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\title{Explicit positive lower density for the Collatz Conjecture}
\author{Kenny Lau}
\address{Axiom Math, 124 University Avenue, Palo Alto, CA 94301}
\email{kenny@axiommath.ai}
\date{}
\subjclass[2020]{Primary 11B83; Secondary 37P99, 60G50}
\keywords{Collatz conjecture, arithmetic dynamics}
\begin{document}
\begin{abstract}
Lech Mazur proved that a positive proportion of integers $n$ reach $1$ under the
Collatz map in at most $10.46\log n$ steps. We give a quantitative refinement
of his construction. For any $X\ge 2^{2^{2^{2^{87640}}}}$, at least
$X/2^{2^{2^{87640}}}$ positive integers $n<X$ reach $1$ within
$\floor{10.46\log n}$ ordinary steps, and $\floor{3.4881\log n}$ accelerated steps. We also obtain results for paths that include any prescribed integer $t$ not divisible by $3$. AxiomProver was used to obtain and formalize this result in Lean.
\end{abstract}
\maketitle


\section{Introduction}\label{sec:intro}

For a positive integer $n$, define the Collatz map $T(n)$ with the rule
\[
T(n)=
\begin{cases}n/2,&n~\mathrm{even}
\\
3n+1,&n~\mathrm{odd}
\end{cases}.
\]

\begin{conjecture}[Collatz]\label{conj:collatz}
Every positive integer $n$ reaches $1$ under finitely many iterations of $T$.
\end{conjecture}

For example, using the starting value $17$ we obtain the orbit
\[
17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1.
\]

The Collatz map can be accelerated by focusing on only the odd values in an orbit. Given an odd number $n$, the next step $3n+1$ is even, and then we keep dividing by $2$ until we reach an odd value. In other words, the next odd number in the orbit after $n$ is
\[S(n) := \frac{3n+1}{2^a}\]
where $a = \ord_2(3n+1)$ is the exponent in the highest power of $2$ that divides $3n+1$. $S$ here is called the Syracuse map, or the accelerated Collatz map. For example, using the same starting value $17$, the accelerated orbit is now
\[17 \to 13 \to 5 \to 1\]
which picks out precisely the odd numbers from the original unaccelerated orbit.

Our main result establishes an \emph{explicit positive density} of starting values that reach $1$, with explicit control on number of steps:
\begin{corollary}
Let $(X_0, c) = (2^{D_A}, D_A^{-1})$ where $D_A = 2^{2^{2^{87640}}}$. Then, for every $X \ge X_0$, at least $c X$ odd starting values $n < X$ reach $1$ within $10.46 \log(n)$ ordinary steps and $3.4881 \log(n)$ accelerated steps.
\end{corollary}

We obtain this as a corollary of a more general theorem, which we also prove in this paper:
\begin{theorem}\label{thm:main}
Let $t$ be a positive integer not divisible by $3$, and let $(X_0(t), c(t)) = (2^{t^3D_A}, t^{-4} D_A^{-1})$. Then, for every $X \ge X_0(t)$, at least $c(t)X$ odd starting values $n < X$ reach $t$ within $10.46 \log(n/t)$ ordinary steps using the Collatz map and reach the odd part of $t$ within $3.4881 \log(n/t)$ accelerated steps using the Syracuse map.
\end{theorem}

\begin{remark}
If $t$ is additionally odd, i.e. $\gcd(t,6)=1$, then we can replace ``the odd part of $t$'' with just $t$.
\end{remark}

\begin{remark}\label{rem:three-nmid}
We require $t$ to be not divisible by $3$ because otherwise the only starting values that reach $t$ are $2^k t$, which has density $0$.
\end{remark}

As a corollary of the main result, we obtain a conditional theorem:
\begin{corollary}
If the Collatz conjecture holds for almost all positive integers, then it is universally true. More precisely, if we have
\[\lim_{X \to \infty} \frac{\#\{1 \le n \le X \mid \text{Collatz is true for $n$}\}}{X} = 1,\]
then the Collatz Conjecture is true.
\end{corollary}
\begin{proof}
Assume that almost all starting values reach $1$. Now let $t$ be a starting value that does not reach $1$. If $t$ is divisible by $3$, then after consecutive even steps followed by one odd step, $t$ will not be divisible by $3$. By the main theorem, a positive density of starting values reach $t$, and those do not reach $1$, and hence it is not true that almost all starting values reach $1$.
\end{proof}

We briefly describe some historical notes about the Collatz conjecture. Krasikov and Lagarias \cite{KL} proved that at least $X^{0.84}$ integers below $X$ reach $1$, for sufficiently large $X$. Tao \cite{Tao} proved that, for any function $f$ tending to infinity, for almost every starting value $n$, the Collatz orbit of $n$ eventually falls under $f(n)$. Here ``almost every'' is in the sense of logarithmic density.

From Tao's proof it follows that if $t$ is sufficiently large then a positive logarithmic density, namely $1-C_T (\log t)^{-c_t}$, of starting values eventually reach some value below $t$. We remark that using the proofs in this paper, the constants can be determined explicitly, but this is out of the scope of this paper. Then Mazur \cite{Mazur} proved:
\begin{theorem}
There exist $X_0$ and $c > 0$ with the property that for every $X \ge X_0$, at least $cX$ starting values $n < X$ reach $1$ within $10.46 \log n$ steps.
\end{theorem}

The pair of constants $(X_0, c)$ is not explicitly specified in Mazur's proof, but Appendix~\ref{app:mazur} reconstructs their values to be
\begin{equation}\label{eq:mazur-bounds}
(X_0, c) = \left(D_M, D_M^{-1}\right), \quad D_M = 2^{2^{2^{2^{2^{2^{2^{173}}}}}}}.
\end{equation}

Mazur also proved \cite{MazurPred} that, more generally, for any prescribed target $t$ with $3 \nmid t$, a positive proportion of integers eventually reach $t$, without a bound on the number of steps. Shaik \cite{Shaik}, building on Mazur's construction, obtains a positive proportion reaching $1$ within $c\log n$ steps for every $c>3/\log(4/3)\approx10.428$.

The paper is organized as follows: Section~\ref{sec:outline} provides a summary of the main argument in Sections~\ref{sec:words}--\ref{sec:steps}. Then Sections~\ref{sec:fb-black}--\ref{sec:fb-cases} establishes a key Fourier bound, while Sections~\ref{sec:est-families} and~\ref{sec:est-histogram} provide key estimates. Then Appendices~\ref{app:constants} and~\ref{app:mazur} contain numerical computations, Appendix~\ref{app:lean} produces the formal Lean certificate, and Appendix~\ref{app:ai} declares the AI usage in this paper.



\section{Outline of the proof}\label{sec:outline}

We provide a high-level overview of the argument in this paper. Recall that we require $3 \nmid t$ (Remark~\ref{rem:three-nmid}). We now further assume that:
\begin{equation}\label{hyp:odd}
\text{$t$ is odd}
\tag{H1}
\end{equation}
We will explain the minor modification to the argument for even $t$ in Section~\ref{sec:outline-seed}.

\subsection{Preliminaries}

We outline some basic facts about the Collatz problem.

\subsubsection{The random model}

Recall that the Syracuse map $S$, which is the accelerated Collatz map, sends an odd number $n$ to $\frac{3n+1}{2^a}$ where $a = \ord_2(3n+1)$. Given a random even number, there is a $\frac12$ chance that it is exactly divisible by $2$ once, then a $\frac14$ chance that it is exactly divisible by $2$ twice, and so on. Therefore, $a$ can be modeled by a geometric random variable $A$ where $\Prob(A=a) = 2^{-a}$. The average value of $a$ is
\[\E(A) = \sum_{a=1}^{\infty} a 2^{-a} = 2.\]

\subsubsection{The heuristic number of steps}

Since the expected value of $a$ is $2$, heuristically the Syracuse map is expected to send $n$ to $\frac{3n+1}{2^2} \approx \frac34 n$. This means that $n$ should reach $1$ in about
\[\log_{4/3}(n) \approx 3.476 \log n\]
accelerated steps using the Syracuse map. Since each Syracuse application with $a$ corresponds to $a+1$ ordinary steps using the Collatz map, and the expected value of $a$ is $2$, this means that $n$ should reach $1$ in about
\[(2+1)\log_{4/3}(n) \approx 10.428 \log n\]
ordinary steps. More generally, if we want $n$ to reach $t$ instead, then heuristically that would take $3.476 \log \frac{n}{t}$ accelerated steps and $10.428 \log \frac{n}{t}$ ordinary steps.

\subsection{First reduction: lower bound for total weight of words landing in a window}

In this section we explain the backward tree model and reduce the goal to obtaining a lower bound for the total weight of the words that result in a starting value in the window $[X, 32X)$.

\subsubsection{Walking backwards}\label{sec:backwards}

Instead of following the Syracuse map $S$ forwards, we analyze the preimages under $S$, in order to construct a large set of sources that will reach the given target $t$.

Suppose $S(n') = n$ for some odd $n$ and $n'$.  We know that for some $a \in \N_{>0}$ we have
\[n = \frac{3n'+1}{2^a},\]
which is equivalent to
\[n' = \frac{2^a n - 1}{3}.\]
This parametrizes the solutions $n'$ given $n$ in terms of $a$, provided the expression gives an integer.

The set of valid solutions depends on the residue class of $n$ modulo $6$. If $n \equiv 1 \pmod 6$, then the valid solutions are given by $a = 2, 4, 6, \cdots$. For example,
\[S^{-1}(7) = \{9, 37, 149, 597, 2389, \cdots\}.\]
If $n \equiv 5 \pmod 6$, then the valid solutions are given by $a = 1, 3, 5, \cdots$. For example,
\[S^{-1}(11) = \{7, 29, 117, 469, 1877, \cdots\}.\]
Finally, if $n \equiv 3 \pmod 6$ then there are no valid solutions.

\subsubsection{Letters and words}\label{sec:word}
Consider the Syracuse orbit $11 \to 17 \to 13 \to 5 \to 1$. The values of $a$ in each step are $(1, 2, 3, 4)$. So if we want to start from the target $1$ and describe the path backwards to reach the source $11$, we would write $w = (4, 3, 2, 1)$, which we call a \emph{word}. Each component is called a \emph{letter}. The example word here has $4$ letters.

As we have seen in Section~\ref{sec:backwards}, not all words correspond to valid integer-valued sources, but if the resulting source is an integer, then all of the intermediate steps are integers (Remark~\ref{rem:admissible}).

\subsubsection{Number of steps}\label{sec:outline-steps}

Given a word $w = (a_1, \cdots, a_d)$, we define its \emph{degree} to be
\[\deg(w) := a_1 + \cdots + a_d.\]
We also denote by $|w|$ the \emph{length} of the word, which is $d$. The constructed source then takes $|w|$ accelerated steps to reach the target $t$, and $|w| + \deg(w)$ ordinary steps (Section~\ref{sec:words}).

\subsubsection{Weight}

A backward step dictated by the letter $a$ causes the value to be multiplied by approximately $\frac{2^a}{3}$. As a result, if we go backwards through the whole word $w$, then the value will be multiplied by approximately $3^{-|w|} 2^{\deg(w)}$. For the example in Section~\ref{sec:word}, we verify that $11 \approx 3^{-4} 2^{4+3+2+1} \approx 12.642$. In fact this will always be an overestimate.

We now define the \emph{weight} of a word $w$ to be
\[\wt(w) := 3^{|w|} 2^{-\deg(w)},\]
which is the reciprocal of the previous value, so that the source is approximately the target divided by $\wt(w)$. We always have
\[\wt(w) \le \frac{t}{s}\]
where $t$ is the target and $s$ is the source.

\subsubsection{Window and reduction}\label{sec:outline-reduction}

To obtain the bound in the main theorem, we consider the window $[X, 32X)$ and try to construct many sources that land in the window. If $t$ is not part of a cycle, then two different words cannot produce the same source (Lemma~\ref{lem_distinct_sources}), so the problem reduces to a count of words. We record this constraint as an explicit hypothesis:
\begin{equation}\label{hyp:not-cycle}
\text{$t$ is not part of a cycle.}
\tag{H2}
\end{equation}

Given the target $t$ and a valid word $w$ with source $s \in [X, 32X)$, we have $\wt(w) \le \frac{t}{s} \le \frac{t}{X}$, which implies
\[\frac{X}{t} \wt(w) \le 1.\]
Summing this over all valid words whose source lands in $[X, 32X)$, we obtain that the count of such words is at least $X/t$ times the total weight $\beta_X(t)$ of these words, which is \eqref{eq:charge}.

Therefore, the main theorem reduces to obtaining an (explicit) constant lower bound of $\beta_X(t)$, with control over the lengths and degrees of the words, in view of Section~\ref{sec:outline-steps}.

\subsection{A mixing lemma}

In this section we explain an important lemma that relates to the stability of the set of valid words across different lengths.

\subsubsection{Residue and offset}

Consider words of length $3$, with target $t$. Say the word is $(a_1, a_2, a_3)$. This word is valid iff the source is an integer, where we recall the source is given by
\[\frac{2^{a_3} \frac{2^{a_2} \frac{2^{a_1} t - 1}{3} - 1}{3} - 1}{3} =
\frac{2^{a_1+a_2+a_3} t - 2^{a_2+a_3} - 2^{a_3} \cdot 3 - 9}{27}.\]
Noting that $2$ is invertible modulo any power of $3$, we conclude that this is an integer iff
\[t \equiv 2^{-a_1} + 2^{-a_1-a_2} \cdot 3 + 2^{-a_1-a_2-a_3} \cdot 9 \pmod{27}.\]

Generalizing this, we can define the offset $\off(w)$ of a word $w = (a_1, \cdots, a_d)$ to be
\[\off(w) := 2^{-a_1} + 2^{-a_1-a_2} \cdot 3 + \cdots + 2^{-a_1-\cdots-a_d} 3^{d-1}\]
so that $w$ is valid iff $t \equiv \off(w) \pmod{3^{|w|}}$, which is Remark~\ref{rem:admissible}. Furthermore, given any $q \in \N$, we denote by $\off_q(w)$ the residue of $\off(w)$ in $\Z/3^q\Z$.

\subsubsection{Residue density}

Recall our probabilistic model where each letter is modeled by a random variable $A$ with $\Prob(A=a) = 2^{-a}$. We form random words $W$ by combining independent letters. The probability of a particular word being chosen is $\Prob(W=w) = 2^{-\deg(w)}$, so $\wt(w) = 3^{|w|} \Prob(W=w)$.

Consider the distribution of $\off_d(W) \in \Z/3^d\Z$, i.e. the residue of the offset of a random word. Define the \emph{residue density} $\rho_d(y) := 3^d \Prob(\off_d(W) = y)$ for $y \in \Z/3^d\Z$. This value coincides with the total weight of length-$d$ words that are valid for $t$ (see \eqref{eq:rho-meaning}):
\[\sum_{w} \wt(w) \cdot 1_{\off_d(w)=t} = \sum_w 3^d \Prob(W=w) \cdot 1_{\off_d(w)=t} = 3^d \Prob(\off_d(W)=t) = \rho_d(t).\]

\subsubsection{Mixing}

The residue density $\rho_d$ depends on $d$ and could differ wildly across different $d$. However, we have a \emph{fine-scale mixing} lemma, in the sense of Tao \cite{Tao}, which states that for large enough $d$, the residue density $\rho_d$ is, on average over residues, well approximated by $\rho_{d_0}$ for much smaller $d_0$. More precisely,
\[3^{-d} \sum_{y \in \Z/3^d\Z} |\rho_d(y) - \rho_{d_0}(y \bmod 3^{d_0})| \le 2^{87533} d_0^{-3/2}\]
for any $d \ge d_0 \ge 2^{80000}$. This is Proposition~\ref{prop:mix}, proved in Section~\ref{sec:mix}.

\subsection{Constructing a large family of words}

Recall we need a constant lower bound of the total weight $\beta_X(t)$ of words whose source lands in $[X, 32X)$, so we need a controlled way to construct many such words to satisfy this constraint.

\subsubsection{Growing the family generation by generation}

We construct a family $L_j$ of words generation by generation (Section~\ref{sec:physical}). $L_0$ consists of only the empty word. At generation $j$, we take every word in the previous generation, and extend it by every element of a yet to be constructed set $\FC(\delta_j, R_j)$. We then keep the words $w$ that are valid for the target $t$ and that additionally satisfy an unspecified filter that among other things has the effect of forcing the inequality $\frac{85}{256} \le \off(w) \le \frac{637}{256}$, whenever $j \ge 5$.

We choose $R_j = 16 + \frac{1}{32}j$ explicitly, but leave $\delta_j$ undefined here except that we know for large $j$ it behaves like $200 \cdot 1.01^j$.

We now describe $\FC(\delta, R)$, the set of \emph{first crossings} with average length $\delta$ and allowed overshoot $R$. For any word $w = (a_1, \cdots, a_d)$, draw the word-graph $(i, \deg_i(w))$ where $\deg_i(w) := a_1 + \cdots + a_i$ is the partial degree. Then draw a straight line passing through $(\delta, 2\delta)$ with slope $\log_2(3) \approx 1.585$. Chop each word off at the \emph{first crossing} of the line, and keep the words with length in $(\delta - \lfloor 3\delta/5 \rfloor, \delta + \lfloor 3\delta/5 \rfloor]$ and overshoot $\le R$. Note that by the first crossing construction, these words are prefix-disjoint, meaning no word is a prefix of any other word. Recall that the average value of a letter is $2$, so almost all words eventually cross the line, and the expected place of crossing is $(\delta, 2\delta)$, so these words have expected length approximately $\delta$ and expected degree approximately $2\delta$.

\subsubsection{Weighted density}

We define the \emph{weighted density} $Z_j(t)$ at generation $j$ (Section~\ref{sec:physical}) to be
\[Z_j(t) := \sum_{w \in L_j} \wt(w) \rho_{\lfloor \frac14 \delta_j \rfloor}(s_w)\]
where $s_w$ is the source of the word $w$ from target $t$. Recall the residue density $\rho$ represents how well words can continue from $s_w$.

\subsubsection{Terminal extension}

We now construct a family $L_{j,X}$ of words that will finally have sources landing near $X$: take every word in $L_j$ and add every extension in $\FC^\term_X(w, \delta_j, R_j)$ so that the resulting word is valid, where $\FC^\term$ is a slight modification of $\FC$ that among other things forces the sources to be $\ge X$.

We will use a notion of $X$ being \emph{reachable} from $j$, which we do not define here (see Section~\ref{sec:physical}), but it roughly means that $X$ is not far away from a typical source of words in $L_j$.

\subsubsection{Terminal weighted density}

We then define the \emph{terminal weighted density} $Z^\term_{j,X}(t)$ to be
\[Z^{\term}_{j,X}(t) := \sum_{w \in L_{j,X}} \wt(w) \rho_{\lfloor \frac14 \delta_j \rfloor}(s_w),\]
with a similar interpretation as the weighted density.


\subsection{Key estimates and the final proof}

We now state several key estimates for the quantities above that are proved in this paper (Section~\ref{sec:census}), and explain how they lead to a proof of the main theorem.

\subsubsection{Key estimates}

The weighted density enjoys a propagation error bound from one generation to the next. Precisely, for any $j \ge 6 \times 10^6$, Proposition~\ref{prop:central-propagation} states that
\begin{equation}\label{eqn:propagation-error}
|Z_{j+1}(t) - Z_j(t)| \le 2^{87534.2} j^{\frac52} 1.01^{-j}.
\end{equation}
As a consequence, if $j$ is sufficiently large, then the sum of all these errors beyond $j$ is negligible, provided that the main term $Z_j(t)$ is large enough.

We indeed have a large enough main term. Precisely, let $j_0 := 6103914$ and $d_0 := 2^{87639}$. This $d_0$ is enough to dominate the subscript of every $\rho$ that appears in the definition of $Z_{j_0}$, so that $Z_{j_0}$ is periodic with period $3^{d_0}$. Then, the estimate we need is that the average value of $Z_{j_0}(t)$ across unit residues $t \in (\Z/3^{d_0}\Z)^\times$ is greater than $2^{-23}$, which is supplied by Proposition~\ref{prop:average} which gives us
\begin{equation}\label{eqn:average-bound}
\frac{1}{|(\Z/3^{d_0}\Z)^\times|} \sum_{t \in (\Z/3^{d_0}\Z)^\times} Z_{j_0}(t) > 2^{-23}.
\end{equation}

Now we bring $X$ into the picture, and seek a large enough bound for $X$ such that it is reachable from some $j$ that is large enough for the estimates below. Let $J(t) := 2 \times 10^{10} + 200 \log_2 t$. Proposition~\ref{prop:coverage} then tells us the following:
\begin{equation}\label{eqn:reachable}
\text{
Let $t \ge 16^9$. Then every $X \ge 2^{2^{2^{29}} t^3}$ is reachable from some $j \ge J(t)$.
}
\end{equation}
This requires another hypothesis on $t$:
\begin{equation}\label{hyp:bound}
t \ge 16^9.
\tag{H3}
\end{equation}

We now provide an estimate to connect the weighted density to the terminal weighted density. Precisely, for $j \ge 6 \times 10^6$ and $X$ reachable from $j$, according to Proposition~\ref{prop:terminal-errors},
\begin{equation}\label{eqn:terminal-error}
|Z^\term_{j,X}(t) - Z_j(t)| \le 2^{87536} j^{\frac92} 1.01^{-j}.
\end{equation}

And finally an inequality to connect the terminal weighted density to the total weight: for any $j$, any $X$ reachable from $j$, and any integer $m$ with $2^{80000} \le m \le \frac14 \delta_j$, \eqref{eq:census} provides us with the bound
\begin{equation}\label{eqn:terminal-total-ineq}
Z^\term_{j,X}(t) \le \tfrac{32}{3} m^{\frac{4097}{2048}} \beta_X(t) + 2^{87538} m^{-\frac32} t.
\end{equation}


\subsubsection{Final proof}\label{sec:outline-final-fake}

We now combine the five estimates above. Firstly from the estimate on the average weighted density \eqref{eqn:average-bound}, we know that there is some unit residue $y_0 \in (\Z/3^{d_0}\Z)^\times$ with
\[Z_{j_0}(y_0) > 2^{-23},\]
which leads to our final hypothesis on $t$:
\begin{equation}\label{hyp:residue}
t \equiv y_0 \pmod{3^{d_0}}
\tag{H4}
\end{equation}

Now following the reachability estimate \eqref{eqn:reachable}, we let $X \ge 2^{2^{2^{29}} t^3}$, with $j \ge J(t)$ so that $X$ is reachable from $j$. This lets us use the estimate on propagation error \eqref{eqn:propagation-error}, which tells us that the errors from generation $j_0$ onwards (Lemma~\ref{lem_Nstar_tail}) sum to less than $2^{-23} - 2^{-25}$, so that we can obtain
\[Z_j(t) \ge 2^{-25},\]
at which point the estimate on terminal weighted density \eqref{eqn:terminal-error} gives us
\[Z^\term_{j,X}(t) \ge 2^{-26}.\]

Now the inequality linking terminal weighted density with total weight \eqref{eqn:terminal-total-ineq} has an additional parameter $m$. After numerical optimization we can choose $m = \ceil{2^{80000} t^{\frac23}}$, and this gives us
\[\beta_X(t) \ge 2^{-160071} t^{-1.34},\]
which we can substitute back into the reduction in Section~\ref{sec:outline-reduction} to obtain the count we desired, namely that
\[\#\{s \in [X, 32X) ~\text{maps to}~t\} \ge 2^{-160071} t^{-2.34} X,\]
subject to the hypotheses \eqref{hyp:odd}, \eqref{hyp:not-cycle}, \eqref{hyp:bound}, \eqref{hyp:residue}. This is Proposition~\ref{prop:core}.

\subsubsection{Discharging the hypotheses: choosing a seed}
\label{sec:outline-seed}

Recall the hypotheses require that $t \ge 16^9$ is odd, $t$ is not part of a cycle, and $t$ has a specific residue modulo $3^{d_0}$. We now describe the process of removing those hypotheses, which mirrors Proposition~\ref{prop:seed}. Starting with an arbitrary $t$ with $3 \nmid t$, we will obtain a seed $t_1$ that satisfies the hypotheses above, with some constraints between the seed $t_1$ and the original target $t$.

The seed $t_1$ will be chosen from $\frac{2^a t - 1}{3}$, so that $t_1$ will reach $t$ with only one odd step at the beginning, and the rest of the steps will be the even step where the value is divided by $2$. This discharges $t_1$ being odd (Hypothesis~\eqref{hyp:odd}). If $t$ is odd, then $t_1$ will reach $t$ in one Syracuse step; otherwise $t_1$ will still reach the odd part of $t$ in one Syracuse step.

Valid choices of $a$ that make $t_1$ an integer will always be either all the even numbers or all the odd numbers. By modular arithmetic, these choices of $t_1$ meet every residue class modulo any power of $3$ (Lemma~\ref{lem_repunit_residues}). This discharges the residue requirement (Hypothesis~\eqref{hyp:residue}).

The bound requirement (Hypothesis~\eqref{hyp:bound}) is discharged by simply choosing a larger $a$, which leaves us with the requirement that $t_1$ cannot be part of a cycle (Hypothesis~\eqref{hyp:not-cycle}), which is discharged by choosing \emph{two} potential candidates for $t_1$. They both map to (the odd part of) $t$ in one step under the Syracuse map, so they cannot both be part of a cycle.

The resulting seed $t_1$ can be bounded in terms of $t$ using Proposition~\ref{prop:seed}, which says that
\[t_1 \le D_A t.\]

\subsubsection{Final proof for general $t$}

Using Section~\ref{sec:outline-final-fake} for the seed $t_1$, we deduce that for $X \ge 2^{2^{2^{29}} t_1^3}$, we have
\[\#\{s \in [X, 32X) ~\text{maps to}~ t_1\} \ge 2^{-160071} t_1^{-2.34} X.\]
Translating this back to $t$, using the bound $t_1 \le D_A t$, gives us that for $X \ge 2^{t^3 D_A}$, we have
\[\#\{s \in [X, 32X) ~\text{maps to}~ t\} \ge t^{-2.34} D_A^{-1} X.\]

The exact version of this bound is \eqref{eq:bridge}. Note that there is some abuse of notation here considering that $D_A$ is a triple exponential and hence absorbs many smaller operations. This also covers replacing $X$ with $\frac1{32}X$ to align with the main theorem counting numbers $< X$.


\subsection{The bound for the number of steps}

What remains is control on the length of the words involved, as well as their degrees (Section~\ref{sec:steps}). This follows from the construction since at each generation we pick the words at their first crossing with the line, which puts constraints on their expected lengths and degrees.

Recall that the heuristic number of steps is $3.476 \log \frac{n}{t}$ accelerated steps and $10.428 \log \frac{n}{t}$ ordinary steps. The extra slack in the main theorem allows the bounds on step count to hold. We also use the fact that the seed $t_1$ is chosen to reach (the odd part of) the target $t$ in one accelerated step, which provides a simple bound for also the number of ordinary steps for this part.



\section{Words and residue densities}\label{sec:words}
This section makes the objects of Section~\ref{sec:outline} precise: valid words, their weights and offsets, the reduction to a lower bound for the total weight, and the residue density with its branch transfers.

A \emph{word} $w=(a_1,\ldots,a_d)$ is a finite sequence of positive integers, of length $|w|=d$. It describes a walk backwards along the Syracuse map $S$: each letter $a$ picks the preimage $(2^an-1)/3$ of $n$ under $S$, as in Section~\ref{sec:backwards}. Starting from a positive odd $n$, the \emph{inverse trajectory} of $w$ is
\[
x_0=n,\qquad x_i=\frac{2^{a_i}x_{i-1}-1}{3}\quad(1\le i\le|w|).
\]
The word is \emph{valid at $n$} if all $x_i$ are positive odd integers. Then $3x_i+1=2^{a_i}x_{i-1}$ with $x_{i-1}$ odd, so $S(x_i)=x_{i-1}$: the Collatz map $T$ takes $x_i$ to $x_{i-1}$ in one step $x\mapsto3x+1$ followed by $a_i$ halvings. Hence the \emph{source} $\src(w,n)=x_{|w|}$ reaches $n$ in $|w|$ accelerated steps and $|w|+\deg(w)$ ordinary steps, where
\begin{gather*}
\deg_i(w)=a_1+\cdots+a_i,\qquad \deg(w)=\deg_{|w|}(w),\\
\wt(w)=3^{|w|}2^{-\deg(w)},\qquad\off(w)=\sum_{i=1}^{|w|}3^{i-1}2^{-\deg_i(w)}
\end{gather*}
are the \emph{partial degrees}, the \emph{degree}, the \emph{weight} and the \emph{offset} of $w$. Iterating $x_{i-1}=(3x_i+1)/2^{a_i}$ gives the basic identity
\begin{equation}\label{eq:affine}
n=\wt(w)\src(w,n)+\off(w),
\end{equation}
which defines $\src(w,n)=(n-\off(w))/\wt(w)$ for every word. For a concatenation $vw$,
\begin{equation}\label{eq:concat}
\wt(vw)=\wt(v)\wt(w),\qquad \off(vw)=\off(v)+\wt(v)\off(w).
\end{equation}
For example, $w=(3,2)$ at $n=5$ gives $x_1=13$ and $x_2=17$, and indeed $\frac9{32}\cdot17+\frac7{32}=5$.

\begin{remark}[Validity is a congruence; cf.~{\cite[Section~2]{Mazur}, \cite[Lemma~2.1]{Tao}}]\label{rem:admissible}
A word $w$ is valid at a positive odd $n$ if and only if $n\equiv\off(w)\pmod{3^{|w|}}$. Indeed $2^{\deg(w)}\off(w)$ is an integer, so validity and \eqref{eq:affine} give the congruence. Conversely, if it holds then $x_{|w|}$ is an integer; the forward relations put every $x_i$ in $\Z[1/2]$ and the backward relations put it in $\Z[1/3]$, so every $x_i$ is an integer, odd because $2^{a_i}x_{i-1}-1$ is odd, and positive.
\end{remark}

\subsection{Distinct sources and the reduction}
To count sources through the weights of their words, distinct words must have distinct sources; this holds unless the target lies on a cycle. A positive odd integer $y$ is \emph{$S$-periodic} if $S^{\circ j}(y)=y$ for some $j\ge1$, where $f^{\circ k}$ denotes the $k$-fold iterate of a map $f$.

\begin{lemma}[Periodic points]\label{lem_syr_periodic_fiber}\label{lem_orbit_hits_seed_once}
Let $y,y'$ be positive odd integers.
\begin{enumerate}
\item If $y$ and $y'$ are $S$-periodic and $S(y)=S(y')$, then $y=y'$.
\item If $y$ is not $S$-periodic, then for every positive odd $n$ there is at most one $r\in\N$ with $S^{\circ r}(n)=y$.
\end{enumerate}
\end{lemma}
\begin{proof}
For (1), let $S^{\circ j}(y)=y$ and $S^{\circ j'}(y')=y'$ with $j,j'\ge1$; then $r=jj'$ satisfies $S^{\circ r}(y)=y$ and $S^{\circ r}(y')=y'$, so $y=S^{\circ(r-1)}(S(y))=S^{\circ(r-1)}(S(y'))=y'$. For (2), if $S^{\circ r}(n)=y=S^{\circ r'}(n)$ with $r<r'$, then $S^{\circ(r'-r)}(y)=y$.
\end{proof}

\begin{lemma}[Distinct sources]\label{lem_distinct_sources}
Let $n$ be a positive odd integer that is not $S$-periodic. Then distinct words valid at $n$ have distinct sources.
\end{lemma}
\begin{proof}
Let $w$ be valid at $n$ with source $s$. Reading its inverse trajectory backwards from $s$, the value before a value $x'$ is $(3x'+1)/2^a$, where $a$ is the letter between them, and this is an odd integer only for $a=\ord_2(3x'+1)$, where $\ord_p(y)$ is the exponent of the prime $p$ in $y$. So $|w|$ and $s$ determine $w$, and $S^{\circ|w|}(s)=n$. By Lemma~\ref{lem_syr_periodic_fiber}(2), $s$ determines $|w|$.
\end{proof}

For real $X>0$ and a positive odd $n$, let $\beta_X(n)$ be the total weight of the words valid at $n$ whose source lies in $[X,32X)$. If $n$ is not $S$-periodic, these sources are distinct by Lemma~\ref{lem_distinct_sources}, and each such word has $\wt(w)\le n/\src(w,n)\le n/X$ by \eqref{eq:affine}, since $\off(w)\ge0$. Hence
\begin{equation}\label{eq:charge}
\#\{\src(w,n):w\text{ valid at }n,\ \src(w,n)\in[X,32X)\}\ge\frac Xn\,\beta_X(n).
\end{equation}

\subsection{The random model and the residue density}
By Remark~\ref{rem:admissible}, the total weight of the words of length $d$ valid at $n$ depends only on $n$ modulo $3^d$; the random model computes it as a function on residues. Write $G_d=\Z/3^d\Z$, call $d$ its \emph{level}, and write $[x]_d\in G_d$ for the reduction of $x\in\Z[1/2]$; a function on $G_d$ is evaluated at such $x$ through $[x]_d$. Take independent letters with $\Prob(a_i=a)=2^{-a}$ for $a\ge1$, and call $W$ the resulting random word; a word $w$ occurs as a prefix of $W$ with probability $2^{-\deg(w)}$. Let the \emph{offset law} $\mu_d$ be the probability mass function on $G_d$ of the offset residue $[\off(W_{\le d})]_d$ of the first $d$ letters, and put the \emph{residue density}
\begin{equation}\label{eq:rho}
\rho_d(y)=3^d\mu_d(y).
\end{equation}
Write $\av df=3^{-d}\sum_{y\in G_d}f(y)$ for $f:G_d\to\C$, and $\pr_{d,m}:G_d\to G_m$ ($m\le d$) for reduction; then $\av d{f\circ\pr_{d,m}}=\av mf$ and $\av d{\rho_d}=1$. Reduction modulo $3^m$ sends $\mu_d$ to $\mu_m$ (\emph{projectivity}). For $d\ge1$ the law lives on units, since $\off(W_{\le d})\equiv2^{-a_1}\pmod3$. By Remark~\ref{rem:admissible},
\begin{equation}\label{eq:rho-meaning}
\sum_{|w|=d}\wt(w)\,\mathbf 1[w\text{ valid at }n]=\rho_d(n),
\end{equation}
where $\mathbf 1[E]$ is $1$ if the statement $E$ holds and $0$ otherwise.

To average quantities such as $\rho_k(\src(w,n))$ over the residue of $n$, we transport functions along words as in \cite[Section~2]{Mazur}. For a word $w$, the map $\iota_w:G_k\to G_{k+|w|}$, $\iota_w(y)=\wt(w)y+\off(w)$, is well-defined and injective, since its linear coefficient is $3^{|w|}$ times a unit. For $f$ on $G_k$, the \emph{branch transfer} $\mathcal T_wf$ on $G_{k+|w|}$ equals $\wt(w)f(y)$ at $\iota_w(y)$ and zero off the image. By \cite[Lemma~2.1]{Mazur},
\begin{equation}\label{eq:transfer-mean}
\av{k+|w|}{\mathcal T_wf}=2^{-\deg(w)}\av kf,\qquad \av{k+|w|}{|\mathcal T_wf|}=2^{-\deg(w)}\av k{|f|}.
\end{equation}
For positive odd $n$, Remark~\ref{rem:admissible} and injectivity of $\iota_w$ give
\begin{equation}\label{eq:fiber-source}
(\mathcal T_wf)(n)=\begin{cases}\wt(w)f(\src(w,n)),&w\text{ valid at }n,\\0,&\text{otherwise}.\end{cases}
\end{equation}
By \eqref{eq:concat}, $\mathcal T_w\rho_k$ is $3^{k+|w|}$ times the law of the offset residue of $W_{\le k+|w|}$ restricted to the event that $W$ begins with $w$.

\section{Mixing and a bounded subdensity}\label{sec:mix}
This section proves the mixing lemma of Section~\ref{sec:outline} and the bounded subdensity used by the census (Proposition~\ref{prop:mix}), assuming the Fourier bound \eqref{eq:fourier} below, and derives the transfer estimate (Lemma~\ref{lem:finite-transfer}). Here $n$ denotes a level. Set $n_0=2^{80000}$ and
\begin{equation}\label{eq:exponents}
A_*=\frac{10241}{4096},\qquad\alpha=\frac{4097}{2048},\qquad A_*-\frac\alpha2=\frac32.
\end{equation}
The Fourier bound below decays with exponent $A_*$ and the bounded subdensity is at most $8m^\alpha$; the relation $A_*-\alpha/2=3/2$ gives the exponent $3/2$ of mixing in \eqref{eq:one-step-mixing}.
Theorem~\ref{lem_ch_primitive_decay}, an explicit form of \cite[Proposition~1.17]{Tao} proved in Sections~\ref{sec:fb-black}--\ref{sec:fb-conclusion}, gives for every $n\ge1$ and every frequency $\xi\in G_n$ prime to $3$ the \emph{Fourier bound}
\begin{equation}\label{eq:fourier}
|\widehat\mu_n(\xi)|\le(160D_*)^{A_*}n^{-A_*},\qquad \widehat\mu_n(\xi)=\sum_{y\in G_n}\mu_n(y)e^{-2\pi\mathrm i\xi y/3^n},
\end{equation}
where $D_*$ is defined in Appendix~\ref{app:analytic}. Put
\begin{equation}\label{eq:C}
C=\frac{477}{20}(3200D_*)^{A_*}+\frac{159}{10}.
\end{equation}

\begin{proposition}[Fine-scale mixing and subdensity]\label{prop:mix}
The \emph{mixing constant} $C$ of \eqref{eq:C} satisfies
\begin{equation}\label{eq:Clog}
87532.002162790<\log_2C<\log_2(C+1)<87532.002162791.
\end{equation}
For integers $n_0\le m\le d$,
\begin{equation}\label{eq:mix}
\av d{|\rho_d-\rho_m\circ\pr_{d,m}|}\le Cm^{-3/2}.
\end{equation}
For every integer $m\ge n_0$ there is a function $\underline\rho_m:G_m\to\R$, the \emph{bounded subdensity}, with
\begin{equation}\label{eq:subdensity}
0\le\underline\rho_m\le\rho_m,\qquad\underline\rho_m\le8m^\alpha,\qquad\av m{\rho_m-\underline\rho_m}\le m^{-3/2}.
\end{equation}
\end{proposition}

\subsection{Restricting at the cut}\label{sec:first-crossing}
Following \cite[Section~6]{Tao}, we \emph{cut} a random word of length $n$ at the first letter where its partial degree exceeds a \emph{cut height} $V$ with $2^V$ slightly below $3^n$: the position of this letter is the \emph{cut point}, the prefix ending at it is the \emph{head}, and the rest is the \emph{tail word}. A head $w'$ with degree $l$ is kept only if $2^l\off(w')<3^n$; then it has probability $2^{-l}\le2^{-V}$, which is $3^{-n}$ up to a polynomial factor, and it is determined by its offset residue (Lemma~\ref{lem:heads}), which bounds the subdensity pointwise. The tail word is fresh, and its offset law inherits the Fourier bound, which gives mixing.

For $n\ge n_0$ set
\begin{align*}
V&=n\log_23-\frac{6145}{4096}\log_2n-2,\\
k^{\mathrm{lo}}&=\floor{V/2-\tfrac14n^{2049/4096}},&k^{\mathrm{hi}}&=\ceil{V/2+\tfrac14n^{2049/4096}}.
\end{align*}
For integers $1\le k\le n$ and $l\ge1$, let the \emph{gate} $\mathcal G_{k,l}$ consist of the heads $w'$ of length $k$ satisfying
\begin{equation}\label{eq:head-gate}
k^{\mathrm{lo}}<k\le k^{\mathrm{hi}},\qquad\deg_{k-1}(w')\le V<\deg(w')=l,\qquad2^l\off(w')<3^n.
\end{equation}
The corresponding \emph{slice} consists of a head followed by any word of length $n-k$. These slices are disjoint, since the partial degrees increase strictly and so pass $V$ only once. Let $\mathcal R_n$ be their union.

Since letters have mean $2$, the cut point lies near $V/2$, and Lemma~\ref{lem:gates} below shows that the slices carry almost all the probability. The cut point leaves $(k^{\mathrm{lo}},k^{\mathrm{hi}}]$ only if a partial degree deviates from twice its length by about $n^{2049/4096}$, which the following Chernoff bound controls.

\begin{lemma}[Geometric tail]\label{lem:geometric-tail}
For real $\mathfrak t<\log2$ put $\mathsf M_{\mathrm{geo}}(\mathfrak t)=\E e^{\mathfrak t(a_1-2)}=e^{-\mathfrak t}/(2-e^{\mathfrak t})$. Then $0\le\log\mathsf M_{\mathrm{geo}}(\mathfrak t)\le8\mathfrak t^2$ for $|\mathfrak t|\le1/4$, and $\log\mathsf M_{\mathrm{geo}}(\mathfrak t)\le9\mathfrak t^2/8$ for $|\mathfrak t|\le1/32$. For integers $d\ge1$ and real $x\ge0$,
\begin{equation}\label{eq:geometric-tail}
\Prob(|\deg_d(W)-2d|\ge x)\le2\exp\left(-\min\{x^2/(32d),x/8\}\right).
\end{equation}
\end{lemma}
\begin{proof}
The formula for $\mathsf M_{\mathrm{geo}}$ is a geometric series. Its logarithm has value and derivative zero at $0$, and its second derivative $2e^{\mathfrak t}/(2-e^{\mathfrak t})^2$ is nonnegative and increasing, at most $6$ on $|\mathfrak t|\le1/4$ (as $e^{1/4}\le4/3$) and at most $\frac{496}{225}<\frac94$ on $|\mathfrak t|\le1/32$ (as $e^{1/32}\le32/31$); Taylor's formula gives the bounds. Since $\E e^{\mathfrak t(\deg_d(W)-2d)}=\mathsf M_{\mathrm{geo}}(\mathfrak t)^d$, the Cram\'er--Chernoff method \cite[Section~2.2]{BLM} for $\pm(\deg_d(W)-2d)$ with $\mathfrak t=\min(x/(16d),1/4)$ gives \eqref{eq:geometric-tail}, which \cite[Section~3.1]{Mazur} states with strict inequality.
\end{proof}

The product condition in \eqref{eq:head-gate} fails only if the offset of the word before the cut point, scaled by the overshoot, is large. The offset has a $1/x$ tail, and the overshoot is geometric and independent of what precedes the crossing.

\begin{lemma}[Offset tail]\label{lem_mx_offset_tail}
For every $n\in\N$ and real $x\ge0$, $\Prob(\off(W_{\le n})\ge x)\le7/(x+2)$.
\end{lemma}
\begin{proof}
First, for real $x\ge5$,
\begin{equation}\label{eq:tail-step}
3\sum_{a\ge1}\frac{2^{-a}}{2^ax+5}\le\frac1{x+2}.
\end{equation}
Indeed $\frac{3\cdot2^{-a}}{2^ax+5}=\frac{3\cdot4^{-a}}x-\frac{15\cdot4^{-a}}{x(2^ax+5)}$; summing with $\sum_a4^{-a}=\frac13$ and keeping the terms $a=1,2$ of the subtracted series, it suffices that $\frac2{x+2}\le\frac{15(18x+25)}{16(2x+5)(4x+5)}$, that is $15(18x+25)(x+2)-32(2x+5)(4x+5)=14x^2-45x-50\ge0$, which holds for $x\ge5$.

Now induct on $n$. For $n=0$ the offset is $0$, and $1\le7/2$. For $x<5$ the bound exceeds $1$. For $x\ge5$, write $W_{\le n}=(a)v$; then $\off(W_{\le n})=2^{-a}(1+3\off(v))$, so the event is $\off(v)\ge x_a=(2^ax-1)/3\ge0$, where $v$ is a random word of length $n-1$ independent of $a$. By the induction hypothesis and \eqref{eq:tail-step}, the probability is at most $\sum_a2^{-a}\frac7{x_a+2}=21\sum_a\frac{2^{-a}}{2^ax+5}\le\frac7{x+2}$.
\end{proof}

\begin{lemma}[Product tail]\label{lem_mx_product_tail}
Let $n\ge n_0$. The probability that the cut point $i$ of $W_{\le n}$ exists and $2^{\deg_i(W)}\off(W_{\le i})\ge3^n$ is at most $\frac45n^{-3/2}$.
\end{lemma}
\begin{proof}
Put $c=\frac{6145}{4096}$, $R=n^c$, $L=\log_2n$ and $H=\floor V+1$, so $2^V=3^n/(4R)$ and $2^H\le3^n/(2R)$. Since $\off(W_{\le i})=\off(W_{\le i-1})+3^{i-1}2^{-\deg_i(W)}$ and $i\le n$, the event forces $2^{\deg_i(W)}\off(W_{\le i-1})\ge\frac23\cdot3^n\ge R2^H$, that is, with the overshoot $r=\deg_i(W)-H\ge0$, $\off(W_{\le i-1})\ge R2^{-r}$.

Fix $i$ and $r$. Given the first $i-1$ letters, with $\deg_{i-1}(W)\le V$ and $\gamma=H-\deg_{i-1}(W)\ge1$, the letter $a_i$ equals $\gamma+r$ with probability $2^{-\gamma-r}=2^{-r-1}\Prob(a_i\ge\gamma)$. So the probability that $i$ is the cut point with overshoot $r$ and $\off(W_{\le i-1})\ge R2^{-r}$ is $2^{-r-1}$ times the probability that $i$ is the cut point and $\off(W_{\le i-1})\ge R2^{-r}$. The latter events are disjoint in $i$ and imply $\off(W_{\le n})\ge R2^{-r}$, so by Lemma~\ref{lem_mx_offset_tail} the probability in question is at most
\[
\sum_{r\ge0}2^{-r-1}\frac7{R2^{-r}+2}=\sum_{r\ge0}\frac7{2R+2^{r+2}}\le\frac7{2R}(\log_2R+2).
\]
For the last step let $r_0=\floor{\log_2R}\ge1$: the $r_0+1$ terms with $r\le r_0$ are each below $\frac7{2R}$, and the terms with $r>r_0$ are below $7\cdot2^{-r-2}$ and sum to $7\cdot2^{-r_0-2}<\frac7{2R}$.

It remains to show $\frac72(cL+2)n^{-c}\le\frac45n^{-3/2}$, that is $\frac{35}8(cL+2)\le2^{L/4096}$, as $c-\frac32=\frac1{4096}$. Put $s=L/80000\ge1$, $K=\frac{35}8(80000c+3)=\frac{35\cdot491612288}{32768}$ and $\beta_0=2^{625/32}$, so $2^{L/4096}=\beta_0^s$. Then $\frac{35}8(cL+2)\le sK$, $K<\beta_0$ by the integer comparison $(35\cdot491612288)^{32}<2^{1105}$, and $s\beta_0\le\beta_0^s$ since $\log\beta_0>1$.
\end{proof}

\begin{lemma}[Mass retained by the gates]\label{lem:gates}
The probability of $\mathcal R_n$ is at least $1-n^{-3/2}$.
\end{lemma}
\begin{proof}
The elementary bounds
\begin{equation}\label{eq:gate-window}
0.33n<k^{\mathrm{lo}}<k^{\mathrm{hi}}<13n/16,\qquad k^{\mathrm{hi}}-k^{\mathrm{lo}}\le n^{2049/4096}
\end{equation}
follow from $79/50<\log_23<317/200$, $\tfrac14n^{2049/4096}\le n/100$, and $(6145/4096)\log_2n+2<5\log n\le0.3n$. At $d=k^{\mathrm{lo}}$ and at $d=k^{\mathrm{hi}}$, with $x=\tfrac12n^{2049/4096}$, the exponent in \eqref{eq:geometric-tail} is at least $n^{1/2048}/128>10\log n$; the latter inequality holds at $n_0$ and persists since $n^{1/2048}/\log n$ increases in $n$ thereon. Thus the two events $\deg_{k^{\mathrm{lo}}}(W)>V$ and $\deg_{k^{\mathrm{hi}}}(W)\le V$ have total probability at most $4n^{-10}$, and outside them the cut point lies in the required range. By Lemma~\ref{lem_mx_product_tail} the product condition fails with probability at most $\frac45n^{-3/2}$, and $\frac45+4n^{-17/2}\le1$.
\end{proof}

For each gate, let $\Gamma_{k,l}(y)$ be the probability mass of its heads with offset residue $y\in G_n$; sums over gates include only nonempty gates. For functions on $G_n$, $\|\cdot\|_1$ and $\|\cdot\|_\infty$ are the unnormalized $\ell^1$ and supremum norms, and $*$ is convolution in counting measure.

\begin{lemma}[Head bounds]\label{lem:heads}
The head masses satisfy
\begin{equation}\label{eq:head-bounds}
\|\Gamma_{k,l}\|_\infty\le2^{-l},\qquad\sum_y\Gamma_{k,l}(y)^2\le2^{-l}\|\Gamma_{k,l}\|_1,\qquad\sum_{k,l}\|\Gamma_{k,l}\|_1\le1,
\end{equation}
and
\begin{equation}\label{eq:head-coefficients}
\sum_{k,l}3^n2^{-l}<8n^\alpha.
\end{equation}
\end{lemma}
\begin{proof}
By \eqref{eq:head-gate}, $2^l\off(w')$ is an integer in $[0,3^n)$ for every head $w'$ in $\mathcal G_{k,l}$, so two heads in the same gate with equal offset residues have equal offsets, and offsets determine words of a fixed length \cite[Lemma~6.2]{Tao}. As in \cite[Corollary~6.3]{Tao}, there is therefore at most one head per residue, and its probability is $2^{-l}$. This proves the first two bounds. Each slice has probability $\|\Gamma_{k,l}\|_1$, since its tail word is unrestricted, so disjointness proves the third. There are at most $n^{2049/4096}$ head lengths by \eqref{eq:gate-window}. For each length, sum the geometric series over $l\ge\floor V+1>V$:
\[
\sum_{k,l}3^n2^{-l}\le n^{2049/4096}3^n2^{1-(\floor V+1)}<8n^{(2049+6145)/4096}=8n^\alpha.
\]
\end{proof}

The \emph{transported tail law} $\mathcal L_{k,l}$ is the law on $G_n$ obtained by sending $\mu_{n-k}$, the offset law of the tail word, through $y\mapsto3^k2^{-l}y$. By \eqref{eq:concat} and independence of head and tail word, the offset law restricted to the slice of $\mathcal G_{k,l}$ is $\Gamma_{k,l}*\mathcal L_{k,l}$. Define
\[
\underline\rho_n=3^n\sum_{k,l}\Gamma_{k,l}*\mathcal L_{k,l}.
\]
This is a restriction of the law, so $0\le\underline\rho_n\le\rho_n$, and Lemma~\ref{lem:gates} gives $\av n{\rho_n-\underline\rho_n}\le n^{-3/2}$. Since convolution with a probability measure cannot increase the supremum norm, \eqref{eq:head-bounds} and \eqref{eq:head-coefficients} give $\underline\rho_n\le8n^\alpha$. This proves \eqref{eq:subdensity} at every level $n\ge n_0$.

\subsection{From the Fourier bound to mixing}\label{sec:fourier-to-mixing}
For $m\le n$, let $\operatorname{Av}_{n,m}$ average a function on each fiber of $\pr_{n,m}$, and put $\Osc_{n,m}(f)=\sum_y|f(y)-\operatorname{Av}_{n,m}f(y)|$. A frequency $\xi\in G_n$ \emph{factors through} $G_m$ if $3^{n-m}\mid\xi$. By projectivity, $\operatorname{Av}_{n,m}\mu_n=3^{-n}\rho_m\circ\pr_{n,m}$, so $\Osc_{n,m}(\mu_n)=\av n{|\rho_n-\rho_m\circ\pr_{n,m}|}$. We bound it first for $n\le\frac{10}9m$ (Tao's regime \cite[(6.1)]{Tao}, which leaves the tail word a fixed fraction of the level), and then along a chain of levels.

\begin{lemma}[Convolution oscillation]\label{lem:oscillation}
Let $0\le m\le n$ be integers and $\Lambda\ge0$. Suppose $f,\psi$ are complex-valued functions on $G_n$ and $|\widehat\psi(\xi)|\le\Lambda$ for all frequencies that do not factor through $G_m$. Then
\[
\Osc_{n,m}(f*\psi)^2\le\Lambda^23^n\sum_y|f(y)|^2.
\]
\end{lemma}
\begin{proof}
This is the computation of \cite[(6.10)--(6.11)]{Tao}. Fiber averaging keeps exactly the Fourier coefficients at frequencies factoring through $G_m$, so Cauchy--Schwarz and Parseval give
\[
\|f*\psi-\operatorname{Av}_{n,m}(f*\psi)\|_1^2\le\sum_{\xi\text{ not factoring through }G_m}|\widehat f(\xi)\widehat\psi(\xi)|^2\le\Lambda^23^n\sum_y|f(y)|^2.
\]
\end{proof}

Suppose $n_0\le m\le n$ and $9n\le10m$. Let $\xi\in G_n$ not factor through $G_m$, and let $3^r$ be its order in $G_n$, so $r\ge m+1$. By \eqref{eq:gate-window}, $20(k-1)\le17n$, and consequently $r-k\ge m+1-k\ge n/20$. Since $\mathcal L_{k,l}$ is the image of $\mu_{n-k}$ under $y\mapsto3^k2^{-l}y$, $\widehat{\mathcal L_{k,l}}(\xi)=\widehat\mu_{n-k}(2^{-l}\xi)$, where $2^{-l}\xi$ is reduced into $G_{n-k}$ and has order $3^{r-k}$ there; by projectivity this is $\widehat\mu_{r-k}(\xi')$ for some $\xi'\in G_{r-k}$ prime to $3$. Since the bound \eqref{eq:fourier} decreases with the level, it gives the hypothesis of Lemma~\ref{lem:oscillation} for $\psi=\mathcal L_{k,l}$ with $\Lambda=(160D_*)^{A_*}(n/20)^{-A_*}=(3200D_*)^{A_*}n^{-A_*}$. Summing Lemma~\ref{lem:oscillation} over the gates and using \eqref{eq:head-bounds} and \eqref{eq:head-coefficients} gives
\[
\sum_{k,l}\Osc_{n,m}(\Gamma_{k,l}*\mathcal L_{k,l})\le\Lambda\Bigl(\sum_{k,l}3^n2^{-l}\Bigr)^{1/2}\Bigl(\sum_{k,l}\|\Gamma_{k,l}\|_1\Bigr)^{1/2}\le3(3200D_*)^{A_*}n^{-3/2}.
\]
Fiber averaging is an $\ell^1$ contraction, so the part of $\mu_n$ outside the slices adds at most twice its probability. Thus
\begin{equation}\label{eq:one-step-mixing}
\Osc_{n,m}(\mu_n)\le B_{\mathrm{osc}}n^{-3/2},\qquad B_{\mathrm{osc}}=3(3200D_*)^{A_*}+2.
\end{equation}

Telescoping, as in \cite[Section~6]{Tao}, reaches all levels; by \eqref{eq:C}, $C=\frac{159}{20}B_{\mathrm{osc}}$. We show $\Osc_{n,m}(\mu_n)\le Cm^{-3/2}$ for $n_0\le m\le n$, which is \eqref{eq:mix}, by induction on $n-m$. For $m\ge n_0$, $\bar n=\floor{10m/9}\ge\frac{99999}{90000}m>m$, and the integer inequality $90000^3\,179^2<159^2\,99999^3$ gives $(\bar n/m)^{-3/2}<159/179$. If $n\le\bar n$, use \eqref{eq:one-step-mixing}. Otherwise $\operatorname{Av}_{n,m}=\operatorname{Av}_{n,m}\operatorname{Av}_{n,\bar n}$, and projectivity identifies $\Osc_{n,m}(\operatorname{Av}_{n,\bar n}\mu_n)$ with $\Osc_{\bar n,m}(\mu_{\bar n})$. The triangle inequality, the induction hypothesis for $(n,\bar n)$ and \eqref{eq:one-step-mixing} for $(\bar n,m)$ give
\[
\Osc_{n,m}(\mu_n)\le\Bigl(\frac{159}{20}+1\Bigr)B_{\mathrm{osc}}\bar n^{-3/2}<\frac{159}{20}B_{\mathrm{osc}}m^{-3/2}=Cm^{-3/2}.
\]
The finite arithmetic in Appendix~\ref{app:constants} gives \eqref{eq:Clog}, completing the proof of Proposition~\ref{prop:mix} from the Fourier bound.

\subsection{Transfer through a prefix-disjoint family}
Each generation of Section~\ref{sec:physical} appends one word from a family and then continues randomly; if the family captures almost all of the probability, the residue density after the appended word is close to the residue density of the unconditioned random word. For a finite family $\mathcal V$ of words write $\mass(\mathcal V)=\sum_{w\in\mathcal V}2^{-\deg(w)}$ for its \emph{mass} and call $1-\mass(\mathcal V)$ its \emph{deficit}. The family is \emph{prefix-disjoint} if no word is a proper prefix of another; then the events that $W$ begins with $w$ ($w\in\mathcal V$) are disjoint and $\mass(\mathcal V)\le1$. For $k\in\N$ and $Q\ge k+|w|$, write $\mathcal T_{w,Q}\rho_k=\mathcal T_w(\rho_k\circ\pr_{Q-|w|,k})$.

\begin{lemma}[Finite transfer]\label{lem:finite-transfer}
Let $d,k,m\in\N$, put $Q=d+k$, and suppose $k\ge n_0$ and $n_0\le m\le Q$. If $\mathcal V$ is a finite prefix-disjoint family whose words have length at most $d$, then
\[
\av Q{\Bigl|\sum_{w\in\mathcal V}\mathcal T_{w,Q}\rho_k-\rho_m\circ\pr_{Q,m}\Bigr|}\le\frac C{k^{3/2}}+\frac C{m^{3/2}}+1-\mass(\mathcal V).
\]
\end{lemma}
\begin{proof}
This is \cite[(3.7)]{Mazur} with $h=d$, $\ell=m$ and exponent $3/2$, in our normalization of $\rho$, and his proof applies verbatim: besides \eqref{eq:transfer-mean} and the disjointness of the events, it uses mixing only from the levels $Q-|w|\ge k$ and $Q$ down to the levels $k$ and $m$, where \eqref{eq:mix} holds since $k,m\ge n_0$.
\end{proof}

\section{The family of words}\label{sec:physical}
This section constructs the families $L_j$ and $L_{j,X}$ of Section~\ref{sec:outline} and the weighted densities $Z_j$ and $Z^\term_{j,X}$. The estimates on scales, first-crossing families and the histogram are proved in Sections~\ref{sec:est-families}--\ref{sec:est-histogram}.

\subsection{Scales and first crossings}
Generation $j$ appends to each word a \emph{block}, a word of length about the \emph{scale} $\delta_j$ whose overshoot (defined below) is at most the \emph{cap} $R_j$. The scales are $\delta_0=9$ and $\delta_{j+1}=\delta_j+e_{\delta_j}$, where
\[
e_\delta=\begin{cases}1,&9\le\delta\le12,\\\max\{i:\ceil{\delta/100}\le i\le\delta,\ 9\cdot3^\delta\cdot16^i<4^{\delta+1}\},&13\le\delta\le255,\\\ceil{\delta/100},&\delta\ge256.\end{cases}
\]
The increments make the sources of generation $j$ exceed $16^{\delta_j}$ (Lemma~\ref{lem_endpoint_large}). The finite maxima are nonempty, and with $g=101/100$ one has (Lemma~\ref{lem_scales_growth})
\begin{equation}\label{eq:scalegrowth}
g^j\delta_i\le\delta_{i+j},\qquad\delta_j<201g^j.
\end{equation}
The integer comparisons $101^{70}>2\cdot100^{70}$ and $101^{69}<2\cdot100^{69}$ give $1/70<\log_2g<1/69$. The caps are $R_j=16+\floor{j/32}$; they grow with $j$ so that the losses $2^{-(R_j+1)}$ decay geometrically (Lemma~\ref{lem_central_deficit_decay}).

For an integer scale $\delta\ge9$, put $\ell_\delta=\floor{3\delta/5}$, $d^{\mathrm{lo}}_\delta=\delta-\ell_\delta$, $d^{\mathrm{hi}}_\delta=\delta+\ell_\delta$, and, with $B_0(n)=\ceil{n\log_23}$, the \emph{shift radius} $r_\delta=2\ell_\delta-B_0(\ell_\delta)-2e_\delta$. For an integer shift $u$ define the \emph{line}
\[
H_{\delta,u}(d)=2\delta+B_0((d-\delta)_+)-B_0((\delta-d)_+)+u-r_\delta,
\]
where $x_+=\max(x,0)$. For $n\in\N$, $n\log_23\le B_0(n)<n\log_23+1$, and $B_0(n)=\bl(3^n)$ for $n\ge1$, where the \emph{bit length} $\bl(x)$ of $x\in\N$ is the least $k\in\N$ with $x<2^k$. Hence, for all integers $\delta\ge9$, $u$ and $d\ge0$,
\begin{equation}\label{eq:hist-barrier}
\bigl|H_{\delta,u}(d)-\bigl(2\delta+(d-\delta)\log_23+u-r_\delta\bigr)\bigr|<1,
\end{equation}
so $H_{\delta,r_\delta}$ is the line of Section~\ref{sec:outline} through $(\delta,2\delta)$ of slope $\log_23$, rounded to integers, and $H_{\delta,u}$ is that line raised by $u-r_\delta$. A word $w$ of length $d$ and degree $H_{\delta,u}(d)+o$ has $2^{\deg(w)}3^{-d}\approx(4/3)^\delta2^{u-r_\delta+o}$: by \eqref{eq:affine} it enlarges the source by a factor that does not depend on its length. The shift radius is chosen so that, for every shift $0\le u\le2r_\delta$, most random words cross the line at a length in $(d^{\mathrm{lo}}_\delta,d^{\mathrm{hi}}_\delta]$ (Lemma~\ref{lem_deficit_family}); blocks use $u=r_\delta$, the terminal extension all $u\in[0,2r_\delta]$.

\begin{lemma}[Shift radius]\label{lem_rb_bounds}
For $\delta\ge9$, $0\le r_\delta\le\frac{23}{100}\delta<\frac3{10}\delta$, with $r_\delta=0$ exactly for $9\le\delta\le13$. For $\delta\ge256$, $r_\delta>\frac{229}{1000}\delta-4$.
\end{lemma}
\begin{proof}
Since $B_0(\ell_\delta)\ge\ell_\delta\log_23\ge\frac{19}{12}\ell_\delta$ (as $3^{12}>2^{19}$), $\ell_\delta\le3\delta/5$ and $e_\delta\ge\delta/100$ (Lemma~\ref{lem_scales_growth}), $r_\delta\le\frac5{12}\cdot\frac{3\delta}5-\frac\delta{50}=\frac{23\delta}{100}$. For $\delta\ge256$, $B_0(\ell_\delta)<\ell_\delta\log_23+1$, $2-\log_23>\frac{83}{200}$ (Lemma~\ref{lem_s03_log3_bounds}(1)), $\ell_\delta\ge3\delta/5-1$ and $e_\delta\le\delta/100+1$ give
\[
r_\delta>\frac{83}{200}\Bigl(\frac{3\delta}5-1\Bigr)-1-\frac\delta{50}-2=\frac{229}{1000}\delta-\frac{683}{200},
\]
which is positive. For the $247$ integers $9\le\delta\le255$, $r_\delta$ is an exact integer computation from the definitions; it is nonnegative, and zero exactly for $9\le\delta\le13$.
\end{proof}

A word $w$ belongs to the \emph{first-crossing family} $\FC_u(\delta,R)$, where $R\in\N$, if $d^{\mathrm{lo}}_\delta<|w|\le d^{\mathrm{hi}}_\delta$, its partial degrees stay strictly below the line at lengths $d^{\mathrm{lo}}_\delta,\ldots,|w|-1$ (so $|w|$ is the \emph{first crossing}), and $H_{\delta,u}(|w|)\le\deg(w)\le H_{\delta,u}(|w|)+R$. Its \emph{overshoot} is $\deg(w)-H_{\delta,u}(|w|)\in[0,R]$. Such a family is finite and prefix-disjoint (Lemma~\ref{lem_first_crossing_prefix_disjoint}). The \emph{block family} $\FC(\delta,R)$ of Section~\ref{sec:outline}, from which the blocks are drawn, consists of the words of $\FC_{r_\delta}(\delta,R)$ that satisfy two further restrictions. For $\delta\ge256$, the \emph{length restriction} puts the length in $[\delta-\ell^{\mathrm c}_\delta,\delta+\ell^{\mathrm c}_\delta]$, where
\[
\ell^{\mathrm c}_\delta=\min\left(\ceil{\sqrt{32\delta\ceil{\log_2\delta}}},\ell_\delta-1\right),
\]
and $\ell^{\mathrm c}_\delta=\ell_\delta$ for $\delta<256$; it narrows the length spread for the histogram estimate \eqref{eq:histogram} and Lemma~\ref{lem_width_sum} at a deficit polynomially small in $\delta$ (Lemma~\ref{lem_deficit_central}). At $\delta=9,10,11$, the overshoot is at least $o_\delta(d)$, the least nonnegative integer $o$ with
\[
16\cdot3^d\cdot2^{3\delta}<(2^{3\delta}-1)\cdot2^{H_{\delta,r_\delta}(d)+o},
\]
and $o_\delta(d)=0$ for $\delta\ge12$; with the increments $e_\delta$, this makes sources grow fast enough at the first scales.

\begin{lemma}[Weight, offset and source of a first-crossing word]\label{lem_s05_block_weight}
Let $\delta\ge9$, $u$ and $R\ge0$ be integers, and let $w\in\FC_u(\delta,R)$. Then
\begin{enumerate}
\item $2^{-R-1+r_\delta-u}(3/4)^\delta<\wt(w)<2^{1+r_\delta-u}(3/4)^\delta$;
\item $0<\off(w)<2^\delta$;
\item if $w\in\FC(\delta,R)$, then $16^{e_\delta}\wt(w)<1-2^{-3\delta}$;
\item for every odd $n\ge16^\delta$,
\begin{equation}\label{eq:word-source}
2^{u-r_\delta-2}(4/3)^\delta n<\src(w,n)<2^{R+1+u-r_\delta}(4/3)^\delta n.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}
For (1), $\wt(w)=3^{|w|}2^{-\deg(w)}$ and $H_{\delta,u}(|w|)\le\deg(w)\le H_{\delta,u}(|w|)+R$; insert \eqref{eq:hist-barrier}. For (2), as in \cite[proof of Proposition~3.1]{Mazur}: every summand of $\off(w)$ is positive and, since the letters are at least $1$, the $i$th summand is at most $3^{i-1}2^{-i}$; as $|w|\le d^{\mathrm{hi}}_\delta\le8\delta/5$ and $(3/2)^8<2^5$, $0<\off(w)<(3/2)^{|w|}<2^\delta$.

For (3), $u=r_\delta$, and we distinguish four ranges. If $\delta\in\{9,10,11\}$, then $e_\delta=1$ and $\deg(w)\ge H_{\delta,r_\delta}(|w|)+o_\delta(|w|)$, so the definition of $o_\delta(|w|)$ gives $16\,\wt(w)\le16\cdot3^{|w|}\cdot2^{-H_{\delta,r_\delta}(|w|)-o_\delta(|w|)}<1-2^{-3\delta}$. If $\delta=12$, then $e_\delta=1$, $d^{\mathrm{lo}}_{12}=5$, $d^{\mathrm{hi}}_{12}=19$, and for $d=6,\ldots,19$ the values $H_{12,r_{12}}(d)$ are
\[
14,16,17,19,20,22,24,26,28,29,31,32,34,36.
\]
The fourteen exact integer comparisons $9\cdot3^d\cdot4^{12}\le16\cdot3^{12}\cdot2^{H_{12,r_{12}}(d)}$ hold (with equality at $d=10$). Hence $16\,\wt(w)\le\frac{256}9(3/4)^{12}=3^{10}/2^{16}<1-2^{-36}$. If $13\le\delta\le255$, then $e_\delta$ belongs to the set defining it, so $16^{e_\delta}<\frac49(4/3)^\delta$, and (1) gives $16^{e_\delta}\wt(w)<\frac89<1-2^{-3\delta}$. Finally let $\delta\ge256$, so $e_\delta=\ceil{\delta/100}\ge3$ and $\delta>100(e_\delta-1)$. The integer inequality $4^{100}>4096\cdot3^{100}$ says $(4/3)^{100}>16^3$, so $(4/3)^\delta>16^{3(e_\delta-1)}\ge4\cdot16^{e_\delta}$, and with (1), $16^{e_\delta}\wt(w)<\frac14(4/3)^\delta\cdot2(3/4)^\delta=\frac12$.

For (4), which is \cite[(3.5)]{Mazur} for general $u$, (2) gives $0<\off(w)<2^\delta\le n/2$, so $\wt(w)\src(w,n)=n-\off(w)\in(n/2,n)$; insert (1).
\end{proof}

\subsection{The selector and the family \texorpdfstring{$L_j$}{Lj}}
Consider the words $w=v_0\cdots v_{j-1}$ with $v_i\in\FC(\delta_i,R_i)$. Since each $\FC(\delta_i,R_i)$ is prefix-disjoint, such a word determines its \emph{blocks} $v_i$: if $v_0\cdots v_{j-1}w=v'_0\cdots v'_{j-1}w'$ with $v_i,v'_i\in\FC(\delta_i,R_i)$, then one of $v_0,v'_0$ is a prefix of the other, so $v_0=v'_0$, and cancelling and inducting gives $v_i=v'_i$ and $w=w'$. For $i\le j$ write $w_{<i}=v_0\cdots v_{i-1}$.

By \eqref{eq:concat}, the offset of a long word is dominated by its first few blocks. The \emph{five-block selector} bounds those offsets, which improves the histogram estimate \eqref{eq:histogram}. For a word $v$ and an integer $\mathfrak r$, let
\[
\operatorname{rd}_{\mathfrak r}(v)=2^{-\mathfrak r}\sum_{i=1}^{|v|}\ceil{2^{\mathfrak r}\cdot3^{i-1}\cdot2^{-\deg_i(v)}},
\]
the offset with each summand rounded up to a multiple of $2^{-\mathfrak r}$; rounding makes the selector a condition on integer bins, which Appendix~\ref{app:enumeration} enumerates exactly. The selector retains a word with first five blocks $v_0,\ldots,v_4$ if
\begin{align*}
\frac38&\le\operatorname{rd}_9(v_0)+\wt(v_0)\operatorname{rd}_6(v_1)\le\frac{19}{8},\\
\operatorname{rd}_6(v_1)&\le16,\quad\operatorname{rd}_3(v_2)\le16,\quad\operatorname{rd}_0(v_3)\le64,\quad\operatorname{rd}_{-3}(v_4)\le256.
\end{align*}
The \emph{selected words} form the set $\bar L_j$ of words $v_0\cdots v_{j-1}$ as above that pass the selector when $j\ge5$; $\bar L_0$ consists of the empty word. The selector involves only the first five blocks, so for $j\ge5$ every word of $\bar L_j$ extended by a word of $\FC(\delta_j,R_j)$ lies in $\bar L_{j+1}$, and every word of $\bar L_j$ has $\frac{85}{256}\le\off(w)\le\frac{637}{256}$ (Lemma~\ref{lem_s05_selected_offset}). Its retained fraction of the mass of all five-block words $v_0\cdots v_4$ is
\begin{equation}\label{eq:five}
\frac{17295610054959855512665734818947672447921469210720928769975}{32309186679205268427697461585468266195406151955473216860286}>\frac12,
\end{equation}
computed in Appendix~\ref{app:enumeration}.

\begin{lemma}[Selected mass]\label{lem_s05_tuple_mass}\label{lem_survival_product}
For every $j\ge5$, $\mass(\bar L_j)>2^{-23}$.
\end{lemma}
\begin{proof}
The degree is additive under concatenation, the families are finite, and the selector constrains only the first five blocks, so
\[
\mass(\bar L_j)=\mass(\bar L_5)\prod_{5\le i<j}\mass(\FC(\delta_i,R_i)),
\]
and $\mass(\bar L_5)$ is $\prod_{i<5}\mass(\FC(\delta_i,R_i))$ times the retained fraction \eqref{eq:five}, which exceeds $\frac12$. All factors lie in $[0,1]$ (Lemma~\ref{lem_first_crossing_prefix_disjoint}). For $j\le444$ the product $\prod_{i<j}\mass(\FC(\delta_i,R_i))$ is at least the product over $i\le443$, which exceeds $\frac{72}5\,2^{-25}>2^{-22}$ (Lemma~\ref{lem_fc_startup_product}); for $j>444$, Lemmas~\ref{lem_fc_startup_product} and~\ref{lem_fc_late_survival} bound it below by $\frac{72}5\,2^{-25}\cdot\frac59=2^{-22}$. Hence $\mass(\bar L_j)>\frac12\cdot2^{-22}$.
\end{proof}

For a positive odd integer $t$, let $L_j=L_j(t)$ be the set of words of $\bar L_j$ valid at $t$, and for $w\in L_j$ write $s_w=\src(w,t)$ for its source; the empty word has $s_w=t$. A prefix of a valid word is valid, so $w_{<i}\in L_i$ for $w\in L_j$ and $i\le j$; the block $v_i$ is valid at $s_{w_{<i}}$, with source $s_{w_{<i+1}}$ from there, so by \eqref{eq:affine}
\begin{equation}\label{eq:hist-step}
s_{w_{<i+1}}=\frac{s_{w_{<i}}-\off(v_i)}{\wt(v_i)}\qquad(i<j).
\end{equation}

\begin{lemma}[Growth of sources]\label{lem_endpoint_large}
Let $t\ge16^9$ be odd, $j\in\N$ and $w\in L_j$. For every $i<j$,
\[
s_{w_{<i}}\ge16^{\delta_i},\qquad16^{e_{\delta_i}}s_{w_{<i}}<s_{w_{<i+1}},\qquad\tfrac14(4/3)^{\delta_i}s_{w_{<i}}\le s_{w_{<i+1}}\le2^{R_i+1}(4/3)^{\delta_i}s_{w_{<i}}.
\]
In particular $s_w\ge16^{\delta_j}$.
\end{lemma}
\begin{proof}
We prove $s_{w_{<i}}\ge16^{\delta_i}$ by induction on $i$; the case $i=0$ is $t\ge16^9$. If $s_{w_{<i}}\ge16^{\delta_i}$, then Lemma~\ref{lem_s05_block_weight}(2) gives $0<\off(v_i)<2^{\delta_i}\le2^{-3\delta_i}s_{w_{<i}}$, and by \eqref{eq:hist-step} and Lemma~\ref{lem_s05_block_weight}(3),
\[
s_{w_{<i+1}}>\frac{(1-2^{-3\delta_i})s_{w_{<i}}}{\wt(v_i)}>16^{e_{\delta_i}}s_{w_{<i}}\ge16^{\delta_i+e_{\delta_i}}=16^{\delta_{i+1}}.
\]
The third inequality is \eqref{eq:word-source} with $u=r_{\delta_i}$ and $R=R_i$, and the induction up to $i=j$ gives $s_w\ge16^{\delta_j}$.
\end{proof}

The \emph{weighted density} is
\[
Z_j(t)=\sum_{w\in L_j}\wt(w)\rho_{k_j}(s_w),\qquad k_j=\floor{\delta_j/4}.
\]
By \eqref{eq:rho-meaning}, it weighs each word of $L_j$ by the total weight of its valid continuations of length $k_j$. The \emph{histogram estimate} (Proposition~\ref{lem_histogram_bound}) says that the weighted sources do not concentrate on few residues: for $j\ge10000$, $0\le q\le2\delta_j$, any $f:G_q\to\R$, and any subset of $L_j$,
\begin{equation}\label{eq:histogram}
\sum_w\wt(w)|f(s_w)|\le\frac{4075}{56}\frac{277}{128}j^{5/2}g^{j/2}\av q{|f|}.
\end{equation}

\subsection{The terminal extension}
To bring the sources into $[X,32X)$, each $w\in L_j$ receives a last word whose line is shifted according to $s_w$. Let $X>0$ be real. For an integer scale $\delta\ge9$ and a real $n>0$, set, as in \cite[Section~5]{Mazur},
\[
X^{\mathrm{lo}}_\delta(n)=\frac{(4/3)^\delta n}{4\cdot2^{r_\delta}},\qquad X^{\mathrm{hi}}_\delta(n)=2^{2r_\delta}X^{\mathrm{lo}}_\delta(n).
\]
By \eqref{eq:word-source}, every $v\in\FC_u(\delta,R)$ and every odd $n\ge16^\delta$ satisfy $2^uX^{\mathrm{lo}}_\delta(n)<\src(v,n)<2^{R+3+u}X^{\mathrm{lo}}_\delta(n)$. The real $X$ is \emph{reachable} at generation $j$ if $X^{\mathrm{lo}}_{\delta_j}(s_w)<X\le X^{\mathrm{hi}}_{\delta_j}(s_w)$ for every $w\in L_j$. For $w\in L_j$ let the \emph{terminal shift} $u_w$ be the least integer $u\ge0$ with $2^uX^{\mathrm{lo}}_{\delta_j}(s_w)\ge X$. If $X$ is reachable at $j$, then
\[
X\le2^{u_w}X^{\mathrm{lo}}_{\delta_j}(s_w)<2X,\qquad0\le u_w\le2r_{\delta_j}.
\]
For $R\in\N$, let $\FC^\term_X(w,\delta_j,R)$ be the words $v\in\FC_{u_w}(\delta_j,R)$ that are valid at $s_w$ and satisfy the \emph{mask}
\begin{equation}\label{eq:mask}
2|v|\le H_{\delta_j,u_w}(|v|)+\floor{\delta_j/8},
\end{equation}
Mazur's terminal restriction \cite[(5.2)]{Mazur}, which bounds the length of $v$ by its degree and is used only in Lemma~\ref{lem_orbit_large}(3). The \emph{terminal extension} $L_{j,X}$ consists of the words $wv$ with $w\in L_j$ and $v\in\FC^\term_X(w,\delta_j,R_j)$; its sources are at least $X$. Define the \emph{terminal weighted density} and the \emph{terminal weight}, which uses only the cap $1$ (the census in Section~\ref{sec:census} reduces the cap $R_j$ to $1$),
\begin{align*}
Z^\term_{j,X}(t)&=\sum_{w\in L_j}\wt(w)\sum_{v\in\FC^\term_X(w,\delta_j,R_j)}\wt(v)\rho_{k_j}(\src(v,s_w)),\\
\beta^\term_{j,X}(t)&=\sum_{w\in L_j}\wt(w)\sum_{v\in\FC^\term_X(w,\delta_j,1)}\wt(v),
\end{align*}
and let $\Src_{j,X}(t)$ be the set of sources $\src(v,s_w)$ of the words $wv$, $w\in L_j$, $v\in\FC^\term_X(w,\delta_j,1)$. If $t\ge16^9$ and $X$ is reachable at $j$, these sources lie in $[X,32X)$ \cite[Lemma~6.1]{Mazur}, since $s_w\ge16^{\delta_j}$ (Lemma~\ref{lem_endpoint_large}): the factor $2$ comes from the rounding of $u_w$, and the factor $16$ from overshoot at most $1$ and the slack $3$ above. If moreover $t$ is not $S$-periodic, \eqref{eq:charge} gives $\beta^\term_{j,X}(t)\le\beta_X(t)$ and
\begin{equation}\label{eq:charge-term}
X\beta^\term_{j,X}(t)\le t\,\#\Src_{j,X}(t).
\end{equation}

\section{Key estimates}\label{sec:census}
This section proves the key estimates of Section~\ref{sec:outline}: propagation (Proposition~\ref{prop:central-propagation}), the average at generation $j_0$ (Proposition~\ref{prop:average}), reachability (Proposition~\ref{prop:coverage}), the terminal error (Proposition~\ref{prop:terminal-errors}) and the census inequality \eqref{eq:census}. Apart from the census, which uses the bounded subdensity, the arguments are Mazur's \cite[Proposition~3.1, Sections~4--6 and~8]{Mazur}. Throughout, $t$ is a positive odd integer.

\subsection{Propagation}
Passing from $Z_j$ to $Z_{j+1}$ appends a block and moves the level from $k_j$ to $k_{j+1}$; the resulting transfer errors (Lemma~\ref{lem:finite-transfer}) are summed over $L_j$ by the histogram estimate. By Lemmas~\ref{lem_level_late_lower} and~\ref{lem_central_deficit_decay}, for $j\ge6000000$ the level $k_j$ is large enough for Proposition~\ref{prop:mix} and the block family has small deficit:
\begin{equation}\label{eq:physical-deficits}
k_j\ge\max(n_0,47g^j),\qquad1-\mass(\FC(\delta_j,R_j))\le(1024+2^{-16})g^{-3j/2}.
\end{equation}
Hence the three errors in Lemma~\ref{lem:finite-transfer} are small: every level $m\ge k_j$ has $Cm^{-3/2}\le C\,47^{-3/2}g^{-3j/2}$, and
\begin{equation}\label{eq:loss-bracket}
2C\cdot47^{-3/2}+2^{121}+1024+2^{-16}<\frac{28}{1000}(C+1),
\end{equation}
because $2000^2<27^2\cdot47^3$ gives $2\cdot47^{-3/2}<27/1000$, and $C>2^{87532}$ gives $C/1000>2^{121}+1024+2^{-16}$. The term $2^{121}$ covers the terminal deficit of Proposition~\ref{lem_full_loss_rate}. Put
\begin{equation}\label{eq:thresholds}
F=\frac92(C+1),\qquad j_0=\ceil{\frac{23+\log_2(3(C+1))+535/8}{\log_2g}}=6103914.
\end{equation}
By \eqref{eq:Clog}, $\log_2F<87534.2$. The generation $j_0$ is chosen so that the propagation errors from $j_0$ on are small compared with $2^{-23}$ (Lemma~\ref{lem_Nstar_tail}).

\begin{proposition}[Propagation]\label{prop:central-propagation}
For every $j\ge6000000$,
\begin{equation}\label{eq:variation}
|Z_{j+1}(t)-Z_j(t)|\le Fj^{5/2}g^{-j}.
\end{equation}
\end{proposition}
\begin{proof}
This is \cite[Proposition~3.1]{Mazur} with explicit constants. Put $Q=d^{\mathrm{hi}}_{\delta_j}+k_{j+1}$. Since $\delta_j\ge256$ (Lemma~\ref{lem_scales_growth}), $\delta_{j+1}\le(51/50)\delta_j$, so $Q\le(8/5+51/200)\delta_j<2\delta_j$. Define on $G_Q$
\[
\mathcal D_j=\sum_{v\in\FC(\delta_j,R_j)}\mathcal T_{v,Q}\rho_{k_{j+1}}-\rho_{k_j}\circ\pr_{Q,k_j}.
\]
Since $j\ge5$, the words of $L_{j+1}$ are exactly the words $wv$ with $w\in L_j$ and $v\in\FC(\delta_j,R_j)$ valid at $s_w$, so \eqref{eq:fiber-source} gives
\[
Z_{j+1}(t)-Z_j(t)=\sum_{w\in L_j}\wt(w)\mathcal D_j(s_w).
\]
Lemma~\ref{lem:finite-transfer} at the levels $k_{j+1},k_j\ge k_j$, \eqref{eq:physical-deficits} and \eqref{eq:loss-bracket} give $\av Q{|\mathcal D_j|}<\frac{28}{1000}(C+1)g^{-3j/2}$. Apply \eqref{eq:histogram}, using $j/2-3j/2=-j$ and $\frac{4075}{56}\frac{277}{128}\frac{28}{1000}=\frac{45151}{10240}<\frac92$.
\end{proof}

\begin{lemma}[Polynomial-geometric tails]\label{lem_moment_tail}
For real $0\le\mathfrak e\le9/2$ and integers $i\ge6000000$, the terms $j^{\mathfrak e}g^{-j}$ decrease in $j\ge i$, and
\[
\sum_{j\ge i}j^{\mathfrak e}g^{-j}<162\,i^{\mathfrak e}g^{-i}.
\]
\end{lemma}
\begin{proof}
For $j\ge i$ the ratio of consecutive terms is
\[
\Bigl(\frac{j+1}j\Bigr)^{\mathfrak e}g^{-1}\le\Bigl(\frac{6000001}{6000000}\Bigr)^{9/2}\frac{100}{101}<\frac{161}{162}.
\]
Squared, the last inequality is the integer comparison $6000001^9\cdot100^2\cdot162^2<161^2\cdot6000000^9\cdot101^2$. Hence the term at $j$ is at most $(161/162)^{j-i}i^{\mathfrak e}g^{-i}$, strictly for $j>i$, and the geometric series sums to $162\,i^{\mathfrak e}g^{-i}$.
\end{proof}

\begin{lemma}[Propagation tail]\label{lem_Nstar_tail}
$F\sum_{j\ge j_0}j^{5/2}g^{-j}<\frac54\cdot2^{-24}$. Consequently, if $Z_{j_0}(t)>2^{-23}$, then $Z_j(t)>\frac34\cdot2^{-24}\ge2^{-25}$ for every $j\ge j_0$.
\end{lemma}
\begin{proof}
By the ceiling in \eqref{eq:thresholds}, $j_0\log_2g\ge23+\log_2(3(C+1))+\frac{535}8$, so $g^{-j_0}\le2^{-23-535/8}/(3(C+1))$, and $F=\frac32\cdot3(C+1)$. Lemma~\ref{lem_moment_tail} with $\mathfrak e=5/2$ gives
\[
F\sum_{j\ge j_0}j^{5/2}g^{-j}<\frac32\cdot162\,j_0^{5/2}2^{-535/8}\cdot2^{-23}.
\]
Raising to the eighth power, $243\,j_0^{5/2}2^{-535/8}<5/8$ is the integer comparison $243^8j_0^{20}8^8<5^82^{535}$, so the bound is less than $\frac58\cdot2^{-23}=\frac54\cdot2^{-24}$. The consequence follows by summing \eqref{eq:variation} from $j_0$ to $j-1$.
\end{proof}

\subsection{The average at generation \texorpdfstring{$j_0$}{j0}}
At generation $j_0$ the words are short enough that $Z_{j_0}(t)$ depends only on $t$ modulo $3^{d_0}$, where $d_0=2^{87639}$, so its average over residues can be computed exactly. The comparison of lengths uses the exact numerical enclosure of Appendix~\ref{app:constants},
\begin{equation}\label{eq:endpoint}
87639.473776839504865<16+j_0\log_2g<87639.473776839504878.
\end{equation}

\begin{lemma}[Period of $Z_{j_0}$]\label{lem_ph_conductor}\label{lem_s04_qstar_ge}\label{lem_ph_h0_large}
Every word of $\bar L_{j_0}$ has length at most $d_0-k_{j_0}$. Consequently, as a function of $t$, $Z_{j_0}(t)$ depends only on $t\bmod3^{d_0}$, and on $G_{d_0}$
\[
Z_{j_0}=\sum_{w\in\bar L_{j_0}}\mathcal T_{w,d_0}\rho_{k_{j_0}}.
\]
\end{lemma}
\begin{proof}
Block $j$ has length at most $d^{\mathrm{hi}}_{\delta_j}\le\frac85\delta_j$. By \eqref{eq:scalegrowth}, $\delta_j\le g^{-(j_0-j)}\delta_{j_0}$ for $j<j_0$, so $\sum_{j<j_0}\delta_j<100\,\delta_{j_0}$, and with $k_{j_0}\le\delta_{j_0}/4$,
\[
k_{j_0}+\sum_{j<j_0}d^{\mathrm{hi}}_{\delta_j}<\Bigl(\frac14+\frac85\cdot100\Bigr)\delta_{j_0}=\frac{641}4\delta_{j_0}<\frac{641\cdot201}4g^{j_0}<2^{15}g^{j_0},
\]
by $\delta_{j_0}<201g^{j_0}$ and $641\cdot201=128841<2^{17}$. By \eqref{eq:endpoint}, $15+j_0\log_2g<87638.48<\log_2d_0$. For $w\in\bar L_{j_0}$, \eqref{eq:fiber-source} shows that the term of $w$ in the displayed sum, at a positive odd $t$, is $\wt(w)\rho_{k_{j_0}}(s_w)$ if $w$ is valid at $t$ and zero otherwise, so the sum is $Z_{j_0}(t)$.
\end{proof}

\begin{proposition}[Average at generation $j_0$]\label{prop:average}
The average of $Z_{j_0}$ over the units of $G_{d_0}$ equals $\frac32\mass(\bar L_{j_0})$, and in particular exceeds $2^{-23}$.
\end{proposition}
\begin{proof}
We follow \cite[Proposition~4.2]{Mazur}. By Lemma~\ref{lem_ph_conductor}, \eqref{eq:transfer-mean} and $\av k{\rho_k}=1$, the average of $Z_{j_0}$ over all of $G_{d_0}$ is $\sum_{w\in\bar L_{j_0}}2^{-\deg(w)}=\mass(\bar L_{j_0})$. Every $w\in\bar L_{j_0}$ is nonempty, so the image of $\iota_w$ consists of units: its offset is $2^{-a_1}$ modulo $3$ and its linear coefficient $\wt(w)$ vanishes modulo $3$. Thus $Z_{j_0}$ vanishes off the units, which are $2/3$ of $G_{d_0}$, and Lemma~\ref{lem_s05_tuple_mass} gives $\frac32\mass(\bar L_{j_0})>2^{-23}$.
\end{proof}

\subsection{Reachability}
Let $J(t)=2\cdot10^{10}+200\log_2t$, a generation large enough for Lemmas~\ref{lem_overlap} and~\ref{lem_ph_m_log}. The \emph{generation-$j$ ranges} $(X^{\mathrm{lo}}_{\delta_j}(s_w),X^{\mathrm{hi}}_{\delta_j}(s_w)]$, $w\in L_j$, with \emph{lower ends} $X^{\mathrm{lo}}_{\delta_j}(s_w)$ and \emph{upper ends} $X^{\mathrm{hi}}_{\delta_j}(s_w)$, intersect in the set of $X$ reachable at $j$; consecutive intersections overlap and move to infinity. The overlap holds because the sources of $L_j$ differ by a factor at most $2^{R_\Sigma}$ (Lemma~\ref{lem_endpoint_ratio}), where $R_\Sigma=\sum_{i<j}(R_i+3)$, suppressing $j$. Since $R_i+3=19+\floor{i/32}$, $\frac1{56}-\frac1{64}=\frac1{448}$ and $448(19j+1)<j^2$ for $j\ge10000$,
\begin{equation}\label{eq:Rsum}
R_\Sigma+1\le19j+\frac{j(j-1)}{64}+1<\frac{j^2}{56}\qquad(j\ge10000).
\end{equation}

\begin{lemma}[Consecutive generations overlap]\label{lem_overlap}
Let $t\ge16^9$ be odd and $j\ge2\cdot10^{10}$. For all $w'\in L_j$ and $w\in L_{j+1}$, $X^{\mathrm{lo}}_{\delta_{j+1}}(s_w)<X^{\mathrm{hi}}_{\delta_j}(s_{w'})$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem_scales_growth}, $\delta_j\ge8000j^2$ and $\delta_j\ge256$, so $e_{\delta_j}\le\delta_j/100+1$ and $\delta_j\le\delta_{j+1}\le\frac{101}{100}\delta_j+1$. Lemmas~\ref{lem_endpoint_large} and~\ref{lem_endpoint_ratio} and \eqref{eq:Rsum} give $s_w\le2^{R_j+1}(4/3)^{\delta_j}s_{w_{<j}}\le2^{R_j+1+R_\Sigma}(4/3)^{\delta_j}s_{w'}$. Since $X^{\mathrm{hi}}_{\delta_j}(s_{w'})=2^{r_{\delta_j}}(4/3)^{\delta_j}s_{w'}/4$, it suffices that $(4/3)^{\delta_{j+1}}2^{R_j+1+R_\Sigma}<2^{r_{\delta_j}+r_{\delta_{j+1}}}$. The inequality $2^{19}\le3^{12}$ gives $(4/3)^\delta\le2^{5\delta/12}$, so $(4/3)^{\delta_{j+1}}\le2^{505\delta_j/1200+5/12}$. By Lemma~\ref{lem_rb_bounds}, $r_{\delta_j}+r_{\delta_{j+1}}\ge\frac{458}{1000}\delta_j-8$. As $\frac{458}{1000}-\frac{505}{1200}=\frac{223}{6000}>\frac{37}{1000}$, it suffices that $R_j+R_\Sigma+10<\frac{37}{1000}\delta_j$, and indeed $R_j+R_\Sigma+10<16+\frac j{32}+\frac{j^2}{56}+10<296j^2\le\frac{37}{1000}\delta_j$.
\end{proof}

\begin{proposition}[Reachability]\label{prop:coverage}
Let $t\ge16^9$ be odd. Every real $X\ge2^{2^{2^{29}}t^3}$ is reachable at some generation $j\ge J(t)$.
\end{proposition}
\begin{proof}
We follow \cite[Lemma~6.2]{Mazur} in the effective form of \cite[Section~8.5]{Mazur}. Put $\operatorname{gr}(\delta)=2\delta-\floor{19\delta/12}$, $\mathfrak l_0=0$ and $\mathfrak l_{j+1}=\mathfrak l_j+R_j+1+\operatorname{gr}(\delta_j)$, Mazur's $t(u)$ and $E(j)$ \cite[Section~8.2]{Mazur} with this paper's caps and scales. The inequality $2^{19}\le3^{12}$ gives $(4/3)^\delta\le2^{\operatorname{gr}(\delta)}$, so Lemma~\ref{lem_endpoint_large} and induction give $s_w\le2^{\mathfrak l_j}t$ for $w\in L_j$. Let $J'=\ceil{J(t)}$ and $\delta=\delta_{J'}$. By Lemma~\ref{lem_rb_bounds} and $\operatorname{gr}(\delta)\le\frac{5\delta}{12}+1$, $X^{\mathrm{hi}}_\delta(n)=2^{r_\delta}(4/3)^\delta n/4\le2^{r_\delta+\operatorname{gr}(\delta)-2}n$ with $r_\delta+\operatorname{gr}(\delta)-2\le\frac{97\delta}{150}-1<\floor{2\delta/3}$. Hence every upper end at generation $J'$ is at most $2^Bt$, where $B=\mathfrak l_{J'}+\floor{2\delta_{J'}/3}$. Using $R_j+1=17+\floor{j/32}$, $\operatorname{gr}(\delta)\le2\delta$ and \eqref{eq:scalegrowth},
\[
B<17J'+\frac{{J'}^2}{64}+2\sum_{j<J'}\delta_j+\frac23\delta_{J'}<17J'+\frac{{J'}^2}{64}+40334g^{J'}.
\]
Since $\log_2g<1/69$ and $J'<J(t)+1$, $g^{J'}<2^{(2\cdot10^{10}+1)/69}t^{200/69}<2^{2.9\cdot10^8}t^3$; also $17J'+{J'}^2/64<{J'}^2<2^{72}t^2$. Hence $B+\log_2t<2^{2.9\cdot10^8+17}t^3<2^{2^{29}}t^3$, and every upper end at generation $J'$ is below $X$.

For $i\ge J'$ and $w\in L_i$, Lemma~\ref{lem_endpoint_large} gives $s_w\ge16^{\delta_i}$, so $X^{\mathrm{hi}}_{\delta_i}(s_w)\ge2^{\delta_i}\to\infty$. Let $j\ge J'$ be least such that every upper end at generation $j$ is at least $X$. If $j=J'$, then $L_{J'}$ is empty and $X$ is reachable vacuously. Otherwise some upper end at generation $j-1\ge J'\ge2\cdot10^{10}$ is below $X$, and by Lemma~\ref{lem_overlap} every lower end at generation $j$ is below it, hence below $X$.
\end{proof}

\subsection{The terminal error}
The terminal extension appends to $w$ words of $\FC_{u_w}(\delta_j,R_j)$, a family that depends on $w$ only through its terminal shift; grouping $L_j$ by shift reduces the terminal error to transfer errors, at the cost of a factor equal to the number of shifts.

\begin{lemma}[Few terminal shifts]\label{lem_shift_count}
Let $t\ge16^9$ be odd, $j\ge10000$ and $X>0$. Then the terminal shifts $u_w$, $w\in L_j$, take fewer than $j^2/56$ values.
\end{lemma}
\begin{proof}
We follow \cite[proof of Proposition~5.2]{Mazur}. Assume $L_j\ne\emptyset$ and let $w_{\min}\in L_j$ have the least source. Since $X^{\mathrm{lo}}_{\delta_j}(n)$ is positive and linear in $n$, $u_w$ is nonincreasing in $s_w$, so $u_w\le u_{w_{\min}}$. By Lemma~\ref{lem_endpoint_ratio}, $s_w\le2^{R_\Sigma}s_{w_{\min}}$, so $2^{u_w+R_\Sigma}X^{\mathrm{lo}}_{\delta_j}(s_{w_{\min}})\ge X$ and $u_{w_{\min}}\le u_w+R_\Sigma$. Thus all shifts lie in a set of at most $R_\Sigma+1<j^2/56$ integers, by \eqref{eq:Rsum}.
\end{proof}

\begin{proposition}[Terminal error]\label{prop:terminal-errors}
Let $t\ge16^9$ be odd. For $j\ge6000000$ and every $X$ reachable at $j$,
\[
|Z^\term_{j,X}(t)-Z_j(t)|\le2Fj^{9/2}g^{-j},
\]
and the right side is less than $2^{-1500}$ for $j\ge10^7$.
\end{proposition}
\begin{proof}
We follow \cite[Proposition~5.2]{Mazur}. Put $Q=d^{\mathrm{hi}}_{\delta_j}+k_j<2\delta_j$, and group the words of $L_j$ by their terminal shift; by Lemma~\ref{lem_shift_count} there are fewer than $j^2/56$ groups. For each shift $0\le u\le2r_{\delta_j}$, Proposition~\ref{lem_full_loss_rate} gives
\begin{gather*}
1-\mass(\FC_u(\delta_j,R_j))\le(2^{121}+2^{-16})g^{-3j/2},\\
\mass(\{v\in\FC_u(\delta_j,R_j):v\text{ fails \eqref{eq:mask}}\})\le2^{95}g^{-3j/2}.
\end{gather*}
For each shift define on $G_Q$
\[
\mathcal D^\term_u=\sum_{v\in\FC_u(\delta_j,R_j)}\mathcal T_{v,Q}\rho_{k_j}-\rho_{k_j}\circ\pr_{Q,k_j},\qquad\tilde{\mathcal D}^\term_u=\sum_{\substack{v\in\FC_u(\delta_j,R_j)\\v\text{ fails \eqref{eq:mask}}}}\mathcal T_{v,Q}\rho_{k_j}.
\]
By \eqref{eq:fiber-source}, $Z^\term_{j,X}(t)-Z_j(t)=\sum_u\sum_{w:u_w=u}\wt(w)\bigl(\mathcal D^\term_u-\tilde{\mathcal D}^\term_u\bigr)(s_w)$. Lemma~\ref{lem:finite-transfer} and \eqref{eq:loss-bracket} give $\av Q{|\mathcal D^\term_u|}<\frac{28}{1000}(C+1)g^{-3j/2}$; also $\tilde{\mathcal D}^\term_u\ge0$ and, by \eqref{eq:transfer-mean}, its mean is the mass of the words failing the mask, so it obeys the same bound. Apply \eqref{eq:histogram} to each group and sum over fewer than $j^2/56$ shifts: the power of $j$ is $5/2+2=9/2$, and each of the two parts is at most $\frac{45151}{10240}(C+1)j^{9/2}g^{-j}/56<Fj^{9/2}g^{-j}$.

By Lemma~\ref{lem_moment_tail}, $j^{9/2}g^{-j}$ decreases for $j\ge6000000$. At $j=10^7$, $2F<2^{87536}$, $j^{9/2}=10^{63/2}<2^{105}$ and $g^{-10^7}<2^{-10^7/70}<2^{-142857}$ give $2Fj^{9/2}g^{-j}<2^{-55216}$.
\end{proof}

\subsection{The census}
The terminal weighted density weighs each source by $\rho_{k_j}$, which might be large on a few residues; the census bounds it by the terminal weight, splitting $\rho_{k_j}$ into the bounded subdensity of a lower level and a remainder $\Phi$ of small average, on which distinct sources cannot concentrate. As in \cite[Proposition~5.3]{Mazur}, it is applied through the maps $\lift_i(s)=4^is+(4^i-1)/3$, $i\ge0$.

\begin{proposition}[Census with a bounded subdensity]\label{prop:census}
Let $0\le m\le q$ be integers and let $X\ge3^q$ be real. Let $\mathcal S$ be a set of integers in $[X,32X)$ with nonnegative weights $\upsilon_s\le t/X$. Suppose $\mathfrak w,\Phi\ge0$ on $G_q$, $0\le\underline{\mathfrak w}\le U$ on $G_m$, and $\mathfrak w\le\underline{\mathfrak w}\circ\pr_{q,m}+\Phi$. Then
\begin{equation}\label{eq:census-unfold}
\sum_{s\in\mathcal S}\upsilon_s\sum_{i\ge0}4^{-i}\mathfrak w(\lift_i(s))\le\frac43U\sum_{s\in\mathcal S}\upsilon_s+44t\av q\Phi.
\end{equation}
\end{proposition}
\begin{proof}
The contribution from $\underline{\mathfrak w}$ is at most $U\sum_s\upsilon_s\sum_{i\ge0}4^{-i}=\frac43U\sum_s\upsilon_s$. For each fixed $i$, $\lift_i$ is a permutation of $G_q$ because $4^i$ is a unit. Each residue class contains at most $31X/3^q+1\le33X/3^q$ integers of $[X,32X)$, since $X\ge3^q$. As the elements of $\mathcal S$ are distinct and $\Phi\ge0$,
\[
\sum_{s\in\mathcal S}\upsilon_s\Phi(\lift_i(s))\le\frac tX\frac{33X}{3^q}\sum_{y\in G_q}\Phi(y)=33t\av q\Phi.
\]
Summing against $4^{-i}$ gives $44t\av q\Phi$. This counting is that of \cite[Proposition~5.3]{Mazur}.
\end{proof}

Let $t\ge16^9$ be odd and not $S$-periodic, and let $X$ be reachable at $j$. A word $v\in\FC^\term_X(w,\delta_j,R_j)$ with overshoot $o$ gives, after subtracting $2i$ from its last letter, $i=\floor{o/2}$, a word $v'\in\FC^\term_X(w,\delta_j,1)$ with the same length, the same earlier letters and the mask, and with $\src(v,s_w)=\lift_i(\src(v',s_w))$ and $\wt(v)=4^{-i}\wt(v')$, since adding $2$ to the last letter turns a source $s'$ into $4s'+1$ \cite[proof of Proposition~5.3]{Mazur}; and $(v',i)$ determines $v$. So $Z^\term_{j,X}(t)$ is at most the left side of \eqref{eq:census-unfold} for $\mathfrak w=\rho_{k_j}$, $\mathcal S=\Src_{j,X}(t)$ and the weights $\upsilon_s=\wt(wv')\le t/X$ of the sources $s=\src(v',s_w)$, whose total is $\beta^\term_{j,X}(t)$.

If $L_j$ is empty, $Z^\term_{j,X}(t)=0$ and \eqref{eq:census} below is trivial; otherwise reachability, $s_w\ge16^{\delta_j}$ and $r_{\delta_j}\le3\delta_j/10$ give $X>16^{\delta_j}/(4\cdot3^{\delta_j})>3^{k_j}$. For an integer $m$ with $n_0\le m\le k_j$, apply Proposition~\ref{prop:census} with $q=k_j$, $\underline{\mathfrak w}=\underline\rho_m$, $U=8m^\alpha$ and $\Phi=|\rho_{k_j}-\rho_m\circ\pr_{k_j,m}|+(\rho_m-\underline\rho_m)\circ\pr_{k_j,m}$, which is nonnegative with $\rho_{k_j}\le\underline\rho_m\circ\pr_{k_j,m}+\Phi$. By \eqref{eq:mix} and \eqref{eq:subdensity}, $\av{k_j}\Phi\le(C+1)m^{-3/2}$, so
\begin{equation}\label{eq:census}
Z^\term_{j,X}(t)\le\frac{32}3m^\alpha\beta^\term_{j,X}(t)+44(C+1)m^{-3/2}t.
\end{equation}

\section{Proof of the main theorem}\label{sec:finish}
This section combines the key estimates into a count for a target $t$ satisfying the hypotheses of Section~\ref{sec:outline} (Proposition~\ref{prop:core}), chooses a seed that satisfies them (Proposition~\ref{prop:seed}), and deduces Theorem~\ref{thm:main}, using the step bounds of Section~\ref{sec:steps}.

\subsection{The count under the hypotheses}
The census inequality \eqref{eq:census} holds at every level $m$ with $n_0\le m\le k_j$; we take $m$ large enough that its error term $44(C+1)m^{-3/2}t$ is negligible.

\begin{lemma}[The census level is available]\label{lem_ph_m_log}\label{lem_m_gate}\label{lem_m_available}\label{lem_mixing_error_small}\label{lem_s04_J_le}
Let $t\ge1$ and $m=\ceil{2^{80000}t^{2/3}}$. Then $m\ge n_0$, $m^{3/2}\ge2^{120000}t$, $m<2^{80001}t^{2/3}$, and $m\le k_j$ for every integer $j\ge J(t)$.
\end{lemma}
\begin{proof}
The first three claims are immediate, as $2^{80000}t^{2/3}\ge1$. Let $j\ge J(t)$. Then $j\ge200\log_2m$, as $200\log_2m<200(80001+\log_2t)\le J(t)$, so $\floor{j/200}\ge\floor{\log_2m}\ge\ceil{\log_2m}-1$. Bernoulli's inequality gives $g^{200}\ge3>2$, so \eqref{eq:scalegrowth} yields $\delta_j\ge9g^j\ge9\cdot2^{\floor{j/200}}\ge9\cdot2^{\ceil{\log_2m}-1}\ge\frac92m$. Hence $m\le\delta_j/4$, and $m\le k_j$ because $m$ is an integer.
\end{proof}

\begin{proposition}[Count under the hypotheses]\label{prop:core}\label{prop:unmarked}
Let $t\ge16^9$ be odd and not $S$-periodic, with $Z_{j_0}(t)>2^{-23}$. Then every real $X\ge2^{2^{2^{29}}t^3}$ is reachable at some generation $j\ge J(t)$, and at every such $j$
\begin{equation}\label{eq:unmarked}
\beta^\term_{j,X}(t)>2^{-160071}t^{-4097/3072},\qquad\#\Src_{j,X}(t)>2^{-160071}t^{-7169/3072}X.
\end{equation}
\end{proposition}
\begin{proof}
Proposition~\ref{prop:coverage} gives $j$. As $j\ge J(t)>10^7>j_0$, Lemma~\ref{lem_Nstar_tail} gives $Z_j(t)>\frac34\cdot2^{-24}$, and Proposition~\ref{prop:terminal-errors} gives $Z^\term_{j,X}(t)>\frac34\cdot2^{-24}-2^{-1500}>2^{-26}$. Take $m$ as in Lemma~\ref{lem_ph_m_log}, so \eqref{eq:census} applies and its last term is at most $44(C+1)2^{-120000}<2^{-32462}$. Hence
\[
\beta^\term_{j,X}(t)>\frac3{32}(2^{-26}-2^{-32462})m^{-\alpha}>2^{-29.42}\cdot2^{-80001\alpha}t^{-2\alpha/3}.
\]
Here $80001\alpha=327764097/2048<160041.07$ and $2\alpha/3=4097/3072$, which gives the first bound; the second follows by \eqref{eq:charge-term}.
\end{proof}

\subsection{Choosing a seed}
A general target need not satisfy the hypotheses of Proposition~\ref{prop:core}, so we apply it to a \emph{seed} $t_1$ that does, chosen among the integers that reach the target in one odd step. Let $t$ be a positive integer with $3\nmid t$. As $2^{38}=4^{19}\equiv1\pmod3$, exactly one $a^\circ\in\{38,39\}$ has $2^{a^\circ}t\equiv1\pmod3$, and the \emph{candidate seeds} are the integers $(2^at-1)/3$ with $a\equiv a^\circ\pmod2$ and $a\ge a^\circ$. They increase with $a$, and $3\cdot(2^at-1)/3+1=2^at$, so a candidate reaches $t$ after one odd step and $a$ halvings.

\begin{lemma}[Candidate residues]\label{lem_repunit_residues}
Let $t'$ be an integer with $t'\equiv1\pmod3$, and let $q,i_0\in\N$. Then the integers $(4^it'-1)/3$, $i_0\le i<i_0+3^q$, meet every residue class modulo $3^q$ exactly once.
\end{lemma}
\begin{proof}
For $t'=1$ this is \cite[Lemma~4.1]{Mazur}. First, $\ord_3(4^e-1)=1+\ord_3(e)$ for every integer $e\ge1$, by induction on $\ord_3(e)$. If $3\nmid e$, then $(4^e-1)/3=\sum_{i<e}4^i\equiv e\not\equiv0\pmod3$. If $e=3e'$, put $x=4^{e'}=1+3z$; then $4^e-1=(x-1)(x^2+x+1)$ with $x^2+x+1=3+9z+9z^2$, so $\ord_3(4^e-1)=\ord_3(4^{e'}-1)+1$. Now let $i_0\le i<i'<i_0+3^q$ and $e=i'-i$, so $\ord_3(e)<q$. The difference of the two integers is $4^it'(4^e-1)/3$, and since $3\nmid4^it'$ its $3$-adic order is $\ord_3(e)<q$. So the $3^q$ integers are pairwise incongruent modulo $3^q$.
\end{proof}

\begin{proposition}[The seed]\label{prop:seed}
Let $t$ be a positive integer with $3\nmid t$. There is a candidate seed $t_1=(2^at-1)/3$ with $a<38+4\cdot3^{d_0}$ that is odd, not $S$-periodic, satisfies
\[
2^{36}t<t_1<2^{36+4\cdot3^{d_0}}t\qquad\text{and}\qquad Z_{j_0}(t_1)>2^{-23},
\]
and reaches $t$ in $1+a$ ordinary steps; $S(t_1)$ is the odd part of $t$.
\end{proposition}
\begin{proof}
By Proposition~\ref{prop:average} there is a unit $y_0\in G_{d_0}$ with $Z_{j_0}(y_0)>2^{-23}$. Apply Lemma~\ref{lem_repunit_residues} with $t'=2^{a^\circ}t$ and $q=d_0$, once with $i_0=0$ and once with $i_0=3^{d_0}$: this gives two candidate seeds $(4^it'-1)/3$, $i<2\cdot3^{d_0}$, congruent to $y_0$ modulo $3^{d_0}$, with $a=a^\circ+2i\le37+4\cdot3^{d_0}$. Each is odd, as $2^at-1$ is odd, satisfies $Z_{j_0}=Z_{j_0}(y_0)$ by Lemma~\ref{lem_ph_conductor}, and lies between $(2^{38}t-1)/3>2^{36}t$ and $2^at/3<2^{36+4\cdot3^{d_0}}t$. Since $3t_1+1=2^at$, $S(t_1)$ is the odd part of $t$, so the two candidates have the same image under $S$, and by Lemma~\ref{lem_syr_periodic_fiber}(1) at most one of them is $S$-periodic.
\end{proof}

\subsection{General targets}\label{sec:tower}
Every integer that reaches the seed reaches $t$, so Proposition~\ref{prop:core} for the seed counts integers reaching $t$. Let $t$ be a positive integer with $3\nmid t$, and fix $t_1$ as in Proposition~\ref{prop:seed}. Put the \emph{internal density} and the \emph{internal cutoff}
\begin{equation}\label{eq:internal}
\cint=2^{-160076}t_1^{-7169/3072},\qquad\Lint=32\cdot2^{2^{2^{29}}t_1^3}.
\end{equation}

\begin{proposition}\label{prop:internal}
For every real $X\ge\Lint$, at least $\cint X$ odd $n<X$ reach $t$ within $\floor{10.46\log(n/t)}$ ordinary steps and reach the odd part of $t$ within $\floor{3.4881\log(n/t)}$ accelerated steps.
\end{proposition}
\begin{proof}
The seed $t_1$ is odd, not $S$-periodic, at least $2^{36}\ge16^9$, and has $Z_{j_0}(t_1)>2^{-23}$, so Proposition~\ref{prop:core} applies to it at $X/32\ge2^{2^{2^{29}}t_1^3}$: at some $j\ge J(t_1)$ with $X/32$ reachable, $\#\Src_{j,X/32}(t_1)>2^{-160071}t_1^{-7169/3072}X/32=\cint X$. Every source in $\Src_{j,X/32}(t_1)$ is odd, lies in $[X/32,X)$, and satisfies both step bounds by Proposition~\ref{prop:clocks}.
\end{proof}

The theorem's constants are rounded versions of $\cint$ and $\Lint$:
\begin{equation}\label{eq:bridge}
1/\cint<t^4D_A,\qquad\Lint<2^{t^3D_A}.
\end{equation}
Indeed $\log_2D_A=2^{2^{87640}}=4^{d_0}$, and $(4/3)^{d_0}\ge(4/3)^{2^{20}}>2^{10^5}$ gives $2^{10^5}\cdot3^{d_0}<4^{d_0}$. With $\log_2(t_1/t)<36+4\cdot3^{d_0}$ (Proposition~\ref{prop:seed}) and $t\ge1$,
\[
\log_2(1/\cint)<160076+\tfrac{7169}{3072}(36+4\cdot3^{d_0})+4\log_2t<16\cdot3^{d_0}+4\log_2t<4^{d_0}+\log_2t^4,
\]
and $\log_2\Lint=5+2^{2^{29}}t_1^3<t^3(5+2^{2^{29}+108+12\cdot3^{d_0}})<t^3\cdot2^{4^{d_0}}=t^3D_A$, since $2^{29}+109+12\cdot3^{d_0}<4^{d_0}$.

\begin{proof}[Proof of Theorem~\ref{thm:main}]
For $X\ge X_0(t)=2^{t^3D_A}$, \eqref{eq:bridge} gives $X>\Lint$ and $\cint>c(t)$, so Proposition~\ref{prop:internal} gives at least $\cint X\ge c(t)X$ odd $n<X$ with both step bounds.
\end{proof}

\section{The number of steps}\label{sec:steps}
This section proves the step bounds used in Proposition~\ref{prop:internal} (Proposition~\ref{prop:clocks}), following \cite[Section~7]{Mazur}: the steps beyond the heuristic $\log(s/t)/\log(4/3)$ of Section~\ref{sec:outline} are paid by the deviation of the degree from twice the length, which the lines, caps and the mask bound, and by the steps from the seed to the target.

Throughout, $t$ and $t_1$ are as in Section~\ref{sec:tower}. For a generation $j$ define, suppressing $j$,
\[
\delta_\Sigma=\sum_{i\le j}\delta_i,\qquad\ell^{\mathrm c}_\Sigma=\sum_{i<j}\ell^{\mathrm c}_{\delta_i},\qquad c_{\mathrm{step}}=\log\Bigl(3+\frac1{4096}\Bigr),\qquad c_{\mathrm{src}}=\frac{28559}{100000}.
\]
The first three are Mazur's $B_*$, $W_*$ (with $\ell^{\mathrm c}_\delta$ in place of his width) and $h=\log(12289/4096)$; $c_{\mathrm{step}}$ bounds each step $\log(3+1/y)$ of an inverse trajectory with values $y\ge4096$, and $c_{\mathrm{src}}$ is the rate in Lemma~\ref{lem_log_source_lower}. We use $(2-\log_23)\log2=\log(4/3)$ and
\begin{equation}\label{eq:stop-logs}
\frac{287682}{10^6}<\log\frac43,\qquad\log2<\frac{347}{500}<\frac7{10},\qquad c_{\mathrm{step}}-\log3<\frac1{12288};
\end{equation}
the first two are Lemma~\ref{lem_s03_log3_bounds}(3), and the last is $\log(1+1/12288)<1/12288$.

\begin{lemma}[Scale sums]\label{lem_width_sum}
Let $j\ge2\cdot10^{10}$. Then (1) $\delta_j\ge2^{40}j$, and hence $2^{22}(j+1)\le\delta_\Sigma$; (2) $\delta_j\le9+\delta_\Sigma/100+2j$; and (3) $9000\,\ell^{\mathrm c}_\Sigma\le\delta_\Sigma$.
\end{lemma}
\begin{proof}
These are this paper's forms of Mazur's bounds $n+1\le B_*/32^5$, $100b_n\le100b_0+B_*$ and \cite[(7.1)]{Mazur}. By Lemma~\ref{lem_scales_growth}, $\delta_i\ge256$, hence $e_{\delta_i}=\ceil{\delta_i/100}$, for $i\ge42$, $2^{40}<\delta_{2254}=1100658881854<2^{41}$, and $\delta_j\ge2^{40}j$; then $2^{22}(j+1)\le2^{23}j\le\delta_j\le\delta_\Sigma$, proving (1). For (2), $e_{\delta_i}<\delta_i/100+1$ for $i\ge42$, so
\[
\delta_j=256+\sum_{42\le i<j}e_{\delta_i}\le256+\frac{\delta_\Sigma}{100}+(j-42)\le9+\frac{\delta_\Sigma}{100}+2j.
\]
For (3), $\ell^{\mathrm c}_\delta\le\ell_\delta\le\delta$ for every $\delta\ge9$. Let $\delta\ge2^{40}$. Then $40\le\floor{\log_2\delta}$, $\ceil{\log_2\delta}\le\floor{\log_2\delta}+1$ and $2^{\floor{\log_2\delta}}\le\delta$, and $(y+1)/2^y$ is nonincreasing in $y\ge1$, so
\[
\frac{\ell^{\mathrm c}_\delta}\delta<\sqrt{\frac{32\ceil{\log_2\delta}}\delta}+\frac1\delta\le\sqrt{\frac{32(\floor{\log_2\delta}+1)}{2^{\floor{\log_2\delta}}}}+\frac1{2^{40}}\le\sqrt{\frac{1312}{2^{40}}}+\frac1{2^{40}}\le\frac1{20000},
\]
the last being the rational comparison $1312/2^{40}\le(1/20000-2^{-40})^2$. Split $\ell^{\mathrm c}_\Sigma$ at $i=2254$. The terms with $2254\le i<j$ contribute at most $\delta_\Sigma/20000$. For $i<2254$, \eqref{eq:scalegrowth} gives $\delta_i\le g^{i-2254}\delta_{2254}$, so these terms total at most $100\,\delta_{2254}<100\cdot2^{41}$. By (1), $\delta_\Sigma\ge\delta_j\ge2^{41}\cdot10^{10}$, hence $9000\,\ell^{\mathrm c}_\Sigma\le9\cdot10^5\cdot2^{41}+\frac9{20}\delta_\Sigma\le\delta_\Sigma$.
\end{proof}

Let $j\ge J(t_1)$, let $X$ be reachable at $j$ for the target $t_1$, and let $s\in\Src_{j,X}(t_1)$ be the source of $wv$, with $w=v_0\cdots v_{j-1}\in L_j(t_1)$ and $v\in\FC^\term_X(w,\delta_j,1)$; we call $s$ a \emph{counted source}. The word $wv$ is valid at $t_1$. Its inverse trajectory from $t_1$ has one value after each number $k=0,1,\ldots,|wv|$ of letters, the first being $t_1$ and the last $s$; for $i\le j$ the value after $|w_{<i}|$ letters is $s_{w_{<i}}\ge16^{\delta_i}$ (Lemma~\ref{lem_endpoint_large}).

\begin{lemma}[Inverse trajectory of a counted source]\label{lem_orbit_large}
In this notation, (1) every value of the inverse trajectory of $wv$ is at least $4096$; (2) $\deg(wv)\log2\le\log s-\log t_1+|wv|\,c_{\mathrm{step}}$; and (3) $\deg(wv)\ge2|wv|-(2-\log_23)\ell^{\mathrm c}_\Sigma-j-\delta_j/8$.
\end{lemma}
\begin{proof}
For (1), we follow \cite[Section~7]{Mazur}. The first value is $t_1>2^{36}$. Let $y$ be the value after $k$ letters, where either $|w_{<i}|<k\le|w_{<i+1}|$ with $i<j$, or $|w|<k\le |wv|$ and $i=j$. Then $S^{\circ(k-|w_{<i}|)}(y)=s_{w_{<i}}$, and $S$ at most doubles every positive odd integer, so $s_{w_{<i}}\le2^{k-|w_{<i}|}y$, where $k-|w_{<i}|\le d^{\mathrm{hi}}_{\delta_i}\le8\delta_i/5$; hence $y\ge16^{\delta_i}2^{-8\delta_i/5}=2^{12\delta_i/5}\ge2^{108/5}>4096$.

For (2), write $wv=(a_1,\ldots,a_{|wv|})$. If $y$ and $y'$ are the values after $k$ and $k-1$ letters, then $3y+1=2^{a_k}y'$, so $a_k\log2=\log y-\log y'+\log(3+1/y)$. Summing over $k$, the first two terms telescope to $\log s-\log t_1$, and by (1) each last term is at most $c_{\mathrm{step}}$; this is \cite[(7.4)]{Mazur}.

For (3), which is \cite[(7.3)]{Mazur} with $\ell^{\mathrm c}_\delta$ in place of his width, let $i<j$. By \eqref{eq:hist-barrier} with $u=r_{\delta_i}$, $\deg(v_i)\ge H_{\delta_i,r_{\delta_i}}(|v_i|)>2|v_i|-(2-\log_23)(|v_i|-\delta_i)-1$, and $|v_i|-\delta_i\le\ell^{\mathrm c}_{\delta_i}$, by the length restriction for $\delta_i\ge256$ and by $|v_i|\le d^{\mathrm{hi}}_{\delta_i}$ otherwise. As $2-\log_23>0$, $\deg(v_i)\ge2|v_i|-(2-\log_23)\ell^{\mathrm c}_{\delta_i}-1$. For $v$, the first crossing and the mask \eqref{eq:mask} give $\deg(v)\ge H_{\delta_j,u_w}(|v|)\ge2|v|-\delta_j/8$. Summing proves (3).
\end{proof}

\begin{lemma}[Size of a counted source]\label{lem_log_source_lower}
Every $s\in\Src_{j,X}(t_1)$ satisfies $\log(s/t)\ge c_{\mathrm{src}}\delta_\Sigma$.
\end{lemma}
\begin{proof}
This is \cite[(7.2)]{Mazur} with this paper's scales. Apply \eqref{eq:word-source} to $v_i$ at $s_{w_{<i}}\ge16^{\delta_i}$ with $u=r_{\delta_i}$ for $i<j$, and to $v$ at $s_w$ with $u=u_w\ge0$: each word of scale $\delta$ multiplies the value by more than $2^{u-r_\delta-2}(4/3)^\delta$. Hence $\log(s/t)\ge\log(t_1/t)+\delta_\Sigma\log\frac43-(r_{\delta_j}+2j+2)\log2$. By Lemmas~\ref{lem_rb_bounds} and~\ref{lem_width_sum}(2), $r_{\delta_j}\le\frac3{10}\delta_j\le\frac{27}{10}+\frac3{1000}\delta_\Sigma+\frac35j$, and $2j+2+\frac35j\le\frac{26}{10}(j+1)$. Since $\log(t_1/t)>36\log2>\frac{27}{10}\log2$ and $j+1\le\delta_\Sigma/2^{22}$ by Lemma~\ref{lem_width_sum}(1),
\[
\log(s/t)\ge\delta_\Sigma\Bigl(\log\frac43-\log2\Bigl(\frac3{1000}+\frac{26}{10\cdot2^{22}}\Bigr)\Bigr),
\]
and by \eqref{eq:stop-logs} the bracket exceeds $\frac{287682}{10^6}-\frac{347}{500}\bigl(\frac3{1000}+\frac{26}{10\cdot2^{22}}\bigr)=\frac{598945709}{2097152000}=c_{\mathrm{src}}+\frac{501733}{52428800000}$.
\end{proof}

\begin{proposition}[Step bounds]\label{prop:clocks}
Let $t_{\mathrm{odd}}$ be the odd part of $t$. Each $s\in\Src_{j,X}(t_1)$ reaches $t$ within $\floor{\frac{523}{50}\log(s/t)}$ ordinary steps and $t_{\mathrm{odd}}$ within $\floor{\frac{34881}{10000}\log(s/t)}$ accelerated steps.
\end{proposition}
\begin{proof}
As $j\ge J(t_1)\ge2\cdot10^{10}$, Lemmas~\ref{lem_width_sum} and~\ref{lem_log_source_lower} apply. For the accelerated steps we follow the proof of \cite[Lemma~7.2]{Mazur}. Put $c_{\mathrm{gain}}=2\log2-c_{\mathrm{step}}$; by \eqref{eq:stop-logs}, $\log(4/3)-1/12288<c_{\mathrm{gain}}<\log(4/3)$. Lemma~\ref{lem_orbit_large}(2) and (3), multiplied out with $(2-\log_23)\log2=\log(4/3)$, give
\[
c_{\mathrm{gain}}(|wv|+1)\le\log s-\log t_1+c_{\mathrm{gain}}+\log\frac43\,\ell^{\mathrm c}_\Sigma+\Bigl(j+\frac{\delta_j}8\Bigr)\log2.
\]
By Lemma~\ref{lem_width_sum}(3), $\log\frac43\,\ell^{\mathrm c}_\Sigma\le\log\frac43\,\delta_\Sigma/9000$. By Lemma~\ref{lem_width_sum}(2), $j+\delta_j/8\le\frac98+\frac{\delta_\Sigma}{800}+\frac54j$, and by $\log2<7/10$ and Lemma~\ref{lem_width_sum}(1), $\frac54j\log2<\delta_\Sigma/10^6$ and $\frac{\delta_\Sigma}{800}\log2<\frac{7\delta_\Sigma}{8000}$. Finally $c_{\mathrm{gain}}+\frac98\log2<36\log2<\log(t_1/t)$. As $1/10^6+7/8000=219/250000$, we obtain, as in \cite[(7.6)]{Mazur},
\begin{equation}\label{eq:accclock}
c_{\mathrm{gain}}(|wv|+1)\le\log(s/t)+\Bigl(\frac{\log(4/3)}{9000}+\frac{219}{250000}\Bigr)\delta_\Sigma.
\end{equation}
Put $c_{\mathrm{acc}}=34881/10000$, the constant of \cite[(7.7)]{Mazur}. Since $0\le\delta_\Sigma\le\log(s/t)/c_{\mathrm{src}}$, it suffices for $|wv|+1\le c_{\mathrm{acc}}\log(s/t)$ that $c_{\mathrm{src}}+\frac{\log(4/3)}{9000}+\frac{219}{250000}\le c_{\mathrm{src}}c_{\mathrm{acc}}c_{\mathrm{gain}}$. Using $c_{\mathrm{gain}}>\log(4/3)-1/12288$ and $\log(4/3)>287682/10^6$, the right side minus the left side exceeds the rational residual
\begin{equation}\label{eq:accres}
\Bigl(c_{\mathrm{src}}c_{\mathrm{acc}}-\frac1{9000}\Bigr)\frac{287682}{10^6}-\frac{c_{\mathrm{src}}c_{\mathrm{acc}}}{12288}-c_{\mathrm{src}}-\frac{219}{250000}=\frac{25365007801}{192000000000000000}>0.
\end{equation}
Hence $|wv|+1\le c_{\mathrm{acc}}\log(s/t)$. Since $s$ reaches $t_1$ in $|wv|$ accelerated steps and $S(t_1)=t_{\mathrm{odd}}$, $s$ reaches $t_{\mathrm{odd}}$ within $|wv|+1$ accelerated steps.

For the ordinary steps we follow the proof of \cite[Lemma~7.3]{Mazur}. By Section~\ref{sec:words}, $s$ reaches $t_1$ in $|wv|+\deg(wv)$ ordinary steps, and $t_1$ reaches $t$ in $1+a$ more, where $3t_1+1=2^at$; so $s$ reaches $t$ within $N_{\mathrm{ord}}=|wv|+1+\deg(wv)+a$ ordinary steps. As $t_1>4096$, $a\log2=\log(3t_1+1)-\log t\le\log t_1+c_{\mathrm{step}}-\log t$, and adding Lemma~\ref{lem_orbit_large}(2) gives $(\deg(wv)+a)\log2\le\log(s/t)+(|wv|+1)c_{\mathrm{step}}$. With $|wv|+1\le c_{\mathrm{acc}}\log(s/t)$,
\[
N_{\mathrm{ord}}\log2\le(|wv|+1)(\log2+c_{\mathrm{step}})+\log(s/t)\le\bigl(c_{\mathrm{acc}}(\log2+c_{\mathrm{step}})+1\bigr)\log(s/t).
\]
It remains to prove Mazur's inequality \cite[(7.9)]{Mazur},
\begin{equation}\label{eq:ordres}
c_{\mathrm{acc}}(\log2+c_{\mathrm{step}})+1\le\frac{523}{50}\log2.
\end{equation}
By \eqref{eq:stop-logs} and Lemma~\ref{lem_s03_log3_bounds}(1), $c_{\mathrm{step}}<(317/200)\log2+1/12288$, so it suffices that
\[
\Bigl(\frac{523}{50}-\frac{517}{200}c_{\mathrm{acc}}\Bigr)\log2\ge\frac{c_{\mathrm{acc}}}{12288}+1.
\] The bracket is $2886523/2000000>0$, and by $\log2>693147180/10^9$ (Lemma~\ref{lem_s03_log3_bounds}(3)) the difference of the two sides exceeds
\[
\frac{693147180}{10^9}\cdot\frac{2886523}{2000000}-\frac{c_{\mathrm{acc}}}{12288}-1=\frac{43510569153}{400000000000000}>0.
\]
Both step counts are integers, so both bounds may be floored.
\end{proof}

\section{The Fourier bound: black points and triangles}\label{sec:fb-black}
This section and the next five prove the Fourier bound \eqref{eq:fourier}, following Tao's argument \cite[Section 7]{Tao} with explicit constants. Section~\ref{sec:fb-charsum} bounds $|\widehat\mu_n(\xi)|$ by the expected penalty of a random walk, which gains a factor below $1$ at the white points of Definition~\ref{def_bk_points}, and reduces its decay to a descent step proved in three cases in Sections~\ref{sec:fb-trace}--\ref{sec:fb-cases}; Section~\ref{sec:fb-renewal} supplies the laws of the walk. Here we describe the geometry of the points at which the character sum gains no cancellation: these \emph{black} points form disjoint triangles, and every point bordering a triangle is \emph{white}.
In Sections~\ref{sec:fb-black}--\ref{sec:fb-conclusion} the letter $j$ denotes the first
coordinate of a point and $J$ is as in Definition~\ref{lem_bk_eps_star_range}, and from Section~\ref{sec:fb-renewal} on $b$ (also $b^k$) denotes a hold
word and $\mathbf b$ a list of hold words; the generations $j$, scales $\delta_j$, families $L_j$ and
the bound $J(t)$ of Sections~\ref{sec:physical}--\ref{sec:finish} do not occur there.

\begin{definition}[Standing conventions]\label{lem_bk_eps_star_range}
From now on, $n\ge1$, $J=\floor{n/2}$, $\xi\in G_n$ is a frequency prime
to $3$, and $\varepsilon=8/217$ is the constant of
Appendix~\ref{app:analytic}. The objects defined below depend on $n$
and $\xi$; this dependence is suppressed from the notation.
\end{definition}
Apart from Lemma~\ref{lem_bk_exit_room}, the arguments of this section use
only $0<\varepsilon<1/27$.

\subsection{Points, angles, white and black points}
In the character sum, the cancellation contributed by the $j$th pair
$(a_{2j-1},a_{2j})$ of letters of the random word $W$ is governed by the angle, defined next, of the point
$(j,\deg_{2j}(W))$ (see Lemma~\ref{lem_ch_front_pascal}).

\begin{definition}[Points and angles]\label{def_bk_points}\label{def_bk_signed_frac}\label{def_bk_theta_residue}\label{def_bk_theta}\label{def_bk_black}\label{def_bk_white}
A \emph{point} is an element $p=(j,l)$ of $\mathcal P=\Z_{\ge1}\times\Z$; we
write $j(p)=j$ and $l(p)=l$, the \emph{raw length} and \emph{height} of $p$, and add points coordinatewise. Points $p\ne v$ are
\emph{cardinal neighbours} if $|j(p)-j(v)|+|l(p)-l(v)|=1$.
For $y\in G_n$, let $\tilde y\in\Z$ be the representative of $y$ with $|\tilde y|<3^n/2$
(it is unique, as $3^n$ is odd) and put $\operatorname{sfr}(y)=\tilde y/3^n$, the
\emph{signed fractional part}. The \emph{angle} of $p=(j,l)$ is
\[
\vartheta(p)=\operatorname{sfr}\bigl(\operatorname{res}(p)\bigr),\qquad
\operatorname{res}(p)=\bigl[3^{2(j-1)}2^{1-l}\bigr]_n\,\xi .
\]
A point $p$ is \emph{white} if $j(p)\le J$ and $|\vartheta(p)|>\varepsilon$, and
\emph{black} if $j(p)\le J$ and $|\vartheta(p)|\le\varepsilon$. Points with
$j(p)>J$ are neither.
\end{definition}

Since the numerator of $\operatorname{sfr}(y)$ is the representative of least modulus,
\begin{equation}\label{eq:fb-sfr}
y'\in\Z,\ y'\equiv y,\ |y'|<3^n/2\ \Longrightarrow\ \operatorname{sfr}(y)=\frac{y'}{3^n};
\qquad
|\operatorname{sfr}(ky)|\le k|\operatorname{sfr}(y)|\quad(k\in\N),
\end{equation}
the second because $k\cdot3^n\operatorname{sfr}(y)$ represents $ky$.

Moving a point one step right multiplies its residue by $9$, and one step down by $2$; the southeast distance measures the resulting growth of the angle (Lemma~\ref{lem_bk_se_bound}).

\begin{definition}[Southeast distance and the constant $\lambda$]\label{def_bk_se_weight}\label{def_bk_slope}\label{def_bk_drift}\label{lem_bk_alpha_bounds}
For points $u,v$ with $j(u)\le j(v)$ and $l(v)\le l(u)$ put the \emph{southeast distance}
\[
\operatorname{se}(u,v)=(j(v)-j(u))\log9+(l(u)-l(v))\log2,
\]
so that $e^{\operatorname{se}(u,v)}=9^{j(v)-j(u)}2^{l(u)-l(v)}$ is a positive integer.
Let $\lambda=\log2/\log9$, the slope of the hypotenuse of a triangle
(Definition~\ref{def_bk_triangle}).
\end{definition}

\begin{lemma}[Southeast transport]\label{lem_bk_se_bound}\label{lem_bk_se_exact}\label{lem_bk_residue_shift}
Let $u,v\in\mathcal P$ with $j(u)\le j(v)$ and $l(v)\le l(u)$, and put
$k=e^{\operatorname{se}(u,v)}$. Then $\operatorname{res}(v)=k\,\operatorname{res}(u)$ and
$|\vartheta(v)|\le k|\vartheta(u)|$. If moreover $k|\vartheta(u)|<1/2$, then
$\vartheta(v)=k\vartheta(u)$.
\end{lemma}
\begin{proof}
As in \cite[(7.12)]{Tao}, $3^{2(j(v)-1)}2^{1-l(v)}=k\cdot3^{2(j(u)-1)}2^{1-l(u)}$ in $\Z[1/2]$; apply
$[\cdot]_n$ and multiply by $\xi$. The bound follows from the second part of
\eqref{eq:fb-sfr}, and the equality, as in \cite[(7.18)]{Tao}, from its first part with
$y'=k\cdot3^n\vartheta(u)$.
\end{proof}

\begin{lemma}[Angles are bounded below]\label{lem_bk_strip}
If $j(p)\le J$, then $3^{\,n+1-2j(p)}|\vartheta(p)|\ge1/3$; in particular
$\vartheta(p)\ne0$.
\end{lemma}
\begin{proof}
This is the proof of \cite[(7.16)]{Tao} with $\varepsilon$ replaced by $|\vartheta(p)|$: for
$p=(j,l)$ and $k=n+1-2j\ge1$, $3^k\operatorname{res}(p)=3^{n-1}[2^{1-l}]_n\xi$ is represented by
$\pm3^{n-1}$, so $1/3=|\operatorname{sfr}(3^k\operatorname{res}(p))|\le3^k|\vartheta(p)|$ by \eqref{eq:fb-sfr}.
\end{proof}

\subsection{Angles of neighbouring points}
The proof of \cite[Lemma 7.4]{Tao} assumes $0<\varepsilon<1/100$; our
$\varepsilon=8/217$ is larger, which is why we reprove its ingredients with explicit
thresholds. The next lemma is a quantitative form of the claims (i)--(iii) of that
proof, which concern points with $|\vartheta|\le1/100$: it recovers the angle of a
point exactly from the angles of two cardinal neighbours.

\begin{lemma}[Quantitative claims (i)--(iii)]\label{lem_bk_claim_i}\label{lem_bk_claim_ii}\label{lem_bk_claim_iii}
Let $p=(j,l)\in\mathcal P$ and let $\bar\vartheta_1,\bar\vartheta_2$ be real numbers with
$|\vartheta(j,l-1)|\le \bar\vartheta_2$. Each of the following implies
$2\vartheta(p)=\vartheta(j,l-1)$:
\begin{enumerate}
\item[(i)] $|\vartheta(j,l)|\le \bar\vartheta_1$ and $2\bar\vartheta_1+\bar\vartheta_2<1$;
\item[(ii)] $|\vartheta(j+1,l)|\le \bar\vartheta_1$ and $2\bar\vartheta_1+9\bar\vartheta_2<1$;
\item[(iii)] $j\ge2$, $|\vartheta(j-1,l)|\le \bar\vartheta_1$ and $18\bar\vartheta_1+\bar\vartheta_2<1$.
\end{enumerate}
\end{lemma}
\begin{proof}
Write $y_2=3^n\vartheta(j,l-1)$; by Lemma~\ref{lem_bk_se_bound} it represents
$2\operatorname{res}(p)$. In each case $\bar\vartheta_1\ge0$, so $\bar\vartheta_2<1$.

(i) The integer $2\cdot3^n\vartheta(p)-y_2$ is divisible by $3^n$ and has
modulus at most $(2\bar\vartheta_1+\bar\vartheta_2)3^n<3^n$, so it vanishes.

(ii) Let $y_1=3^n\vartheta(j+1,l)$, which represents $9\operatorname{res}(p)$. Then
$2y_1-9y_2\equiv18\operatorname{res}(p)-18\operatorname{res}(p)=0$ and $|2y_1-9y_2|<3^n$, so $2y_1=9y_2$. The
integer $y_1-4y_2$ represents $9\operatorname{res}(p)-8\operatorname{res}(p)=\operatorname{res}(p)$ and equals $y_2/2$, of
modulus at most $\bar\vartheta_2\,3^n/2<3^n/2$. By \eqref{eq:fb-sfr},
$\vartheta(p)=y_2/(2\cdot3^n)$.

(iii) Let $y_3=3^n\vartheta(j-1,l)$, which represents $\operatorname{res}(j-1,l)$; then
$\operatorname{res}(p)=9\operatorname{res}(j-1,l)$ and $y_2$ represents $18\operatorname{res}(j-1,l)$. As before $|18y_3-y_2|<3^n$
forces $y_2=18y_3$, and $9y_3=y_2/2$ represents $\operatorname{res}(p)$ with modulus less than
$3^n/2$, so $\vartheta(p)=y_2/(2\cdot3^n)$.
\end{proof}

\begin{corollary}[Small angles along a row or column]\label{lem_bk_prop_left}\label{lem_bk_prop_right}\label{lem_bk_prop_up}
Let $(j,l)\in\mathcal P$ and $m\in\N$.
\begin{enumerate}
\item[(a)] If $|\vartheta(j+m,l)|\le\varepsilon$ and $|\vartheta(j+k,l-1)|\le\varepsilon$
for $0\le k<m$, then $|\vartheta(j+k,l)|\le\varepsilon$ for $0\le k\le m$.
\item[(b)] If $|\vartheta(j,l)|\le\varepsilon$ and $|\vartheta(j+k,l-1)|\le\varepsilon$
for $1\le k\le m$, then $|\vartheta(j+k,l)|\le\varepsilon$ for $0\le k\le m$.
\item[(c)] If $|\vartheta(j,l)|\le\varepsilon$ and $|\vartheta(j+1,l+k)|\le\varepsilon$
for $1\le k\le m$, then $|\vartheta(j,l+k)|\le\varepsilon$ for $0\le k\le m$.
\end{enumerate}
\end{corollary}
\begin{proof}
Take $\bar\vartheta_1=\bar\vartheta_2=\varepsilon$, so that $2\bar\vartheta_1+9\bar\vartheta_2=11\varepsilon<1$
and $18\bar\vartheta_1+\bar\vartheta_2=19\varepsilon<1$. In (a), induct downwards on $k$ using
Lemma~\ref{lem_bk_claim_i}(ii) at $(j+k,l)$: $|\vartheta(j+k,l)|=|\vartheta(j+k,l-1)|/2\le\varepsilon$.
In (b), induct upwards on $k$ using Lemma~\ref{lem_bk_claim_i}(iii) at $(j+k,l)$, where
$j+k\ge2$. In (c), induct on $k$ using Lemma~\ref{lem_bk_claim_i}(ii) at
$(j,l+k)$, whose right neighbour is $(j+1,l+k)$ and lower neighbour
$(j,l+k-1)$.
\end{proof}

\subsection{Column tops and canonical triangles}
To make the decomposition of the black points canonical, each black point is
assigned the triangle whose corner is reached by moving up through black points
of its column as far as possible and then left through black points of that row,
as in the proof of \cite[Lemma 7.4]{Tao}. Moving up stops: if $p=(j,l)$ is black,
then $(j,l+k)$ is white for some $1\le k\le2n$. Otherwise
$(j,l+k)$ is black for $0\le k\le2n$, since $j\le J$. Moving down one
step doubles the angle exactly, by Lemma~\ref{lem_bk_se_bound} and
$2\varepsilon<1/2$; hence $|\vartheta(p)|=4^n|\vartheta(j,l+2n)|$. By
Lemma~\ref{lem_bk_strip} and $3^{n+1-2j}\le3^{n-1}<4^n$, this is at least
$1/3>\varepsilon$, contradicting blackness of $p$. As every point $v$ with
$j(v)\le J$ is either black or white, the following objects are well defined.

\begin{definition}[Column top, gap, row start]\label{def_bk_col_top}\label{def_bk_row_start}\label{lem_bk_coltop_spec}\label{lem_bk_rowstart_spec}\label{lem_bk_column_ends}
Let $p=(j,l)$ be black. Its \emph{column top} $l_*(p)$ is the unique $l''\ge l$
such that $(j,l')$ is black for $l\le l'\le l''$ and $(j,l''+1)$ is white. Its
\emph{gap} is $\operatorname{gap}(p)=l_*(p)-l(p)\in\N$. Its \emph{row start} $j_*(p)$
is the unique $j'\in[1,j]$ such that $(i,l_*(p))$ is black for $j'\le i\le j$ and
either $j'=1$ or $(j'-1,l_*(p))$ is white.
\end{definition}

\begin{definition}[Triangles and the canonical family]\label{def_bk_triangle}\label{def_bk_canon_triangle}\label{def_bk_family}
A \emph{triangle} is a triple $\Delta=(j_\Delta,l_\Delta,S_\Delta)\in
\Z_{\ge1}\times\Z\times\R$, with \emph{corner} $p_\Delta=(j_\Delta,l_\Delta)$
and \emph{size} $S_\Delta$. A point $v$ \emph{lies in} $\Delta$, written
$v\in\Delta$, if
\[
j(v)\ge j_\Delta,\qquad l(v)\le l_\Delta,\qquad
\operatorname{se}(p_\Delta,v)\le S_\Delta .
\]
The \emph{top edge} of $\Delta$ is the row $l=l_\Delta$, and its \emph{left
edge} is the column $j=j_\Delta$.
For a black point $p$, the \emph{canonical triangle} is
\[
\Delta(p)=\Bigl(j_*(p),\ l_*(p),\
\log\frac{\varepsilon}{|\vartheta(j_*(p),l_*(p))|}\Bigr),
\]
and the \emph{canonical family} is
$\mathfrak T=\{\Delta(p):p\text{ black}\}$.
\end{definition}

Thus $v\in\Delta$ means that $v$ lies weakly southeast of the corner and
weakly left of the hypotenuse $j-j_\Delta=\lambda(l-l_\Delta)+S_\Delta/\log9$. Membership is
\emph{closed to the northwest}: if $v\in\Delta$ and $j_\Delta\le j(v')\le j(v)$,
$l(v)\le l(v')\le l_\Delta$, then $v'\in\Delta$, since $\operatorname{se}(p_\Delta,\cdot)$
decreases. Also every $v\in\Delta$ satisfies
$j(v)\le j_\Delta+S_\Delta/\log9$.

\begin{lemma}[Canonical triangles]\label{lem_bk_corner_value}\label{lem_bk_canon_edge}\label{lem_bk_size_nonneg}\label{lem_bk_black_on}\label{lem_bk_right_edge}
Let $p$ be black and $\Delta=\Delta(p)$. Then
\begin{enumerate}
\item[(a)] $p_\Delta$ is black, $|\vartheta(p_\Delta)|=\varepsilon e^{-S_\Delta}$ and $S_\Delta\ge0$;
\item[(b)] (the \emph{right-end bound})
\begin{equation}\label{eq:fb-right-edge}
j_\Delta+S_\Delta/\log9\le n/2+1-\log(1/\varepsilon)/\log9<n/2-1/2\le J;
\end{equation}
\item[(c)] every $v\in\Delta$ is black.
\end{enumerate}
\end{lemma}
\begin{proof}
(a) The corner is black by Definition~\ref{def_bk_col_top}, so $j_\Delta\le J$,
$\vartheta(p_\Delta)\ne0$ by Lemma~\ref{lem_bk_strip}, and the size is defined by
the stated formula; $S_\Delta\ge0$ as $|\vartheta(p_\Delta)|\le\varepsilon$.

(b) Lemma~\ref{lem_bk_strip} gives $\log(1/|\vartheta(p_\Delta)|)\le(n+2-2j_\Delta)\log3=
(n/2+1-j_\Delta)\log9$. Subtract $\log(1/\varepsilon)$ to obtain the first inequality.
The second follows from $\log(1/\varepsilon)>\log27=\frac32\log9$, and the third
from $J=\floor{n/2}$.

(c) By Lemma~\ref{lem_bk_se_bound} and (a),
$|\vartheta(v)|\le e^{\operatorname{se}(p_\Delta,v)}\varepsilon e^{-S_\Delta}\le\varepsilon$. By (b),
$j(v)\le j_\Delta+S_\Delta/\log9<n/2-1/2$, so $j(v)\le J$.
\end{proof}

\subsection{The neighbourhood of a triangle and the partition of the black points}
The \emph{cardinal neighbourhood} of a triangle $\Delta$ consists of the points
not in $\Delta$ that are cardinal neighbours of a point of $\Delta$. Tao's
claim $(*)$ in the proof of \cite[Lemma 7.4]{Tao} makes every point within distance
$\frac1{10}\log\frac1\varepsilon$ of a triangle white, for small $\varepsilon$. The next
lemma is its cardinal-neighbour case, proved at $\varepsilon=8/217$ by Tao's three cases.

\begin{lemma}[White neighbourhood]\label{lem_bk_claim_star}\label{lem_bk_star_case1}\label{lem_bk_star_case2}\label{lem_bk_star_case3}\label{lem_bk_white_near}
Let $p$ be black. Every point of the cardinal neighbourhood of $\Delta(p)$ is white.
\end{lemma}
\begin{proof}
We follow the proof of claim $(*)$ in \cite[Lemma 7.4]{Tao}, with distance $1$ in place
of $\frac1{10}\log\frac1\varepsilon$; since $v$ below is itself black, Corollary~\ref{lem_bk_prop_left}
replaces Tao's claims (i)--(iii) and weakly black points. Write $\Delta=\Delta(p)$, and let
$v\notin\Delta$ be a cardinal neighbour of $x\in\Delta$. As $j(x)\le j_\Delta+S_\Delta/\log9<J$
by Lemma~\ref{lem_bk_corner_value}(b), $j(v)\le J$, so it suffices to show that $v$ is not black.
Recall $j_\Delta\le j(p)$, that $(i,l_\Delta)$ is black for $j_\Delta\le i\le j(p)$, and that
$p'=(j(p),l_\Delta+1)$ is white. Tao's three cases are exhaustive.

\emph{Case 1:} $j(v)\ge j_\Delta$, $l(v)\le l_\Delta$. Then $S_\Delta<\operatorname{se}(p_\Delta,v)\le S_\Delta+\log9$, as a
cardinal step changes $\operatorname{se}(p_\Delta,\cdot)$ by at most $\log9$. Since $9\varepsilon<1/2$,
Lemma~\ref{lem_bk_se_bound} and Lemma~\ref{lem_bk_corner_value}(a) give
$|\vartheta(v)|=\varepsilon e^{\operatorname{se}(p_\Delta,v)-S_\Delta}>\varepsilon$.

\emph{Case 2:} $j(v)\ge j_\Delta$, $l(v)>l_\Delta$. Then $v=x+(0,1)$ and $l(x)=l_\Delta$; suppose $v$
is black. If $j(v)<j(p)$, Corollary~\ref{lem_bk_prop_left}(b), started at $v$ along the black
points $(j(v)+k,l_\Delta)$, $1\le k\le j(p)-j(v)$, gives $|\vartheta(p')|\le\varepsilon$. If
$j(v)>j(p)$, the points $(j(p)+k,l_\Delta)$, $0\le k<j(v)-j(p)$, lie in $\Delta$ by northwest
closure from $x$ and are black by Lemma~\ref{lem_bk_corner_value}(c), and
Corollary~\ref{lem_bk_prop_left}(a), ending at $v$, gives the same. If $j(v)=j(p)$, then $p'=v$.
So $p'$ is black, as $j(p')\le J$; but it is white.

\emph{Case 3:} $j(v)<j_\Delta$. Then $v=x-(1,0)$, $j_\Delta\ge2$ and $l(v)\le l_\Delta$, so
$v'=(j_\Delta-1,l_\Delta)$ is white; suppose $v$ is black. The points $(j_\Delta,l(v)+k)$,
$1\le k\le l_\Delta-l(v)$, lie between $x$ and $p_\Delta$, hence in $\Delta$, and are black;
Corollary~\ref{lem_bk_prop_left}(c), started at $v$, gives $|\vartheta(v')|\le\varepsilon$, so
$v'$ is black, a contradiction.
\end{proof}

\begin{lemma}[Membership and coherence]\label{lem_bk_source_mem}\label{lem_bk_coherent_top}\label{lem_bk_coherent}
Let $p$ be black. Then $p\in\Delta(p)$. If $v\in\Delta(p)$ is black, then
$l_*(v)=l_*(p)$ and $\Delta(v)=\Delta(p)$.
\end{lemma}
\begin{proof}
As in the last paragraph of the proof of \cite[Lemma 7.4]{Tao}, with
Lemma~\ref{lem_bk_claim_star} in place of claim $(*)$. Write $\Delta=\Delta(p)$. By
Lemma~\ref{lem_bk_claim_star}, a black cardinal neighbour of a point of $\Delta$ lies in $\Delta$. The corner
lies in $\Delta$ because $S_\Delta\ge0$. Walking right through the black points
$(i,l_\Delta)$, $j_\Delta\le i\le j(p)$, and then down through the black points
$(j(p),l')$, $l(p)\le l'\le l_\Delta$, every point lies in $\Delta$; the last one is $p$.

Let $v\in\Delta$ be black. By northwest closure and
Lemma~\ref{lem_bk_corner_value}(c), the points $(j(v),l')$ with
$l(v)\le l'\le l_\Delta$ are black. The point $(j(v),l_\Delta+1)$ is not in $\Delta$ and is
a cardinal neighbour of $(j(v),l_\Delta)\in\Delta$, so it is white by
Lemma~\ref{lem_bk_claim_star}. Hence $l_*(v)=l_\Delta$ by
Definition~\ref{def_bk_col_top}. Similarly the points $(i,l_\Delta)$,
$j_\Delta\le i\le j(v)$, lie in $\Delta$ and are black, and either $j_\Delta=1$ or
$(j_\Delta-1,l_\Delta)$ is white, so $j_*(v)=j_\Delta$. The corners agree, and the size
depends only on the corner.
\end{proof}

\begin{proposition}[Partition of the black points]\label{lem_bk_cover}\label{lem_bk_disjoint}
A point is black if and only if it lies in some $\Delta\in\mathfrak T$.
Two members of $\mathfrak T$ with a common point are equal.
\end{proposition}
\begin{proof}
A black $p$ lies in $\Delta(p)$, and points of canonical triangles are black,
by Lemmas~\ref{lem_bk_source_mem} and~\ref{lem_bk_corner_value}(c). If
$x\in\Delta(p)\cap\Delta(p')$, then $x$ is black and
Lemma~\ref{lem_bk_source_mem} gives $\Delta(p)=\Delta(x)=\Delta(p')$.
\end{proof}

\subsection{White exit points above a triangle}
In Sections~\ref{sec:fb-trace} and~\ref{sec:fb-cases} a random walk leaves a
triangle $\Delta\in\mathfrak T$ from a point $p\in\Delta$ at height $l_\Delta-l(p)$ below the
top edge and first lands above that edge at an exit point
$(j(p)+r,l_\Delta+o)$ with $1\le o\le4$. Such an exit point is white
when $r$ is not too large.

\begin{lemma}[White exit points]\label{lem_bk_exit_room}\label{lem_bk_exit_white}\label{lem_bk_exit_strip}\label{lem_bk_exit_left}
For $\omega\in\N$ put $f_{\omega}=\floor{(5\omega+16)/16}$. Let $\Delta\in\mathfrak T$, $p\in\Delta$ and
$\omega=l_\Delta-l(p)$, which is $\operatorname{gap}(p)$ by
Lemmas~\ref{lem_bk_corner_value}(c) and~\ref{lem_bk_source_mem}. Then $(j(p)+r,\,l_\Delta+o)$ is white for all integers
$1\le r\le f_{\omega}$ and $1\le o\le4$.
\end{lemma}
\begin{proof}
Since $f_{\omega}\le1+5\omega/16$ and $9^5\le2^{16}$, $9^{f_{\omega}}\le9\cdot2^{\omega}$. Hence
$3\varepsilon\,9^r\le3\varepsilon\,9^{f_{\omega}}\le27\varepsilon\,2^{\omega}=\frac{216}{217}2^{\omega}<2^{\omega}$.
Put $p'=(j(p)+r,l_\Delta+o)$.
Membership of $p$ gives
$j(p)\le j_\Delta+S_\Delta/\log9-\lambda \omega$, so Lemma~\ref{lem_bk_corner_value}(b) gives
$j(p)\le n/2+1-\log(1/\varepsilon)/\log9-\lambda \omega$. Taking logarithms in
$9^r<2^{\omega}/(3\varepsilon)$ gives $r<\lambda \omega+\log(1/\varepsilon)/\log9-1/2$. Adding,
$2j(p')<n+1$, so $j(p')\le J$.

Suppose $p'$ is not white; then it is black. Put $p''=(j(p'),l_\Delta+1)$. Lemma~\ref{lem_bk_se_bound}
gives $|\vartheta(p'')|\le2^{o-1}\varepsilon\le8\varepsilon$. By northwest closure,
$(j(p),l_\Delta)\in\Delta$. We use that points of $\Delta$ are black and points of its
cardinal neighbourhood are white
(Lemmas~\ref{lem_bk_corner_value}(c) and~\ref{lem_bk_claim_star}).

\emph{Case 1.} Suppose $(j(p'),l_\Delta)\in\Delta$. Then $|\vartheta(j(p'),l_\Delta)|\le\varepsilon$, and
Lemma~\ref{lem_bk_claim_i}(i) at $p''$ with $\bar\vartheta_1=8\varepsilon$, $\bar\vartheta_2=\varepsilon$ ($17\varepsilon<1$) gives
$|\vartheta(p'')|\le\varepsilon/2$. So $p''$ is black; but $p''\notin\Delta$ is a
cardinal neighbour of $(j(p'),l_\Delta)$, hence white.

\emph{Case 2.} Suppose $(j(p'),l_\Delta)\notin\Delta$. Let $i$ be the largest
integer in $[j(p),j(p')]$ with $(i,l_\Delta)\in\Delta$; then $i<j(p')$, and put
$m=j(p')-i\ge1$. For $0<k<m$ the point $x=(i+k,l_\Delta)$ satisfies, by
Lemma~\ref{lem_bk_se_bound}, Lemma~\ref{lem_bk_corner_value}(a) and
$S_\Delta\ge(j(p)-j_\Delta)\log9+\omega\log2$,
\[
|\vartheta(x)|\le9^{j(x)-j_\Delta}\varepsilon e^{-S_\Delta}\le\varepsilon\,9^{j(x)-j(p)}2^{-\omega}
\le\varepsilon\,9^{r-1}2^{-\omega}<\frac1{27}.
\]
Also $|\vartheta(i,l_\Delta)|\le\varepsilon<1/27$. Put $\bar\vartheta_1=\max(8\varepsilon,1/54)$ and
$\bar\vartheta_2=1/27$, so $2\bar\vartheta_1+9\bar\vartheta_2\le25/27<1$. Starting from
$|\vartheta(i+m,l_\Delta+1)|=|\vartheta(p'')|\le \bar\vartheta_1$, Lemma~\ref{lem_bk_claim_i}(ii) at
$(i+k,l_\Delta+1)$ for $k=m-1,\ldots,0$ gives successively
$2\vartheta(i+k,l_\Delta+1)=\vartheta(i+k,l_\Delta)$, hence
$|\vartheta(i+k,l_\Delta+1)|\le1/54\le \bar\vartheta_1$. At $k=0$ this gives
$|\vartheta(i,l_\Delta+1)|\le\varepsilon/2$, so $(i,l_\Delta+1)$ is black, as $i<j(p')\le J$.
But it is not in $\Delta$ and is a cardinal neighbour of $(i,l_\Delta)\in\Delta$,
hence white.
\end{proof}
The range $1\le r\le f_{\omega}$, $1\le o\le4$ of exit points is the
\emph{exit window}.

\section{The renewal process}\label{sec:fb-renewal}
This section splits the raw letters (sums of consecutive pairs of letters) of a
random word after each raw letter $4$ or $5$, where \cite[\S7.3]{Tao} splits after
each raw letter $3$; the pieces are independent, so their hold points form a random walk with independent steps, the \emph{renewal process} or hold walk of Lemma~\ref{lem_rn_hold_mass}, along which the penalty of Section~\ref{sec:fb-charsum} factorizes. In place of
the local limit bounds of \cite[Lemmas~2.2, 7.6, 7.7]{Tao}, we prove exact laws,
exponential moments and explicit tail bounds for it, used in Sections~\ref{sec:fb-charsum}--\ref{sec:fb-cases} and
Appendix~\ref{app:constants}.

Nothing here depends on $n$ or $\xi$. Throughout, $a_1,a_2,\ldots$ are the
independent random letters of the random word $W$, and $\deg_i=\deg_i(W)$ are its partial degrees
(Section~\ref{sec:words}), and $\binom Nk=0$ unless $0\le k\le N$.

\subsection{Raw letters and holds}
\begin{definition}[Raw letters]\label{def_rn_pascal}\label{def_rn_raw_mass}
The \emph{raw letters} of a word are the pair sums $c_i=a_{2i-1}+a_{2i}$ $(i\ge1)$,
so that $c_1+\cdots+c_k=\deg_{2k}$ ($\deg_0=0$). For
$c\in\Z$ put $\mathfrak q(c)=(c-1)2^{-c}$ if $c\ge2$ and $\mathfrak q(c)=0$
otherwise. Since exactly $c-1$ pairs of positive integers have sum $c$,
each of probability $2^{-c}$, the raw letters are independent with law
$\Prob(c_i=c)=\mathfrak q(c)$. Below, a \emph{word of raw letters} is a deterministic word
$\mathbf c=(\mathbf c_1,\ldots,\mathbf c_k)\in\Z^k$. For $k\in\N$ and $l\in\Z$
put
\[
\mathsf R(k,l)=\Prob(\deg_{2k}=l)
=\sum_{\substack{\mathbf c\in\Z^k,\ \mathbf c_i\ge2\\ \mathbf c_1+\cdots+\mathbf c_k=l}}\
 \prod_{i=1}^k\mathfrak q(\mathbf c_i).
\]
The raw letters $4$ and $5$ are called \emph{closing}.
\end{definition}

All exact formulas below come from one classical observation
\cite[Section~VI.8]{Feller}: the letters are the waiting times between
successes in a sequence of independent fair Bernoulli trials, so the set
$\mathcal A=\{\deg_1,\deg_2,\ldots\}$ of success times contains each positive
integer independently with probability $1/2$, and $|\mathcal A\cap[1,l]|$
is binomial with parameters $l$ and $1/2$.

\begin{lemma}[Sums of raw letters]\label{lem_rn_raw_mass_formula}
\label{lem_rn_pascal_total}\label{lem_rn_pascal_tail}
\label{lem_rn_raw_odd_sum}\label{lem_rn_straddle}
\begin{enumerate}
\item $\sum_c\mathfrak q(c)=1$, and $\sum_{c>m}\mathfrak q(c)=(m+1)2^{-m}$ for
$m\in\N$.
\item $\mathsf R(0,l)=\mathbf 1[l=0]$; for $k\ge1$,
$\mathsf R(k,l)=\binom{l-1}{2k-1}2^{-l}$ if $l\ge1$ and
$\mathsf R(k,l)=0$ if $l\le0$.
\item $\sum_{k\ge1}\mathsf R(k,l)=\frac14$ for every integer $l\ge2$.
\item For $l,k\in\N$,
$\Prob(\deg_{2k}\le l<\deg_{2k+2})=\binom{l+1}{2k+1}2^{-l}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The raw letters are almost surely finite, so $\sum_c\mathfrak q(c)=1$, and
$c_1>m$ means $|\mathcal A\cap[1,m]|\le1$, of probability
$(1+m)2^{-m}$. Part (2) is the negative binomial law \cite[Section~VI.8]{Feller}:
for $k,l\ge1$, $\deg_{2k}=l$ means $l\in\mathcal A$ and
$|\mathcal A\cap[1,l-1]|=2k-1$. Summing over
$k\ge1$, $\sum_k\mathsf R(k,l)$ is the probability that $l\in\mathcal A$
and $|\mathcal A\cap[1,l-1]|$ is odd, which is $\frac12\cdot\frac12$ when
$l-1\ge1$. Finally $\deg_{2k}\le l<\deg_{2k+2}$ means
$|\mathcal A\cap[1,l]|\in\{2k,2k+1\}$, of probability
$(\binom l{2k}+\binom l{2k+1})2^{-l}=\binom{l+1}{2k+1}2^{-l}$.
\end{proof}

We split the sequence of raw letters after each closing letter; the possible pieces are the hold words below.

\begin{definition}[Hold words]\label{def_rn_hold_words}\label{def_rn_hold}
\label{def_rn_geom4}\label{def_rn_hold_list}\label{def_ch_block_weight}
\label{def_ch_block_point}\label{def_ch_block_path}\label{def_ch_raw}
For $(j,l)\in\mathcal P$, let $\mathfrak B_{j,l}$ be the set of
\emph{hold words} $b=(\mathbf c_1,\ldots,\mathbf c_j)\in\Z^j$ with all $\mathbf c_i\ge2$, with
$\mathbf c_i\notin\{4,5\}$ for $i<j$, with $\mathbf c_j\in\{4,5\}$, and with
$\mathbf c_1+\cdots+\mathbf c_j=l$; such a $b$ has \emph{length} $|b|=j$ and \emph{hold point}
$\operatorname{pt}(b)=(j,l)$, its raw length and height. Let
$\mathfrak B=\bigcup_{(j,l)\in\mathcal P}\mathfrak B_{j,l}$, a countable set. For a list
$\mathbf b=(b^1,\ldots,b^K)\in\mathfrak B^K$ put
$\operatorname{pt}_k(\mathbf b)=\operatorname{pt}(b^1)+\cdots+\operatorname{pt}(b^k)$ for $0\le k\le K$,
so that $\operatorname{pt}_0(\mathbf b)=(0,0)\notin\mathcal P$ and $j(\operatorname{pt}_k(\mathbf b))\ge k$; the
\emph{raw word} of $\mathbf b$ is the concatenation $b^1b^2\cdots b^K$. For a word of raw
letters $\mathbf c=(\mathbf c_1,\ldots,\mathbf c_k)$ put
$\mathfrak q(\mathbf c)=\prod_{i=1}^k\mathfrak q(\mathbf c_i)$, and for a list let
$\mathfrak q$ be $\mathfrak q$ of its raw word. Put
\begin{equation}\label{eq:fbc-eta}
\eta(y)=\sum_{b\in\mathfrak B:\ \operatorname{pt}(b)=y}\mathfrak q(b)\quad(y\in\mathcal P),\qquad
\eta^{\otimes N}(\mathbf y)=\prod_{i=1}^N\eta(y_i)\quad(\mathbf y\in\mathcal P^N),
\end{equation}
with $\eta^{\otimes0}=1$, and $\nu_{45}(r)=\frac5{16}(\frac{11}{16})^{r-1}$
for $r\ge1$, $\nu_{45}(r)=0$ for $r\le0$.
\end{definition}

The pieces of the random raw letters are the \emph{random holds}. A hold word is also called a \emph{hold}.

\begin{lemma}[Holds are independent]\label{lem_rn_hold_mass}
\label{lem_rn_hold_jmarginal}\label{lem_rn_hold_support}
\label{lem_rn_hold_finite}\label{lem_rn_list_marginal}
\label{lem_ch_raw_law}\label{lem_ch_block_hold}\label{lem_ch_block_list_mass}
Almost surely infinitely many raw letters are closing. The successive
random holds are independent, each equal to a given hold word $b$ with
probability $\mathfrak q(b)$. Consequently:
\begin{enumerate}
\item the hold points $\mathsf h_1,\mathsf h_2,\ldots$ are independent
with law $\eta$, the \emph{hold law}, so $\sum_{y\in\mathcal P}\eta(y)=1$ and
$\eta^{\otimes N}$ is the law of $(\mathsf h_1,\ldots,\mathsf h_N)$;
\item the raw length $j(\mathsf h_1)$ of the first hold point has the law
\begin{equation}\label{eq:hold-law}
\Prob(j(\mathsf h_1)=r)=\sum_l\eta(r,l)=\nu_{45}(r)\quad(r\ge1);
\end{equation}
\item each $\mathfrak B_{j,l}$ is finite, and $\eta(j,l)\ne0$ implies
$l\ge2j+2\ge4$;
\item $\mathsf W_N=\mathsf h_1+\cdots+\mathsf h_N$, the \emph{hold walk},
equals $(i,\deg_{2i})$, where $i$ is the index of the $N$th closing raw letter;
it is $\operatorname{pt}_N(\mathbf b)$ for the list $\mathbf b$ of the first $N$ random holds;
\item for every $K\in\N$, $\sum_{\mathbf b\in\mathfrak B^K}\mathfrak q(\mathbf b)=1$, and if
$0\le m\le K$ and $\mathbf c\in\Z^m$, then
\begin{equation}\label{eq:fbc-raw-law}
\sum_{\mathbf b\in\mathfrak B^K}\mathfrak q(\mathbf b)\,
\mathbf 1[\text{the raw word of }\mathbf b\text{ begins with }\mathbf c]
=\prod_{i=1}^m\mathfrak q(\mathbf c_i).
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}
Each raw letter is closing with probability
$\mathfrak q(4)+\mathfrak q(5)=\frac3{16}+\frac18=\frac5{16}$, independently, so
almost surely infinitely many are. Since a hold word has exactly one closing
letter, at its end, the first $N$ holds are given hold words $b^1,\ldots,b^N$ if
and only if the raw letters begin with their concatenation, an event of
probability $\prod_{i=1}^N\mathfrak q(b^i)$. Summing $\mathfrak q(b)$ over the hold
words $b$ of length $r$ factorizes as
\[
\Bigl(\sum_{c\notin\{4,5\}}\mathfrak q(c)\Bigr)^{r-1}
\bigl(\mathfrak q(4)+\mathfrak q(5)\bigr)
=\left(\frac{11}{16}\right)^{r-1}\frac5{16}=\nu_{45}(r),
\]
by Lemma~\ref{lem_rn_raw_mass_formula}(1). Summing over $r\ge1$ gives
$\sum_b\mathfrak q(b)=1$, the sum running over all hold words, so the holds are
independent with law $\mathfrak q$ on hold words. This proves (1) and (2). In (3), the raw letters of
a word in $\mathfrak B_{j,l}$ lie in $[2,l]$, and $l\ge2(j-1)+4$. Part (4)
holds because the $N$th hold ends at the $N$th closing letter. In (5), the
first $K$ random holds equal $\mathbf b$ with probability $\mathfrak q(\mathbf b)$, which gives
the total mass $1$; their concatenation, the raw word of $\mathbf b$, has length at least
$K\ge m$, so the event in \eqref{eq:fbc-raw-law} is that the first $m$ raw letters
equal $\mathbf c$, of probability $\prod_{i=1}^m\mathfrak q(\mathbf c_i)$.
\end{proof}

\subsection{First passage over a height}
The exit from a triangle in Section~\ref{sec:fb-black} is modelled by the first time
the hold walk, started at gap $\omega$ below the top edge, rises above that edge.

\begin{definition}[First passage]\label{def_rn_fp_word}\label{def_rn_fp_law}
Let $\omega\in\N$. The \emph{first-passage index} $N_{\omega}$ is the least $N\ge1$
with $l(\mathsf W_N)>\omega$. Since every $l(\mathsf h_i)\ge4$, it exists and
$N_{\omega}\le\floor{\omega/4}+1$. The \emph{first-passage law} is
$\mathsf F_{\omega}(x)=\Prob(\mathsf W_{N_{\omega}}=x)$ for $x\in\Z\times\Z$, where, as for points, $j(x)$ and $l(x)$ denote the coordinates of $x$.
Equivalently, call $\mathbf y\in\mathcal P^K$ a \emph{first-passage list over
height $\omega$} if $l(y_1)+\cdots+l(y_k)\le \omega$ for $k<K$ and
$l(y_1)+\cdots+l(y_K)>\omega$; then $\mathsf F_{\omega}(x)$ is the sum of
$\eta^{\otimes K}(\mathbf y)$ over $K\ge1$ and the first-passage lists $\mathbf y$ with
$y_1+\cdots+y_K=x$.
\end{definition}

The \emph{first-passage overshoot} over height $\omega$ is $l(\mathsf W_{N_{\omega}})-\omega\ge1$.
The \emph{horizontal first-passage coordinate} over height $\omega$ is
$j(\mathsf W_{N_{\omega}})$, and its law $r\mapsto\sum_{x:\ j(x)=r}\mathsf F_{\omega}(x)$ is the
\emph{horizontal first-passage law}.

\begin{lemma}[First-passage law]\label{lem_rn_fp_mass}
\label{lem_rn_fp_support}
\label{lem_rn_fp_finite}\label{lem_rn_fp_raw}
Let $\omega\in\N$.
\begin{enumerate}
\item $\mathsf F_{\omega}$ is a probability on $\Z\times\Z$, and
$\mathsf F_{\omega}(x)\ne0$ implies $x\in\mathcal P$ (that is, $j(x)\ge1$) and $l(x)>\omega$.
\item $\mathsf W_{N_{\omega}}=(i,\deg_{2i})$ for the least index $i$ with $c_i$
closing and $\deg_{2i}>\omega$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) holds since $N_{\omega}$ is finite, $j(\mathsf W_{N_{\omega}})\ge N_{\omega}\ge1$,
and $l(\mathsf W_{N_{\omega}})>\omega$. Part (2) follows from
Lemma~\ref{lem_rn_hold_mass}(4): the points $\mathsf W_N$, $N\ge1$, are
the points $(i,\deg_{2i})$ at the closing indices $i$, in increasing order.
\end{proof}

\subsection{Exponential moments}
Chernoff bounds for sums of hold points need the exponential moments of one hold point.
\begin{definition}\label{def_rn_mgf}\label{def_rn_vpgf}
For real $\mathfrak t$ and $y>0$ put
\[
\mathsf M_{45}(\mathfrak t)=\sum_{r\ge1}\nu_{45}(r)e^{\mathfrak t r}=\E e^{\mathfrak t j(\mathsf h_1)},\qquad
\mathsf V(y)=\sum_{(j,l)\in\mathcal P}\eta(j,l)y^l=\E y^{l(\mathsf h_1)},
\]
with values in $[0,\infty]$.
\end{definition}

\begin{lemma}[Moments of one hold point]\label{lem_rn_mgf_bound}
\label{lem_rn_mgf_summable}\label{lem_rn_mgf_finite}
\label{lem_rn_mgf_sixteenth}\label{lem_rn_hold_vpgf}
\label{lem_rn_vpgf_tilt}\label{lem_rn_vpgf_low}
\begin{enumerate}
\item If $11e^{\mathfrak t}<16$, then $\mathsf M_{45}(\mathfrak t)=5e^{\mathfrak t}/(16-11e^{\mathfrak t})<\infty$.
\item If $0\le \mathfrak t\le1/1024$, then $\mathsf M_{45}(\mathfrak t)\le e^{13\mathfrak t/4}$.
\item $\mathsf M_{45}(1/16)\le5/4$.
\item If $0<y<2$ and
$\mathsf V_{\mathrm{nc}}(y)=\frac{y^2}{(2-y)^2}-\frac3{16}y^4-\frac18y^5<1$, then
$\mathsf V(y)=(\frac3{16}y^4+\frac18y^5)/(1-\mathsf V_{\mathrm{nc}}(y))$.
\item $\mathsf V(27/25)<8$ and
$\mathsf V(4095/4096)^{10}<(4095/4096)^{127}$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) is a geometric series. For (2), $e^{-\mathfrak t}\ge1-\mathfrak t$ gives
$16e^{-\mathfrak t}-11\ge5-16\mathfrak t>0$, so $\mathsf M_{45}(\mathfrak t)=5/(16e^{-\mathfrak t}-11)$, and it suffices
that $5-16\mathfrak t\ge5e^{-13\mathfrak t/4}$. Since $e^{-13\mathfrak t/4}\le1-\frac{13}4\mathfrak t+\frac12(\frac{13}4\mathfrak t)^2$ for $\mathfrak t\ge0$, the
right side is at most $5-\frac{65}4\mathfrak t+\frac{845}{32}\mathfrak t^2$, which is at most
$5-16\mathfrak t$ because $\mathfrak t\le1/1024<8/845$. For (3), $e^{1/16}\le16/15$ by
$e^{-1/16}\ge15/16$, and by (1) $\mathsf M_{45}(\mathfrak t)=5e^{\mathfrak t}/(16-11e^{\mathfrak t})$ increases with $e^{\mathfrak t}$ on
$(0,16/11)$, with value $5/4$ at $e^{\mathfrak t}=16/15$.

For (4), $y^l=\prod_{i=1}^jy^{\mathbf c_i}$ for $b=(\mathbf c_1,\ldots,\mathbf c_j)\in\mathfrak B_{j,l}$, so the
factorization in the proof of Lemma~\ref{lem_rn_hold_mass} gives
$\mathsf V(y)=\sum_{j\ge1}\mathsf V_{\mathrm{nc}}(y)^{j-1}(\frac3{16}y^4+\frac18y^5)$. Here
$\sum_c\mathfrak q(c)y^c=\sum_{c\ge2}(c-1)(y/2)^c=y^2/(2-y)^2$, so $\mathsf V_{\mathrm{nc}}(y)$
is the sum of $\mathfrak q(c)y^c$ over $c\notin\{4,5\}$, a nonnegative
number. Sum the geometric series.

For (5), exact evaluation gives $\mathsf V_{\mathrm{nc}}(27/25)=77640184719/82656250000<1$
and $\mathsf V(27/25)=36266065281/5016065281<8$. Similarly
$\mathsf V_{\mathrm{nc}}(4095/4096)<1$ and
\[
\mathsf V(4095/4096)=\frac{48328698012322974813669375}{48479813739774803460507647};
\]
the claim is the exact integer comparison
\[
48328698012322974813669375^{10}\,4096^{127}<48479813739774803460507647^{10}\,4095^{127}
\]
(the ratio of the two sides is $0.99979\ldots$).
\end{proof}

\begin{lemma}[Tails of sums of hold points]\label{lem_rn_list_hor_tail}
\label{lem_rn_list_hor_mgf}\label{lem_rn_list_ver_tail}
\label{lem_rn_list_ver_low}\label{lem_rn_fp_mgf}\label{lem_rn_fp_mgf_summable}
Let $N\in\N$, and let $\mathsf m$, $\mathsf k$ and $\mathfrak t$ be real. Then
$\E e^{\mathfrak t(j(\mathsf h_1)+\cdots+j(\mathsf h_N))}=\mathsf M_{45}(\mathfrak t)^N$ in $[0,\infty]$, and
\begin{align*}
\Prob\Bigl(\mathsf m\le10\sum_{i\le N}j(\mathsf h_i)\Bigr)
 &\le\exp\left(\frac N2-\frac{\mathsf m}{160}\right),\\
\Prob\Bigl(\mathsf k\le\sum_{i\le N}l(\mathsf h_i)\Bigr)
 &\le8^N\left(\frac{25}{27}\right)^{\mathsf k},\\
\Prob\Bigl(\sum_{i\le N}l(\mathsf h_i)\le\mathsf k\Bigr)
 &\le\left(\frac{4095}{4096}\right)^{\frac{127}{10}N-\mathsf k}.
\end{align*}
For $\omega\in\N$ and real $\mathfrak t\ge0$ with $11e^{\mathfrak t}<16$,
$\sum_{x\in\mathcal P}\mathsf F_{\omega}(x)e^{\mathfrak t j(x)}\le \mathsf M_{45}(\mathfrak t)^{\omega+1}<\infty$.
\end{lemma}
\begin{proof}
The identity is independence. Each bound is the exponential Markov
inequality: the three probabilities are at most
\[
e^{-\mathsf m/160}\mathsf M_{45}(1/16)^N,\qquad
(25/27)^{\mathsf k}\mathsf V(27/25)^N,\qquad
y^{-\mathsf k}\mathsf V(y)^N\quad(y=4095/4096).
\]
Apply Lemma~\ref{lem_rn_mgf_bound}, with
$5/4<1+\frac12+\frac18\le e^{1/2}$ in the first case and
$0<\mathsf V(y)<y^{127/10}$ in the last. In the final claim the sum is
$\E e^{\mathfrak t j(\mathsf W_{N_{\omega}})}$; since $N_{\omega}\le \omega+1$ and $\mathfrak t\ge0$,
$\mathfrak t j(\mathsf W_{N_{\omega}})\le \mathfrak t\sum_{i\le \omega+1}j(\mathsf h_i)$, so the identity and
Lemma~\ref{lem_rn_mgf_bound}(1) apply.
\end{proof}

\subsection{First-passage overshoot}
Section~\ref{sec:fb-case3} needs the first passage over a height not to overshoot it by much (Lemma~\ref{lem_c3_eprime_mass}).
\begin{lemma}[First-passage overshoot tail]\label{lem_rn_overshoot_tail}
\label{def_rn_occupation}\label{lem_rn_occupation_value}\label{def_rn_green}\label{lem_rn_green_formula}
For integers $\omega\ge64$ and $\mathsf k\ge1$,
\[
\sum_{x:\ l(x)\ge \omega+\mathsf k}\mathsf F_{\omega}(x)
 <9\left(\frac{25}{27}\right)^{\mathsf k}.
\]
\end{lemma}
\begin{proof}
Put $\mathsf W_0=(0,0)$ and, for $m\in\Z$, let $\mathsf u(m)=\sum_{N\ge0}\Prob(l(\mathsf W_N)=m)$ be
the expected number of $N\ge0$ with $l(\mathsf W_N)=m$.
By Lemma~\ref{lem_rn_hold_mass}(4), the heights $l(\mathsf W_N)$, $N\ge1$,
are the $\deg_{2i}$ at the closing indices $i$. Since $c_{k+1}$ is independent
of $\deg_{2k}$,
\[
\mathsf u(m)=\mathbf 1[m=0]+\sum_{k\ge0}\bigl(\mathfrak q(4)\mathsf R(k,m-4)
+\mathfrak q(5)\mathsf R(k,m-5)\bigr).
\]
By Lemma~\ref{lem_rn_raw_mass_formula}(2),(3), $\sum_{k\ge0}\mathsf R(k,l)$ is
$1$ at $l=0$, is $1/4$ for $l\ge2$, and vanishes otherwise. With
$\mathfrak q(4)=3/16$ and $\mathfrak q(5)=1/8$, for $m\in\N$,
\[
\mathsf u(m)=\mathbf 1[m=0]+\tfrac3{16}\mathbf 1[m=4]+\tfrac18\mathbf 1[m=5]
+\tfrac3{64}\mathbf 1[m\ge6]+\tfrac1{32}\mathbf 1[m\ge7],
\]
so $\mathsf u(m)\le\frac5{64}+\mathbf 1[m=0]+\frac3{16}\mathbf 1[m=4]+\frac18\mathbf 1[m=5]$.

The left side of the claim is $\Prob(l(\mathsf W_{N_{\omega}})\ge \omega+\mathsf k)$. On this event,
$N=N_{\omega}-1$ is the unique $N\ge0$ with $l(\mathsf W_N)\le \omega$ and
$l(\mathsf h_{N+1})\ge \omega+\mathsf k-l(\mathsf W_N)$; conversely, any such $N$
equals $N_{\omega}-1$. Since $\mathsf h_{N+1}$ is independent of $\mathsf W_N$,
with $y=27/25$,
\[
\Prob(l(\mathsf W_{N_{\omega}})\ge \omega+\mathsf k)
=\sum_{m=0}^{\omega}\mathsf u(m)\Prob(l(\mathsf h_1)\ge \omega+\mathsf k-m)
\le\mathsf V(y)y^{-\mathsf k}\sum_{m=0}^{\omega}\mathsf u(m)y^{m-\omega},
\]
and
\[
\sum_{m=0}^{\omega}\mathsf u(m)y^{m-\omega}
\le\frac5{64}\sum_{i\ge0}y^{-i}
+y^{-\omega}\left(1+\frac3{16}y^4+\frac18y^5\right)
\le\frac{135}{128}+\frac9{128}=\frac98.
\]
The last step uses the exact rational comparison
$(25/27)^{64}\cdot224805889/156250000<9/128$ (the left side is
$0.0104\ldots$), where $224805889/156250000=1+\frac3{16}y^4+\frac18y^5$.
Lemma~\ref{lem_rn_mgf_bound}(5) gives $\mathsf V(y)<8$.
\end{proof}

\subsection{The horizontal first-passage law}
Section~\ref{sec:fb-case3} needs a packing bound and a tail bound for the horizontal first-passage law (Lemmas~\ref{lem_rn_fp_packing} and~\ref{lem_rn_hor_tail}). Both follow from its exact form, a mixture of two binomial rows (Lemma~\ref{lem_rn_cross_law}); we first bound binomial rows.
\begin{lemma}[Binomial bounds]\label{lem_rn_binom_row}
\label{lem_rn_binom_central}\label{lem_rn_hoeffding}
For $N\in\N$ and $k\in\Z$, $(\binom Nk2^{-N})^2\le2/(3N+2)$. For $N\ge1$
and real $u\ge0$,
\[
\sum_{|k-N/2|\ge u}\binom Nk2^{-N}\le2e^{-2u^2/N}.
\]
\end{lemma}
\begin{proof}
We have $(3m+1)\bigl(\binom{2m}m4^{-m}\bigr)^2\le1$ by induction on $m$:
it holds at $m=0$, and $\binom{2m+2}{m+1}4^{-m-1}=\binom{2m}m4^{-m}(2m+1)/(2m+2)$ with
$(3m+4)(2m+1)^2=12m^3+28m^2+19m+4\le(3m+1)(2m+2)^2$. The largest
coefficient of row $N$ is the middle one. For $N=2m$ this gives the claim;
for $N=2m+1$, $\binom{2m+1}m2^{-N}=\binom{2m+2}{m+1}4^{-m-1}$, whose square is at most
$1/(3m+4)=2/(3N+5)$.

The second bound is Hoeffding's inequality \cite[Theorem~2]{Hoeffding} for a
sum of $N$ independent fair Bernoulli variables, applied to each of the two tails.
\end{proof}

\begin{definition}[Binomial masses]\label{def_rn_cross_p}\label{def_rn_cross_e}
For $l\in\N$ and $i\in\Z$ put
\[
\mathsf p_l(i)=\binom{l+1}{2i+1}2^{-l},\qquad
\mathsf p^{\mathrm{cl}}_l(i)=\frac18\binom{l-1}{2i-1}2^{-l}\quad(l,i\ge1),
\]
and $\mathsf p^{\mathrm{cl}}_l(i)=0$ otherwise.
\end{definition}

By Lemma~\ref{lem_rn_raw_mass_formula}, $\mathsf p_l(i)$ is the
probability that $\deg_{2i}\le l<\deg_{2i+2}$, and for $l\ge2$,
$\mathsf p^{\mathrm{cl}}_l(i)=\mathfrak q(5)\mathsf R(i,l)$ is the probability that $\deg_{2i}=l$
and $c_{i+1}=5$. Odd-index sums of binomial rows give
$\sum_i\mathsf p_l(i)=1$ and, for $l\ge2$, $\sum_i\mathsf p^{\mathrm{cl}}_l(i)=1/32$.

\begin{lemma}[Exact horizontal first-passage law]\label{lem_rn_cross_law}
Let $\omega\ge6$ and $r\ge1$ be integers, and $l=\omega-4$. Then
\[
\sum_{x:\ j(x)=r}\mathsf F_{\omega}(x)=\mathsf p^{\mathrm{cl}}_l(r-1)
+\sum_{k=0}^{r-2}\bigl(\mathsf p_l(k)-\mathsf p^{\mathrm{cl}}_l(k)\bigr)\nu_{45}(r-1-k).
\]
\end{lemma}
\begin{proof}
By Lemma~\ref{lem_rn_fp_mass}(2), the left side is $\Prob(i_1=r)$, where $i_1=j(\mathsf W_{N_{\omega}})$. Let
$i_0$ be the largest $i\ge0$ with $\deg_{2i}\le l$; then $i_0<i_1$. Closing
indices $i\le i_0$ have $\deg_{2i}\le l\le \omega$. If $c_{i_0+1}=5$ and
$\deg_{2i_0}=l$, then $\deg_{2i_0+2}=\omega+1$ and $i_1=i_0+1$. Otherwise a closing
letter at $i_0+1$ gives $\deg_{2i_0+2}\le \omega$, while every closing index
$i\ge i_0+2$ has $\deg_{2i}\ge \deg_{2i_0+2}+4>\omega$; so $i_1$ is the first closing
index after $i_0+1$. Therefore $\{i_1=r\}$ is the disjoint union of
$\{\deg_{2r-2}=l,\ c_r=5\}$, of probability $\mathsf p^{\mathrm{cl}}_l(r-1)$ (zero for
$r=1$, since $l\ne0$), and, for $0\le k\le r-2$, of the events
\[
\bigl(\{\deg_{2k}\le l<\deg_{2k+2}\}\setminus\{\deg_{2k}=l,\ c_{k+1}=5\}\bigr)
\ \cap\ \{c_{k+2},\ldots,c_{r-1}\text{ not closing},\ c_r\text{ closing}\}.
\]
The two parts are independent, with probabilities
$\mathsf p_l(k)-\mathsf p^{\mathrm{cl}}_l(k)$ and $\nu_{45}(r-1-k)$.
\end{proof}

\begin{lemma}[Packing of the horizontal first-passage coordinate]
\label{lem_rn_fp_packing}\label{lem_rn_cross_max}\label{lem_rn_unimodal_pack}
Let $\omega\ge1024$ be an integer, $\mathsf d>0$, and $\mathcal K\subseteq\Z$ with
$|\zeta-\zeta'|\ge\mathsf d$ for distinct $\zeta,\zeta'\in\mathcal K$. Then
\[
\sum_{\zeta\in\mathcal K}\sum_{x:\ j(x)=\zeta}\mathsf F_{\omega}(x)
<\frac{17}{10\sqrt \omega}+\frac{33}{32\mathsf d}.
\]
\end{lemma}
\begin{proof}
Put $l=\omega-4$. By Lemma~\ref{lem_rn_cross_law}, with $\mathsf p^{\mathrm{cl}}_l\ge0$,
\[
\sum_{x:\ j(x)=r}\mathsf F_{\omega}(x)\le\mathsf p^{\mathrm{cl}}_l(r-1)
+\sum_{k\ge1}\nu_{45}(k)\mathsf p_l(r-1-k)
\]
for $r\ge1$, and the left side vanishes for $r\le0$.
Let $\mathsf g:\Z\to[0,\infty)$ have finite support, with every set
$\{\mathsf g\ge y\}$, $y>0$, consisting of consecutive integers. Such a set $I$
contains at most $1+|I|/\mathsf d$ points of $\mathcal K$, so
\[
\sum_{\zeta\in\mathcal K}\mathsf g(\zeta)=\int_0^{\max\mathsf g}\bigl|\mathcal K\cap\{\mathsf g\ge y\}\bigr|\,dy
\le\int_0^{\max\mathsf g}\Bigl(1+\frac{|\{\mathsf g\ge y\}|}{\mathsf d}\Bigr)dy
=\max\mathsf g+\frac1{\mathsf d}\sum_x\mathsf g(x).
\]
Binomial rows are unimodal, so every superlevel set of $i\mapsto\binom Ni$ is an
interval, and so are its preimages under $i\mapsto2i\pm1$ and their translates.
Thus this bound applies to
$i\mapsto\mathsf p^{\mathrm{cl}}_l(i-1)$ and to each $i\mapsto\mathsf p_l(i-1-k)$, of
total masses $1/32$ and $1$. Since $\sum_k\nu_{45}(k)=1$, the left side of the claim is
at most $\max\mathsf p^{\mathrm{cl}}_l+\max\mathsf p_l+\frac{33}{32\mathsf d}$.

Finally, Lemma~\ref{lem_rn_binom_row} with rows $l+1$ and $l-1$ gives, for all $i\in\Z$,
\[
\mathsf p_l(i)\le2\sqrt{\frac2{3\omega-7}},\qquad
\mathsf p^{\mathrm{cl}}_l(i)\le\frac1{16}\sqrt{\frac2{3\omega-13}}.
\]
Both $\omega/(3\omega-7)$ and $\omega/(3\omega-13)$ decrease in $\omega>13/3$. At $\omega=1024$ the exact
comparisons $8192/3065<(41/25)^2$ and $8/3059<(3/50)^2$ give
$\max\mathsf p_l<\frac{41}{25\sqrt \omega}$ and
$\max\mathsf p^{\mathrm{cl}}_l<\frac3{50\sqrt \omega}$, and $\frac{41}{25}+\frac3{50}=\frac{17}{10}$.
\end{proof}

\begin{lemma}[Horizontal tail of the first passage]\label{lem_rn_hor_tail}\label{lem_rn_threefifths_tail}
For $\omega\in\N$ and real $u\ge0$,
\[
\sum_{x:\ |j(x)-\omega/4|\ge u}\mathsf F_{\omega}(x)
\le2^{54}\left[\exp\left(-\frac{u^2}{2^{15}(1+\omega)}\right)
+\exp\left(-\frac u{256}\right)\right].
\]
\end{lemma}
\begin{proof}
The left side is at most $1$. If $u\le4096$, then
$e^{-u/256}\ge e^{-16}>3^{-16}>2^{-26}$, so the right side exceeds $1$.
Let $u>4096$.

\emph{Case 1.} Suppose $\omega\le5$. A nonzero term has $j(x)\ge1$, so $j(x)-\omega/4>-u$; hence
$j(x)-\omega/4\ge u$ and $j(x)\ge u$. Markov's inequality, Lemma~\ref{lem_rn_fp_mgf} with exponent $1/16$ and
Lemma~\ref{lem_rn_mgf_bound}(3) bound the left side by
$(5/4)^{\omega+1}e^{-u/16}\le(5/4)^6e^{-u/16}\le4e^{-u/256}$.

\emph{Case 2.} Suppose $\omega\ge6$. Put $l=\omega-4\ge2$; then $r-\omega/4=(r-1)-l/4$ for $r=j(x)$. Let
$\mathcal J=\{i\ge0:|i-l/4|\ge u/2\}$. By Pascal's rule
$\mathsf p_l(i)=(\binom l{2i}+\binom l{2i+1})2^{-l}\ge\mathsf p^{\mathrm{cl}}_l(i)$.
For $i\in\mathcal J$ and $i'\in\{2i,2i+1\}$ we have
$|i'-l/2|\ge2|i-l/4|-1\ge u/2$, so Lemma~\ref{lem_rn_hoeffding} gives
$\sum_{i\in\mathcal J}\mathsf p_l(i)\le2e^{-u^2/(2l)}$. Sum
Lemma~\ref{lem_rn_cross_law} over the $r$ with $|(r-1)-l/4|\ge u$. The
$\mathsf p^{\mathrm{cl}}_l$-part is at most $\sum_{i\in\mathcal J}\mathsf p_l(i)$. In the remaining
part, with $k=r-1-i\ge1$, $|i+k-l/4|\ge u$ forces $i\in\mathcal J$ or $k\ge u/2$, so
it is at most
\[
\sum_{i\in\mathcal J}\mathsf p_l(i)+\sum_{k\ge u/2}\nu_{45}(k)
\le2e^{-u^2/(2l)}+\left(\frac{11}{16}\right)^{u/2-1}
\le2e^{-u^2/(2l)}+2e^{-5u/32},
\]
using $\log(16/11)\ge5/16$. The total is at most
$4e^{-u^2/(2l)}+2e^{-u/256}$, and $2l<2^{15}(1+\omega)$.
\end{proof}

\subsection{The exit window}
The exit mass is the probability that the first passage lands in the exit window (Lemma~\ref{lem_rn_p45_mass}), whose points are white (Lemma~\ref{lem_bk_exit_room}); it bounds below the reward of one passage (Lemma~\ref{lem_tr_passage_bound}) and the gain at black points near the top of a triangle (Lemma~\ref{lem_mo_case2}).
\begin{definition}[Exit mass]\label{def_rn_p45}
For $\omega\in\N$, with $f_{\omega}$ as in
Lemma~\ref{lem_bk_exit_room}, put
\[
\mathsf q_{45}(\omega)=\sum_{c=4}^5\mathfrak q(c)\sum_{o=1}^4\sum_{r=1}^{f_{\omega}}
 \mathsf R(r-1,\omega+o-c).
\]
\end{definition}

\begin{lemma}[Exit masses as first-passage probabilities]\label{lem_rn_p45_mass}
Let $\omega\in\N$. Then
\[
\sum_{x:\ j(x)\le f_{\omega},\ \omega<l(x)\le \omega+4}\mathsf F_{\omega}(x)=\mathsf q_{45}(\omega).
\]
\end{lemma}
\begin{proof}
If $c_r$ is closing and $\omega<\deg_{2r}\le \omega+4$, then $\deg_{2r-2}=\deg_{2r}-c_r\le \omega$, so
every earlier closing index $i$ has $\deg_{2i}\le \omega$ and $r=j(\mathsf W_{N_{\omega}})$. Hence, by
Lemma~\ref{lem_rn_fp_mass}(2), the left side is the probability that some
$r\le f_{\omega}$ has $c_r\in\{4,5\}$ and $\deg_{2r}=\omega+o$ with $1\le o\le4$. These
events are disjoint in $(r,c_r,o)$, and $c_r$ is independent of
$\deg_{2r-2}$, which gives $\mathsf q_{45}(\omega)$.
\end{proof}

\begin{lemma}[Lower bounds for $\mathsf q_{45}$]\label{lem_rn_p45_floor}
\label{lem_rn_p45_large}
For every integer $\omega\ge256$, $\mathsf q_{45}(\omega)\ge\frac5{16}-\frac{125}{8\omega}\ge\frac{515}{2048}$.
For every $\omega\in\N$, $\mathsf q_{45}(\omega)\ge\frac3{16}$.
\end{lemma}
\begin{proof}
Let $\omega\ge256$, fix $c\in\{4,5\}$ and $o\in\{1,\ldots,4\}$, and put
$l=\omega+o-c\in[\omega-4,\omega]$. By Lemma~\ref{lem_rn_raw_mass_formula}(2),(3),
\[
\sum_{r=1}^{f_{\omega}}\mathsf R(r-1,l)\ge\frac14-\sum_{k\ge f_{\omega}}\mathsf R(k,l),
\qquad
\sum_{k\ge f_{\omega}}\mathsf R(k,l)\le\frac12\Prob\bigl(|\mathcal A\cap[1,l-1]|\ge2f_{\omega}-1\bigr),
\]
where $|\mathcal A\cap[1,l-1]|$ is binomial with parameters $l-1$ and $1/2$. Since
$16f_{\omega}\ge5\omega+1$ and $l\le \omega$,
$2f_{\omega}-1-\frac{l-1}2\ge\frac{5\omega+1}8-1-\frac{\omega-1}2=\frac{\omega-3}8\ge\frac \omega{10}$,
so Chebyshev's inequality bounds the probability by
$\frac{l-1}4(10/\omega)^2\le25/\omega$. Summing over the eight pairs $(c,o)$, with
$\mathfrak q(4)+\mathfrak q(5)=5/16$, gives
$\mathsf q_{45}(\omega)\ge\frac54(\frac14-\frac{25}{2\omega})=\frac5{16}-\frac{125}{8\omega}$.
For $\omega\ge256$ this is at least $515/2048>3/16$.

For $\omega\le255$, $\mathsf q_{45}(\omega)$ is an explicit rational computed from
the definition with $\mathsf R$ given by
Lemma~\ref{lem_rn_raw_mass_formula}(2). At $\omega=0$ only $r=1$, $c=o=4$
contributes, and $\mathsf q_{45}(0)=\mathfrak q(4)=3/16$. For $1\le \omega\le255$,
exact evaluation of these $255$ rationals shows $\mathsf q_{45}(\omega)>3/16$;
the minimum is $\mathsf q_{45}(6)=227/1024$.
\end{proof}

\section{The character sum and the weighted maxima}\label{sec:fb-charsum}
This section bounds $|\widehat\mu_n(\xi)|$ by the expectation
$\sum_y\eta(y)\mathsf Q(y)$ of the expected penalty $\mathsf Q$
(Proposition~\ref{lem_ch_front_end}, used in Theorem~\ref{lem_ch_primitive_decay}),
following \cite[\S\S7.1, 7.3]{Tao}
but with cancellation at the raw letters $4$ and $5$ as well as $3$
(Lemma~\ref{lem_ch_white_cancel}). It then introduces the weighted maxima
$\mathsf Q_m$, whose descent step $\mathsf Q_m\le\mathsf Q_{m-1}$ is proved in
Sections~\ref{sec:fb-trace}--\ref{sec:fb-conclusion}.

The constants $z=\frac{21}{500}$ and $\kappa=\frac4{25}$ are those of
Appendix~\ref{app:analytic}. Every series below is indexed by a countable set
and has nonnegative terms or converges absolutely, so it may be
rearranged and regrouped freely.

\subsection{Pairing}\label{sec:fb-characters}
Grouping the letters in pairs, as in \cite[\S7.1]{Tao}, bounds $|\widehat\mu_n(\xi)|$ by an
average of products of the following factors, one for each raw letter $\mathbf c_i$ at its point $(i,\mathbf c_1+\cdots+\mathbf c_i)$.
For $p\in\mathcal P$ and an integer $c\ge2$ put
\begin{equation}\label{eq:fbc-pair-factor}
\varphi_c(p)=\frac1{c-1}\left|\sum_{k=1}^{c-1}
 e^{-2\pi\mathrm i(2^{k-1}-1)\vartheta(p)}\right|,
\end{equation}
and put $\varphi_c(p)=1$ for $c\le1$. By the triangle inequality,
$0\le\varphi_c(p)\le1$.

\begin{lemma}[Pairing]\label{lem_ch_front_pascal}\label{lem_fx_character_sum}\label{lem_fx_offset_law}
\begin{equation}\label{eq:fbc-pascal}
|\widehat\mu_n(\xi)|\le\sum_{\mathbf c\in\Z^J}\ \prod_{i=1}^J
 \mathfrak q(\mathbf c_i)\,\varphi_{\mathbf c_i}\bigl(i,\,\mathbf c_1+\cdots+\mathbf c_i\bigr).
\end{equation}
\end{lemma}
\begin{proof}
Let $\chi(y)=e^{-2\pi\mathrm i\xi y/3^n}$ for $y\in G_n$, evaluated at elements of $\Z[1/2]$
after reduction (Section~\ref{sec:words}). Then $\widehat\mu_n(\xi)=\E\chi(\off(W_{\le n}))$ is the sum
$S_\chi(n)$ of \cite[(7.2)]{Tao}, since the expression in \cite[(1.29)]{Tao} is $\off(W_{\le n})$ letter
for letter. Tao's pair
sums $a_{2i-1}+a_{2i}$ are the raw letters, with law $\mathfrak q$
(Definition~\ref{def_rn_pascal}), and given $a_{2i-1}+a_{2i}=c$ the letter $a_{2i}$ is uniform
on $\{1,\ldots,c-1\}$. So, with $p_i=(i,\mathbf c_1+\cdots+\mathbf c_i)$ and
$x_i=3^{2(i-1)}2^{-l(p_i)}$, \cite[(7.4)--(7.5)]{Tao} read
\[
|\widehat\mu_n(\xi)|\le\sum_{\mathbf c\in\Z^J}\ \prod_{i=1}^J\mathfrak q(\mathbf c_i)\,
|f(x_i,\mathbf c_i)|,\qquad f(x,c)=\frac1{c-1}\sum_{k=1}^{c-1}\chi\bigl((2^k+3)x\bigr),
\]
where the terms with some $\mathbf c_i\le1$ vanish. Since $2^k+3=5+2(2^{k-1}-1)$ and
$\chi(2x_i)=e^{-2\pi\mathrm i\vartheta(p_i)}$ (Definition~\ref{def_bk_points}), we have
$\chi((2^k+3)x_i)=\chi(5x_i)\,e^{-2\pi\mathrm i(2^{k-1}-1)\vartheta(p_i)}$, so
$|f(x_i,\mathbf c_i)|=\varphi_{\mathbf c_i}(p_i)$.
\end{proof}

\subsection{Cancellation at white points}
At a white point the factors $\varphi_3$, $\varphi_4$, $\varphi_5$ are bounded away from $1$.

\begin{lemma}[White cancellation]\label{lem_ch_white_cancel}
\label{lem_ch_white_cancel_four}\label{lem_ch_white_cancel_five}
If $p\in\mathcal P$ is white, then
\[
\varphi_3(p)<e^{-\kappa z}=e^{-21/3125},\qquad
\varphi_4(p)<e^{-z},\qquad\varphi_5(p)<e^{-z}=e^{-21/500}.
\]
\end{lemma}
\begin{proof}
Put $x=\pi|\vartheta(p)|\in(\pi\varepsilon,\frac\pi2]$,
$\mathfrak a=e^{-2\mathrm ix}$ and $x_\circ=3.1415\,\varepsilon=\frac{6283}{54250}$. Replacing
$\vartheta(p)$ by $-\vartheta(p)$ conjugates the sum in
\eqref{eq:fbc-pair-factor}, so
$\varphi_c(p)=\frac1{c-1}\bigl|\sum_{k=1}^{c-1}\mathfrak a^{2^{k-1}-1}\bigr|$. Since
$\pi>3.1415$,
\[
0<x_\circ<\pi\varepsilon<x\le\frac\pi2.
\]
We use the Taylor bounds
$\sin y\ge y-y^3/6$ for $y\ge0$, $\cos y\le1-y^2/2+y^4/24$, and
$e^{-y}\ge1-y+y^2/2-y^3/6$ for $y\ge0$. The exponents $2^{k-1}-1$ in
\eqref{eq:fbc-pair-factor} are $0,1,3,7$ for $k=1,2,3,4$.

\emph{The raw letter $3$.} As in \cite[Lemma~7.2]{Tao},
$\varphi_3(p)=\frac12|1+\mathfrak a|=\cos x$.
As $\cos$ decreases on $[0,\pi]$, $\cos x<\cos x_\circ\le
1-x_\circ^2/2+x_\circ^4/24$. Since $\kappa z=\frac{21}{3125}$, the exact rational
comparison
\[
1-\frac{x_\circ^2}2+\frac{x_\circ^4}{24}
 <1-\kappa z+\frac{(\kappa z)^2}2-\frac{(\kappa z)^3}6\le e^{-\kappa z}
\]
(decimal values $0.9933009$ and $0.9933025$) proves the first bound.

\emph{The raw letter $4$.} Here $\varphi_4(p)=\frac13|1+\mathfrak a+\mathfrak a^3|$, and since
$|\mathfrak a|=1$,
\[
|1+\mathfrak a+\mathfrak a^3|^2=3+2\cos2x+2\cos4x+2\cos6x=9-4\mathfrak h(\mathfrak u),
\qquad \mathfrak h(\mathfrak u)=14\mathfrak u-28\mathfrak u^2+16\mathfrak u^3,
\]
by the multiple-angle formulas in $\mathfrak u=\sin^2x$. As $\sin$ increases on
$[0,\frac\pi2]$, $\sin x>\sin x_\circ\ge x_\circ-x_\circ^3/6>0$, so
$\mathfrak u>\mathfrak u_\circ=(x_\circ-x_\circ^3/6)^2$. If $\mathfrak u\le\frac14$, then $\mathfrak h$ is
nondecreasing on $[0,\frac14]$ (as $\mathfrak h'(y)\ge14-56y$), so $\mathfrak h(\mathfrak u)\ge
\mathfrak h(\mathfrak u_\circ)$. If $\mathfrak u>\frac14$, then $\mathfrak h(\mathfrak u)-\frac74\mathfrak u=16\mathfrak u(\mathfrak u-\frac78)^2\ge0$
gives $\mathfrak h(\mathfrak u)>\frac7{16}$. Exact rational arithmetic gives
$\frac{18199}{100000}<\mathfrak h(\mathfrak u_\circ)<\frac7{16}$ and
\[
9-4\cdot\frac{18199}{100000}<9\Bigl(1-2z+2z^2-\frac43z^3\Bigr)\le9e^{-2z}
\]
(decimal values $8.27204$ and $8.27486$; the last step is the Taylor bound
at $y=2z$). In both cases $\mathfrak h(\mathfrak u)>\frac{18199}{100000}$, hence
$9\varphi_4(p)^2=9-4\mathfrak h(\mathfrak u)<9e^{-2z}$.

\emph{The raw letter $5$.} By the triangle inequality and
$|1+e^{-2\mathrm iy}|=2|\cos y|$,
\[
\varphi_5(p)=\tfrac14|1+\mathfrak a+\mathfrak a^3+\mathfrak a^7|\le\tfrac14\bigl(|1+\mathfrak a|+|1+\mathfrak a^4|\bigr)
=\tfrac12\bigl(|\cos x|+|\cos4x|\bigr).
\]
If $x\le\frac\pi8$, then $4x_\circ<4x\le\frac\pi2$, so
$0\le\cos4x<\cos4x_\circ$, and $\varphi_5(p)<
1-(4x_\circ)^2/4+(4x_\circ)^4/48$, which is less than $0.94731$. If
$\frac\pi8<x<\frac\pi6$, then $|\cos4x|<\frac12$ and
$\varphi_5(p)<\frac34$. If $x\ge\frac\pi6$, then
$0\le\cos x\le\frac{\sqrt3}2<\frac{87}{100}$ and
$\varphi_5(p)<\frac{187}{200}$. All three bounds are below
$1-z=\frac{479}{500}\le e^{-z}$.
\end{proof}

The penalty below records these bounds along a word of raw letters.

\begin{definition}[White factors and penalty]\label{def_ch_white_factor}
\label{def_ch_raw3_factor}\label{def_ch_penalty}
For $p\in\mathcal P$ the \emph{white factors} are $\mathsf w(p)=e^{-z}$ and
$\mathsf w_3(p)=e^{-\kappa z}$ if $p$ is white, and
$\mathsf w(p)=\mathsf w_3(p)=1$ otherwise. For a word of raw letters
$\mathbf c=(\mathbf c_1,\ldots,\mathbf c_j)\in\Z^j$, the
\emph{penalty} is
\[
\operatorname{Pen}(\mathbf c)=\prod_{i:\ \mathbf c_i\in\{4,5\}}\mathsf w(i,\mathbf c_1+\cdots+\mathbf c_i)
 \prod_{i:\ \mathbf c_i=3}\mathsf w_3(i,\mathbf c_1+\cdots+\mathbf c_i)\in(0,1].
\]
\end{definition}

\subsection{Splitting the penalty into holds}
Splitting the raw letters into holds turns the penalty into a product along
the hold walk: the closing letters carry $\mathsf w$ at the hold points, and
the raw letters $3$ inside a hold carry the internal factor
(Lemma~\ref{lem_ch_raw_penalty}).

\begin{definition}[Internal factors]\label{def_ch_internal_weight}
For $x\in\N\times\Z$ and a hold $b=(\mathbf c_1,\ldots,\mathbf c_{|b|})$ (Definition~\ref{def_rn_hold_words}), the \emph{internal
factor}
\[
\mathsf I(x;b)=\prod_{i:\ \mathbf c_i=3}\mathsf w_3\bigl(x+(i,\,\mathbf c_1+\cdots+\mathbf c_i)\bigr)
 \in(0,1]
\]
records the raw letters $3$ inside the hold.
\end{definition}

\begin{lemma}[Penalty along holds]\label{lem_ch_raw_penalty}
Let $K\ge J$, let $\mathbf b\in\mathfrak B^K$, and let $\mathbf c$ be the
first $J$ raw letters of its raw word. Then
\[
\operatorname{Pen}(\mathbf c)=\prod_{k=1}^K\mathsf w\bigl(\operatorname{pt}_k(\mathbf b)\bigr)
\prod_{k=1}^K\mathsf I\bigl(\operatorname{pt}_{k-1}(\mathbf b);b^k\bigr).
\]
\end{lemma}
\begin{proof}
The raw letters $\mathbf c^k_1,\ldots,\mathbf c^k_{|b^k|}$ of $b^k$ occupy positions
$j(\operatorname{pt}_{k-1}(\mathbf b))+1,\ldots,j(\operatorname{pt}_k(\mathbf b))$ of the raw
word, and the sum of the raw letters of the raw word up to $\mathbf c^k_i$ is
$l(\operatorname{pt}_{k-1}(\mathbf b))+\mathbf c^k_1+\cdots+\mathbf c^k_i$. Hence the point attached
to $\mathbf c^k_i$ in Definition~\ref{def_ch_white_factor} is
$\operatorname{pt}_{k-1}(\mathbf b)+(i,\mathbf c^k_1+\cdots+\mathbf c^k_i)$, which
for the closing letter $i=|b^k|$ is $\operatorname{pt}_k(\mathbf b)$.
The raw letters in $\{4,5\}$ are exactly the closing letters, and the raw
letters equal to $3$ are among the others. So the right side is the
penalty of the entire raw word, whose length is at least $K\ge J$. The raw
letters beyond position $J$ have points of raw length greater than
$J$, which are not white, so their factors are $1$.
\end{proof}

\subsection{Reduction to the expected penalty}
The character sum is bounded by the average penalty of the hold walk (Proposition~\ref{lem_ch_front_end}), the analogue of \cite[(7.34)]{Tao}. We truncate Tao's infinite product after $K$ holds, so that $\mathsf U_K$ below is a sum over finite lists of holds, with no infinite product of random factors.

\begin{definition}[The expected penalty]\label{def_ch_Qfinite}\label{def_ch_Q}
For $K\in\N$ and $p\in\mathcal P$ let
\[
\mathsf U_K(p)=\sum_{\mathbf b\in\mathfrak B^K}
 \mathfrak q(\mathbf b)
 \prod_{k=0}^K\mathsf w\bigl(p+\operatorname{pt}_k(\mathbf b)\bigr)
 \prod_{k=1}^K\mathsf I\bigl(p+\operatorname{pt}_{k-1}(\mathbf b);b^k\bigr),
\]
and call $\mathsf Q(p)=\mathsf U_J(p)$ the \emph{expected penalty} of a hold
walk started at $p$, counting the closing letter at $p$ itself.
\end{definition}

\begin{lemma}[Properties of the expected penalty]\label{lem_ch_Q_range}
\label{lem_ch_Qfinite_range}\label{lem_ch_Qfinite_step}
\label{lem_ch_Qfinite_beyond}\label{lem_ch_Qfinite_stable}
\label{lem_ch_Q_horizon}\label{lem_ch_Q_beyond}
\label{lem_ch_Q_recursion}\label{lem_ch_Q_recursion_le}
Let $p\in\mathcal P$.
\begin{enumerate}
\item $0\le\mathsf U_K(p)\le1$ for all $K$, and
$\mathsf U_K(p)=\mathsf Q(p)$ whenever $j(p)+K\ge J$. If $j(p)>J$, then
$\mathsf Q(p)=1$.
\item $\displaystyle\mathsf Q(p)=\mathsf w(p)\sum_{b\in\mathfrak B}
\mathfrak q(b)\,\mathsf I(p;b)\,\mathsf Q\bigl(p+\operatorname{pt}(b)\bigr)$.
\item $\displaystyle\mathsf Q(p)\le\mathsf w(p)\sum_{y\in\mathcal P}\eta(y)\,\mathsf Q(p+y)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (2) is the analogue of \cite[(7.35)]{Tao}, with internal factors and truncated sums.
All factors in Definition~\ref{def_ch_Qfinite} lie in $(0,1]$ and the
lists have total mass $1$ (Lemma~\ref{lem_rn_hold_mass}(5)), so
$0\le\mathsf U_K\le1$. Writing a list in $\mathfrak B^{K+1}$ as
$(b,\mathbf b')$ with
$\operatorname{pt}_i(b,\mathbf b')=\operatorname{pt}(b)
+\operatorname{pt}_{i-1}(\mathbf b')$ for $i\ge1$ gives the one-step
identity
\begin{equation}\label{eq:fbc-step}
\mathsf U_{K+1}(p)=\mathsf w(p)\sum_{b\in\mathfrak B}\mathfrak q(b)\,
\mathsf I(p;b)\,\mathsf U_K\bigl(p+\operatorname{pt}(b)\bigr).
\end{equation}
Splitting off the last hold instead shows that
$\mathsf U_{K+1}(p)=\mathsf U_K(p)$ when $j(p)+K\ge J$: the new
factors $\mathsf w(p+\operatorname{pt}_{K+1}(\mathbf b))$ and
$\mathsf I(p+\operatorname{pt}_K(\mathbf b);b^{K+1})$ are evaluated at points of
raw length at least $j(p)+K+1>J$, which are not white, so they equal
$1$, and the last hold then sums to mass $1$. By induction,
$\mathsf U_K(p)$ does not depend on $K$ once $j(p)+K\ge J$; as $K=J$
qualifies, the common value is $\mathsf Q(p)$. If $j(p)>J$, the same
reasoning makes every factor equal to $1$, and $\mathsf Q(p)=1$. This
proves (1). For (2), apply \eqref{eq:fbc-step} with $K=J$, noting
$\mathsf U_{J+1}(p)=\mathsf Q(p)$ by (1). For (3), drop
$\mathsf I\le1$ in (2) and group the holds by their point using
\eqref{eq:fbc-eta}.
\end{proof}

\begin{proposition}[Character sum and expected penalty]\label{lem_ch_front_end}
\begin{equation}\label{eq:fbc-front}
|\widehat\mu_n(\xi)|\le\sum_{y\in\mathcal P}\eta(y)\,\mathsf Q(y).
\end{equation}
\end{proposition}
\begin{proof}
If $J=0$, then $\mathsf Q(y)=1$ for every $y\in\mathcal P$
(Lemma~\ref{lem_ch_Q_range}(1)), so the right side is $1$, while
$|\widehat\mu_n(\xi)|\le\sum_x\mu_n(x)=1$.

Let $J\ge1$ and $\mathbf c\in\Z^J$. The factor of $\operatorname{Pen}(\mathbf c)$ at $i$ is
$1\ge\varphi_{\mathbf c_i}(i,\mathbf c_1+\cdots+\mathbf c_i)$ unless $\mathbf c_i\in\{3,4,5\}$ and
$(i,\mathbf c_1+\cdots+\mathbf c_i)$ is white, and then Lemma~\ref{lem_ch_white_cancel}
applies. So \eqref{eq:fbc-pascal} gives
\[
|\widehat\mu_n(\xi)|\le\sum_{\mathbf c\in\Z^J}\ \prod_{i=1}^J\mathfrak q(\mathbf c_i)\,
 \operatorname{Pen}(\mathbf c).
\]
By \eqref{eq:fbc-raw-law} with
$m=K=J$, the mass $\prod_i\mathfrak q(\mathbf c_i)$ is the total mass of the lists
$\mathbf b\in\mathfrak B^J$ whose raw word begins with $\mathbf c$. Summing
over $\mathbf c$ first and applying Lemma~\ref{lem_ch_raw_penalty}, the right side
becomes
\[
\sum_{\mathbf b\in\mathfrak B^J}\mathfrak q(\mathbf b)
 \prod_{k=1}^J\mathsf w\bigl(\operatorname{pt}_k(\mathbf b)\bigr)
 \prod_{k=1}^J\mathsf I\bigl(\operatorname{pt}_{k-1}(\mathbf b);b^k\bigr).
\]
Write $\mathbf b=(b^1,\mathbf b')$ and
$y=\operatorname{pt}(b^1)$, so that
$\operatorname{pt}_k(\mathbf b)=y+\operatorname{pt}_{k-1}(\mathbf b')$
for $k\ge1$. Summing over $\mathbf b'\in\mathfrak B^{J-1}$ turns the
summand into $\mathfrak q(b^1)\,\mathsf I((0,0);b^1)\,\mathsf
U_{J-1}(y)$, by Definition~\ref{def_ch_Qfinite}. Since $j(y)\ge1$,
Lemma~\ref{lem_ch_Q_range}(1) gives $\mathsf U_{J-1}(y)=\mathsf Q(y)$.
Finally drop $\mathsf I\le1$ and group the holds $b^1$ by their point,
using \eqref{eq:fbc-eta}.
\end{proof}

Two cases of the descent step halt the hold walk started at a black point $p$ when its height first exceeds $l(p)+\omega$, the top edge of a triangle at gap $\omega$ above $p$, and continue from there (Lemmas~\ref{lem_mo_case2} and~\ref{lem_c3_continuation}). The expected penalty at $p$ is at most its average at the halting point.

\begin{lemma}[Halting at a first passage]\label{lem_ch_Q_stopped}
For every $\omega\in\N$ and $p\in\mathcal P$,
\[
\mathsf Q(p)\le\sum_{x\in\mathcal P}\mathsf F_\omega(x)\,\mathsf Q(p+x).
\]
\end{lemma}
\begin{proof}
Compare \cite[(7.45)]{Tao}. Put $\mathsf W_0=(0,0)$. The hold points are
independent with law $\eta$ (Lemma~\ref{lem_rn_hold_mass}(1)), so
Lemma~\ref{lem_ch_Q_range}(3) and $\mathsf w\le1$ give
$\E[\mathsf Q(p+\mathsf W_{k+1})\mid\mathsf h_1,\ldots,\mathsf h_k]
=\sum_y\eta(y)\mathsf Q(p+\mathsf W_k+y)\ge\mathsf Q(p+\mathsf W_k)$:
the process $k\mapsto\mathsf Q(p+\mathsf W_k)\in[0,1]$ is a submartingale. The first-passage
index $N_\omega$ is a stopping time in the sense of \cite{Durrett}, with $N_\omega\le\floor{\omega/4}+1$
(Definition~\ref{def_rn_fp_law}), so optional stopping \cite[Theorem~4.4.1]{Durrett} gives
$\mathsf Q(p)\le\E\,\mathsf Q(p+\mathsf W_{N_\omega})=\sum_x\mathsf F_\omega(x)\mathsf Q(p+x)$, where the sum
may be taken over $x\in\mathcal P$ by Lemma~\ref{lem_rn_fp_mass}(1).
\end{proof}

\subsection{Weighted maxima near the horizon}
To show that $\mathsf Q(p)$ decays like a power of the distance to the
horizon $j=J$, we follow \cite[(7.38)--(7.41)]{Tao} and track the largest
weighted value $d_J^{A_*}\mathsf Q$ within distance $m$ of the horizon; a
bound on it that does not grow with $m$ is the decay we need.

\begin{definition}[Weighted maxima]\label{def_ch_Qm}\label{def_ch_dist}
For $p\in\mathcal P$ let $d_J(p)=\max(J-j(p),1)$, the \emph{distance to
the horizon} $j=J$. For an integer $m\ge1$, a \emph{boundary point} for $m$
is a $p\in\mathcal P$ with $j(p)+m=J$. With the exponent $A_*=\frac{10241}{4096}$ of
\eqref{eq:exponents}, for an integer $m\ge0$ the \emph{weighted maxima} are
\[
\mathsf Q_m=\sup\bigl\{d_J(p)^{A_*}\,\mathsf Q(p):\ p\in\mathcal P,\
 j(p)\ge J-m\bigr\}.
\]
\end{definition}

\begin{lemma}[Basic properties of the weighted maxima]\label{lem_ch_Qm_bounded}
\label{lem_ch_Qm_nonempty}\label{lem_ch_Qm_set_bound}\label{lem_ch_Qm_le_pow}
\label{lem_ch_Qm_dominate}\label{lem_ch_Qm_mono}
Let $m\ge0$ be an integer.
\begin{enumerate}
\item $0\le\mathsf Q_m\le\max(m,1)^{A_*}$.
\item If $j(p)\ge J-m$, then $d_J(p)^{A_*}\,\mathsf Q(p)\le\mathsf Q_m$.
\item If $0\le m\le m'$, then $\mathsf Q_m\le\mathsf Q_{m'}$.
\end{enumerate}
\end{lemma}
\begin{proof}
As in \cite[(7.39)]{Tao}: if $j(p)\ge J-m$, then $1\le d_J(p)\le\max(m,1)$ and
$0\le\mathsf Q(p)\le1$ (Lemma~\ref{lem_ch_Q_range}(1)), and the set in
Definition~\ref{def_ch_Qm} contains the point $(J+1,0)$. This gives (1) and (2); for (3),
the sets increase with $m$.
\end{proof}

\begin{lemma}[Boundary reduction]\label{lem_ch_boundary_suffices}
\label{lem_ch_boundary_step}
Let $m\ge1$ be an integer.
\begin{enumerate}
\item If $r\ge1$ and $j(p)+m=J+r$, then
$\mathsf Q(p)\le\max(m-r,1)^{-A_*}\mathsf Q_{m-1}$.
\item If $\mathsf Q(p)\le m^{-A_*}\mathsf Q_{m-1}$ for every boundary point
$p$ for $m$, then $\mathsf Q_m\le\mathsf Q_{m-1}$.
\end{enumerate}
\end{lemma}
\begin{proof}
As in \cite[\S7.4]{Tao} (the bound opening Case~1, and the reduction of (7.40) to
(7.41)), by Lemma~\ref{lem_ch_Qm_bounded}(2): in (1), $j(p)\ge J-(m-1)$ and
$d_J(p)=\max(m-r,1)$; in (2), a point with $j(p)\ge J-m$ either has $j(p)\ge J-(m-1)$ or
is a boundary point for $m$, where $d_J(p)=m$.
\end{proof}

By Lemma~\ref{lem_ch_boundary_suffices}(2), the descent step $\mathsf Q_m\le\mathsf Q_{m-1}$ (Proposition~\ref{lem_mo_monotonicity}) reduces to the bound $\mathsf Q(p)\le m^{-A_*}\mathsf Q_{m-1}$ at every boundary point $p$ for $m$, which is white or black as $j(p)<J$. We prove it in three cases: \emph{Case~1}, $p$ is white; \emph{Case~2}, $p$ is black with $\operatorname{gap}(p)\le m/(\log m)^2$; and \emph{Case~3}, $p$ is black with $\operatorname{gap}(p)>m/(\log m)^2$. Cases~1 and~2 need one hold or one passage over the top edge of a triangle (Section~\ref{sec:fb-cases}); Case~3 needs the trace (Section~\ref{sec:fb-trace}) and a packing estimate (Section~\ref{sec:fb-case3}).


\section{The trace}\label{sec:fb-trace}
In Case~3 a hold walk with low reward must enter many triangles (Section~\ref{sec:fb-case3}). Each entry is preceded by a passage over the top edge of the previous triangle, which earns a reward with probability bounded below, so many entries are unlikely. This section proves this in the explicit form of \cite[Lemma~7.9]{Tao}: from any starting point, the average of the
discounted reward $e^{-\gamma\operatorname{Rw}}$ at the $(k+1)$st trace time is at most
$\mathsf r_2^{\floor{k/2}}(1-d_{\mathrm{tr}})^{k-2\floor{k/2}}$
(Lemma~\ref{lem_tr_lemma79}). The proof rests on bounds for the powers of the entry
kernel (Proposition~\ref{lem_tr_kernel_two}). The constants are those of
Definitions~\ref{def_eps_star} and~\ref{def_s02_d}. All sums below have nonnegative terms
and values in $[0,\infty]$ unless stated otherwise, and are rearranged
freely.

\subsection{Paths and rewards}
A path follows the hold walk for a fixed number of holds, and its reward counts the white points
where the penalty of Section~\ref{sec:fb-charsum} receives a factor below $1$.

\begin{definition}[Paths and rewards]
\label{def_tr_list_weight}\label{def_tr_path}\label{def_tr_reward}\label{def_tr_count}
For a list $\mathbf b=(b^1,\ldots,b^L)\in\mathfrak B^L$, $L\in\N$, write
$|\mathbf b|=L$ and $\mathbf b_{[i,i']}=(b^i,\ldots,b^{i'})$, and write $\mathbf w\mathbf w'$
for the concatenation of lists $\mathbf w$ and $\mathbf w'$. The \emph{path} of $\mathbf b$
from $\bar o\in\mathcal P$ is
$\mathsf y_u(\bar o,\mathbf b)=\bar o+\operatorname{pt}_{\min(u,L)}(\mathbf b)$, $u\in\N$.
The \emph{reward} of a hold $b=(\mathbf c_1,\ldots,\mathbf c_{|b|})$ at a point $x$ is
\[
\operatorname{rw}(x,b)=\mathbf 1[x+\operatorname{pt}(b)\text{ white}]
+\kappa\,\#\{i<|b|:\ \mathbf c_i=3,\ x+(i,\mathbf c_1+\cdots+\mathbf c_i)\text{ white}\},
\]
and the \emph{accumulated reward} is
$\operatorname{Rw}(\bar o,\mathbf b;u)=\sum_{i=1}^{\min(u,L)}\operatorname{rw}(\mathsf y_{i-1}(\bar o,\mathbf b),b^i)$.
\end{definition}

In the notation of Definitions~\ref{def_ch_white_factor}
and~\ref{def_ch_internal_weight},
$\mathsf w(x+\operatorname{pt}(b))\,\mathsf I(x;b)=e^{-z\,\operatorname{rw}(x,b)}$.

\begin{lemma}[Elementary properties]\label{lem_tr_concat}
Let $\bar o\in\mathcal P$, $\mathbf w\in\mathfrak B^{L_1}$, $\mathbf w'\in\mathfrak B^{L_2}$, and
$\mathbf b\in\mathfrak B^L$.
\begin{enumerate}
\item For $\Psi\ge0$ on $\mathfrak B^{L_1}\times\mathfrak B^{L_2}$,
\[\sum_{\mathbf b\in\mathfrak B^{L_1+L_2}}\mathfrak q(\mathbf b)\Psi(\mathbf b_{[1,L_1]},\mathbf b_{[L_1+1,L_1+L_2]})
=\sum_{\mathbf w,\mathbf w'}\mathfrak q(\mathbf w)\mathfrak q(\mathbf w')\Psi(\mathbf w,\mathbf w').\]
\item $\mathsf y_u(\bar o,\mathbf w\mathbf w')=\mathsf y_u(\bar o,\mathbf w)$ for $u\le L_1$, $\mathsf y_{L_1+u}(\bar o,\mathbf w\mathbf w')=\mathsf y_u(\mathsf y_{L_1}(\bar o,\mathbf w),\mathbf w')$ and
$\operatorname{Rw}(\bar o,\mathbf w\mathbf w';L_1+u)=\operatorname{Rw}(\bar o,\mathbf w;L_1)+\operatorname{Rw}(\mathsf y_{L_1}(\bar o,\mathbf w),\mathbf w';u)$ for $u\in\N$.
\item For $u\le u'$, with $\mathsf y_u=\mathsf y_u(\bar o,\mathbf b)$ and
$k=\min(u',L)-\min(u,L)$, we have $j(\mathsf y_{u'})-j(\mathsf y_u)\ge k$ and
$l(\mathsf y_{u'})-l(\mathsf y_u)\ge4k$.
\item For $L'\in\N$, the masses of the lists in a prefix-disjoint set of lists of length at
most $L'$ sum to at most $1$.
\item For $u\le L$, $\operatorname{Rw}(\bar o,\mathbf b;u)\ge\#\{1\le i\le u:\mathsf y_i(\bar o,\mathbf b)\text{ white}\}$,
and $u\mapsto\operatorname{Rw}(\bar o,\mathbf b;u)$ is nondecreasing.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) Concatenation is a bijection $\mathfrak B^{L_1}\times\mathfrak B^{L_2}\to\mathfrak B^{L_1+L_2}$
and the mass is multiplicative. (2) is immediate from
$\operatorname{pt}_{L_1+u}(\mathbf w\mathbf w')=\operatorname{pt}_{L_1}(\mathbf w)+\operatorname{pt}_u(\mathbf w')$.
(3) Every $\operatorname{pt}(b)=(j,l)$ has $j\ge1$ and $l\ge4$
(Lemma~\ref{lem_rn_hold_mass}(3)). (4) Extend
every list $\mathbf w$ of the set by all lists of length $L'-|\mathbf w|$. By (1) the extensions of
$\mathbf w$ carry mass $\mathfrak q(\mathbf w)$, and the extensions of distinct lists of
the set are disjoint subsets of $\mathfrak B^{L'}$, which has total mass $1$
(Lemma~\ref{lem_rn_hold_mass}(5)).
(5) The first term of $\operatorname{rw}(\mathsf y_{i-1},b^i)$ is the indicator that
$\mathsf y_i$ is white, and all rewards are nonnegative.
\end{proof}

\subsection{Gaps and the trace times}
The trace times mark the successive entries of a path into triangles; each new entry lies above the top edge of the triangle entered before, so between two entries the path makes a full passage over that edge (Lemma~\ref{lem_tr_passage_bound}).
A black point $p$ lies in $\Delta(p)$ and $l_{\Delta(p)}=l_*(p)$
(Lemma~\ref{lem_bk_source_mem} and Definition~\ref{def_bk_triangle}), so its gap (Definition~\ref{def_bk_col_top}) is the
distance from $p$ up to the top edge of $\Delta(p)$. If $(j,l)$ and $(j,l+1)$ are black,
they have the same column top, so
\begin{equation}\label{eq:tr-gap-adjacent}
\operatorname{gap}((j,l))=\operatorname{gap}((j,l+1))+1.
\end{equation}

Call the path $\mathsf y_u=\mathsf y_u(\bar o,\mathbf b)$ of $\mathbf b\in\mathfrak B^L$ \emph{complete}
if $j(\bar o)+L>J$, so that it passes the horizon and its trace times below are finite. If $\mathsf y_u$ is black on a complete path, then $u<L$ and $j(\bar o)+u\le J$: otherwise
$j(\mathsf y_u)\ge j(\bar o)+\min(u,L)>J$ by Lemma~\ref{lem_tr_concat}(3). A time $u$
is a \emph{hit} if $\mathsf y_u$ is black.

\begin{definition}[Trace times]\label{def_tr_trace}
The \emph{trace times} $u_1<u_2<\cdots$ of a complete path are defined as
follows: $u_1$ is the least hit, if there is one; and once $u_i$ is
defined, $u_{i+1}$ is the least hit $u>u_i$ with
$l(\mathsf y_u)>l_*(\mathsf y_{u_i})$, if there is one. The sequence ends at the first
index where no such time exists. Since hits are less than $L$, it is
finite. The points $\mathsf y_{u_i}$ for $i\ge2$ are the \emph{entries}; their
sequence is the \emph{trace}.
\end{definition}

For a point $p\in\mathcal P$ and a list $\mathbf w$ write $p_{\mathbf w}=\mathsf y_{|\mathbf w|}(p,\mathbf w)$, and $p_b=p+\operatorname{pt}(b)$ for a hold $b$.
The first two trace times are read off from prefixes of the list, as follows. For $\bar o\in\mathcal P$ let $\mathcal U(\bar o)$, the \emph{bridge lists}, be the set of lists $\mathbf w$ such
that $\bar o_{\mathbf w}$ is black and $\mathsf y_u(\bar o,\mathbf w)$ is not black for $u<|\mathbf w|$. For
a black $v$ let $\mathcal E(v)$, the \emph{entry lists}, be the set of
nonempty lists $\mathbf w$ such that $v_{\mathbf w}$ is black with
$l(v_{\mathbf w})>l_*(v)$, and no $\mathsf y_u(v,\mathbf w)$ with $1\le u<|\mathbf w|$ has these two
properties. The sets $\mathcal U(\bar o)$ and $\mathcal E(v)$ are
prefix-disjoint. If $\mathbf w\in\mathcal U(\bar o)$, then
$j(\bar o)+|\mathbf w|\le j(\bar o_{\mathbf w})\le J$, and similarly for $\mathcal E(v)$.
The path of a list agrees with the path of each of its prefixes up to the prefix's
length. So, for the complete path of $\mathbf b\in\mathfrak B^L$ from $\bar o$, $u_1$ exists
and equals $\tilde u$ if and only if $\tilde u\le L$ and $\mathbf b_{[1,\tilde u]}\in\mathcal U(\bar o)$; and if
$\bar o=v$ is black, then $u_1=0$, and $u_2$ exists and equals $\tilde u$ if and only if
$1\le \tilde u\le L$ and $\mathbf b_{[1,\tilde u]}\in\mathcal E(v)$.

\begin{lemma}[Restart at a trace time]
\label{lem_tr_restart}\label{lem_tr_first_stop_prefix}\label{lem_tr_entry_prefix}
Let $\mathsf y_u=\mathsf y_u(\bar o,\mathbf b)$ be the complete path of $\mathbf b\in\mathfrak B^L$
from $\bar o$. Suppose $u_i=\tilde u$ exists, and let $\mathsf y'_u=\mathsf y_u(\mathsf y_{\tilde u},\mathbf b_{[\tilde u+1,L]})$
be the path of $\mathbf b_{[\tilde u+1,L]}$ from $\mathsf y_{\tilde u}$. Then $j(\mathsf y_{\tilde u})+(L-\tilde u)>J$, and for every
$k\ge1$, $u_{i-1+k}$ exists if and only if $u'_k$ (the trace
times of the path $\mathsf y'_u$) does, and then $u_{i-1+k}=\tilde u+u'_k$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem_tr_concat}(3),(2), $j(\mathsf y_{\tilde u})\ge j(\bar o)+\tilde u$ and
$\mathsf y'_u=\mathsf y_{\tilde u+u}$. Since $\mathsf y'_0=\mathsf y_{\tilde u}$ is
black, $u'_1=0$, matching $u_i=\tilde u$, and both sequences are continued
by the same rule applied to the same points.
\end{proof}

\subsection{One passage}
Between two trace times the path climbs from a black point $v$ above its column top.
This subsection shows that such a passage earns a reward with probability bounded
below, because it leaves through the white exit points of Lemma~\ref{lem_bk_exit_room}.
For $\omega\in\N$, let $\Pi_\omega$ be the set of nonempty lists $\mathbf f$ with
$l(\operatorname{pt}_i(\mathbf f))\le \omega$ for $i<|\mathbf f|$ and
$l(\operatorname{pt}_{|\mathbf f|}(\mathbf f))>\omega$, that is, the lists whose hold points
form a first-passage list over height $\omega$ (Definition~\ref{def_rn_fp_word}).
Grouping lists by their hold points and using \eqref{eq:fbc-eta} gives,
for $\Psi\ge0$,
\begin{equation}\label{eq:tr-passage-law}
\sum_{\mathbf f\in\Pi_\omega}\mathfrak q(\mathbf f)\,\Psi(\operatorname{pt}_{|\mathbf f|}(\mathbf f))
=\sum_{x\in\mathcal P}\mathsf F_\omega(x)\Psi(x).
\end{equation}
In particular $\sum_{\mathbf f\in\Pi_\omega}\mathfrak q(\mathbf f)=1$. Let $\mathbf c$ be the raw word of
$\mathbf f$. Its raw letters are at least $2$, its raw letters in $\{4,5\}$ are exactly
the closing letters of the holds, $\mathfrak q(\mathbf f)=\prod_i\mathfrak q(\mathbf c_i)$, and the
closing letter of the $i$th hold is $\mathbf c_r$ with $r=j(\operatorname{pt}_i(\mathbf f))$.

For $\omega\in\N$, with $f_\omega$ as in Lemma~\ref{lem_bk_exit_room} and the exit mass
$\mathsf q_{45}$ of Definition~\ref{def_rn_p45}, put
\begin{equation}\label{def_rn_p3}
\mathsf q_3(\omega)=\mathfrak q(3)\sum_{o=1}^3\sum_{r=1}^{f_\omega}\mathsf R(r-1,\omega+o-3),
\end{equation}
the mass of the passages over height $\omega$ that first exceed it at a raw letter $3$ at a white
exit point (Lemma~\ref{lem_tr_passage_bound}), and define the \emph{passage surplus}
\[
\delta_{\mathrm{tr}}(\omega)=E_8(\gamma)\mathsf q_{45}(\omega)+E_8(\kappa\gamma)\mathsf q_3(\omega)-d_{\mathrm{tr}},
\]
the excess of the one-passage gain at gap $\omega$ over its value $d_{\mathrm{tr}}$ at gap $0$.

\begin{lemma}[Reward over one passage]\label{lem_tr_passage_bound}
\label{lem_tr_passage_raw3}
Let $v$ be black and $\omega=\operatorname{gap}(v)$. Then
\[
\sum_{\mathbf f\in\Pi_\omega}\mathfrak q(\mathbf f)e^{-\gamma\operatorname{Rw}(v,\mathbf f;|\mathbf f|)}
\le1-d_{\mathrm{tr}}-\delta_{\mathrm{tr}}(\omega).
\]
\end{lemma}
\begin{proof}
Let $\Pi^{45}_\omega$ be the set of $\mathbf f\in\Pi_\omega$ with
$\operatorname{pt}_{|\mathbf f|}(\mathbf f)=(r,\omega+o)$, $1\le r\le f_\omega$, $1\le o\le4$. By
\eqref{eq:tr-passage-law} and Lemma~\ref{lem_rn_p45_mass}, its mass is
$\mathsf q_{45}(\omega)$. Let $\Pi^3_\omega$ be the set of $\mathbf f\in\Pi_\omega$ whose
raw word $\mathbf c$ first exceeds $\omega$ at a position $r\le f_\omega$ with $\mathbf c_r=3$, that
is, $\mathbf c_1+\cdots+\mathbf c_{r-1}\le \omega<\mathbf c_1+\cdots+\mathbf c_r$. The two sets are disjoint.
Indeed, for $\mathbf f\in\Pi^{45}_\omega$, the raw word has length $r$ and ends
with a closing letter $k\ge4\ge o$, so $\mathbf c_1+\cdots+\mathbf c_{r-1}=\omega+o-k\le \omega$ and
the raw word first exceeds $\omega$ at the closing letter.

\emph{The mass of $\Pi^3_\omega$ is $\mathsf q_3(\omega)$.} Consider the choices of
$(r,o,\mathbf c^-,\mathbf c^+,k)$ with $1\le r\le f_\omega$, $o\in\{1,2,3\}$, a word $\mathbf c^-$ of $r-1$
raw letters $\ge2$ with sum $\omega+o-3$, a word $\mathbf c^+$ of raw letters $\ge2$ outside
$\{4,5\}$, and $k\in\{4,5\}$. Splitting the raw word $\mathbf c^-$, then $3$, then $\mathbf c^+$,
then $k$ after each closing letter gives a list $\mathbf f$. Every hold of $\mathbf f$ except the last
ends inside $\mathbf c^-$, at height at most $\omega+o-3\le \omega$, and the last hold ends at
height at least $\omega+o>\omega$. So $\mathbf f\in\Pi_\omega$, and its raw word first exceeds
$\omega$ at position $r$, with raw letter $3$. Hence $\mathbf f\in\Pi^3_\omega$.
Conversely, let $\mathbf f\in\Pi^3_\omega$, and let $r$ be the position at which
its raw word first exceeds $\omega$. The raw letter $3$ at $r$ is not closing, so it lies in some hold of
$\mathbf f$ before that hold's closing letter. Earlier holds end before $r$ and at height
at most $\omega$, and this hold ends after $r$, at height greater than $\omega$. By
the definition of $\Pi_\omega$, it is the last hold of $\mathbf f$. Thus $\mathbf c$ is $\mathbf c^-$, then $3$,
then $\mathbf c^+$, then $k$, as above, the two maps are inverse, and summing the product masses
over the length $N$ of $\mathbf c^+$ gives
\[
\mathfrak q(3)\sum_{o=1}^3\sum_{r=1}^{f_\omega}\mathsf R(r-1,\omega+o-3)
\sum_{N\ge0}\Bigl(\frac{11}{16}\Bigr)^N\frac5{16}=\mathsf q_3(\omega),
\]
since $\sum_{c\notin\{4,5\}}\mathfrak q(c)=1-\frac5{16}$. The argument also
shows that, for $\mathbf f\in\Pi^3_\omega$, the raw letter $\mathbf c_r=3$ is the $i$th
raw letter of the last hold of $\mathbf f$ for some $i$, and
\[
\mathsf y_{|\mathbf f|-1}(v,\mathbf f)+(i,\mathbf c_{r-i+1}+\cdots+\mathbf c_r)=v+(r,\omega+o).
\]

\emph{Rewards.} Since $v\in\Delta(v)$ and $l_{\Delta(v)}=l(v)+\omega$
(Lemma~\ref{lem_bk_source_mem} and Definition~\ref{def_bk_triangle}), Lemma~\ref{lem_bk_exit_room} shows that $v+(r,\omega+o)$ is
white for $1\le r\le f_\omega$, $1\le o\le4$. If $\mathbf f\in\Pi^{45}_\omega$, this is its last point, so the last hold has reward
at least $1$. If $\mathbf f\in\Pi^3_\omega$, it is the point of the raw letter $\mathbf c_r=3$
displayed above, so the last hold has reward at least $\kappa$. Since rewards are nonnegative,
\[
e^{-\gamma\operatorname{Rw}(v,\mathbf f;|\mathbf f|)}\le1-(1-e^{-\gamma})\mathbf 1[\mathbf f\in\Pi^{45}_\omega]
-(1-e^{-\kappa\gamma})\mathbf 1[\mathbf f\in\Pi^3_\omega].
\]
Summing against $\mathfrak q$ and using $E_8(y)<1-e^{-y}$ for $y>0$
(Definition~\ref{def_s02_d}) gives the bound
$1-E_8(\gamma)\mathsf q_{45}(\omega)-E_8(\kappa\gamma)\mathsf q_3(\omega)=1-d_{\mathrm{tr}}-\delta_{\mathrm{tr}}(\omega)$.
\end{proof}

\begin{lemma}[Entry lists split]\label{lem_tr_entry_split}
Let $v$ be black and $\omega=\operatorname{gap}(v)$. The map $(\mathbf f,\mathbf w)\mapsto\mathbf f\mathbf w$ is a
bijection from the pairs with $\mathbf f\in\Pi_\omega$ and
$\mathbf w\in\mathcal U(v_{\mathbf f})$ onto $\mathcal E(v)$.
\end{lemma}
\begin{proof}
For a list $\mathbf b$, $l(\mathsf y_u(v,\mathbf b))>l_*(v)$ holds if and only if
$l(\operatorname{pt}_u(\mathbf b))>\omega$. Given $\mathbf b\in\mathcal E(v)$, let $k$ be the least
index with $l(\operatorname{pt}_k(\mathbf b))>\omega$; it exists because the last point lies
above $l_*(v)$. Then $\mathbf f=\mathbf b_{[1,k]}\in\Pi_\omega$, and $\mathbf f$ is the only
prefix of $\mathbf b$ in $\Pi_\omega$. By Lemma~\ref{lem_tr_concat}(3), every
$\mathsf y_u(v,\mathbf b)$ with $u\ge k$ lies above $l_*(v)$. So the defining conditions
of $\mathcal E(v)$ say exactly that
$\mathbf b_{[k+1,|\mathbf b|]}\in\mathcal U(\mathsf y_k(v,\mathbf b))$. The same two facts show that every
$\mathbf f\mathbf w$ as in the statement lies in $\mathcal E(v)$.
\end{proof}

\subsection{The entry kernel}
By Lemma~\ref{lem_tr_lemma79}, the $\mathfrak q$-average of $e^{-\gamma\operatorname{Rw}}$ at the
$(k+1)$st trace time from a black point $v$ is $\mathsf K^k\mathbf 1(v)$, for the entry
kernel $\mathsf K$ below and the constant function $\mathbf 1$, so it suffices to bound the powers of $\mathsf K$. The bridge loss
measures how far a bridge list followed by one passage falls below $1-d_{\mathrm{tr}}$; it
produces the improvement of two passages over one.
For a point $p\in\mathcal P$ and a list $\mathbf w$ write
$\mathfrak q^p(\mathbf w)=\mathfrak q(\mathbf w)e^{-\gamma\operatorname{Rw}(p,\mathbf w;|\mathbf w|)}$. The \emph{entry kernel} acts on
nonnegative functions on black points by
\[
\mathsf K\Psi(v)=\sum_{\mathbf w\in\mathcal E(v)}\mathfrak q^v(\mathbf w)\,\Psi(v_{\mathbf w}).
\]
It is positive: $0\le\Psi\le\Psi'$ implies $\mathsf K\Psi\le\mathsf K\Psi'$. For
$p\in\mathcal P$ define the \emph{bridge mass}, \emph{bridge gain} and
\emph{bridge loss}
\begin{gather*}
\operatorname{Br}(p)=\sum_{\mathbf w\in\mathcal U(p)}\mathfrak q^p(\mathbf w),\qquad
\operatorname{Gain}(p)=\sum_{\mathbf w\in\mathcal U(p)}\mathfrak q^p(\mathbf w)\,\delta_{\mathrm{tr}}(\operatorname{gap}(p_{\mathbf w})),\\
\operatorname{Loss}(p)=(1-d_{\mathrm{tr}})(1-\operatorname{Br}(p))+\operatorname{Gain}(p).
\end{gather*}

\begin{lemma}[One kernel step and the bridge loss]\label{lem_tr_kernel_one}
\begin{enumerate}
\item $\operatorname{Br}(p)\le1$ for every $p$, and
$\mathsf K\mathbf 1(v)\le1-d_{\mathrm{tr}}-\delta_{\mathrm{tr}}(\operatorname{gap}(v))\le1-d_{\mathrm{tr}}$ for
every black $v$.
\item $0\le\operatorname{Loss}(p)\le1-d_{\mathrm{tr}}$ for every $p$; $\operatorname{Loss}(p)=\delta_{\mathrm{tr}}(\operatorname{gap}(p))$ if $p$ is
black, and $\operatorname{Loss}(p)=1-d_{\mathrm{tr}}$ if $j(p)>J$.
\item For every black $v$, with $\omega=\operatorname{gap}(v)$,
\[
\mathsf K^2\mathbf 1(v)\le(1-d_{\mathrm{tr}})^2-(1-d_{\mathrm{tr}})\delta_{\mathrm{tr}}(\omega)
-\sum_{\mathbf f\in\Pi_\omega}\mathfrak q^v(\mathbf f)\,\operatorname{Loss}(v_{\mathbf f}).
\]
\end{enumerate}
\end{lemma}
\begin{proof}
(1) Rewards are nonnegative, so $\mathfrak q^p(\mathbf w)\le\mathfrak q(\mathbf w)$. Lists in
$\mathcal U(p)$ have length less than $J$ and form a prefix-disjoint family,
so $\operatorname{Br}(p)\le1$ by Lemma~\ref{lem_tr_concat}(4). By
Lemma~\ref{lem_tr_entry_split} with $\omega=\operatorname{gap}(v)$ and Lemma~\ref{lem_tr_concat}(1),(2),
$\mathfrak q^v(\mathbf f\mathbf w)=\mathfrak q^v(\mathbf f)\mathfrak q^{v_{\mathbf f}}(\mathbf w)$, and therefore
\[
\mathsf K\mathbf 1(v)=\sum_{\mathbf f\in\Pi_\omega}\mathfrak q^v(\mathbf f)\operatorname{Br}(v_{\mathbf f})\le\sum_{\mathbf f\in\Pi_\omega}\mathfrak q^v(\mathbf f).
\]
Lemma~\ref{lem_tr_passage_bound} and
Lemma~\ref{lem_tr_delta_floor}(1) conclude.

(2) By (1), $0\le\delta_{\mathrm{tr}}(\operatorname{gap}(v))\le1-d_{\mathrm{tr}}$ for every
black $v$, so $0\le\operatorname{Gain}(p)\le(1-d_{\mathrm{tr}})\operatorname{Br}(p)$, which gives the two bounds. If $p$
is black, then $\mathcal U(p)$ consists of the empty list only, so $\operatorname{Br}(p)=1$
and $\operatorname{Gain}(p)=\delta_{\mathrm{tr}}(\operatorname{gap}(p))$. If $j(p)>J$, no point of
a path from $p$ is black, so $\mathcal U(p)=\emptyset$.

(3) Apply (1) at each $v_{\mathbf w}$ and Lemma~\ref{lem_tr_entry_split}:
\[
\mathsf K^2\mathbf 1(v)\le\sum_{\mathbf f\in\Pi_\omega}\mathfrak q^v(\mathbf f)
\sum_{\mathbf w\in\mathcal U(v_{\mathbf f})}\mathfrak q^{v_{\mathbf f}}(\mathbf w)\bigl(1-d_{\mathrm{tr}}-\delta_{\mathrm{tr}}(\operatorname{gap}((v_{\mathbf f})_{\mathbf w}))\bigr).
\]
The inner sum is $(1-d_{\mathrm{tr}})\operatorname{Br}(v_{\mathbf f})-\operatorname{Gain}(v_{\mathbf f})=(1-d_{\mathrm{tr}})-\operatorname{Loss}(v_{\mathbf f})$, a
difference of finite numbers. Since $1-d_{\mathrm{tr}}>0$,
Lemma~\ref{lem_tr_passage_bound} bounds $(1-d_{\mathrm{tr}})\sum_{\mathbf f}\mathfrak q^v(\mathbf f)$ by
$(1-d_{\mathrm{tr}})(1-d_{\mathrm{tr}}-\delta_{\mathrm{tr}}(\omega))$.
\end{proof}

\subsection{Two passages}
Squaring the one-passage bound would give only $(1-d_{\mathrm{tr}})^2$; Proposition~\ref{lem_tr_kernel_two} improves it to $\mathsf r_2=(1-d_{\mathrm{tr}})^2-\Omega$ (Definition~\ref{def_s02_d}). The correction
$\Omega$ comes from the loss after the first of the two passages. If that passage ends
at a black point, the loss is the surplus of the next passage. If it ends
at a white point, the path earns a reward before its next trace time. Put
\[
d_{\mathrm{pair}}=\frac3{16}\delta_{\mathrm{tr}}(6)+\frac18\delta_{\mathrm{tr}}(5),\qquad
d_{\mathrm{wh}}=\frac{10000}{3299}d_{\mathrm{pair}}:
\]
$d_{\mathrm{pair}}$ bounds the surplus after the two closing letters $4,5$ of a hold
(Lemma~\ref{lem_tr_delta_floor}(2)), and $d_{\mathrm{wh}}$ the resulting loss at a white point
(Lemma~\ref{lem_tr_bridge_white}).

In this subsection every hold is written $(\mathbf d,k)=(\mathbf c_1,\ldots,\mathbf c_N,k)$, where
$\mathbf d=(\mathbf c_1,\ldots,\mathbf c_N)$, $N\in\N$, is the word of its raw letters before the
closing letter $k\in\{4,5\}$ (these are at least $2$ and outside $\{4,5\}$), and
$\mathfrak q((\mathbf d,k))=\mathfrak q(\mathbf d)\mathfrak q(k)$. For
such a word $\mathbf d$ let $\#_3(\mathbf d)$ be the number of its raw letters equal to $3$; for
$p\in\mathcal P$ let $\#^{\mathrm w}_3(p,\mathbf d)$ be the number of $i$ with $\mathbf c_i=3$ and
$p+(i,\mathbf c_1+\cdots+\mathbf c_i)$ white, and put
$\operatorname{disc}(p,\mathbf d)=e^{-\kappa\gamma\#^{\mathrm w}_3(p,\mathbf d)}$.
Since $1-\kappa\gamma=1163/1250$, summing over the words of length $N$
factorizes:
\begin{equation}\label{eq:tr-word-sum}
\sum_{|\mathbf d|=N}\mathfrak q(\mathbf d)(1-\kappa\gamma)^{\#_3(\mathbf d)}
=\Bigl(1-\frac14-\frac5{16}+\frac14\cdot\frac{1163}{1250}\Bigr)^N
=\Bigl(\frac{6701}{10000}\Bigr)^N.
\end{equation}

\begin{lemma}[A white point loses at least $d_{\mathrm{wh}}$]\label{lem_tr_bridge_white}
\label{lem_tr_word_sum}
For every white $p$, $\operatorname{Loss}(p)\ge d_{\mathrm{wh}}$.
\end{lemma}
\begin{proof}
Since $p$ is not black, every $\mathbf w\in\mathcal U(p)$ is nonempty, and
$\mathbf w\mapsto(b,\mathbf w')$ with $\mathbf w=(b)\mathbf w'$ is a bijection onto the pairs with
$b\in\mathfrak B$ and $\mathbf w'\in\mathcal U(p_b)$. Since
$\mathfrak q^p((b)\mathbf w')=\mathfrak q(b)e^{-\gamma\operatorname{rw}(p,b)}\mathfrak q^{p_b}(\mathbf w')$ and
$\sum_b\mathfrak q(b)=1$,
\begin{equation}\label{eq:tr-loss-step}
\operatorname{Loss}(p)=\sum_{b\in\mathfrak B}\mathfrak q(b)\mathsf L(p,b),\qquad
\mathsf L(p,b)=(1-d_{\mathrm{tr}})\bigl(1-e^{-\gamma\operatorname{rw}(p,b)}\bigr)+e^{-\gamma\operatorname{rw}(p,b)}\operatorname{Loss}(p_b)\ge0.
\end{equation}

Fix $\mathbf d$ as above. Then $\operatorname{rw}(p,(\mathbf d,k))=\mathbf 1[p_{(\mathbf d,k)}\text{ white}]+\kappa\#^{\mathrm w}_3(p,\mathbf d)$. We claim
\begin{equation}\label{eq:tr-pair}
\tfrac3{16}\mathsf L(p,(\mathbf d,4))+\tfrac18\mathsf L(p,(\mathbf d,5))\ge \operatorname{disc}(p,\mathbf d)\,d_{\mathrm{pair}}.
\end{equation}
The points $p_{(\mathbf d,4)}$ and $p_{(\mathbf d,5)}=p_{(\mathbf d,4)}+(0,1)$ lie in the same column. If
$j(p_{(\mathbf d,4)})>J$, then $\mathsf L(p,(\mathbf d,k))=1-d_{\mathrm{tr}}$ by Lemma~\ref{lem_tr_kernel_one}(2),
and $\frac5{16}(1-d_{\mathrm{tr}})>d_{\mathrm{pair}}$. If both points are black, then
$\mathsf L(p,(\mathbf d,k))\ge \operatorname{disc}(p,\mathbf d)\,\delta_{\mathrm{tr}}(\operatorname{gap}(p_{(\mathbf d,k)}))$. By
\eqref{eq:tr-gap-adjacent} the two gaps are $\omega$ and $\omega-1$ for some
$\omega\ge1$, and Lemma~\ref{lem_tr_delta_floor}(2) applies. Otherwise some
$p_{(\mathbf d,k')}$ is white. Then $e^{-\gamma\operatorname{rw}(p,(\mathbf d,k'))}=\operatorname{disc}(p,\mathbf d)\,e^{-\gamma}$, so
$\mathsf L(p,(\mathbf d,k'))\ge(1-d_{\mathrm{tr}})(1-\operatorname{disc}(p,\mathbf d)\,e^{-\gamma})\ge(1-d_{\mathrm{tr}})\operatorname{disc}(p,\mathbf d)(1-e^{-\gamma})>(1-d_{\mathrm{tr}})\operatorname{disc}(p,\mathbf d)E_8(\gamma)$.
Since $\mathfrak q(k')\ge1/8$, Lemma~\ref{lem_tr_delta_floor}(3) gives
\eqref{eq:tr-pair}.

Finally $\operatorname{disc}(p,\mathbf d)\ge e^{-\kappa\gamma\#_3(\mathbf d)}\ge(1-\kappa\gamma)^{\#_3(\mathbf d)}$.
Multiplying \eqref{eq:tr-pair} by $\mathfrak q(\mathbf d)$, summing over $\mathbf d$ and
using \eqref{eq:tr-word-sum} gives
$\operatorname{Loss}(p)\ge d_{\mathrm{pair}}\sum_{N\ge0}(6701/10000)^N=d_{\mathrm{wh}}$.
\end{proof}

\begin{lemma}[A black point of gap zero]\label{lem_tr_gap_zero_loss}
If $v$ is black with $\operatorname{gap}(v)=0$, then
\[
\sum_{\mathbf f\in\Pi_0}\mathfrak q^v(\mathbf f)\,\operatorname{Loss}(v_{\mathbf f})
\ge e^{-\gamma}d_{\mathrm{wh}}\Bigl(\frac3{16}+\frac{6701}{26392}\Bigr).
\]
\end{lemma}
\begin{proof}
Holds climb at least $4$ (Lemma~\ref{lem_rn_hold_mass}(3)), so $\Pi_0$ consists of the one-hold lists
$(b)$, $b\in\mathfrak B$. Their terms are
$\mathfrak q(b)e^{-\gamma\operatorname{rw}(v,b)}\operatorname{Loss}(v_b)\ge0$.
For the hold $b=(4)$ the point $v+(1,4)$ is white by
Lemma~\ref{lem_bk_exit_room} ($\omega=0$, $r=1$, $o=4$). So its reward is
$1$, and by Lemma~\ref{lem_tr_bridge_white} its term is at least
$\frac3{16}e^{-\gamma}d_{\mathrm{wh}}$. For a nonempty $\mathbf d$, the two holds $(\mathbf d,4)$, $(\mathbf d,5)$ contribute at least
$\mathfrak q(\mathbf d)\operatorname{disc}(v,\mathbf d)\cdot\frac5{16}(1-d_{\mathrm{tr}})$ if $j(v_{(\mathbf d,4)})>J$, at least
$\mathfrak q(\mathbf d)\operatorname{disc}(v,\mathbf d)d_{\mathrm{pair}}$ if $v_{(\mathbf d,4)},v_{(\mathbf d,5)}$ are black, and otherwise at least
$\frac18\mathfrak q(\mathbf d)\operatorname{disc}(v,\mathbf d)\,e^{-\gamma}d_{\mathrm{wh}}$, by
Lemma~\ref{lem_tr_kernel_one}(2), \eqref{eq:tr-gap-adjacent}
and Lemma~\ref{lem_tr_delta_floor}(2), and Lemma~\ref{lem_tr_bridge_white}
respectively. Since $\frac5{16}(1-d_{\mathrm{tr}})>d_{\mathrm{pair}}>d_{\mathrm{wh}}/8$, each case
gives at least $\frac18\mathfrak q(\mathbf d)(1-\kappa\gamma)^{\#_3(\mathbf d)}e^{-\gamma}d_{\mathrm{wh}}$.
Summing over $N\ge1$ by \eqref{eq:tr-word-sum} gives the factor
$\frac18\cdot\frac{6701}{3299}=\frac{6701}{26392}$.
\end{proof}

\begin{proposition}[The entry kernel]\label{lem_tr_kernel_two}
\label{lem_tr_kernel_powers}\label{lem_tr_kernel_two_bound}
For every black point $v$,
\[
\mathsf K\mathbf 1(v)\le1-d_{\mathrm{tr}},\qquad \mathsf K^2\mathbf 1(v)\le\mathsf r_2,
\]
and hence $\mathsf K^i\mathbf 1(v)\le\mathsf r_2^{\floor{i/2}}(1-d_{\mathrm{tr}})^{i-2\floor{i/2}}$ for
every $i\in\N$.
\end{proposition}
\begin{proof}
The first bound is Lemma~\ref{lem_tr_kernel_one}(1). Let
$\omega=\operatorname{gap}(v)$. Losses are nonnegative, so if $\omega\ge1$,
Lemma~\ref{lem_tr_kernel_one}(3) and Lemma~\ref{lem_tr_delta_floor}(1),(3) give
$\mathsf K^2\mathbf 1(v)\le(1-d_{\mathrm{tr}})^2-(1-d_{\mathrm{tr}})\delta_{\mathrm{tr}}(6)<\mathsf r_2$. If
$\omega=0$, then $\delta_{\mathrm{tr}}(0)=0$, and Lemma~\ref{lem_tr_gap_zero_loss}
and $e^{-\gamma}>1-\gamma>0$ give
\[
\mathsf K^2\mathbf 1(v)\le(1-d_{\mathrm{tr}})^2-e^{-\gamma}d_{\mathrm{wh}}\Bigl(\frac3{16}+\frac{6701}{26392}\Bigr)
<(1-d_{\mathrm{tr}})^2-\Omega=\mathsf r_2
\]
by Lemma~\ref{lem_tr_delta_floor}(3) and the definition of $\mathsf r_2$. For the powers, induct on $i$
using positivity: if $\mathsf K^i\mathbf 1$ is at most the constant of the statement pointwise, then
$\mathsf K^{i+2}\mathbf 1=\mathsf K^2(\mathsf K^i\mathbf 1)$ is at most $\mathsf r_2$ times that constant, which is the constant for $i+2$.
\end{proof}

\subsection{Accumulated rewards at the trace times}
For $\mathbf b\in\mathfrak B^L$ whose path from $\bar o$ is complete,
and $k\in\N$, put
$\mathcal Z_{k+1}(\bar o,\mathbf b)=e^{-\gamma\operatorname{Rw}(\bar o,\mathbf b;u_{k+1})}$ if $u_{k+1}$
exists, and $\mathcal Z_{k+1}(\bar o,\mathbf b)=0$ otherwise.

\begin{lemma}[Explicit form of Tao's Lemma 7.9]\label{lem_tr_lemma79}\label{lem_tr_moment_black}
Let $\bar o\in\mathcal P$ and $L\in\N$ with $j(\bar o)+L>J$, so that the paths
from $\bar o$ of all $\mathbf b\in\mathfrak B^L$ are complete. Then
\begin{enumerate}
\item if $\bar o=v$ is black, then
$\sum_{\mathbf b\in\mathfrak B^L}\mathfrak q(\mathbf b)\mathcal Z_{k+1}(v,\mathbf b)=\mathsf K^k\mathbf 1(v)$ for every
$k\in\N$;
\item for every $k\in\N$,
$\sum_{\mathbf b\in\mathfrak B^L}\mathfrak q(\mathbf b)\mathcal Z_{k+1}(\bar o,\mathbf b)
\le\mathsf r_2^{\floor{k/2}}(1-d_{\mathrm{tr}})^{k-2\floor{k/2}}$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) We induct on $k$, for all black $v$ and all $L$ with $j(v)+L>J$ at
once. For $k=0$, $u_1=0$ and $\mathcal Z_1(v,\mathbf b)=1$, and the total mass is
$1$. For the step, by the description of $u_2$ before Lemma~\ref{lem_tr_restart}, that
lemma with $i=2$, and Lemma~\ref{lem_tr_concat}(2),
\[
\mathcal Z_{k+2}(v,\mathbf b)=\sum_{u=1}^L\mathbf 1[\mathbf b_{[1,u]}\in\mathcal E(v)]\,
e^{-\gamma\operatorname{Rw}(v,\mathbf b_{[1,u]};u)}\,\mathcal Z_{k+1}(\mathsf y_{u}(v,\mathbf b),\mathbf b_{[u+1,L]}),
\]
with at most one nonzero term. Split $\mathbf b$ at $u$ by
Lemma~\ref{lem_tr_concat}(1). For $\mathbf w\in\mathcal E(v)$, the point $v_{\mathbf w}$ is
black and $j(v_{\mathbf w})+L-|\mathbf w|>J$, so the induction hypothesis evaluates the sum
over the remainder as $\mathsf K^k\mathbf 1(v_{\mathbf w})$. Every $\mathbf w\in\mathcal E(v)$
has $|\mathbf w|\le J-j(v)<L$. Hence the total is
$\sum_{\mathbf w\in\mathcal E(v)}\mathfrak q^v(\mathbf w)\mathsf K^k\mathbf 1(v_{\mathbf w})=\mathsf K^{k+1}\mathbf 1(v)$.

(2) In the same way, the description of $u_1$ and Lemma~\ref{lem_tr_restart} with $i=1$ give
\[
\sum_\mathbf b\mathfrak q(\mathbf b)\mathcal Z_{k+1}(\bar o,\mathbf b)
=\sum_{\mathbf w\in\mathcal U(\bar o)}\mathfrak q^{\bar o}(\mathbf w)\sum_{\mathbf w'\in\mathfrak B^{L-|\mathbf w|}}\mathfrak q(\mathbf w')\mathcal Z_{k+1}(\bar o_{\mathbf w},\mathbf w').
\]
By (1) and Proposition~\ref{lem_tr_kernel_two}, the inner sum is at most
the claimed bound. Finally
$\sum_{\mathbf w\in\mathcal U(\bar o)}\mathfrak q^{\bar o}(\mathbf w)=\operatorname{Br}(\bar o)\le1$.
\end{proof}

\section{Boundary points far below the top edge}\label{sec:fb-case3}
This section proves Case~3 of the descent step (Section~\ref{sec:fb-charsum}): the boundary point $p$ for $m$ is black
and lies in a triangle $\Delta\in\mathfrak T$ at gap $\operatorname{gap}(p)=l_\Delta-l(p)$ greater
than $m/(\log m)^2$ below its top edge (Lemma~\ref{lem_c3_case3}). Following Tao \cite[Case 3 of
Proposition 7.8 and Lemma 7.10]{Tao}, we restart the hold walk after its first passage above the
top edge; outside rare events, a low reward then forces many entries into triangles
(Lemma~\ref{lem_tr_survival_time}), which Lemma~\ref{lem_tr_lemma79} makes unlikely, and the main
rare event, a visit to a large triangle, is bounded by the packing estimate
(Lemma~\ref{lem_c3_lemma710}). The constants are those of Definitions~\ref{def_Dstar},
\ref{def_eps_star}, \ref{def_s02_schedule}, \ref{def_Pstar} and~\ref{def_s02_d}, and
$Q_{\mathrm{pk}}$ of Lemma~\ref{lem_c3_pack_const}(2).

\subsection{The fresh law and the Case~3 events}
In Case~3 the path is restarted at a boundary point $p\in\mathcal P$ after a
first passage over height $\omega\in\N$, with the law $\mathsf F_\omega$.

\begin{definition}[Fresh law]\label{def_tr_fresh_law}\label{def_tr_atoms}
\label{def_tr_fresh_path}\label{def_tr_white_count}
The \emph{atoms} are the pairs $a=(x,\mathbf b)$ with \emph{displacement}
$x\in\mathcal P$ and $\mathbf b=(b^1,\ldots,b^J)\in\mathfrak B^J$.
The \emph{fresh law} is
$\Prob_{p,\omega}(a)=\mathsf F_\omega(x)\,\mathfrak q(\mathbf b)$ (independent of $p$), and for a set
$\mathcal Q$ of atoms, $\Prob_{p,\omega}(\mathcal Q)=\sum_{a\in\mathcal Q}\Prob_{p,\omega}(a)$.
The \emph{fresh path} and its accumulated reward are
\[
y^p_u(a)=p+x+\operatorname{pt}_{\min(u,J)}(\mathbf b),\qquad
\operatorname{Rw}^p_u(a)=\operatorname{Rw}(p+x,\mathbf b;u).
\]
Thus $y^p_u(a)=\mathsf y_u(p+x,\mathbf b)$ is the path of $\mathbf b$ from $p+x$.
\end{definition}

By Lemmas~\ref{lem_rn_fp_mass}(1) and~\ref{lem_rn_hold_mass}(5), $\Prob_{p,\omega}$ is a probability on the
atoms. Since $j(p+x)\ge1$, every fresh path is complete.

\begin{lemma}[Displacement marginal]\label{lem_tr_fresh_marginal}
For $0\le k\le J$ and $\Psi\ge0$ on $\mathcal P\times\mathcal P^k$,
\[
\sum_{a=(x,\mathbf b)}\Prob_{p,\omega}(a)\,\Psi\bigl(x,(\operatorname{pt}(b^1),\ldots,\operatorname{pt}(b^k))\bigr)
=\sum_{x\in\mathcal P}\mathsf F_\omega(x)\sum_{\mathbf h\in\mathcal P^k}\eta^{\otimes k}(\mathbf h)\,\Psi(x,\mathbf h).
\]
\end{lemma}
\begin{proof}
Split $\mathbf b$ after $k$ holds and sum out the remainder, which has
total mass $1$ (Lemmas~\ref{lem_tr_concat}(1) and~\ref{lem_rn_hold_mass}(5)). Then group the first $k$
holds by their points, using \eqref{eq:fbc-eta}.
\end{proof}

\begin{definition}[Case~3 events]\label{def_tr_low}\label{def_tr_bad}
\label{def_tr_estar}\label{def_tr_clear}
For $p\in\mathcal P$ and $m\in\N$, define the sets of atoms
\begin{align*}
\mathrm{Low}^p&=\{a:\ \operatorname{Rw}^p_{P-1}(a)\le T_{\mathrm{tr}}\},\\
\mathrm{Bad}^p_m&=\{a:\ 400\,(j(y^p_P(a))-j(p))\ge317m\},\\
\mathrm E^{*,p}&=\{a:\ y^p_u(a)\in\Delta\text{ and }S_\Delta\ge\beta(u)
\text{ for some }0\le u\le P-1,\ \Delta\in\mathfrak T\},\\
\mathrm{Clr}&=\Bigl\{a:\ \sum_{i=u+1}^{u+h_{\mathrm{tr}}(u)}l(\operatorname{pt}(b^i))\le\frac{13}9\beta(u)
\text{ for some }u\text{ with }u+h_{\mathrm{tr}}(u)\le P-1\Bigr\}.
\end{align*}
We call $\mathrm{Low}^p$ the \emph{low-reward event}, $\mathrm E^{*,p}$ the \emph{oversize event} and $\mathrm{Clr}$ the
\emph{slow-clearing event}.
\end{definition}
Off $\mathrm{Bad}^p_m$ the advance $j(y^p_P)-j(p)$ stays below $\frac{317}{400}m$, so the factor
$\max(m-\cdot,1)^{-A_*}$ of the boundary reduction stays comparable to $m^{-A_*}$; off
$\mathrm E^{*,p}\cup\mathrm{Clr}$ the path climbs out of every triangle it enters on schedule.

\subsection{The Markov bound}
On $\mathrm{Low}^p$ outside $\mathrm{Bad}^p_m\cup\mathrm E^{*,p}\cup\mathrm{Clr}$ the fresh path has
$r_{\mathrm{tr}}+1$ trace times before time $P$ (Lemma~\ref{lem_tr_survival_time}), so
Lemma~\ref{lem_tr_lemma79} and Markov's inequality bound its mass, the \emph{Markov bound}
(Lemma~\ref{lem_tr_survival_export}).

\begin{lemma}[The trace times come on schedule]\label{lem_tr_survival_time}
\label{lem_tr_next_stop}\label{lem_tr_survival_first_hit}
Let $p\in\mathcal P$ and $m\in\N$ with $j(p)+m=J$, suppose $P\le J$, and let
$a=(x,\mathbf b)\in\mathrm{Low}^p\setminus(\mathrm{Bad}^p_m\cup\mathrm E^{*,p}\cup\mathrm{Clr})$.
Then the fresh path
$y^p_u=y^p_u(a)$ has at least $r_{\mathrm{tr}}+1=581$ trace times, and
$u_i\le \mathfrak p_{i-1}$ for $1\le i\le r_{\mathrm{tr}}+1$.
\end{lemma}
\begin{proof}
Since $a\notin\mathrm{Bad}^p_m$, $j(y^p_P)<j(p)+m=J$, so $j(y^p_u)\le J$ for
$u\le P$. Every point $y^p_u$ with $u\le P$ is therefore black or white.
Also $\mathfrak p_i$ increases with $i$ (Lemma~\ref{lem:schedule}(1)), and
$P-1=\mathfrak p_{r_{\mathrm{tr}}}\le J-1$. By Lemma~\ref{lem_tr_concat}(5), for every
$u'\le P-1$ the number of white points among $y^p_1,\ldots,y^p_{u'}$ is at
most $\operatorname{Rw}^p_{P-1}(a)\le T_{\mathrm{tr}}$. In particular, every run of
$T_{\mathrm{tr}}+1$ consecutive times in $[1,P-1]$ contains a hit.

For $i=1$, the run $[1,T_{\mathrm{tr}}+1]$, where
$T_{\mathrm{tr}}+1=102\le \mathfrak p_0$, contains a hit, so
$u_1\le \mathfrak p_0$. Suppose $u_i=\tilde u\le \mathfrak p_{i-1}$ with $i\le r_{\mathrm{tr}}$. Since $\mathfrak g$ is nondecreasing
(Lemma~\ref{lem:schedule}(1)),
$\tilde u+h_{\mathrm{tr}}(\tilde u)+T_{\mathrm{tr}}+1=\mathfrak g(\tilde u)\le\mathfrak g(\mathfrak p_{i-1})=\mathfrak p_i\le P-1$.
Let $\Delta=\Delta(y^p_{\tilde u})$. Then
$y^p_{\tilde u}\in\Delta$ (Lemma~\ref{lem_bk_source_mem}) and $l_\Delta=l_*(y^p_{\tilde u})$. As
$a\notin\mathrm E^{*,p}$, $S_\Delta<\beta(\tilde u)$, and membership gives
$\operatorname{gap}(y^p_{\tilde u})\log2\le S_\Delta$. Since $\log2>9/13$, we get
$\operatorname{gap}(y^p_{\tilde u})<\frac{13}9\beta(\tilde u)$. As $a\notin\mathrm{Clr}$ and
$\tilde u+h_{\mathrm{tr}}(\tilde u)\le P-1\le J$,
\[
l(y^p_{\tilde u+h_{\mathrm{tr}}(\tilde u)})-l(y^p_{\tilde u})=\sum_{k=\tilde u+1}^{\tilde u+h_{\mathrm{tr}}(\tilde u)}l(\operatorname{pt}(b^k))>\frac{13}9\beta(\tilde u).
\]
So $l(y^p_u)>l_*(y^p_{\tilde u})$ for all $u\ge \tilde u+h_{\mathrm{tr}}(\tilde u)$, by Lemma~\ref{lem_tr_concat}(3).
The run $[\tilde u+h_{\mathrm{tr}}(\tilde u)+1,\,\tilde u+h_{\mathrm{tr}}(\tilde u)+T_{\mathrm{tr}}+1]$ contains a hit $u''$; the point
$y^p_{u''}$ lies above $l_*(y^p_{\tilde u})$, so $u_{i+1}$ exists and $u_{i+1}\le u''\le \mathfrak p_i$.
\end{proof}

\begin{lemma}[Slow clearing is rare]\label{lem_tr_clear_mass}
Let $p\in\mathcal P$ and $\omega\in\N$. If $P\le J$, then $\Prob_{p,\omega}(\mathrm{Clr})<2^{-43}$.
\end{lemma}
\begin{proof}
Fix $u$ with $u+h_{\mathrm{tr}}(u)\le P-1$. By Lemma~\ref{lem_tr_fresh_marginal} with
$k=u+h_{\mathrm{tr}}(u)\le J$, the points $\operatorname{pt}(b^i)$, $u<i\le u+h_{\mathrm{tr}}(u)$, are
independent with law $\eta$ under $\Prob_{p,\omega}$. So, by the third bound of
Lemma~\ref{lem_rn_list_ver_low}, the atoms
with $\sum_{i=u+1}^{u+h_{\mathrm{tr}}(u)}l(\operatorname{pt}(b^i))\le\frac{13}9\beta(u)$ have mass at most
$(4095/4096)^{\frac{127}{10}h_{\mathrm{tr}}(u)-\frac{13}9\beta(u)}$. Since
$h_{\mathrm{tr}}(u)\ge\frac4{35}\beta(u)$ and $\beta(u)\ge K_{\mathrm{tr}}\cdot(u+1)$,
\[
\frac{127}{10}h_{\mathrm{tr}}(u)-\frac{13}9\beta(u)\ge\Bigl(\frac{508}{350}-\frac{13}9\Bigr)\beta(u)
=\frac{11}{1575}\beta(u)\ge32\cdot4096\,(u+1),
\]
using the integer comparison $11K_{\mathrm{tr}}=108493309639\ge32\cdot1575\cdot4096=206438400$.
As $\log(1-1/4096)\le-1/4096$, this mass is at most $e^{-32(u+1)}$. Summing
over $u\ge0$ gives at most $2e^{-32}$, and $2e^{-32}<2^{-43}$ because
$44\log2<44\cdot347/500<32$.
\end{proof}

\begin{lemma}[Case~3 low-reward bound]\label{lem_tr_survival_export}\label{lem_tr_trace_event}
Let $p\in\mathcal P$ and $\omega,m\in\N$ with $j(p)+m=J$. If $P\le J$, then
\[
\Prob_{p,\omega}(\mathrm{Low}^p\setminus\mathrm{Bad}^p_m)
\le\Prob_{p,\omega}(\mathrm E^{*,p})+\frac1{P_{\mathrm{tr}}}+2^{-43}.
\]
\end{lemma}
\begin{proof}
Let $\mathcal Q=\mathrm{Low}^p\setminus(\mathrm{Bad}^p_m\cup\mathrm E^{*,p}\cup\mathrm{Clr})$. Since
$\Prob_{p,\omega}(\mathrm{Clr})<2^{-43}$ (Lemma~\ref{lem_tr_clear_mass}), it suffices to prove
the Markov bound
$\Prob_{p,\omega}(\mathcal Q)\le e^{T_{\mathrm{tr}}\gamma}\mathsf r_2^{r_{\mathrm{tr}}/2}=e^{101\gamma}\mathsf r_2^{290}$,
which is less than $1/P_{\mathrm{tr}}$ by Lemma~\ref{lem_s02_trace_term}.
Let $a=(x,\mathbf b)\in\mathcal Q$ and put $\bar o=p+x$. By
Lemma~\ref{lem_tr_survival_time}, $u_{r_{\mathrm{tr}}+1}$ exists and
$u_{r_{\mathrm{tr}}+1}\le P-1$. The accumulated reward is nondecreasing, so
$\operatorname{Rw}^p_{u_{r_{\mathrm{tr}}+1}}(a)\le\operatorname{Rw}^p_{P-1}(a)\le T_{\mathrm{tr}}$, and hence
$\mathbf 1[a\in\mathcal Q]\le e^{T_{\mathrm{tr}}\gamma}\mathcal Z_{r_{\mathrm{tr}}+1}(\bar o,\mathbf b)$.
Sum against $\Prob_{p,\omega}$: Lemma~\ref{lem_tr_lemma79}(2) with $k=r_{\mathrm{tr}}$, which is even, and
$L=J$ bounds the sum over $\mathbf b$ by $\mathsf r_2^{r_{\mathrm{tr}}/2}$, and
$\sum_{x}\mathsf F_\omega(x)=1$.
\end{proof}

\subsection{Row geometry of triangles}
For a triangle $\Delta$ and $l\in\Z$ put
\[
\tip_\Delta=l_\Delta-\frac{S_\Delta}{\log2},\qquad
\row_\Delta(l)=j_\Delta+\lambda\bigl(l-\tip_\Delta\bigr).
\]
Thus $\tip_\Delta$ is the (real) height at which the hypotenuse meets the left edge, and
$\row_\Delta(l)$ is the right end of row $l$. After a first passage over height $\omega$ the fresh path
lies a little above the top edge of $\Delta$, about $\omega/4$ to the right, while the rows of $\Delta$
widen by $\lambda \omega$ over that height; the margin $\lambda_0=\lambda-\frac14$ satisfies
$\frac1{16}\le\lambda_0\le\frac3{20}$ (Lemma~\ref{lem_c3_pack_const}(1)). Making explicit the
geometric steps of the proof of \cite[Lemma 7.10]{Tao}, we show that large triangles met there
have tips at or above $l_\Delta$ (Lemma~\ref{lem_c3_close}) and corners in well-separated columns
(Lemma~\ref{lem_c3_anchor_sep}), so that the packing bound of Lemma~\ref{lem_rn_fp_packing} applies.

\begin{lemma}[Rows of a triangle]\label{lem_c3_row_mem}
Let $\Delta$ be a triangle and $(j,l)\in\mathcal P$ with $l\le l_\Delta$.
Then $(j,l)\in\Delta$ if and only if $j_\Delta\le j\le\row_\Delta(l)$. In
particular every point $(j,l)\in\Delta$ satisfies $l\ge\tip_\Delta$ and
$(j-j_\Delta)\log9\le(l-\tip_\Delta)\log2$.
\end{lemma}
\begin{proof}
Since $S_\Delta-(l_\Delta-l)\log2=(l-\tip_\Delta)\log2$, the third
membership condition reads, as in \cite[(7.65)]{Tao}, $(j-j_\Delta)\log9\le(l-\tip_\Delta)\log2$, that
is, $j\le\row_\Delta(l)$. With $j\ge j_\Delta$ it forces $l\ge\tip_\Delta$.
\end{proof}

\begin{lemma}[Tips of nearby triangles]\label{lem_c3_close}
Let $\Delta\ne\Delta'$ be members of $\mathfrak T$, let
$p\in\Delta$ and $\omega=\operatorname{gap}(p)$. Let $(j,l)\in\Delta'$ with
$l_\Delta\le l\le l_\Delta+W$ and $|j-(j(p)+\omega/4)|\le Y$, where
$W,Y\ge0$ and $Y+\lambda W+1\le \omega\lambda_0$. Then
$\tip_{\Delta'}\ge l_\Delta$.
\end{lemma}
\begin{proof}
Suppose $\tip_{\Delta'}<l_\Delta$; as in the proof of \cite[Lemma 7.10]{Tao}, we find a point
in both triangles, contradicting Proposition~\ref{lem_bk_cover}. Put $\bar\jmath=\row_{\Delta'}(l_\Delta)$, so
$\bar\jmath>j_{\Delta'}$. By Lemma~\ref{lem_c3_row_mem},
$j\le\row_{\Delta'}(l)=\bar\jmath+\lambda(l-l_\Delta)\le\bar\jmath+\lambda W$. Let
$j'=\min(j,\floor{\bar\jmath})$. Then $j_{\Delta'}\le j'\le\bar\jmath$ and
$0\le j-j'\le\lambda W+1$. Since $l_\Delta\le l\le l_{\Delta'}$,
Lemma~\ref{lem_c3_row_mem} gives $(j',l_\Delta)\in\Delta'$.

On the other hand $\omega\ge0$ and $\lambda_0\le1/4$, so
$Y+\lambda W+1\le \omega\lambda_0\le \omega/4$. Hence
$j'\ge j-\lambda W-1\ge j(p)+\omega/4-(Y+\lambda W+1)\ge j(p)\ge j_\Delta$, and
$j'\le j\le j(p)+\omega/4+(Y+\lambda W+1)\le j(p)+(\frac14+\lambda_0)\omega=j(p)+\lambda \omega$.
Multiplying the last inequality by $\log9$ and using the membership of
$p$,
\[
(j'-j_\Delta)\log9\le(j(p)-j_\Delta)\log9+\omega\log2\le S_\Delta.
\]
Thus $(j',l_\Delta)\in\Delta$ as well.
\end{proof}

For $\Delta\in\mathfrak T$ and reals $\mathsf S$ and $W\ge0$, the \emph{anchor set}
$\operatorname{Anc}(\Delta,\mathsf S,W)\subseteq\Z$ consists of the raw lengths
$j_{\Delta'}$ of the corners of the $\Delta'\in\mathfrak T$ with $S_{\Delta'}\ge\mathsf S$ and
$l_\Delta\le\tip_{\Delta'}\le l_\Delta+W$.

\begin{lemma}[Anchors are separated]\label{lem_c3_anchor_sep}
Let $\Delta\in\mathfrak T$, let $\mathsf X\ge1$ be an integer and $\mathsf S\ge4096\mathsf X$.
Distinct elements $\zeta<\zeta'$ of $\operatorname{Anc}(\Delta,\mathsf S,2\mathsf X)$ satisfy
$\zeta'-\zeta>\frac{36}{25}\lambda\mathsf S$.
\end{lemma}
\begin{proof}
Let $\Delta',\Delta''$ be triangles defining $\zeta=j_{\Delta'}$ and
$\zeta'=j_{\Delta''}$; they are distinct. As in the proof of \cite[Lemma 7.10]{Tao}, we compare
their rows at one height, here $\hat l=l_\Delta+\floor{\mathsf S/\log2}$. Since $\mathsf S/\log2>\mathsf S\ge2\mathsf X+1$, we have
$\tip_{\Delta'}\le l_\Delta+2\mathsf X\le \hat l$ and
$l_{\Delta'}=\tip_{\Delta'}+S_{\Delta'}/\log2\ge \hat l$, and likewise for
$\Delta''$. By
Lemma~\ref{lem_c3_row_mem}, $(\zeta',\hat l)\in\Delta''$. If
$\zeta'\le\row_{\Delta'}(\hat l)$, then $\zeta\le\zeta'$ and
Lemma~\ref{lem_c3_row_mem} give $(\zeta',\hat l)\in\Delta'$ as well, contradicting
Proposition~\ref{lem_bk_cover}. Hence
\[
\zeta'-\zeta>\lambda\bigl(\hat l-\tip_{\Delta'}\bigr)\ge
\lambda\Bigl(\frac{\mathsf S}{\log2}-1-2\mathsf X\Bigr).
\]
Finally $\log2<347/500$ gives
$\frac{\mathsf S}{\log2}-1-2\mathsf X-\frac{36}{25}\mathsf S>\frac{8\mathsf S}{8675}-2\mathsf X-1
\ge\frac{32768}{8675}\mathsf X-2\mathsf X-1>0$, so $\zeta'-\zeta>\frac{36}{25}\lambda\mathsf S$.
\end{proof}

\subsection{The packing estimate}
Outside a small deviation event the first passage and the next $v$ holds stay close to their
typical displacements, so a visit to a large triangle pins $j(x)$ near an anchor
(Lemma~\ref{lem_c3_near}) and the packing bound applies; $\Exc(v)$ collects the tail
probabilities. Let $\omega,v\in\N$ with $v\le J$. For real $\mathsf X>0$, the
\emph{deviation event} $\mathrm{Dev}_{\omega,v,\mathsf X}$ is the set of atoms
satisfying at least one of
\[
l(x)-\omega\ge \mathsf X,\qquad l(\operatorname{pt}_v(\mathbf b))\ge \mathsf X,\qquad |j(x)-\omega/4|\ge \omega^{3/5},\qquad
j(\operatorname{pt}_v(\mathbf b))\ge \omega^{3/5}.
\]
For $p\in\mathcal P$ and real $\mathsf S$, the \emph{large-triangle event} $\mathrm{Big}^p_{v,\mathsf S}$
is the set of atoms $a$ with $y^p_u(a)\in\Delta$ for some $u\le v$ and some
$\Delta\in\mathfrak T$ with $S_\Delta\ge\mathsf S$. The \emph{deviation error} is
\[
\Exc(v)=9q_{\mathrm{ex}}^{v+1}
 +\mathbf 1[v\ge1]\bigl(q_{\mathrm{ex}}(8q_{\mathrm{ex}})^v+e^{-14(v+1)}\bigr),
\qquad q_{\mathrm{ex}}=(25/27)^{136}.
\]

\begin{lemma}[Mass of the deviation event]\label{lem_c3_eprime_mass}
Let $p\in\mathcal P$, let $\omega\ge2^{160}$ be an integer, $v\le J$, $\mathsf X=136(v+1)$, and let $\mathsf S$ be real
with $4096\mathsf X\le\mathsf S$ and $\mathsf S^2<2\omega$. Then
$\Prob_{p,\omega}(\mathrm{Dev}_{\omega,v,\mathsf X})\le1/\mathsf S+\Exc(v)$.
\end{lemma}
\begin{proof}
We follow the proof of \cite[(7.61)]{Tao}. Each of the four conditions involves only $x$ or only
the first $v$ hold points, so by Lemma~\ref{lem_tr_fresh_marginal} it suffices to bound four
probabilities under $\mathsf F_\omega$ or $\eta^{\otimes v}$. First, the
first-passage overshoot tail (Lemma~\ref{lem_rn_overshoot_tail}, valid for $\omega\ge64$ and
integer $\mathsf X\ge1$) bounds the probability of $l(x)\ge \omega+\mathsf X$ by
$9(25/27)^{\mathsf X}=9q_{\mathrm{ex}}^{v+1}$. Second, if $v=0$ then $l(\operatorname{pt}_v(\mathbf b))=0<\mathsf X$;
otherwise the second bound of
Lemma~\ref{lem_rn_list_ver_tail} bounds the probability of $l(\operatorname{pt}_v(\mathbf b))\ge \mathsf X$ by
$8^v(25/27)^{\mathsf X}=q_{\mathrm{ex}}(8q_{\mathrm{ex}})^v$. Third, the horizontal tail
of the first passage (Lemma~\ref{lem_rn_hor_tail} with threshold $\omega^{3/5}$, where both exponents are at
least $\omega^{1/5}/65536$ as $1+\omega\le2\omega$) and Lemma~\ref{lem_c3_pack_const}(4) bound the probability of
$|j(x)-\omega/4|\ge \omega^{3/5}$ by
\[
2^{55}e^{-\omega^{1/5}/65536}<(2\omega)^{-1/2}<1/\mathsf S .
\]
Fourth, if $v=0$ then $j(\operatorname{pt}_v(\mathbf b))=0$. Otherwise it is an integer, so the event
is $10\ceil{\omega^{3/5}}\le10\,j(\operatorname{pt}_v(\mathbf b))$, and the first bound of
Lemma~\ref{lem_rn_list_hor_tail} bounds its probability by
$e^{v/2-\omega^{3/5}/16}$. Since $\omega\ge1$,
$\mathsf S<\sqrt{2\omega}\le16\omega^{3/5}$, so $\omega^{3/5}/16>\mathsf S/256\ge16\mathsf X=2176(v+1)$, and the
bound is at most $e^{-14(v+1)}$. Adding the four bounds gives
$1/\mathsf S+\Exc(v)$.
\end{proof}

\begin{lemma}[Outside the deviation event, large triangles sit at anchors]
\label{lem_c3_near}
Let $p\in\Delta\in\mathfrak T$, $\omega=\operatorname{gap}(p)$, $v\le J$,
$\mathsf X\ge1$ an integer and $\mathsf S$ real, and assume
$2\omega^{3/5}+2\lambda \mathsf X+1\le \omega\lambda_0$. If an atom $a=(x,\mathbf b)$
of positive mass lies in $\mathrm{Big}^p_{v,\mathsf S}$ but not in
$\mathrm{Dev}_{\omega,v,\mathsf X}$, then some
$\zeta\in\operatorname{Anc}(\Delta,\mathsf S,2\mathsf X)$ satisfies
$-\mathsf X<j(x)+j(p)-\zeta<2\lambda \mathsf X$.
\end{lemma}
\begin{proof}
We follow \cite[(7.63)--(7.66)]{Tao}, for all $u\le v$ at once. Choose $u\le v$ and $\Delta'\in\mathfrak T$ with $S_{\Delta'}\ge\mathsf S$ and
$(j,l)=y^p_u(a)\in\Delta'$. By Definition~\ref{def_tr_fresh_law}, $j=j(p)+j(x)+j(\operatorname{pt}_u(\mathbf b))$ and
$l=l(p)+l(x)+l(\operatorname{pt}_u(\mathbf b))$. Since $a$ has positive mass and lies outside the
deviation event, Lemmas~\ref{lem_rn_fp_mass}(1) and~\ref{lem_rn_hold_mass}(3) give
\[
\begin{gathered}
0<l(x)-\omega<\mathsf X,\qquad
0\le j(\operatorname{pt}_u(\mathbf b))\le l(\operatorname{pt}_u(\mathbf b))\le l(\operatorname{pt}_v(\mathbf b))<\mathsf X,\\
|j(x)-\omega/4|<\omega^{3/5},\qquad j(\operatorname{pt}_u(\mathbf b))\le j(\operatorname{pt}_v(\mathbf b))<\omega^{3/5}.
\end{gathered}
\]
Hence $0<l-l_\Delta=(l(x)-\omega)+l(\operatorname{pt}_u(\mathbf b))<2\mathsf X$ and
$|j-(j(p)+\omega/4)|<2\omega^{3/5}$. As $l>l_\Delta$, the point is not in
$\Delta$, so $\Delta'\ne\Delta$. Lemma~\ref{lem_c3_close}, with $W=2\mathsf X$
and $Y=2\omega^{3/5}$, gives $\tip_{\Delta'}\ge l_\Delta$; and
$\tip_{\Delta'}\le l<l_\Delta+2\mathsf X$ by Lemma~\ref{lem_c3_row_mem}. Thus
$\zeta=j_{\Delta'}\in\operatorname{Anc}(\Delta,\mathsf S,2\mathsf X)$.
Lemma~\ref{lem_c3_row_mem} also gives
$(j-\zeta)\log9\le(l-\tip_{\Delta'})\log2\le(l-l_\Delta)\log2<2\mathsf X\log2$, so
$0\le j-\zeta<2\lambda \mathsf X$. Since $j(x)+j(p)-\zeta=(j-\zeta)-j(\operatorname{pt}_u(\mathbf b))$ and
$0\le j(\operatorname{pt}_u(\mathbf b))<\mathsf X$, the claim follows.
\end{proof}

\begin{lemma}[Large triangles outside the deviation event]
\label{lem_c3_outside_mass}
Let $p\in\Delta\in\mathfrak T$, let $\omega=\operatorname{gap}(p)\ge1024$,
$v\le J$, let $\mathsf X\ge1$ be an integer and $\mathsf S$ real with $4096\mathsf X\le\mathsf S$ and
$\mathsf S^2<2\omega$, and assume $2\omega^{3/5}+2\lambda \mathsf X+1\le \omega\lambda_0$. Then
\[
\Prob_{p,\omega}\bigl(\mathrm{Big}^p_{v,\mathsf S}\setminus
\mathrm{Dev}_{\omega,v,\mathsf X}\bigr)<Q_{\mathrm{pk}}\frac{\mathsf X}{\mathsf S},
\]
with $Q_{\mathrm{pk}}$ of Lemma~\ref{lem_c3_pack_const}(2).
\end{lemma}
\begin{proof}
We follow the proof of \cite[(7.62)]{Tao}. The integers $i$ with $-\mathsf X<i<2\lambda\mathsf X$ number
$\mathsf X-1+\ceil{2\lambda\mathsf X}<(1+2\lambda)\mathsf X$. For each such $i$ put
$\mathcal K_i=\{\zeta-j(p)+i:\zeta\in\operatorname{Anc}(\Delta,\mathsf S,2\mathsf X)\}$; by
Lemma~\ref{lem_c3_anchor_sep}, distinct elements of $\mathcal K_i$ differ by more
than $\mathsf d=\frac{36}{25}\lambda\mathsf S$. By Lemma~\ref{lem_c3_near}, every
atom of positive mass in the event has $j(x)\in\bigcup_i\mathcal K_i$, and the law
of $j(x)$ is the horizontal first-passage law. The packing bound for the
horizontal first-passage coordinate (Lemma~\ref{lem_rn_fp_packing}, for $\omega\ge1024$,
separation at least $\mathsf d$, applied to the set $\mathcal K_i$) and the union bound over these $i$ bound the
mass of the event by
\[
(1+2\lambda)\mathsf X\Bigl(\frac{17}{10\sqrt\omega}+\frac{33}{32\mathsf d}\Bigr)
<\frac{\mathsf X}{\mathsf S}\Bigl((1+2\lambda)\frac{17}{10}\sqrt2+
\Bigl(\frac1\lambda+2\Bigr)\frac{33}{32}\cdot\frac{25}{36}\Bigr),
\]
using $1/\sqrt\omega<\sqrt2/\mathsf S$ and $(1+2\lambda)/\lambda=1/\lambda+2$.
Lemma~\ref{lem_c3_pack_const}(2) completes the proof.
\end{proof}

\begin{lemma}[Packing estimate; explicit form of Tao's Lemma 7.10]\label{lem_c3_lemma710}
Let $p\in\Delta\in\mathfrak T$, let $\omega=\operatorname{gap}(p)\ge2^{160}$,
let $v\le J$ and $\mathsf X=136(v+1)$, and let $\mathsf S$ be real with $4096\mathsf X\le\mathsf S$ and
$\mathsf S^2<2\omega$. Assume $2\omega^{3/5}+2\lambda \mathsf X+1\le \omega\lambda_0$. Then
\begin{equation}\label{eq:c3-packing}
\Prob_{p,\omega}(\mathrm{Big}^p_{v,\mathsf S})\le Q_{\mathrm{pk}}\frac{\mathsf X}{\mathsf S}
+\frac1{\mathsf S}+\Exc(v).
\end{equation}
\end{lemma}
\begin{proof}
Split $\mathrm{Big}^p_{v,\mathsf S}$ along $\mathrm{Dev}_{\omega,v,\mathsf X}$ and
add Lemmas~\ref{lem_c3_eprime_mass} and~\ref{lem_c3_outside_mass}.
\end{proof}

\begin{corollary}[Mass of the oversize event]\label{lem_c3_estar_mass}\label{lem:dyadic-passage}
Let $p\in\Delta\in\mathfrak T$ and $\omega=\operatorname{gap}(p)$. Let
$m\ge D_{\mathrm{sc}}$ be real with $P\le J$ and $m/(\log m)^2<\omega$. Then the
oversize event $\mathrm E^{*,p}$ of Definition~\ref{def_tr_estar} satisfies
\[
\Prob_{p,\omega}(\mathrm E^{*,p})<\frac{2+272(H_{\mathrm{tr}}+1)Q_{\mathrm{pk}}}{K_{\mathrm{tr}}}+\frac1{3360}.
\]
\end{corollary}
\begin{proof}
Put $\hat u=P-1$, which is at least $2$ by Lemma~\ref{lem_s02_P_bits},
take $\mathsf v(k)$, $\mathsf X(k)$ and $\mathsf S(k)$ from Lemma~\ref{lem:schedule}(3) with this $\hat u$,
and put $\bar k=\floor{\log_2(\hat u+1)}$. A union bound over the dyadic ranges
$0\le k\le\bar k$ of times reduces the oversize event to one packing estimate per range.
Indeed, let $u\le\hat u$ and choose $k$ with $2^k\le u+1<2^{k+1}$. Then
$k\le\bar k$ and $u\le\mathsf v(k)$, and since $u\ge2^k-1$ and $\beta$ is nondecreasing
(Lemma~\ref{lem:schedule}(1)), $\beta(u)\ge\mathsf S(k)$. Hence
$\mathrm E^{*,p}\subseteq\bigcup_{k=0}^{\bar k}\mathrm{Big}^p_{\mathsf v(k),\mathsf S(k)}$.

Fix $0\le k\le\bar k$; we check the hypotheses of Lemma~\ref{lem_c3_lemma710} with
$(v,\mathsf X,\mathsf S)=(\mathsf v(k),\mathsf X(k),\mathsf S(k))$. First $\mathsf v(k)\le P-1<J$, and
$4096\mathsf X(k)<\mathsf S(k)$ by Lemma~\ref{lem:schedule}(3). As $2^k\le\hat u+1$, monotonicity gives
$\mathsf S(k)=\beta(2^k-1)\le\beta(P-1)=B_{\mathrm{sz}}$. By Lemma~\ref{lem_s02_far_B}(2) and
Lemma~\ref{lem_s02_P_bits}, the integer $\omega$ satisfies
$\omega>m/(\log m)^2>\omega_{\mathrm{far}}=2^{36}P>2^{365}$, and
$\mathsf S(k)^2\le B_{\mathrm{sz}}^2<2m/(\log m)^2<2\omega$. Finally, as $2\lambda \mathsf X(k)=272\lambda(\mathsf v(k)+1)$,
Lemma~\ref{lem_c3_pack_const}(3) gives
$2\omega^{3/5}+2\lambda \mathsf X(k)+1\le \omega\lambda_0$.

Sum \eqref{eq:c3-packing} over $0\le k\le\bar k$. The partial sums of
\eqref{eq:schedule-sums} bound the first two terms by
$(272(H_{\mathrm{tr}}+1)Q_{\mathrm{pk}}+2)/K_{\mathrm{tr}}$, with strict inequality, and
Lemma~\ref{lem_c3_exc_sum} bounds $\sum_{k\le\bar k}\Exc(\mathsf v(k))$ by $1/3360$.
\end{proof}

\subsection{The outer horizontal tail and the assembly}
We bound the event $\mathrm{Bad}^p_m$ of Definition~\ref{def_tr_low}, continue $\mathsf Q$
through the first passage, and assemble Case~3.

\begin{lemma}[Outer horizontal tail]\label{lem_c3_bad_mass}
Let $p\in\mathcal P$, $\omega\in\N$, and let $m\ge D_{\mathrm{bad}}$ be an
integer with $P\le J$ and $\omega\log2\le m\log9$. Then
$m^{A_*}\,\Prob_{p,\omega}(\mathrm{Bad}^p_m)<1/8000$.
\end{lemma}
\begin{proof}
As in \cite[proof of (7.54)]{Tao}, we split the advance between the first passage and the next $P$ holds, with explicit thresholds.
By Definition~\ref{def_tr_fresh_law}, $\mathrm{Bad}^p_m$ is the set of atoms with
$400(j(x)+j(\operatorname{pt}_P(\mathbf b)))\ge317m$. Since
$317/400=158500/200000$, it is contained in the union of
$\{j(x)\ge\frac{158499}{200000}m\}$ and $\{j(\operatorname{pt}_P(\mathbf b))\ge\frac m{200000}\}$.

By Lemma~\ref{lem_c3_pack_const}(1), $\omega\le m/\lambda<\frac{158497}{50000}m$.
On the first event, $j(x)-\omega/4\ge m/100000$. The horizontal tail of the first
passage (Lemma~\ref{lem_rn_hor_tail} with threshold $m/100000$) bounds its mass by
$2^{54}(e^{-m^2/(10^{10}2^{15}(1+\omega))}+e^{-m/(256\cdot10^5)})$. As $m\ge1$,
$1+\omega\le\frac{21}5m$, and the integer comparison
$21\cdot2^{15}\cdot10^{10}\le5\cdot2^{51}$ makes both exponents at most $-m/2^{51}$. So the first
event has mass at most $2^{55}e^{-m/2^{51}}$. On the second event,
$10\ceil{m/200000}\le10\,j(\operatorname{pt}_P(\mathbf b))$, and the first bound of
Lemma~\ref{lem_rn_list_hor_tail}, applied via Lemma~\ref{lem_tr_fresh_marginal} with $k=N=P$ and
threshold $10\ceil{m/200000}$, bounds its mass by $e^{P/2-m/3200000}\le e^{P/2}e^{-m/2^{51}}$. Therefore
\[
m^{A_*}\Prob_{p,\omega}(\mathrm{Bad}^p_m)\le m^{A_*}(2^{55}+e^{P/2})e^{-m/2^{51}}.
\]
Since $m\ge D_{\mathrm{bad}}=(2^{53}A_*)^2+2^{53}(P+65)$, we have
$A_*\log m\le A_*\sqrt m\le m/2^{53}$ and $P+56\le m/2^{53}-9$; and
$2^{55}+e^{P/2}\le2e^{P+55}\le e^{P+56}$. The logarithm of the right side
is therefore at most $2m/2^{53}-9-4m/2^{53}<-9$, and $e^9>8000$
(Lemma~\ref{lem_s03_log3_bounds}(4)).
\end{proof}

\begin{lemma}[Continuation through the first passage]\label{lem_c3_continuation}\label{lem_ch_Q_continuation}
Let $p\in\mathcal P$, $\omega\in\N$ and $1\le L\le J$. Then
\[
\mathsf Q(p)\le\sum_a\Prob_{p,\omega}(a)\,e^{-z\operatorname{Rw}^p_{L-1}(a)}\,
\mathsf Q\bigl(y^p_{L}(a)\bigr).
\]
\end{lemma}
\begin{proof}
We adapt \cite[(7.45), (7.53)]{Tao}. Lemma~\ref{lem_ch_Q_stopped} gives
$\mathsf Q(p)\le\sum_{x\in\mathcal P}\mathsf F_\omega(x)\,\mathsf Q(p+x)$. Fix
$x$ with $\mathsf F_\omega(x)\ne0$ and put $\bar o=p+x\in\mathcal P$.
Induction on $K$, substituting Lemma~\ref{lem_ch_Q_range}(2) at the point
$\bar o+\operatorname{pt}_K(\mathbf b)$ in the inductive step, gives with $K=L$ the continuation identity
\[
\mathsf Q(\bar o)=\sum_{\mathbf b\in\mathfrak B^{L}}\mathfrak q(\mathbf b)
\prod_{k=0}^{L-1}\mathsf w(\bar o+\operatorname{pt}_k(\mathbf b))
\prod_{k=1}^{L}\mathsf I(\bar o+\operatorname{pt}_{k-1}(\mathbf b);b^k)\,
\mathsf Q(\bar o+\operatorname{pt}_L(\mathbf b)),
\]
with all factors in $[0,1]$. Drop $\mathsf w(\bar o)$ and
$\mathsf I(\bar o+\operatorname{pt}_{L-1}(\mathbf b);b^{L})$. Since
$\operatorname{pt}_k(\mathbf b)=\operatorname{pt}_{k-1}(\mathbf b)+\operatorname{pt}(b^k)$,
the identity recorded after Definition~\ref{def_tr_reward} gives
\[
\mathsf w(\bar o+\operatorname{pt}_k(\mathbf b))\,\mathsf I(\bar o+\operatorname{pt}_{k-1}(\mathbf b);b^k)
=e^{-z\,\operatorname{rw}(\bar o+\operatorname{pt}_{k-1}(\mathbf b),b^k)}\quad\text{for }1\le k\le L-1,
\]
and the product of these is $e^{-z\operatorname{Rw}(\bar o,\mathbf b;L-1)}$.
Extending $\mathbf b$ by $J-L$ further holds changes neither
$\operatorname{Rw}(\bar o,\mathbf b;L-1)$ nor $\bar o+\operatorname{pt}_L(\mathbf b)$ (Lemma~\ref{lem_tr_concat}(2)), and the extensions of
$\mathbf b$ have total mass $\mathfrak q(\mathbf b)$ (Lemmas~\ref{lem_tr_concat}(1) and~\ref{lem_rn_hold_mass}(5)).
Multiplying by $\mathsf F_\omega(x)$ and summing over $x$ gives the claim, by
Definition~\ref{def_tr_fresh_law}.
\end{proof}

\begin{lemma}[Case~3 of the descent step]\label{lem_c3_case3}\label{def_s02_budget}
Let $m$ be an integer with $m\ge D_{\mathrm{sc}}$, $m\ge D_{\mathrm{bad}}$
and $m\le J$. Let $p\in\mathcal P$ with $j(p)+m=J$, and suppose
$p\in\Delta$ for some $\Delta\in\mathfrak T$ with
$\omega=\operatorname{gap}(p)>m/(\log m)^2$. Then
$\mathsf Q(p)\le m^{-A_*}\mathsf Q_{m-1}$.
\end{lemma}
\begin{proof}
We follow \cite[(7.52)--(7.56)]{Tao}, with $\frac{317}{400}m$ in place of $0.9m$ and an explicit error budget.
By Lemma~\ref{lem_s02_far_B}(2), $P\le m\le J$, and $\omega$ is a positive integer.
The right-end bound \eqref{eq:fb-right-edge} and the membership of $p$ give
\[
\omega\log2\le S_\Delta-(j(p)-j_\Delta)\log9<(J-j(p))\log9=m\log9 .
\]

Lemma~\ref{lem_c3_continuation} with $L=P$ gives
$\mathsf Q(p)\le\sum_a\Prob_{p,\omega}(a)e^{-z\operatorname{Rw}^p_{P-1}(a)}\mathsf Q(y^p_P(a))$.
Let $a$ have positive mass and put
$\operatorname{Adv}(a)=j(y^p_P(a))-j(p)=j(x)+j(\operatorname{pt}_P(\mathbf b))\ge1$. Then
$j(y^p_P(a))+m=J+\operatorname{Adv}(a)$, so the boundary reduction
(Lemma~\ref{lem_ch_boundary_step}(1) with $r=\operatorname{Adv}(a)$) gives
\[
e^{-z\operatorname{Rw}^p_{P-1}(a)}\mathsf Q(y^p_P(a))\le e^{-z\operatorname{Rw}^p_{P-1}(a)}
\max(m-\operatorname{Adv}(a),1)^{-A_*}\mathsf Q_{m-1}.
\]
Since $\mathsf Q_{m-1}\ge0$, the right side is at most
$m^{A_*}\cdot m^{-A_*}\mathsf Q_{m-1}$. If $a\notin\mathrm{Bad}^p_m$, then
$m-\operatorname{Adv}(a)>\frac{83}{400}m$, so the right side is at most
$(400/83)^{A_*}m^{-A_*}\mathsf Q_{m-1}\le c_{\mathrm{cont}}\,m^{-A_*}\mathsf Q_{m-1}$, with
$c_{\mathrm{cont}}=511/10$ of Definition~\ref{def_s02_cont}. If moreover
$a\notin\mathrm{Low}^p$, then $\operatorname{Rw}^p_{P-1}(a)>T_{\mathrm{tr}}=101$, and the right
side is at most $e^{-zT_{\mathrm{tr}}}(400/83)^{A_*}m^{-A_*}\mathsf Q_{m-1}\le
m^{-A_*}\mathsf Q_{m-1}/P_{\mathrm{wc}}$, with $P_{\mathrm{wc}}$ of the same definition. The two numerical bounds are
Lemma~\ref{lem_s02_budget}(1),(2). Splitting the atoms into
$\mathrm{Bad}^p_m$, $\mathrm{Low}^p\setminus\mathrm{Bad}^p_m$ and the rest,
and using that $\Prob_{p,\omega}$ has total mass $1$,
\[
\mathsf Q(p)\le m^{-A_*}\mathsf Q_{m-1}\Bigl(m^{A_*}\Prob_{p,\omega}(\mathrm{Bad}^p_m)
+c_{\mathrm{cont}}\,\Prob_{p,\omega}(\mathrm{Low}^p\setminus\mathrm{Bad}^p_m)+\frac1{P_{\mathrm{wc}}}\Bigr).
\]
Lemma~\ref{lem_c3_bad_mass} bounds the first term by $1/8000$. The
Case~3 low-reward bound (Lemma~\ref{lem_tr_survival_export}, which uses $j(p)+m=J$
and $P\le J$) gives
$\Prob_{p,\omega}(\mathrm{Low}^p\setminus\mathrm{Bad}^p_m)\le
\Prob_{p,\omega}(\mathrm E^{*,p})+1/P_{\mathrm{tr}}+2^{-43}$, and
Corollary~\ref{lem_c3_estar_mass} bounds
$\Prob_{p,\omega}(\mathrm E^{*,p})$. Since $c_{\mathrm{cont}}/3360<1/60$
(Lemma~\ref{lem_s02_budget}(1)), the bracket is less than the \emph{error budget}
\begin{equation}\label{eq:budget}
\mathcal B=\frac1{8000}+\frac{c_{\mathrm{cont}}(2+272(H_{\mathrm{tr}}+1)Q_{\mathrm{pk}})}{K_{\mathrm{tr}}}
+\frac{c_{\mathrm{cont}}}{P_{\mathrm{tr}}}
+\frac1{60}+\frac1{P_{\mathrm{wc}}}+c_{\mathrm{cont}}\,2^{-43},
\end{equation}
and $\mathcal B<1$ (Lemma~\ref{lem_s02_budget}(3)).
\end{proof}

\section{The white and near-top cases and the Fourier bound}\label{sec:fb-cases}\label{sec:fb-conclusion}
This section proves Cases~1 and~2 of the descent step (Section~\ref{sec:fb-charsum}), where the boundary point $p$ for $m$ is white or black with $\operatorname{gap}(p)\le m/(\log m)^2$, following Cases~1 and~2 of
\cite[\S7.4]{Tao} with the constants of Appendix~\ref{app:analytic}. With Case~3
(Lemma~\ref{lem_c3_case3}) this gives the descent step for the weighted maxima beyond the
threshold $D_*$, from which \eqref{eq:fourier} follows.

\subsection{Envelope factors}
To average the bound of Lemma~\ref{lem_ch_boundary_suffices}(1) over the hold
law or a first-passage law, we dominate $\max(m-r,1)^{-A_*}$ by $m^{-A_*}$
times the \emph{envelope factor} $e^{\mathsf t_mr}$ (compare
\cite[(7.42)]{Tao}), whose average is an exponential moment; here, for an
integer $m\ge1$,
\begin{equation}\label{def_mo_rate}
\mathsf t_m=\frac{4A_*(1+\log m)}m.
\end{equation}

\begin{lemma}[Envelope bound]\label{lem_mo_envelope}\label{lem_mo_rate_D1}
Let $m\ge2$ and $r\ge1$ be integers. Then
\begin{equation}\label{eq:F4-envelope}
\max(m-r,1)^{-A_*}\le m^{-A_*}e^{\mathsf t_m r}.
\end{equation}
If $m\ge D_{\mathrm{wc}}$, then $\mathsf t_m\le\chi/32\le1/1024$, and
\begin{equation}\label{eq:F4-invpow}
\sum_{r\ge1}\nu_{45}(r)\max(m-r,1)^{-A_*}\le\mathsf M_{45}(\mathsf t_m)\,m^{-A_*}\le e^{\chi/2}m^{-A_*}.
\end{equation}
\end{lemma}
\begin{proof}
If $2r\le m$, then $(1-r/m)(1+2r/m)\ge1$ gives
$(1-r/m)^{-1}\le1+2r/m\le e^{2r/m}$, so $(m-r)^{-A_*}\le m^{-A_*}e^{2A_*r/m}$, and
$2A_*/m\le \mathsf t_m$. If $2r>m$, then
$\max(m-r,1)^{-A_*}\le1=m^{-A_*}e^{A_*\log m}$ and
$A_*\log m\le A_*(1+\log m)\cdot4r/m=\mathsf t_m r$.

Let $m\ge D_{\mathrm{wc}}$, where $D_{\mathrm{wc}}=(256A_*/\chi)^2+2$ by Definition~\ref{def_Dstar}. Since $\log y\le\sqrt y$ for $y>0$,
$\mathsf t_m\le8A_*/\sqrt m\le8A_*\cdot\chi/(256A_*)=\chi/32=63/80000<1/1024$. Summing
\eqref{eq:F4-envelope} against $\nu_{45}$ gives the first inequality of
\eqref{eq:F4-invpow}, and Lemma~\ref{lem_rn_mgf_bound}(2) gives
$\mathsf M_{45}(\mathsf t_m)\le e^{13\mathsf t_m/4}\le e^{13\chi/128}\le e^{\chi/2}$.
\end{proof}

\subsection{The white case}
At a white boundary point the closing letter at $p$ carries the factor $e^{-z}$, which outweighs the cost $e^{\chi/2}$ of averaging the envelope factor over one hold.
\begin{lemma}[White boundary points]\label{lem_mo_case1}\label{lem_mo_hold_average}
Let $m\ge D_{\mathrm{wc}}$ be an integer, and let $p\in\mathcal P$ be white with
$j(p)+m=J$. Then
\[
\mathsf Q(p)\le e^{-z+\chi/2}m^{-A_*}\mathsf Q_{m-1}\le m^{-A_*}\mathsf Q_{m-1}.
\]
\end{lemma}
\begin{proof}
We follow \cite[(7.43)]{Tao}. The second inequality holds as $\chi/2=63/5000<z$.
By Lemma~\ref{lem_ch_Q_range}(3) and $\mathsf w(p)=e^{-z}$,
$\mathsf Q(p)\le e^{-z}\sum_y\eta(y)\mathsf Q(p+y)$. Every
$y\in\mathcal P$ has $j(y)\ge1$ and $j(p+y)+m=J+j(y)$, so
Lemma~\ref{lem_ch_boundary_suffices}(1) gives
$\mathsf Q(p+y)\le\max(m-j(y),1)^{-A_*}\mathsf Q_{m-1}$. Grouping by
$r=j(y)$ and using \eqref{eq:hold-law},
\[
\sum_y\eta(y)\mathsf Q(p+y)
\le\mathsf Q_{m-1}\sum_{r\ge1}\nu_{45}(r)\max(m-r,1)^{-A_*}
\le e^{\chi/2}m^{-A_*}\mathsf Q_{m-1}
\]
by \eqref{eq:F4-invpow}, since $m\ge D_{\mathrm{wc}}\ge2$ and $\mathsf Q_{m-1}\ge0$.
\end{proof}

\subsection{The near-top case}
Following Case~2 of \cite[\S7.4]{Tao}, we halt the hold walk from a black
boundary point $p$ near the top of its triangle $\Delta$ when its height
first exceeds $l_\Delta$. With probability at least $\frac3{16}$ it halts in
the exit window, at a white point where Case~1 applies; there, after paying $\chi/2$, it
still gains a factor $1-\frac35z$ (Lemma~\ref{lem_mo_rate_tiny}), which more than pays the
cost $\chi/8$ of the envelope factors along the first passage (Lemma~\ref{lem_mo_moment_threshold}).

\begin{lemma}[Moment threshold]\label{lem_mo_moment_threshold}
Let $m$ be an integer with $m\ge\max(D_{\mathrm{wc}},D_{\mathrm{exp}})$, and let $\omega\in\N$
with $\omega\le m/(\log m)^2$. Then, with $\mathsf t_m$ of \eqref{def_mo_rate},
$\mathsf M_{45}(\mathsf t_m)^{\omega+1}\le1+\chi/8$.
\end{lemma}
\begin{proof}
Part (3) of Lemma~\ref{lem_s02_Dstar_eq} gives $\chi\log m>234A_*$; as
$A_*\ge1\ge\chi$, also $\log m>234$. First, because $\log m\ge32$,
\[
8\mathsf t_m\,m/(\log m)^2=32A_*(1+\log m)/(\log m)^2\le A_*/\log m+32A_*/\log m.
\]
Second,
$m\ge(\log m)^4/24$ and $(\log m)^3\ge1024\log m\ge256(1+\log m)$, so
$3m\ge(\log m)^4/8\ge32\log m\,(1+\log m)$, that is $8\mathsf t_m\le3A_*/\log m$. Adding,
$8\mathsf t_m(\omega+1)\le8\mathsf t_m(m/(\log m)^2+1)\le36A_*/\log m$. By Lemma~\ref{lem_mo_envelope},
$\mathsf t_m\le1/1024$, so Lemma~\ref{lem_rn_mgf_bound}(2) gives
\[
\mathsf M_{45}(\mathsf t_m)^{\omega+1}\le\exp\Bigl(\frac{13\mathsf t_m(\omega+1)}4\Bigr)
\le\exp\Bigl(\frac{13}{32}\cdot\frac{36A_*}{\log m}\Bigr)
=\exp\Bigl(\frac{117A_*}{8\log m}\Bigr)<e^{\chi/16}.
\]
Finally $e^y\le(1-y)^{-1}\le1+2y$ for $0\le y\le\tfrac12$; take $y=\chi/16$.
\end{proof}

\begin{lemma}[Black boundary points near the top]\label{lem_mo_case2}\label{lem_mo_discounted_expectation}
Let $m$ be an integer with $m\ge\max(D_{\mathrm{pt}},D_{\mathrm{exp}})$, and let
$p\in\mathcal P$ with $j(p)+m=J$ lie in a triangle
$\Delta\in\mathfrak T$ with $\operatorname{gap}(p)\le m/(\log m)^2$. Then
\[
\mathsf Q(p)\le m^{-A_*}\mathsf Q_{m-1}.
\]
\end{lemma}
\begin{proof}
Put $\omega=\operatorname{gap}(p)\in\N$; every bound below is a multiple of
$m^{-A_*}\mathsf Q_{m-1}\ge0$. Since
$m\ge D_{\mathrm{pt}}\ge D_{\mathrm{wc}}\ge2$, Lemma~\ref{lem_mo_envelope} gives
$0\le \mathsf t_m\le1/1024$, so $11e^{\mathsf t_m}<16$ and
Lemma~\ref{lem_rn_fp_mgf} applies with $\mathfrak t=\mathsf t_m$. Let
\[
\mathcal X_\omega=\{x\in\mathcal P:1\le j(x)\le f_\omega,\ \omega<l(x)\le \omega+4\}.
\]
Let $x\in\mathcal P$. Then $j(x)\ge1$ and $j(p+x)+m=J+j(x)$, so
Lemma~\ref{lem_ch_boundary_suffices}(1) and \eqref{eq:F4-envelope} give
$\mathsf Q(p+x)\le m^{-A_*}\mathsf Q_{m-1}e^{\mathsf t_m j(x)}$.

Let moreover $x=(r,\omega+o)\in\mathcal X_\omega$, so $1\le r\le f_\omega$ and $1\le o\le4$. Since
$m\ge D_{\mathrm{pt}}\ge344>e$, $\log m\ge1$ and $\omega\le m$. Hence
$r\le f_\omega\le5\omega/16+1\le5m/16+1$ and, as $D_{\mathrm{wc}}=(D_{\mathrm{pt}}-344)/2\le(m-344)/2$ by
Definition~\ref{def_Dstar},
\[
r+D_{\mathrm{wc}}\le\frac{13m}{16}-171<m.
\]
Put $m'=m-r$, so that $m'>D_{\mathrm{wc}}\ge2$ and $m'=\max(m-r,1)$. The point
$p'=p+x=(j(p)+r,l_\Delta+o)$ is white by Lemma~\ref{lem_bk_exit_room}, and $j(p')+m'=J$. So
Lemma~\ref{lem_mo_case1} at $(p',m')$, $\mathsf Q_{m'-1}\le\mathsf Q_{m-1}$
(Lemma~\ref{lem_ch_Qm_bounded}(3)), \eqref{eq:F4-envelope} and Lemma~\ref{lem_mo_rate_tiny} give
\[
\mathsf Q(p+x)\le e^{-z+\chi/2}m'^{-A_*}\mathsf Q_{m-1}
\le(1-\tfrac35z)m^{-A_*}\mathsf Q_{m-1}e^{\mathsf t_m j(x)}.
\]
Since $e^{\mathsf t_m j(x)}\ge1$,
Lemmas~\ref{lem_rn_fp_mgf} and~\ref{lem_mo_moment_threshold} and the exit
mass bound $\sum_{x\in\mathcal X_\omega}\mathsf F_\omega(x)=\mathsf q_{45}(\omega)\ge\frac3{16}$
(Lemmas~\ref{lem_rn_p45_mass} and~\ref{lem_rn_p45_floor}, as $\mathsf F_\omega$
vanishes off $\mathcal P$ by Lemma~\ref{lem_rn_fp_mass}(1)) yield
\begin{align*}
\sum_{x\in\mathcal P}\mathsf F_\omega(x)\mathsf Q(p+x)
&\le m^{-A_*}\mathsf Q_{m-1}\Bigl(\sum_{x\in\mathcal P}\mathsf F_\omega(x)e^{\mathsf t_m j(x)}-\frac35z\sum_{x\in\mathcal X_\omega}\mathsf F_\omega(x)\Bigr)\\
&\le m^{-A_*}\mathsf Q_{m-1}\Bigl(1+\frac\chi8-\frac35z\cdot\frac3{16}\Bigr)\le m^{-A_*}\mathsf Q_{m-1},
\end{align*}
the last step by Lemma~\ref{lem_mo_rate_tiny}. Finally Lemma~\ref{lem_ch_Q_stopped} gives
$\mathsf Q(p)\le\sum_{x\in\mathcal P}\mathsf F_\omega(x)\mathsf Q(p+x)$.
\end{proof}

\subsection{The descent step and the Fourier bound}
\begin{proposition}[Descent]\label{lem_mo_monotonicity}
For every integer $m$ with $D_*\le m\le J$, $\mathsf Q_m\le\mathsf Q_{m-1}$.
\end{proposition}
\begin{proof}
By its definition $D_*$ is at least each of $D_{\mathrm{wc}}$, $D_{\mathrm{pt}}$,
$D_{\mathrm{exp}}$, $D_{\mathrm{bad}}$, $D_{\mathrm{sc}}$ and $2$, hence so is $m$. By
Lemma~\ref{lem_ch_boundary_suffices}(2), it suffices to prove
$\mathsf Q(p)\le m^{-A_*}\mathsf Q_{m-1}$ for every $p\in\mathcal P$ with
$j(p)+m=J$. Fix such a point.

\emph{Case~1.} $p$ is white. Apply Lemma~\ref{lem_mo_case1}.

Otherwise $p$ is black, because $j(p)\le J$, so it lies in some
$\Delta\in\mathfrak T$ by Proposition~\ref{lem_bk_cover}.

\emph{Case~2.} $\operatorname{gap}(p)\le m/(\log m)^2$. Apply
Lemma~\ref{lem_mo_case2}.

\emph{Case~3.} $\operatorname{gap}(p)>m/(\log m)^2$. Apply
Lemma~\ref{lem_c3_case3}.
\end{proof}

\begin{theorem}[The Fourier bound]\label{lem_ch_primitive_decay}\label{def_ch_prim_coeff}
\label{lem_ch_Q_pointwise}\label{lem_ch_Qm_threshold}\label{lem_ch_real_moment}
\label{lem_ch_fifth_moment}\label{lem_ch_binomial_series}\label{lem_ch_hold_expectation}\label{lem_ch_gap}
For every integer $n\ge1$ and every frequency $\xi\in G_n$ prime to $3$,
\[
|\widehat\mu_n(\xi)|\le(160D_*)^{A_*}n^{-A_*}.
\]
This is \eqref{eq:fourier}.
\end{theorem}
\begin{proof}
Since $\xi$ is prime to $3$, it is a unit of $G_n$; form $\mathsf Q$ and $\mathsf Q_m$
for $(n,\xi,\varepsilon)$ and the exponent $A_*$. Proposition~\ref{lem_ch_front_end} gives
$|\widehat\mu_n(\xi)|\le\sum_y\eta(y)\mathsf Q(y)$. We bound the right side in three steps,
following \cite[(7.36)--(7.40)]{Tao}.

First, $\mathsf Q_m\le D_*^{A_*}$ for $0\le m\le J$, by induction on $m$: if $m\le D_*$, then
$\mathsf Q_m\le\max(m,1)^{A_*}\le D_*^{A_*}$ by Lemma~\ref{lem_ch_Qm_bounded}(1), as $D_*\ge2$; if
$m>D_*$, then $\mathsf Q_m\le\mathsf Q_{m-1}$ by Proposition~\ref{lem_mo_monotonicity}. So
Lemma~\ref{lem_ch_Qm_bounded}(2) with $m=J$ gives $\mathsf Q(y)\le D_*^{A_*}d_J(y)^{-A_*}$ for every
$y\in\mathcal P$.

Second, let $y\in\mathcal P$ and $r=j(y)\ge1$; then $n\le8r\,d_J(y)$. Indeed
$n\le2J+1$. If $r<J$, then $J\ge2$ and
$r(J-r)-(J-1)=(r-1)(J-r-1)\ge0$, so $n\le2J+1\le8(J-1)\le8r(J-r)$. If
$r\ge J$, then $d_J(y)=1$ and $n\le2r+1\le8r$. Hence
$\mathsf Q(y)\le(8D_*)^{A_*}r^{A_*}n^{-A_*}$.

Third, $\sum_{r\ge1}\nu_{45}(r)r^{A_*}\le20^{A_*}$. For $r\ge1$,
$r^5\le r(r+1)\cdots(r+4)=5!\binom{r+4}5$, and the negative binomial
series $\sum_{k\ge0}\binom{k+5}5x^k=(1-x)^{-6}$ at $x=11/16$ gives
\[
\sum_{r\ge1}\nu_{45}(r)r^5\le\frac5{16}\cdot120\cdot\Bigl(\frac{16}5\Bigr)^6
=\frac{25165824}{625}<20^5.
\]
Since $\nu_{45}$ is a probability law, $1\le A_*\le5$ and $u\mapsto u^{A_*/5}$ is concave,
Jensen's inequality gives
$\sum_r\nu_{45}(r)r^{A_*}\le(\sum_r\nu_{45}(r)r^5)^{A_*/5}\le20^{A_*}$.
Summing the second step against $\eta$ and grouping by $r=j(y)$ with \eqref{eq:hold-law} gives
$\sum_y\eta(y)\mathsf Q(y)\le(8D_*)^{A_*}20^{A_*}n^{-A_*}$.
\end{proof}

\section{Estimates for the first-crossing families}\label{sec:est-families}
This section proves the facts about scales and first-crossing families used in Sections~\ref{sec:physical}--\ref{sec:steps}: the growth of scales and levels (Lemma~\ref{lem_scales_growth}), prefix-disjointness (Lemma~\ref{lem_first_crossing_prefix_disjoint}), the lower bounds on products of masses of block families used in Lemma~\ref{lem_s05_tuple_mass}, the deficit bound in \eqref{eq:physical-deficits}, and the bounds on the terminal deficit and mask failure used in Proposition~\ref{prop:terminal-errors}.

\subsection{Scales and levels}
The finite computations below need the first scales exactly, the deficit bounds need geometric growth of the scales, and propagation needs the levels $k_j=\floor{\delta_j/4}$ to reach $n_0$, where Proposition~\ref{prop:mix} applies.

\begin{lemma}[Scales and levels]\label{lem_scales_growth}\label{lem_level_late_lower}
\leavevmode
\begin{enumerate}
\item For $13\le\delta\le255$ the set defining $e_\delta$ contains $\ceil{\delta/100}$; hence $e_\delta$ is well defined and $9\cdot3^\delta\cdot16^{e_\delta}<4^{\delta+1}$. For every $\delta\ge9$, $e_\delta\ge\max(1,\delta/100)$, and $e_\delta=\ceil{\delta/100}\le(\delta+99)/100$ for $\delta\ge256$.
\item The scales increase; $(\delta_0,\ldots,\delta_6)=(9,\ldots,15)$, $\delta_j\le255$ for $j\le41$, $\delta_j\ge256$ for $j\ge42$, $\delta_{42}=256$, $\delta_{444}=16544$ and $2^{40}<\delta_{2254}=1100658881854<2^{41}$.
\item The scales satisfy \eqref{eq:scalegrowth}; $\delta_j>190g^j$ for $j\ge444$; and $\delta_j\ge2^{40}j$ and $\delta_j\ge8000j^2$ for $j\ge20\cdot10^9$.
\item For $j\ge444$, $k_j\ge47g^j$; for $j\ge6000000$, $k_j\ge\max(n_0,47g^j)$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $i=\ceil{\delta/100}$, so $i\le\delta$. The inequality $9\cdot3^\delta\cdot16^i<4^{\delta+1}$ is equivalent to $(4/3)^\delta>36\cdot16^{i-1}$. If $i=1$, then $(4/3)^\delta\ge(4/3)^{13}>36$, by the integer comparison $4^{13}>36\cdot3^{13}$. If $i\ge2$, then $\delta\ge100(i-1)+1\ge13+88(i-1)$, and $(4/3)^{88}>16$ gives the claim. Thus the maximum is over a nonempty set, all of whose elements are at least $\ceil{\delta/100}\ge\max(1,\delta/100)$; for $\delta\le12$, $e_\delta=1$.

Consequently $\delta_{m+1}\ge(1+1/100)\delta_m=g\delta_m$ for every $m$, and induction on $j$ gives $g^j\delta_m\le\delta_{m+j}$. The scales increase. Running the integer recurrence to $j=2254$ gives the listed values and $\delta_{41}=233$. With these values, the following are exact integer comparisons:
\[
100^j\delta_j<201\cdot101^j\quad(0\le j\le444),\qquad 16644\cdot100^{444}<201\cdot101^{444},\qquad 16544\cdot100^{444}>190\cdot101^{444}.
\]
The first proves $\delta_j<201g^j$ for $j\le444$. For $j\ge444$ we show $\delta_j+100<201g^j$ by induction, the case $j=444$ being the second comparison. Since $\delta_j\ge256$,
\[
\delta_{j+1}+100\le\delta_j+\frac{\delta_j+99}{100}+100<g(\delta_j+100)<201g^{j+1}.
\]
Next $\delta_j\ge g^{j-444}\delta_{444}>190g^j$ for $j\ge444$, by the third comparison. Finally, for $j\ge20\cdot10^9$, \eqref{eq:scalegrowth} and the binomial theorem give $\delta_j\ge9(1+1/100)^j\ge9\binom j4\cdot100^{-4}$. Each of $j,\ldots,j-3$ is at least $j/2$, so $\binom j4\ge j^4/384$ and $\delta_j\ge9j^4/(384\cdot10^8)$. This is at least $2^{40}j$ because $9j^3\ge72\cdot10^{30}>2^{40}\cdot384\cdot10^8=42221246506598400000000$, and at least $8000j^2$ because $9j^2\ge8000\cdot384\cdot10^8$.

For $j\ge444$, $k_j\ge\delta_j/4-1>47.5g^j-1\ge47g^j$, because $g^j\ge g^{444}>2$ (the integer comparison $101^{444}>2\cdot100^{444}$). For $j\ge6000000$, moreover $47g^j>g^j=2^{j\log_2g}>2^{6000000/70}>n_0$, since $\log_2g>1/70$ (Section~\ref{sec:physical}).
\end{proof}


\subsection{The line and the families}
The transfer of Section~\ref{sec:mix} needs finite prefix-disjoint families; the first lemma shows that the first-crossing families are such.

\begin{lemma}[Structure of the families]\label{lem_first_crossing_prefix_disjoint}\label{lem_s03_barrier_mono}
Let $\delta\ge9$, $u$ and $R\ge0$ be integers. The line is nondecreasing: $H_{\delta,u}(d)\le H_{\delta,u}(d+1)$ for integers $d\ge0$. The family $\FC_u(\delta,R)$ is finite and prefix-disjoint, hence so is each of its subfamilies, in particular $\FC(\delta,R)\subseteq\FC_{r_\delta}(\delta,R)$.
\end{lemma}
\begin{proof}
The line is nondecreasing since $B_0$ and $(d-\delta)_+$ are nondecreasing and $(\delta-d)_+$ is nonincreasing in $d$. Words of $\FC_u(\delta,R)$ have length at most $d^{\mathrm{hi}}_\delta$ and letters at most $\max_{d\le d^{\mathrm{hi}}_\delta}H_{\delta,u}(d)+R$, so the family is finite. Suppose $w\in\FC_u(\delta,R)$ of length $d$ is a proper prefix of $w'\in\FC_u(\delta,R)$. Since $d^{\mathrm{lo}}_\delta<d<|w'|$, the definition for $w'$ gives $\deg(w)=\deg_d(w')<H_{\delta,u}(d)$, while the definition for $w$ gives $\deg(w)\ge H_{\delta,u}(d)$.
\end{proof}

All deficit estimates below rest on one decomposition of the random word $W=(a_1,a_2,\ldots)$ of Section~\ref{sec:words} and its partial degrees $\deg_i(W)$. For integers $\delta\ge9$ and $u$, provided $\deg_{d^{\mathrm{lo}}_\delta}(W)<H_{\delta,u}(d^{\mathrm{lo}}_\delta)$ and some $d$ with $d^{\mathrm{lo}}_\delta<d\le d^{\mathrm{hi}}_\delta$ satisfies $\deg_d(W)\ge H_{\delta,u}(d)$, call the least such $d$ the first crossing $\hat d$ of $W$. Then for every $R\in\N$ and every length $d$, $W_{\le d}\in\FC_u(\delta,R)$ if and only if $d=\hat d$ and $\deg_{\hat d}(W)\le H_{\delta,u}(\hat d)+R$.

\begin{lemma}[Covering the deficit]\label{lem_overshoot}
Let $\delta\ge9$, $u$, $R\ge0$ be integers, and let $d^{\mathrm{lo}}_\delta<d'\le d''\le d^{\mathrm{hi}}_\delta$. Let $\mathcal V$ consist of the words of $\FC_u(\delta,R)$ with length in $[d',d'']$. Then
\[
1-\mass(\mathcal V)\le\Prob(\deg_i(W)\ge H_{\delta,u}(i)\text{ for some }d^{\mathrm{lo}}_\delta\le i<d')+\Prob(\deg_{d''}(W)<H_{\delta,u}(d''))+2^{-(R+1)}.
\]
\end{lemma}
\begin{proof}
We follow \cite[proof of (3.6)]{Mazur}. As $\mathcal V$ is prefix-disjoint, the events that $W$ begins with $w\in\mathcal V$ are disjoint (Section~\ref{sec:mix}), so $\mass(\mathcal V)$ is the probability that some prefix of $W$ lies in $\mathcal V$. If neither of the first two events occurs, the first crossing exists and lies in $[d',d'']$, so $W$ has a prefix in $\mathcal V$ unless the event $\mathcal O$ occurs that $\deg_{\hat d}(W)>H_{\delta,u}(\hat d)+R$. Fix $d^{\mathrm{lo}}_\delta<d\le d^{\mathrm{hi}}_\delta$ and a word $(a_1,\ldots,a_{d-1})$ whose partial degrees at lengths $d^{\mathrm{lo}}_\delta,\ldots,d-1$ lie below the line; its degree $l$ satisfies $l<H_{\delta,u}(d-1)\le H_{\delta,u}(d)$ by Lemma~\ref{lem_s03_barrier_mono}. Given that $W$ begins with this word, $\hat d=d$ is the event $a_d\ge H_{\delta,u}(d)-l\ge1$, and $\mathcal O$ adds $a_d\ge H_{\delta,u}(d)-l+R+1$. For the geometric law, $\Prob(a_1\ge c+R+1)=2^{-(R+1)}\Prob(a_1\ge c)$ for $c\ge1$. Summing over the disjoint pairs $(d,\text{prefix})$ gives $\Prob(\mathcal O)=2^{-(R+1)}\Prob(\hat d\text{ exists})\le2^{-(R+1)}$.
\end{proof}

\subsection{Tail bounds for partial degrees}
The sub-Gaussian maximal bound serves the deficits at large scales (Lemma~\ref{lem_deficit_window}); the exact binomial form with L\'evy's reflection serves the finite evaluation at the scales $256\le\delta<16544$ (Lemma~\ref{lem_fc_binomial_lower}).

\begin{lemma}[Maximal inequality]\label{lem_fc_maximal}
Let $d\ge1$ be an integer and $x>0$ real with $4x/(9d)\le1/32$. Then
\[
\Prob\Bigl(\max_{1\le i\le d}(\deg_i(W)-2i)\ge x\Bigr)\le e^{-2x^2/(9d)},\qquad\Prob(\deg_d(W)-2d\le-x)\le e^{-2x^2/(9d)}.
\]
\end{lemma}
\begin{proof}
Put $\mathfrak t=4x/(9d)\le1/32$, so that $1\le\mathsf M_{\mathrm{geo}}(\pm\mathfrak t)\le e^{9\mathfrak t^2/8}$ by Lemma~\ref{lem:geometric-tail}. As $\mathsf M_{\mathrm{geo}}(\mathfrak t)\ge1$, the process $(e^{\mathfrak t(\deg_i(W)-2i)})_{i\le d}$ is a nonnegative submartingale, and Doob's maximal inequality \cite[Theorem 4.4.2]{Durrett} gives
\[
e^{\mathfrak tx}\Prob\Bigl(\max_{1\le i\le d}(\deg_i(W)-2i)\ge x\Bigr)\le\E e^{\mathfrak t(\deg_d(W)-2d)}=\mathsf M_{\mathrm{geo}}(\mathfrak t)^d\le e^{9d\mathfrak t^2/8},
\]
and $-\mathfrak tx+9d\mathfrak t^2/8=-2x^2/(9d)$. The second bound is the exponential Markov inequality, as in Lemma~\ref{lem:geometric-tail}, with $\mathsf M_{\mathrm{geo}}(-\mathfrak t)^d\le e^{9d\mathfrak t^2/8}$.
\end{proof}

\begin{lemma}[Binomial form and reflection]\label{lem_fc_reflection}
For integers $d,N\ge1$,
\[
\Prob(\deg_d(W)\ge N)=2^{-(N-1)}\sum_{c=0}^{d-1}\binom{N-1}c.
\]
For an integer $d\ge1$ and real $x$,
\[
\Prob\Bigl(\max_{1\le i\le d}(\deg_i(W)-2i)\ge x\Bigr)\le2\Prob(\deg_d(W)-2d\ge x).
\]
\end{lemma}
\begin{proof}
The first claim is the duality between the negative binomial and binomial laws \cite[Section~VI.8]{Feller}: the letters are the waiting times between successes in independent fair Bernoulli trials, and $\deg_d(W)\ge N$ if and only if the first $N-1$ trials show fewer than $d$ successes. By the symmetry of $\binom{2d'-1}c$ it gives $\Prob(\deg_{d'}(W)\ge2d')=\frac12$ for $d'\ge1$, so $0$ is a median of $\deg_{d'}(W)-2d'$. The variables $a_i-2$ are independent, and for $i<d$ the difference $(\deg_d(W)-2d)-(\deg_i(W)-2i)$ has the law of $\deg_{d-i}(W)-2(d-i)$, hence median $0$. L\'evy's inequality \cite[Section~18.1]{Loeve} therefore gives the second claim.
\end{proof}

\subsection{Deficits of first-crossing families}
Lemma~\ref{lem_overshoot} is applied in four regimes: large scales (Lemmas~\ref{lem_deficit_window} and~\ref{lem_deficit_central}), the scales $256\le\delta<16544$ (Lemma~\ref{lem_fc_binomial_lower}), the scales $\delta\le255$, where the masses are computed exactly by counting (Lemma~\ref{lem_fc_dp_mass}), and every shift $0\le u\le2r_\delta$, as the terminal extension needs (Lemma~\ref{lem_deficit_family}).

\begin{lemma}[Central length ranges]\label{lem_deficit_window}
Let $\delta,R\ge0$ and $y$ be integers with $1000\le y\le\delta/8$, and let $\mathcal V_y$ consist of the words of $\FC_{r_\delta}(\delta,R)$ with length in $[\delta-y,\delta+y]$. Then
\[
1-\mass(\mathcal V_y)\le2\exp\Bigl(-\frac{2(207y/500)^2}{9(\delta+y)}\Bigr)+2^{-(R+1)}.
\]
\end{lemma}
\begin{proof}
Put $x=207y/500$. Since $y\le\delta/8<3\delta/5-1\le\ell_\delta$, Lemma~\ref{lem_overshoot} applies with $d'=\delta-y$, $d''=\delta+y$. The shift cancels: $H_{\delta,r_\delta}(i)=2\delta+B_0((i-\delta)_+)-B_0((\delta-i)_+)$. If $d^{\mathrm{lo}}_\delta\le i<\delta-y$ and $\deg_i(W)\ge H_{\delta,r_\delta}(i)$, then
\[
\deg_i(W)-2i\ge2(\delta-i)-B_0(\delta-i)>(2-\log_23)(\delta-i)-1>\tfrac{83}{200}y-1\ge x,
\]
by $B_0(n)<n\log_23+1$ (Section~\ref{sec:physical}), $2-\log_23>\frac{83}{200}$ (Lemma~\ref{lem_s03_log3_bounds}(1)) and $(\frac{83}{200}-\frac{207}{500})y=y/1000\ge1$. If $\deg_{\delta+y}(W)<H_{\delta,r_\delta}(\delta+y)=2\delta+B_0(y)$, then $\deg_{\delta+y}(W)-2(\delta+y)<B_0(y)-2y<1-(2-\log_23)y<-x$. Also $4x/(9(\delta+y))\le\frac{4\cdot207}{9\cdot500\cdot8}<\frac1{32}$. Lemma~\ref{lem_fc_maximal} with $d=\delta+y$ bounds the two events of Lemma~\ref{lem_overshoot} by $e^{-2x^2/(9(\delta+y))}$ each.
\end{proof}

\begin{lemma}[Deficit of the block family at large scales]\label{lem_deficit_central}
Let $\delta\ge16384$ and $R\ge0$ be integers. Then
\[
1-\mass(\FC(\delta,R))\le2\exp\bigl(-\min(\ceil{\log_2\delta}^2,32\ceil{\log_2\delta})/30\bigr)+2^{-(R+1)}.
\]
\end{lemma}
\begin{proof}
For $\delta\ge256$, $\FC(\delta,R)=\mathcal V_{\ell^{\mathrm c}_\delta}$ in the notation of Lemma~\ref{lem_deficit_window}, and $\mathcal V_y\subseteq\mathcal V_{y'}$ for $y\le y'$. Note $\frac{16}{81}(\frac{207}{500})^2=\frac{529}{15625}>\frac1{30}$. We choose $1000\le y\le\min(\delta/8,\ell^{\mathrm c}_\delta)$ with $y^2\ge\delta\min(\ceil{\log_2\delta}^2,32\ceil{\log_2\delta})$; then $\delta+y\le9\delta/8$, and Lemma~\ref{lem_deficit_window} gives the bound $2e^{-\frac{529}{15625}\min(\ceil{\log_2\delta}^2,32\ceil{\log_2\delta})}+2^{-(R+1)}$, which suffices.

\emph{Case $\ceil{\log_2\delta}\le32$.} Take $y=\ceil{\sqrt\delta\,\ceil{\log_2\delta}}$. Since $\log_2\delta\ge14$, $\ceil{\log_2\delta}\le\log_2\delta+1$ and $(\log_2x+1)/\sqrt x$ decreases for real $x\ge4$, $\ceil{\log_2\delta}/\sqrt\delta\le15/2^7$, so $y\le\frac{15}{128}\delta+1\le\delta/8$. Also $y\ge128\cdot14>1000$, and $y\le\ell^{\mathrm c}_\delta$ because $\sqrt\delta\,\ceil{\log_2\delta}\le\sqrt{32\delta\ceil{\log_2\delta}}$ and $y\le\delta/8\le\ell_\delta-1$.

\emph{Case $\ceil{\log_2\delta}\ge33$.} Take $y=\ell^{\mathrm c}_\delta$. As $\log_2\delta\ge32$, $2^{\log_2\delta-13}\ge2^{19}(1+(\log_2\delta-32)/2)\ge\log_2\delta+1>\ceil{\log_2\delta}$, so $\delta>8192\ceil{\log_2\delta}$ and $\sqrt{32\delta\ceil{\log_2\delta}}<\delta/16$. Hence $\ceil{\sqrt{32\delta\ceil{\log_2\delta}}}\le\delta/16+1\le\ell_\delta-1$, so $\ell^{\mathrm c}_\delta=\ceil{\sqrt{32\delta\ceil{\log_2\delta}}}\le\delta/8$, $(\ell^{\mathrm c}_\delta)^2\ge32\delta\ceil{\log_2\delta}$, and $\ell^{\mathrm c}_\delta\ge\sqrt{32\cdot2^{32}\cdot33}>1000$.
\end{proof}

\begin{lemma}[Binomial lower bound]\label{lem_fc_binomial_lower}
Let $\delta\ge256$ and $R\ge0$ be integers, $y=\ell^{\mathrm c}_\delta$, $\bar y=\floor{83y/200}$, $d'=\delta-y$, $d''=\delta+y$, $N'=2d'+\bar y-1$ and $N''=2d''-\bar y-1$. Then
\[
\mass(\FC(\delta,R))\ge1-\frac{2}{2^{N'}}\sum_{c=0}^{d'-1}\binom{N'}c-\frac1{2^{N''}}\sum_{c=0}^{d''-\bar y-1}\binom{N''}c-2^{-(R+1)}.
\]
\end{lemma}
\begin{proof}
Apply Lemma~\ref{lem_overshoot} with these $d',d''$; this is legitimate since $1\le y\le\ell_\delta-1<\ell_\delta$. If $d^{\mathrm{lo}}_\delta\le i<d'$ and $\deg_i(W)\ge H_{\delta,r_\delta}(i)$, the integer $\deg_i(W)-2i$ is at least $2(\delta-i)-B_0(\delta-i)>(2-\log_23)(y+1)-1>83y/200-1$, hence at least $\bar y$. Lemma~\ref{lem_fc_reflection} bounds the first event of Lemma~\ref{lem_overshoot} by $2\Prob(\deg_{d'}(W)\ge N'+1)$, which is the first sum. If $\deg_{d''}(W)<H_{\delta,r_\delta}(d'')$, the integer $\deg_{d''}(W)-2d''\le B_0(y)-2y-1$ is less than $-(2-\log_23)y<-83y/200$, hence at most $-(\bar y+1)$, that is, $\deg_{d''}(W)\le N''$. By the first claim of Lemma~\ref{lem_fc_reflection} and the symmetry of binomial coefficients, this has probability $1-2^{-N''}\sum_{c<d''}\binom{N''}c=2^{-N''}\sum_{c\le N''-d''}\binom{N''}c$, and $N''-d''=d''-\bar y-1$.
\end{proof}

For $9\le\delta\le255$ define counts $\mathcal N_i(l)$ for $d^{\mathrm{lo}}_\delta\le i<d^{\mathrm{hi}}_\delta$ and $l\in\N$ by
\begin{equation}\label{eq:fc-counts}
\begin{gathered}
\mathcal N_{d^{\mathrm{lo}}_\delta}(l)=\binom{l-1}{d^{\mathrm{lo}}_\delta-1}\mathbf 1[d^{\mathrm{lo}}_\delta\le l<H_{\delta,r_\delta}(d^{\mathrm{lo}}_\delta)],\\
\mathcal N_i(l)=\mathbf 1[l<H_{\delta,r_\delta}(i)]\sum_{l'<l}\mathcal N_{i-1}(l')\quad(i>d^{\mathrm{lo}}_\delta).
\end{gathered}
\end{equation}
By induction, $\mathcal N_i(l)$ is the number of words of length $i$ and degree $l$ whose partial degrees at lengths $d^{\mathrm{lo}}_\delta,\ldots,i$ lie strictly below the line $H_{\delta,r_\delta}$: there are $\binom{l-1}{d^{\mathrm{lo}}_\delta-1}$ words of length $d^{\mathrm{lo}}_\delta$ and degree $l\ge d^{\mathrm{lo}}_\delta$, and a word of length $i$ is a word of length $i-1$ followed by one positive letter.

\begin{lemma}[Exact masses at small scales]\label{lem_fc_dp_mass}
For $9\le\delta\le255$ and an integer $R\ge1$,
\[
\mass(\FC(\delta,R))=\sum_{d=d^{\mathrm{lo}}_\delta+1}^{d^{\mathrm{hi}}_\delta}\Bigl(\sum_l\mathcal N_{d-1}(l)\Bigr)\frac{2^{R+1-o_\delta(d)}-1}{2^{H_{\delta,r_\delta}(d)+R}}.
\]
\end{lemma}
\begin{proof}
For $\delta<256$ the family $\FC(\delta,R)$ has no length condition beyond that of $\FC_{r_\delta}(\delta,R)$. A word of $\FC(\delta,R)$ of length $d$ is a word of length $d-1$ and degree $l$ counted by $\mathcal N_{d-1}(l)$, followed by a letter $a$ with $H_{\delta,r_\delta}(d)+o_\delta(d)\le l+a\le H_{\delta,r_\delta}(d)+R$. Since $l<H_{\delta,r_\delta}(d-1)\le H_{\delta,r_\delta}(d)$, each total $c$ in this range arises from exactly one positive $a$, so every such word of length $d-1$ contributes $\sum_{c=H_{\delta,r_\delta}(d)+o_\delta(d)}^{H_{\delta,r_\delta}(d)+R}2^{-c}$. The values of $o_\delta$ satisfy $o_\delta(d)\le2\le R+1$ (Appendix~\ref{app:enumeration} lists them).
\end{proof}

\begin{lemma}[First-crossing families and the mask]\label{lem_deficit_family}
Let $\delta\ge9$, $R\ge0$ and $0\le u\le2r_\delta$ be integers. Then
\begin{gather*}
1-\mass(\FC_u(\delta,R))\le2^{116}\delta^{-6}+2^{-(R+1)},\\
\mass(\{w\in\FC_u(\delta,R):2|w|>H_{\delta,u}(|w|)+\floor{\delta/8}\})\le2^{113}\delta^{-6};
\end{gather*}
the words of the second set are those failing the mask \eqref{eq:mask} at scale $\delta$ and shift $u$.
\end{lemma}
\begin{proof}
We follow \cite[proofs of Proposition 5.2 and Lemma 5.1]{Mazur}, with smaller constants. Apply Lemma~\ref{lem_overshoot} with $d'=d^{\mathrm{lo}}_\delta+1$, $d''=d^{\mathrm{hi}}_\delta$. From $\delta-d^{\mathrm{lo}}_\delta=d^{\mathrm{hi}}_\delta-\delta=\ell_\delta$, the definition of $r_\delta$ and $0\le u\le2r_\delta$,
\[
H_{\delta,u}(d^{\mathrm{lo}}_\delta)=2\delta-B_0(\ell_\delta)+u-r_\delta\ge2d^{\mathrm{lo}}_\delta+2e_\delta,\qquad H_{\delta,u}(d^{\mathrm{hi}}_\delta)=2\delta+B_0(\ell_\delta)+u-r_\delta\le2d^{\mathrm{hi}}_\delta-2e_\delta.
\]
So both events of Lemma~\ref{lem_overshoot} lie in $\{|\deg_i(W)-2i|\ge2e_\delta\}$ for $i=d^{\mathrm{lo}}_\delta$ and $i=d^{\mathrm{hi}}_\delta$, respectively. As $e_\delta\ge\delta/100$ and $i\le2\delta$, the exponent in \eqref{eq:geometric-tail} (Lemma~\ref{lem:geometric-tail}) is at least $\delta/160000$. The bound $e^{-\delta/160000}\le6!\,(\delta/160000)^{-6}$ then gives $1-\mass(\FC_u(\delta,R))\le4\cdot6!\cdot160000^6\delta^{-6}+2^{-(R+1)}$, with $4\cdot6!\cdot160000^6<2^{116}$.

For the mask, let $\mathcal V$ be the set of failing words; it is prefix-disjoint. Since $2^{113}>13^6$, assume $\delta\ge14$. If $w\in\mathcal V$ has length $d$, its prefix of length $d-1\ge d^{\mathrm{lo}}_\delta$ has degree at most $H_{\delta,u}(d-1)-1\le H_{\delta,u}(d)-1\le2d-\floor{\delta/8}-2$, so $\deg_{d-1}(W)-2(d-1)\le-\floor{\delta/8}\le-\delta/16$ if $W$ begins with $w$. Hence $\mass(\mathcal V)\le\sum_{i=d^{\mathrm{lo}}_\delta}^{d^{\mathrm{hi}}_\delta-1}\Prob(|\deg_i(W)-2i|\ge\delta/16)$. There are at most $2\delta$ terms, each at most $2e^{-\delta/16384}$ by \eqref{eq:geometric-tail}, and $e^{-\delta/16384}\le7!\,(\delta/16384)^{-7}$ gives $\mass(\mathcal V)\le4\cdot7!\cdot16384^7\delta^{-6}<2^{113}\delta^{-6}$.
\end{proof}

\subsection{Deficits along the scale sequence}
At the scales $\delta_j$ these bounds give the decay of the block deficits used for propagation, the lower bounds on products of block masses used in Lemma~\ref{lem_s05_tuple_mass}, and the terminal bounds used in Proposition~\ref{prop:terminal-errors}.

\begin{lemma}[Deficit of the block families]\label{lem_central_deficit_decay}
For every $j\in\N$, $1-\mass(\FC(\delta_j,R_j))\le(1024+2^{-16})g^{-3j/2}$.
\end{lemma}
\begin{proof}
First, $2^{-(R_j+1)}\le2^{-16}2^{-j/32}\le2^{-16}g^{-3j/2}$, because $g^{48}<2$ (the integer comparison $101^{48}<2\cdot100^{48}$). Next, $g^{666}\le2^{10}$, since $101^{666}\le2^{10}\cdot100^{666}$. For $j<444$ the claim follows from $1-\mass(\FC(\delta_j,R_j))\le1\le1024g^{-3j/2}$.

Let $j=444+m$ with $m\ge0$. By Lemma~\ref{lem_scales_growth} and $101^{70}>2\cdot100^{70}$, $\delta_j\ge16544g^m\ge16544\cdot2^{\floor{m/70}}$, so $x=\ceil{\log_2\delta_j}\ge15+\floor{m/70}>y=14+m/70$. Then $\min(x^2,32x)/30\ge\min(y^2,32y)/30\ge6+\frac{3m}{200}$: for $y\le32$ this is $y^2-\frac{63}2y+261\ge0$, a quadratic with negative discriminant (as $\frac{3m}{200}=\frac{21}{20}(y-14)$), and for $y\ge32$ it follows from $\frac{32}{30}>\frac{21}{20}$ and $32\cdot14/30>6$. So Lemma~\ref{lem_deficit_central} gives
\[
1-\mass(\FC(\delta_j,R_j))\le2e^{-6}e^{-3m/200}+2^{-(R_j+1)}.
\]
Here $2e^{-6}<1$, and $\log g<1/100$ gives $e^{-3m/200}\le g^{-3m/2}=g^{666}g^{-3j/2}\le1024g^{-3j/2}$.
\end{proof}


\begin{lemma}[Startup product]\label{lem_fc_startup_product}
$\prod_{j=0}^{443}\mass(\FC(\delta_j,R_j))>\frac{72}5\,2^{-25}$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem_scales_growth}, $9\le\delta_j\le255$ for $j\le41$ and $256\le\delta_j<16544$ for $42\le j\le443$. All factors lie in $[0,1]$ by Lemma~\ref{lem_first_crossing_prefix_disjoint}, so it suffices to bound each factor below by a nonnegative rational. The following finite computation uses exact integer and rational arithmetic, with $\delta_j$ and $e_\delta$ computed exactly from their definitions (Lemma~\ref{lem_scales_growth}) and $B_0(n)=\bl(3^n)$ for $n\ge1$ (Section~\ref{sec:physical}).
\begin{enumerate}
\item For $j\le41$, evaluate the right side of Lemma~\ref{lem_fc_dp_mass} at $\delta=\delta_j$, $R=R_j$, computing the counts $\mathcal N_i(l)$ by their recurrence and, for $\delta\le11$, each $o_\delta(d)$ as the least $o\ge0$ satisfying the integer inequality of Section~\ref{sec:physical}. This is $\mass(\FC(\delta_j,R_j))$ exactly; for $j\le4$ these are the masses $\mass(\FC(\delta_j,16))$ tabulated in Appendix~\ref{app:enumeration}.
\item For $42\le j\le443$, evaluate the right side of Lemma~\ref{lem_fc_binomial_lower} at $\delta=\delta_j$, $R=R_j$, with the binomial sums as exact integers, $\ceil{\log_2\delta}=\bl(\delta-1)$, and $\ceil{\sqrt{32\delta\ceil{\log_2\delta}}}=\floor{\sqrt{32\delta\ceil{\log_2\delta}-1}}+1$ (an integer square root). Rounding it down to a multiple of $2^{-64}$ gives a rational $\mass(\FC(\delta_j,R_j))^{\mathrm{lo}}\le\mass(\FC(\delta_j,R_j))$.
\item Multiply the $42$ exact masses of step~1 and the $402$ numbers $\mass(\FC(\delta_j,R_j))^{\mathrm{lo}}$, and compare $5\cdot2^{25}$ times this product with $72$.
\end{enumerate}
All these factors lie in $(0,1)$ (the least $\mass(\FC(\delta_j,R_j))^{\mathrm{lo}}$ is about $0.98866$). The masses of step~1 have product $\approx5.7117\cdot10^{-7}$, the $\mass(\FC(\delta_j,R_j))^{\mathrm{lo}}$ have product $\approx0.76732$, and $2^{25}$ times the product of all $444$ factors lies in $(14.7058,14.7059)$; in particular it exceeds $72/5$.
\end{proof}

\begin{lemma}[Late product]\label{lem_fc_late_survival}
For every integer $j_1\ge444$, $\prod_{j=444}^{j_1-1}\mass(\FC(\delta_j,R_j))>5/9$.
\end{lemma}
\begin{proof}
All factors lie in $[0,1]$, so the product is at least $1-\sum_{j=444}^{j_1-1}(1-\mass(\FC(\delta_j,R_j)))$. Let $j=444+m$ and $i=\floor{m/71}$. By Lemma~\ref{lem_scales_growth} and $101^{71}>2\cdot100^{71}$, $\delta_j\ge16544\cdot2^i>2^{14+i}$, so $\ceil{\log_2\delta_j}\ge14+i$, and both $(14+i)^2$ and $32(14+i)$ are at least $196+29i$. By Lemma~\ref{lem_deficit_central} and $e^{98/15}>650$, $e^{29/30}>2$ (Lemma~\ref{lem_s03_log3_bounds}(4) with $k=11$ and $k=2$),
\[
1-\mass(\FC(\delta_j,R_j))\le\frac2{650}2^{-i}+2^{-(R_j+1)}.
\]
Each $i$ occurs for $71$ values of $m$, so the first terms sum to at most $\frac{142}{650}\cdot2$. Since $R_{444+m}+1\ge30+\floor{m/32}$, the cap terms sum to at most $32\cdot2^{-30}\cdot2<1/2048$. Hence the product exceeds $1-\frac{284}{650}-\frac1{2048}=\frac{374459}{665600}>\frac59$.
\end{proof}


\begin{proposition}[Terminal deficit and mask failure]\label{lem_full_loss_rate}
For every $j\in\N$ and every integer $0\le u\le2r_{\delta_j}$,
\begin{align*}
1-\mass(\FC_u(\delta_j,R_j))&\le(2^{121}+2^{-16})g^{-3j/2},\\
\mass(\{v\in\FC_u(\delta_j,R_j):v\text{ fails \eqref{eq:mask}}\})&\le2^{95}g^{-3j/2},
\end{align*}
where the mask is taken at scale $\delta_j$ and shift $u$, as in Lemma~\ref{lem_deficit_family}.
\end{proposition}
\begin{proof}
By \eqref{eq:scalegrowth}, $\delta_j\ge9g^j$, so $\delta_j^{-6}\le9^{-6}g^{-6j}\le9^{-6}g^{-3j/2}$, and $9^6>2^{19}$. By Lemma~\ref{lem_deficit_family}, the two quantities are at most $2^{97}g^{-3j/2}+2^{-(R_j+1)}$ and $2^{94}g^{-3j/2}$. Bound the cap term as in the proof of Lemma~\ref{lem_central_deficit_decay}.
\end{proof}

\section{Sources and the histogram estimate}\label{sec:est-histogram}
This section proves the spread of the sources (Lemma~\ref{lem_endpoint_ratio}), used in Lemmas~\ref{lem_shift_count} and~\ref{lem_overlap}, and the histogram estimate \eqref{eq:histogram} (Proposition~\ref{lem_histogram_bound}) used in Section~\ref{sec:census}. The arguments are those of \cite[proof of Proposition~3.1]{Mazur}, made explicit and extended to the scales $9\le\delta<256$.

\subsection{Sources}
\begin{lemma}[Spread of the sources]\label{lem_endpoint_ratio}
Let $t\ge16^9$ be odd and $j\in\N$. For all $w,w'\in L_j$,
\[
s_{w'}\le2^{R_\Sigma}s_w.
\]
\end{lemma}

\begin{proof}
Multiplying the bounds of Lemma~\ref{lem_endpoint_large} over $i<j$, starting from $s_{w_{<0}}=s_{w'_{<0}}=t$, gives $s_w\ge t\prod_{i<j}\frac14(4/3)^{\delta_i}$ and $s_{w'}\le t\prod_{i<j}2^{R_i+1}(4/3)^{\delta_i}$. Divide.
\end{proof}

\subsection{Offsets of selected words}
The five-block selector pins down the offset of every selected word of at least five blocks; this places the sources of the words of one length and degree in a short interval (Lemma~\ref{lem_s05_hist_fiber_bound}).

\begin{lemma}[Offsets of selected words]\label{lem_s05_selected_offset}
Let $j\ge5$ and $w=v_0\cdots v_{j-1}\in\bar L_j$. Then
\[
\frac{85}{256}\le\off(w)\le\frac{637}{256}.
\]
\end{lemma}

\begin{proof}
For every word $v$ and integer $\mathfrak r$, rounding each of the $|v|$ summands up to a multiple of $2^{-\mathfrak r}$ gives
\begin{equation}\label{eq:hist-rounding}
\operatorname{rd}_{\mathfrak r}(v)-|v|2^{-\mathfrak r}\le\off(v)\le\operatorname{rd}_{\mathfrak r}(v).
\end{equation}
By Lemma~\ref{lem_scales_growth}, $(\delta_0,\ldots,\delta_5)=(9,\ldots,14)$, and Lemma~\ref{lem_s05_block_weight}(3) gives $0<\wt(v_i)<1/16$ for $i<5$. Moreover $|v_0|\le d^{\mathrm{hi}}_9=14$ and $|v_1|\le d^{\mathrm{hi}}_{10}=16$. For $5\le i\le j$ put $\mathfrak o_i=\off(v_i\cdots v_{j-1})$, so $\mathfrak o_j=0$ and $\mathfrak o_i=\off(v_i)+\wt(v_i)\mathfrak o_{i+1}\ge0$ for $i<j$ by \eqref{eq:concat}. If $\mathfrak o_{i+1}\le2^{\delta_{i+1}+1}$, then Lemma~\ref{lem_s05_block_weight}(2),(3), with $e_{\delta_i}\ge1$ and $\delta_{i+1}=\delta_i+e_{\delta_i}$, give
\[
\mathfrak o_i\le2^{\delta_i}+16^{-e_{\delta_i}}\cdot2^{\delta_i+e_{\delta_i}+1}=2^{\delta_i}+2^{\delta_i+1-3e_{\delta_i}}\le2^{\delta_i+1}.
\]
Downward induction from $i=j$, as in \cite[proof of Proposition~3.1]{Mazur}, gives $0\le\off(v_5\cdots v_{j-1})\le2^{\delta_5+1}=2^{15}$. Applying \eqref{eq:concat} repeatedly,
\begin{multline*}
\off(w)=\off(v_0)+\wt(v_0)\off(v_1)\\
+\wt(v_0v_1)\Bigl[\off(v_2)+\wt(v_2)\bigl(\off(v_3)+\wt(v_3)(\off(v_4)+\wt(v_4)\off(v_5\cdots v_{j-1}))\bigr)\Bigr],
\end{multline*}
where all offsets are nonnegative and all weights positive.

For the lower bound, drop the bracket and apply \eqref{eq:hist-rounding} with $\mathfrak r=9$ and $\mathfrak r=6$:
\[
\off(w)\ge\operatorname{rd}_9(v_0)+\wt(v_0)\operatorname{rd}_6(v_1)-\frac{14}{512}-\frac{\wt(v_0)\cdot16}{64}\ge\frac38-\frac{14}{512}-\frac1{64}=\frac{85}{256}.
\]
For the upper bound, \eqref{eq:hist-rounding} and the five-block selector give $\off(v_0)+\wt(v_0)\off(v_1)\le19/8$, $\off(v_2)\le16$, $\off(v_3)\le64$ and $\off(v_4)\le256$. With the weight bounds, the bracketed term multiplied by $\wt(v_0v_1)$ is at most
\[
\frac{16}{16^2}+\frac{64}{16^3}+\frac{256}{16^4}+\frac{2^{15}}{16^5}=\frac{29}{256},
\]
so $\off(w)\le19/8+29/256=637/256$.
\end{proof}

\subsection{The histogram estimate}
The histogram estimate \eqref{eq:histogram} follows from a uniform bound on the \emph{histogram} of $L_j=L_j(t)$ at level $q$, defined for a positive odd integer $t$, integers $j,q\ge0$ and $y\in G_q$ by
\[
\Xi_{j,q}(t;y)=\sum_{\substack{w\in L_j\\ [s_w]_q=y}}\wt(w).
\]
It is bounded by splitting according to the length and the degree of $w$. Words with the same pair share their weight, their sources are distinct, and by Lemma~\ref{lem_s05_selected_offset} those sources lie in a short interval.

\begin{lemma}[Words of one length and degree]\label{lem_s05_hist_fiber_bound}
Let $t$ be a positive odd integer, $j\ge42$, $0\le q\le2\delta_j$, $y\in G_q$, and $d,l\in\Z$. Then
\[
3^q\sum_{\substack{w\in L_j,\ |w|=d,\ \deg(w)=l\\ [s_w]_q=y}}\wt(w)<\frac{277}{128}.
\]
\end{lemma}

\begin{proof}
All words $w$ in the sum have the same weight $\wt(w)=3^d\cdot2^{-l}$, and $t=\wt(w)s_w+\off(w)$ by \eqref{eq:affine}. Since $j\ge5$, Lemma~\ref{lem_s05_selected_offset} places $s_w=(t-\off(w))/\wt(w)$ in a closed interval of length $\frac{637-85}{256}\wt(w)^{-1}=\frac{69}{32}\wt(w)^{-1}$. A word valid at $t$ is determined by its length and its source (proof of Lemma~\ref{lem_distinct_sources}), so the sources in the sum are distinct integers congruent to $y$ modulo $3^q$ in this interval; there are at most $\frac{69}{32}\wt(w)^{-1}3^{-q}+1$ of them. The left side is therefore at most $\frac{69}{32}+3^q\wt(w)$.

By Lemma~\ref{lem_s05_block_weight}(3), $\wt(v_i)<16^{-e_{\delta_i}}$ for every block $v_i$ of $w$, and $\sum_{i<j}e_{\delta_i}=\delta_j-\delta_0$. Thus $\wt(w)<16^{9-\delta_j}$, and with $3^q\le9^{\delta_j}$,
\[
3^q\wt(w)<2^{36}(9/16)^{\delta_j}\le2^{36}(9/16)^{256}<2^{-7},
\]
using $\delta_j\ge\delta_{42}=256$ and the exact integer inequality $2^{43}\cdot9^{256}<16^{256}$. Since $\frac{69}{32}+\frac1{128}=\frac{277}{128}$, the lemma follows.
\end{proof}

\begin{lemma}[Lengths and degrees of words]\label{lem_s05_valuation_count}
Let $t$ be a positive odd integer and $j\ge10000$. The pairs $(|w|,\deg(w))$ with $w\in L_j$ take fewer than $\frac{4075}{56}j^{5/2}g^{j/2}$ values. More precisely, the lengths $|w|$ take fewer than $4075\sqrt j\,g^{j/2}$ values, and for each length the degrees take fewer than $j^2/56$ values.
\end{lemma}

\begin{proof}
First, the lengths. Every $v\in\FC(\delta,R)$ satisfies $\delta-\ell^{\mathrm c}_\delta\le|v|\le\delta+\ell^{\mathrm c}_\delta$: for $\delta\ge256$ this is the length restriction, and for $\delta<256$ it is $d^{\mathrm{lo}}_\delta<|v|\le d^{\mathrm{hi}}_\delta$ with $\ell^{\mathrm c}_\delta=\ell_\delta$. With $\ell^{\mathrm c}_\Sigma=\sum_{i<j}\ell^{\mathrm c}_{\delta_i}$ as in Section~\ref{sec:steps}, the length $|w|$ is one of the $2\ell^{\mathrm c}_\Sigma+1$ integers within $\ell^{\mathrm c}_\Sigma$ of $\sum_{i<j}\delta_i$. We claim $\ell^{\mathrm c}_\delta\le\ceil{\sqrt{32\delta\ceil{\log_2\delta}}}$ for every $\delta\ge9$. For $\delta\ge256$ this is part of the definition of $\ell^{\mathrm c}_\delta$. For $9\le\delta\le255$, $\ceil{\log_2\delta}\ge4$ and $\ell_\delta^2\le\frac{9\cdot255}{25}\delta<128\delta\le32\delta\ceil{\log_2\delta}$.

By \eqref{eq:scalegrowth}, $\delta_i<201g^i$. The exact inequalities $201<2^8$ and $101^{69}<2\cdot100^{69}$ give $\log_2\delta_i<8+i/69$, so $\ceil{\log_2\delta_i}<9+i/69$. Since $32\cdot201<81^2$,
\[
\ell^{\mathrm c}_{\delta_i}<\sqrt{32\cdot201\,g^i(9+i/69)}+1<81\,g^{i/2}\sqrt{9+j/69}+1\qquad(i<j).
\]
The exact inequality $(202/201)^2<101/100$ gives $g^{1/2}-1>1/201$, hence $\sum_{i<j}g^{i/2}<201g^{j/2}$. For $j\ge10000$, $9/j+1/69<1/64$ (exact), so $\sqrt{9+j/69}<\sqrt j/8$. Moreover $g^j\ge\binom j3\cdot100^{-3}\ge j$ by the binomial theorem, because $(j-1)(j-2)\ge6\cdot10^6$; thus $j\le\sqrt j\,g^{j/2}$. Therefore
\[
\ell^{\mathrm c}_\Sigma<\Bigl(\frac{81\cdot201}8+1\Bigr)\sqrt j\,g^{j/2}<2037\sqrt j\,g^{j/2},\qquad2\ell^{\mathrm c}_\Sigma+1<4075\sqrt j\,g^{j/2},
\]
using $\sqrt j\,g^{j/2}\ge1$ in the last step.

Next, the degrees at a fixed length $d$. For a word $v\in\FC(\delta,R)$, the inequalities $H_{\delta,r_\delta}(|v|)\le\deg(v)\le H_{\delta,r_\delta}(|v|)+R$ and \eqref{eq:hist-barrier} give
\[
2\delta+(|v|-\delta)\log_23-1<\deg(v)<2\delta+(|v|-\delta)\log_23+1+R.
\]
Summing over the blocks $v_i\in\FC(\delta_i,R_i)$ of $w$, with $\sum_i|v_i|=d$,
\begin{multline*}
2\sum_{i<j}\delta_i+\Bigl(d-\sum_{i<j}\delta_i\Bigr)\log_23-j<\deg(w)\\
<2\sum_{i<j}\delta_i+\Bigl(d-\sum_{i<j}\delta_i\Bigr)\log_23+j+\sum_{i<j}R_i.
\end{multline*}
This open interval has integer length $2j+\sum_{i<j}R_i<R_\Sigma+1$, so it contains fewer than $j^2/56$ integers by \eqref{eq:Rsum}.
\end{proof}

\begin{proposition}[Histogram estimate]\label{lem_histogram_bound}
Let $t$ be a positive odd integer, $j\ge10000$ and $0\le q\le2\delta_j$. Then
\[
3^q\Xi_{j,q}(t;y)<\frac{4075}{56}\frac{277}{128}j^{5/2}g^{j/2}\qquad(y\in G_q).
\]
Consequently \eqref{eq:histogram} holds for every $f:G_q\to\R$ and every subset of $L_j$.
\end{proposition}

\begin{proof}
Split the sum defining $\Xi_{j,q}(t;y)$ according to the pair $(|w|,\deg(w))$. By Lemma~\ref{lem_s05_valuation_count} fewer than $\frac{4075}{56}j^{5/2}g^{j/2}$ pairs occur, and by Lemma~\ref{lem_s05_hist_fiber_bound} (as $j\ge42$) each contributes less than $\frac{277}{128}3^{-q}$. For \eqref{eq:histogram}, group a subset of $L_j$ by the residue $y=[s_w]_q$; since weights are nonnegative,
\[
\sum_w\wt(w)|f(s_w)|\le\sum_{y\in G_q}\Xi_{j,q}(t;y)|f(y)|\le\max_y\bigl(3^q\Xi_{j,q}(t;y)\bigr)\av q{|f|}.
\]
\end{proof}


\appendix
\section{Numerical constants and finite checks}\label{app:constants}
This appendix defines the constants of the Fourier argument and records the finite computations
behind the numerical claims. Apart from the standard values of $\log_23$, $\log2$, $\log(4/3)$, $e$,
$e^9$ and $e^{10}$ stated in Lemma~\ref{lem_s03_log3_bounds}, every claim is an exact computation:
finite decimals denote exact rational numbers, the integers involved have at most tens of thousands
of bits and are computed exactly, a ceiling $\ceil{x/y}$ of a positive rational is evaluated as
$\floor{(x+y-1)/y}$, and bit lengths by comparison with powers of $2$.

\begin{lemma}[Elementary bounds]\label{lem_s03_log3_bounds}%
\label{lem_s02_atanh_bounds}\label{lem_s02_log2_lower}%
\label{lem_s02_log2_upper}\label{lem_s02_exp_partial}\label{lem_s02_exp9}%
\label{lem_s02_e_upper}
\leavevmode
\begin{enumerate}
\item $79/50<\log_23<317/200$.
\item For real $y\ge1$ and integer $N\ge1$, with $r=(y-1)/(y+1)$,
\begin{equation}\label{eq:logcert}
2\sum_{j=0}^{N-1}\frac{r^{2j+1}}{2j+1}
\le\log y\le
2\sum_{j=0}^{N-1}\frac{r^{2j+1}}{2j+1}
+\frac{2r^{2N+1}}{(2N+1)(1-r^2)}.
\end{equation}
\item $693147180/10^9<\log2<347/500$ and $\log(4/3)>287682/10^6$.
\item For real $x\ge0$ and $k\in\N$, $\sum_{j=0}^kx^j/j!\le e^x$. In
particular $e^9>8000$ and $e^{10}>10^4$.
\item $e<68/25$.
\end{enumerate}
\end{lemma}
\begin{proof}
Parts (1), (3) and (5), and the two numerical claims in (4), follow from the standard
values $\log_23=1.58496\ldots$, $\log2=0.6931471805\ldots$, $\log(4/3)=0.2876820724\ldots$,
$e=2.7182818\ldots$, $e^9=8103.08\ldots$ and $e^{10}=22026.46\ldots$. Part (2) is the series
$\log y=2\sum_{j\ge0}r^{2j+1}/(2j+1)$, valid as $0\le r<1$ and $(1+r)/(1-r)=y$; its tail after
$N$ terms lies between $0$ and
$\frac2{2N+1}\sum_{k\ge0}r^{2N+1+2k}=\frac{2r^{2N+1}}{(2N+1)(1-r^2)}$. Part (4) holds because
the exponential series has nonnegative terms.
\end{proof}

\subsection{The trace schedule, the threshold \texorpdfstring{$D_*$}{D*}, and numerical facts for Cases~2 and~3}\label{app:analytic}
The descent step (Proposition~\ref{lem_mo_monotonicity}), from which the Fourier bound
\eqref{eq:fourier} follows, holds above a threshold $D_*$, which also determines the mixing
constant $C$ of \eqref{eq:C}. This subsection defines $D_*$ together with the parameters and
the trace schedule from which it is built, proves their numerical properties, and ends with the
numerical facts used in Cases~2 and~3 (Sections~\ref{sec:fb-cases} and~\ref{sec:fb-case3}).

\begin{definition}[Analytic parameters]\label{def_eps_star}\label{def_s02_gamma}%
\label{def_s02_kappa}\label{def_s02_z}\label{def_s02_w}\label{def_s02_Aexp}%
\label{def_Tstar}\label{def_s02_pinit}\label{def_Rstar}%
\label{def_s02_trace_reserve}\label{def_s02_K}\label{def_s02_aK}\label{def_s02_H}
Set
\[
\varepsilon=\frac8{217},\quad \gamma=\frac{87}{200},\quad
\kappa=\frac4{25},\quad z=\frac{21}{500},\quad \chi=\frac{63}{2500},
\]
\[
T_{\mathrm{tr}}=101,\quad r_{\mathrm{tr}}=580,\quad \mathfrak p_0=1024,\quad
L_{\mathrm{tr}}=\frac{7046129}{2^{20}},
\]
\[
K_{\mathrm{tr}}=9863028149,\quad H_{\mathrm{tr}}=17451,\quad
a_{\mathrm{tr}}=\ceil{4K_{\mathrm{tr}}/35}+11=1127203229.
\]
\end{definition}
Here $\varepsilon$ is the threshold on $|\vartheta|$ separating white from black points;
$\gamma$ is the rate of the reward discount $e^{-\gamma\operatorname{Rw}}$ of the trace
(Section~\ref{sec:fb-trace}); $\kappa$ weights the raw letter $3$, both in $e^{-\kappa z}$
and in $E_8(\kappa\gamma)$ (Definition~\ref{def_s02_d}); $z$ is the cancellation at a white
point and $\chi$ the allowance for the envelope factors $e^{\mathsf t_mr}$ of
Lemma~\ref{lem_mo_envelope}; $T_{\mathrm{tr}}$ is the reward
bound and $r_{\mathrm{tr}}+1$ the number of trace times used; and
$L_{\mathrm{tr}}$ is the \emph{trace reserve}. The value of $a_{\mathrm{tr}}$ follows from
$35\cdot1127203217<4K_{\mathrm{tr}}\le35\cdot1127203218$.

The \emph{size bound} $\beta(u)$ defined next is the triangle size from which a visit at
time $u$ counts towards the oversize event (Definition~\ref{def_tr_estar}). It is linear in $u$
along the positions $\mathfrak p_0,\ldots,\mathfrak p_{r_{\mathrm{tr}}}$ of the trace schedule, and
it grows faster afterwards so that the dyadic sums below converge.

\begin{definition}[Trace schedule]\label{def_s02_bitlength}%
\label{def_s02_beta}\label{def_s02_h}\label{def_recipe_map}%
\label{def_s02_schedule}
For $u\in\N$ define
\[
\beta(u)=K_{\mathrm{tr}}\cdot(u+1)\cdot2^{(\bl(u+1)-H_{\mathrm{tr}})_+},\qquad
h_{\mathrm{tr}}(u)=\ceil{4\beta(u)/35}+8192,\qquad
\mathfrak g(u)=u+h_{\mathrm{tr}}(u)+T_{\mathrm{tr}}+1,
\]
and $\mathfrak p_{i+1}=\mathfrak g(\mathfrak p_i)$ for $i\in\N$, starting from the
$\mathfrak p_0$ of Definition~\ref{def_s02_pinit}.
\end{definition}

\begin{lemma}[Linear initial schedule and summable continuation]
\label{lem:schedule}\label{lem_s02_beta_lower}\label{lem_s02_beta_mono}%
\label{lem_s02_envelope}\label{lem_s02_plateau}\label{lem_s02_block_gap}%
\label{lem_s02_block_sum_X}\label{lem_s02_block_sum_one}%
\label{lem_s02_bl_upper}\label{lem_s02_bl_lower}
\leavevmode
\begin{enumerate}
\item $\beta(u)\ge K_{\mathrm{tr}}\cdot(u+1)$, and $\beta$ and $\mathfrak g$ are
nondecreasing with $\mathfrak g(u)>u$. Hence $(\mathfrak p_i)$ is increasing.
\item $1025a_{\mathrm{tr}}^{r_{\mathrm{tr}}}<2^{H_{\mathrm{tr}}}$, and for $0\le i\le r_{\mathrm{tr}}$ we have
$\beta(\mathfrak p_i)=K_{\mathrm{tr}}\cdot(\mathfrak p_i+1)$ and
\begin{equation}\label{eq:F0-envelope}
1025\left(\frac{35+4K_{\mathrm{tr}}}{35}\right)^i\le \mathfrak p_i+1\le1025a_{\mathrm{tr}}^i.
\end{equation}
\item For every $\hat u\in\N$ and $k\in\N$ put
$\mathsf v(k)=\min(\hat u,2^{k+1}-2)$, $\mathsf X(k)=136(\mathsf v(k)+1)$ and $\mathsf S(k)=\beta(2^k-1)$;
the notation suppresses the dependence of $\mathsf v(k)$ and $\mathsf X(k)$ on $\hat u$. These describe the dyadic ranges of Case~3 (Section~\ref{sec:fb-case3}). Then
$4096\mathsf X(k)<\mathsf S(k)$ and
\begin{equation}\label{eq:schedule-sums}
\sum_{k\ge0}\frac{\mathsf X(k)}{\mathsf S(k)}<\frac{272(H_{\mathrm{tr}}+1)}{K_{\mathrm{tr}}},
\qquad \sum_{k\ge0}\frac1{\mathsf S(k)}\le\frac2{K_{\mathrm{tr}}}.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}
For (1), the power of $2$ in $\beta(u)$ is at least $1$, and all three
factors of $\beta$ are nondecreasing in $u$ because $\bl$ is. Hence
$h_{\mathrm{tr}}$ and $\mathfrak g$ are nondecreasing, and
$\mathfrak g(u)>u$ is clear.

For (2), the inequality $1025a_{\mathrm{tr}}^{r_{\mathrm{tr}}}<2^{H_{\mathrm{tr}}}$ is an exact integer
comparison, and $\mathfrak p_0+1=1025$. Suppose that
\eqref{eq:F0-envelope} holds at some $i\le r_{\mathrm{tr}}$. Then $\mathfrak p_i+1\le1025a_{\mathrm{tr}}^{r_{\mathrm{tr}}}<2^{H_{\mathrm{tr}}}$,
so $\bl(\mathfrak p_i+1)\le H_{\mathrm{tr}}$ and $\beta(\mathfrak p_i)=K_{\mathrm{tr}}\cdot(\mathfrak p_i+1)$. If moreover
$i<r_{\mathrm{tr}}$, then, as
$h_{\mathrm{tr}}(u)+T_{\mathrm{tr}}+1=\ceil{4\beta(u)/35}+8294$,
$\mathfrak p_{i+1}+1=\mathfrak p_i+1+\ceil{4K_{\mathrm{tr}}\cdot(\mathfrak p_i+1)/35}+8294$. This is at least
$(1+4K_{\mathrm{tr}}/35)(\mathfrak p_i+1)$, and, since $\mathfrak p_i+1\ge1025$ and $8295<10\cdot1025$, it is at
most
\[
\left(1+\frac{4K_{\mathrm{tr}}}{35}\right)(\mathfrak p_i+1)+8295
 <\left(11+\frac{4K_{\mathrm{tr}}}{35}\right)(\mathfrak p_i+1)\le a_{\mathrm{tr}}(\mathfrak p_i+1).
\]
Induction on $i$ proves \eqref{eq:F0-envelope}, and with it
$\beta(\mathfrak p_i)=K_{\mathrm{tr}}\cdot(\mathfrak p_i+1)$, for $0\le i\le r_{\mathrm{tr}}$.

For (3), $\bl(2^k)=k+1$ gives $\mathsf S(k)=K_{\mathrm{tr}}\cdot2^k\cdot2^{(k+1-H_{\mathrm{tr}})_+}$.
Since $\mathsf v(k)\le2^{k+1}-2$, we have
$\mathsf X(k)\le136(2^{k+1}-1)<272\cdot2^k$. Since $K_{\mathrm{tr}}>4096\cdot272=1114112$, it
follows that $4096\mathsf X(k)<K_{\mathrm{tr}}\cdot2^k\le \mathsf S(k)$. Moreover
\[
\sum_{k\ge0}\frac{\mathsf X(k)}{\mathsf S(k)}
 <\frac{272}{K_{\mathrm{tr}}}\sum_{k\ge0}2^{-(k+1-H_{\mathrm{tr}})_+}
 =\frac{272}{K_{\mathrm{tr}}}\left(H_{\mathrm{tr}}+\sum_{j\ge1}2^{-j}\right)=\frac{272(H_{\mathrm{tr}}+1)}{K_{\mathrm{tr}}},
\]
and $\mathsf S(k)\ge K_{\mathrm{tr}}\cdot2^k$ gives the second bound by a geometric series.
\end{proof}

Case~3 (Section~\ref{sec:fb-case3}) needs the last position of the schedule, the size bound there, and a lower bound on the gaps it treats.

\begin{definition}[Schedule-derived quantities]\label{def_Pstar}%
\label{def_s02_B}\label{def_s02_Lb}\label{def_s02_S}
Put
\[
P=\mathfrak p_{r_{\mathrm{tr}}}+1,\qquad B_{\mathrm{sz}}=\beta(\mathfrak p_{r_{\mathrm{tr}}}),\qquad
\omega_{\mathrm{far}}=2^{36}P.
\]
\end{definition}

\begin{lemma}[Size of $P$]\label{lem_s02_P_upper}\label{lem_s02_P_bits}%
\label{lem_s02_Lb_value}\label{lem_s02_S_lower}
We have $B_{\mathrm{sz}}=K_{\mathrm{tr}}P$, $2^{17450}\le P<2^{17451}$, $2^{17483}\le B_{\mathrm{sz}}<2^{17484}$, and
hence $\bl(P)=17451$ and $\bl(B_{\mathrm{sz}})=17484$. Moreover
$\omega_{\mathrm{far}}>2^{365}$ and $P-1\ge2$.
\end{lemma}
\begin{proof}
Lemma~\ref{lem:schedule}(2) gives $B_{\mathrm{sz}}=K_{\mathrm{tr}}\cdot(\mathfrak p_{r_{\mathrm{tr}}}+1)=K_{\mathrm{tr}}P$ and, with $i=r_{\mathrm{tr}}=580$,
$1025((35+4K_{\mathrm{tr}})/35)^{580}\le P\le1025a_{\mathrm{tr}}^{580}$. The claim then reduces to the
three exact integer comparisons
\begin{gather*}
2^{17450}\cdot35^{580}\le1025(35+4K_{\mathrm{tr}})^{580},\qquad
2^{17483}\cdot35^{580}\le1025K_{\mathrm{tr}}\cdot(35+4K_{\mathrm{tr}})^{580},\\
1025K_{\mathrm{tr}}a_{\mathrm{tr}}^{580}<2^{17484},
\end{gather*}
together with $K_{\mathrm{tr}}>2^{33}$, which gives $P=B_{\mathrm{sz}}/K_{\mathrm{tr}}<2^{17484-33}=2^{17451}$. The remaining claims follow from
$P\ge2^{17450}$.
\end{proof}

The integer $P-1=\mathfrak p_{r_{\mathrm{tr}}}$ is obtained by iterating the recurrence of
Definition~\ref{def_s02_schedule}, in the form found in the proof of Lemma~\ref{lem:schedule}(2),
exactly $r_{\mathrm{tr}}$ times from $\mathfrak p_0$. Both $P-1$ and $P$ have $5254$ decimal
digits, beginning $14859748051937132700$.

\begin{definition}[Thresholds]\label{def_s02_D1}\label{def_s02_Dpt}%
\label{def_s02_Dexp}\label{def_s02_Dbad}\label{def_s02_Dcap}%
\label{def_s02_Dsc}\label{def_Dstar}\label{def_s02_X}
Put $D_{\mathrm{cap}}=(B_{\mathrm{sz}}\cdot\bl(B_{\mathrm{sz}}))^2$ and
\begin{align*}
D_{\mathrm{wc}}&=(256A_*/\chi)^2+2,&D_{\mathrm{pt}}&=2D_{\mathrm{wc}}+344,\\
D_{\mathrm{exp}}&=2^{33495},&
D_{\mathrm{bad}}&=(2^{53}A_*)^2+2^{53}(P+65),\\
D_{\mathrm{sc}}&=2+P+D_{\mathrm{cap}}+(100\omega_{\mathrm{far}})^2,\\
D_*&=\max(D_{\mathrm{wc}},D_{\mathrm{pt}},D_{\mathrm{exp}},D_{\mathrm{bad}},D_{\mathrm{sc}},2).
\end{align*}
\end{definition}
Case~1 of the descent step needs $m\ge D_{\mathrm{wc}}$ and Case~2 needs
$m\ge\max(D_{\mathrm{pt}},D_{\mathrm{exp}})$ (Section~\ref{sec:fb-cases}); Case~3 needs
$m\ge\max(D_{\mathrm{bad}},D_{\mathrm{sc}})$ (Section~\ref{sec:fb-case3}), and $D_{\mathrm{cap}}$ enters
only through $D_{\mathrm{sc}}$.

\begin{lemma}[The thresholds]\label{lem_s02_Dstar_eq}\label{lem_s02_X_head}%
\label{lem_s02_far_B}\label{lem_s02_far_S}\label{lem_s02_Dexp_margin}
\leavevmode
\begin{enumerate}
\item $D_{\mathrm{wc}}=836081643217/1296<2^{30}$, $D_{\mathrm{pt}}<2^{31}$,
$D_{\mathrm{bad}}<2^{17506}$, and $D_{\mathrm{sc}}>D_{\mathrm{cap}}\ge2^{34966}$.
Hence $D_*=D_{\mathrm{sc}}$, and
\begin{equation}\label{eq:bits}
\bl(3200D_*)=35008,\quad
\floor{3200D_*/2^{34928}}=886085405399619011114762.
\end{equation}
\item Every real $m\ge D_{\mathrm{sc}}$ satisfies $P\le m$ and
\[
\frac m{(\log m)^2}>\omega_{\mathrm{far}},\qquad
\frac m{(\log m)^2}>\frac{B_{\mathrm{sz}}^2}2.
\]
\item Every real $m\ge D_{\mathrm{exp}}$ satisfies $\chi\log m>234A_*$.
\end{enumerate}
\end{lemma}
\begin{proof}
For (1), $256A_*/\chi=10241\cdot2500/(16\cdot63)$, which gives the value of
$D_{\mathrm{wc}}$; its comparison with $2^{30}$ is exact, and
$D_{\mathrm{pt}}<2^{31}$ follows. Since $A_*<4$, $(2^{53}A_*)^2<2^{110}$. By
Lemma~\ref{lem_s02_P_bits}, $P+65<2^{17452}$, so
$D_{\mathrm{bad}}<2^{110}+2^{17505}<2^{17506}$. Also
$D_{\mathrm{cap}}\ge B_{\mathrm{sz}}^2\ge2^{34966}$ by Lemma~\ref{lem_s02_Lb_value}, and
$D_{\mathrm{exp}}=2^{33495}$. Thus $D_{\mathrm{sc}}$ is the maximum.
Equation \eqref{eq:bits} is an exact integer computation from the integer
$P$ described above, using $B_{\mathrm{sz}}=K_{\mathrm{tr}}P$ and $\bl(B_{\mathrm{sz}})=17484$.

For (2), first $P\le D_{\mathrm{sc}}-2\le m$. The function
$x\mapsto x/(\log x)^2$ has derivative $(\log x-2)/(\log x)^3$, so it increases on
$[e^2,\infty)$, and $e^2<9$ by Lemma~\ref{lem_s03_log3_bounds}(5).
Both $D_{\mathrm{cap}}$ and $m'=(100\omega_{\mathrm{far}})^2$ are at least $9$ and
less than $m$, so $m/(\log m)^2$ exceeds the values of this function at
$D_{\mathrm{cap}}$ and at $m'$.

Since $x\mapsto\log x/x$ decreases for $x>e$ and $\bl(B_{\mathrm{sz}})\ge4096$, we get\linebreak
$\log\bl(B_{\mathrm{sz}})/\bl(B_{\mathrm{sz}})\le12\log2/4096<1/256$. Since $B_{\mathrm{sz}}<2^{\bl(B_{\mathrm{sz}})}$ and
$\log2<347/500$,
\[
\log D_{\mathrm{cap}}=2\log B_{\mathrm{sz}}+2\log\bl(B_{\mathrm{sz}})
 <2\bl(B_{\mathrm{sz}})\left(\frac{347}{500}+\frac1{256}\right)<\frac75\bl(B_{\mathrm{sz}}),
\]
so $D_{\mathrm{cap}}/(\log D_{\mathrm{cap}})^2>\frac{25}{49}B_{\mathrm{sz}}^2>B_{\mathrm{sz}}^2/2$.

For $m'$, use $\log x<\sqrt x$ for $x>0$ and $\log10^4<10$ (as $e^{10}>10^4$). Since
$\omega_{\mathrm{far}}\ge1$, we get
$\log m'<10+2\sqrt{\omega_{\mathrm{far}}}<100\sqrt{\omega_{\mathrm{far}}}$, and therefore
$m'/(\log m')^2>10^4\omega_{\mathrm{far}}^2/(10^4\omega_{\mathrm{far}})=\omega_{\mathrm{far}}$.

For (3), $\log m\ge33495\log2$. The claim reduces, by
Lemma~\ref{lem_s03_log3_bounds}(3), to the exact rational comparison
\[
\frac{63}{2500}\cdot33495\cdot\frac{693147180}{10^9}=585.0675\ldots
>585.0571\ldots=234\cdot\frac{10241}{4096}.
\]
\end{proof}

\begin{lemma}[Numerical facts for Case~2]\label{lem_mo_rate_tiny}
$e^{-z+\chi/2}<1-\frac35z$ and $\frac\chi8<\frac3{16}\cdot\frac35z$.
\end{lemma}
\begin{proof}
Since $z-\chi/2=147/5000$, the Taylor bound $e^{-y}\le1-y+y^2/2$ $(y\ge0)$ gives
$e^{-z+\chi/2}\le\frac{48551609}{50000000}<\frac{48740000}{50000000}=1-\frac35z$; and
$\frac\chi8=\frac{126}{40000}<\frac{189}{40000}=\frac3{16}\cdot\frac35z$.
\end{proof}

\begin{lemma}[Numerical facts for Case~3]\label{lem_c3_pack_const}\label{def_s02_Qpack}
\leavevmode
\begin{enumerate}
\item $\frac5{16}\le\lambda<\frac8{25}$ and $1/\lambda<158497/50000<127/40$.
\item $\displaystyle(1+2\lambda)\frac{17}{10}\sqrt2
+\Bigl(\frac1\lambda+2\Bigr)\frac{33}{32}\cdot\frac{25}{36}<Q_{\mathrm{pk}}
=\frac{34267893}{4480000}$.
\item If $v$ is an integer with $0\le v\le P-1$ and $\omega\ge\omega_{\mathrm{far}}$ is real, then
$2\omega^{3/5}+272\lambda(v+1)+1\le\omega(\lambda-\frac14)$.
\item If $\omega\ge2^{160}$ is real, then $2^{55}e^{-\omega^{1/5}/65536}<(2\omega)^{-1/2}$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1) follows from the integer comparisons $9^5=59049\le2^{16}$, $2^{25}<9^8$ and
$9^{50000}<2^{158497}$ (the integer $9^{50000}$ has $158497$ binary
digits). For (2), use (1), $1+2\lambda<41/25$ and $\sqrt2<99/70$ (as
$99^2>2\cdot70^2$): the left side is less than
\[
\frac{41}{25}\cdot\frac{17}{10}\cdot\frac{99}{70}
+\Bigl(\frac{127}{40}+2\Bigr)\frac{33}{32}\cdot\frac{25}{36}
=\frac{69003}{17500}+\frac{3795}{1024}=\frac{34267893}{4480000},
\]
an exact rational identity ($Q_{\mathrm{pk}}\approx7.6491$). For (3),
$\omega\ge\omega_{\mathrm{far}}=2^{36}P\ge2^{15}$ gives $\omega^{2/5}\ge2^6$, so $2\omega^{3/5}\le\omega/32$; and
$272\lambda(v+1)+1\le(2176/25+1)P\le2^{31}P\le\omega/32$ by (1). Hence the left side is
at most $\omega/16\le\omega(\lambda-\frac14)$. For (4), $\omega^{1/5}\ge2^{32}$, and it suffices
that $\frac{111}2\log2+\frac12\log\omega<2^{-16}\omega^{1/5}$. Now
$\frac12\log\omega=10\log\omega^{1/20}<10\omega^{1/20}\le2^{-17}\omega^{1/5}$ because
$\omega^{3/20}\ge2^{24}>10\cdot2^{17}$, and
$\frac{111}2\log2<2^{15}\le2^{-17}\omega^{1/5}$.
\end{proof}

\begin{lemma}[Deviation errors]\label{lem_c3_exc_sum}
Let $\hat u\ge2$ be an integer, take $\mathsf v(k)$ from Lemma~\ref{lem:schedule}(3) with this
$\hat u$, put $\bar k=\floor{\log_2(\hat u+1)}$, and let $\Exc$ be the deviation error of
Section~\ref{sec:fb-case3}. Then $\sum_{k=0}^{\bar k}\Exc(\mathsf v(k))<1/3360$.
\end{lemma}
\begin{proof}
$\mathsf v(0)=0$. For $1\le k\le\bar k$ we have $\mathsf v(k)\ge2$, since $\hat u\ge2$
and $2^{k+1}-2\ge2$. These indices are distinct: if $1\le k<k'\le\bar k$,
then $\mathsf v(k)\le2^{k+1}-2<2^{k'}-1\le \mathsf v(k')$, because
$\hat u\ge2^{\bar k}-1\ge2^{k'}-1$. Hence the sum is at most
$\Exc(0)+\sum_{v\ge2}\Exc(v)$. Since $e\ge8/3$ (Lemma~\ref{lem_s03_log3_bounds}(4) with $x=1$, $k=3$), we
have $e^{-14(v+1)}\le(3/8)^{14(v+1)}$. Since $8q_{\mathrm{ex}}<1$, this is at most
\[
9\left(q_{\mathrm{ex}}+\frac{q_{\mathrm{ex}}^3}{1-q_{\mathrm{ex}}}\right)
 +\frac{q_{\mathrm{ex}}(8q_{\mathrm{ex}})^2}{1-8q_{\mathrm{ex}}}+\frac{(3/8)^{42}}{1-(3/8)^{14}}<\frac1{3360}.
\]
The last inequality is an exact rational comparison. In fact $3360$ times
the left side is $0.8609\ldots$.
\end{proof}

\subsection{Trace reserve and error budget}\label{app:budget}
This subsection defines the trace constants, which control the entry kernel of the trace
(Proposition~\ref{lem_tr_kernel_two}), proves the numerical facts about the passage surplus,
and proves that the trace reserve suffices and that the error budget \eqref{eq:budget} of
Case~3 closes.

\begin{definition}[Trace constants]\label{def_s02_d}\label{def_s02_Omega}%
\label{def_s02_rho}\label{def_s02_b12}\label{def_s02_PT}
Put
\[
E_8(x)=\sum_{j=1}^8\frac{(-1)^{j+1}x^j}{j!},\qquad
 d_{\mathrm{tr}}=\frac3{16}E_8(\gamma)+\frac14E_8(\kappa\gamma),
\]
\[
\Omega=\frac{17900546471008806829928229500985832153923}
{14265091358720000000000000000000000000000000},\qquad
\mathsf r_2=(1-d_{\mathrm{tr}})^2-\Omega,
\]
\[
b_{\mathrm{tr}}=\sum_{j=1}^{12}\frac{(1-\mathsf r_2)^j}{j},\qquad
P_{\mathrm{tr}}=\sum_{j=0}^{50}\frac{L_{\mathrm{tr}}^j}{j!}.
\]
\end{definition}
Here $d_{\mathrm{tr}}=0.0829461616\ldots$ is
the one-passage decrement, $\Omega=0.0012548497\ldots$ the two-passage correction (its formula is
Lemma~\ref{lem_tr_delta_floor}(3)), and $\mathsf r_2=0.8397328926\ldots$ the two-passage rate.

By Lemma~\ref{lem_rn_raw_mass_formula}(2), the exit mass $\mathsf q_{45}(\omega)$, the mass
$\mathsf q_3(\omega)$ of \eqref{def_rn_p3}, and hence the passage surplus $\delta_{\mathrm{tr}}(\omega)$
(Section~\ref{sec:fb-trace}) are explicit rational numbers, and so are $d_{\mathrm{pair}}$ and
$d_{\mathrm{wh}}$, defined there.

\begin{lemma}[Passage surplus]\label{lem_tr_delta_floor}
\label{lem_tr_delta_nonneg}\label{lem_tr_W}\label{lem_tr_Omega_value}
\begin{enumerate}
\item $\delta_{\mathrm{tr}}(0)=0$, and $\delta_{\mathrm{tr}}(\omega)\ge\delta_{\mathrm{tr}}(6)>0$ for
every $\omega\ge1$.
\item $\frac3{16}\delta_{\mathrm{tr}}(\omega)+\frac18\delta_{\mathrm{tr}}(\omega-1)\ge d_{\mathrm{pair}}>0$ for every
$\omega\ge1$.
\item $(1-d_{\mathrm{tr}})E_8(\gamma)>8d_{\mathrm{pair}}$, $\ 5(1-d_{\mathrm{tr}})>16d_{\mathrm{pair}}$,
$\ (1-d_{\mathrm{tr}})\delta_{\mathrm{tr}}(6)>\Omega$, and
$\Omega=(1-\gamma)d_{\mathrm{wh}}\bigl(\frac3{16}+\frac{6701}{26392}\bigr)$.
\item $0<d_{\mathrm{tr}}<1$.
\end{enumerate}
\end{lemma}
\begin{proof}
Exact rational arithmetic gives
\begin{gather*}
E_8(\gamma)=\frac{4045450969326735650631}{11468800000000000000000}\approx0.35274,\\
E_8(\kappa\gamma)=\frac{1795318739900437013458321431}{26702880859375000000000000000}\approx0.06723,
\end{gather*}
so $d_{\mathrm{tr}}\approx0.082946$; this proves (4). By Lemma~\ref{lem_rn_p45_large} and its
proof, $\mathsf q_{45}(0)=3/16$ and $\mathsf q_{45}(6)=227/1024$. At $\omega=0$, $f_0=1$ and
$\mathsf R(0,l)=\mathbf 1[l=0]$ give $\mathsf q_3(0)=\mathfrak q(3)=1/4$, so
$\delta_{\mathrm{tr}}(0)=0$ by the definition of $d_{\mathrm{tr}}$. Also
$\mathsf q_{45}(5)=61/256$, $\mathsf q_3(5)=9/64$ and $\mathsf q_3(6)=25/256$, which give
\begin{gather*}
\delta_{\mathrm{tr}}(6)=\frac{3250389656929039669352283093088629}
{1792000000000000000000000000000000000}\approx0.0018138,\\
d_{\mathrm{pair}}=\frac{6799086470348268519226291189141329}
{4096000000000000000000000000000000000}\approx0.0016599.
\end{gather*}

The following finite computation is an exact rational computation.
Compute $\delta_{\mathrm{tr}}(\omega)$ for $0\le\omega\le255$ from the definitions.
Then $\delta_{\mathrm{tr}}(\omega)>\delta_{\mathrm{tr}}(6)$ for
$1\le\omega\le255$, $\omega\ne6$, and
$\frac3{16}\delta_{\mathrm{tr}}(\omega)+\frac18\delta_{\mathrm{tr}}(\omega-1)>d_{\mathrm{pair}}$
for $1\le\omega\le255$, $\omega\ne6$. Moreover
$\delta_{\mathrm{tr}}(255)>39886/10^6$.

For $\omega\ge256$, Lemma~\ref{lem_rn_p45_large} gives
$\mathsf q_{45}(\omega)\ge\frac{515}{2048}$. Since
$\mathsf q_3(\omega)\ge0$ and $E_8(\gamma),E_8(\kappa\gamma)>0$, it follows that
$\delta_{\mathrm{tr}}(\omega)\ge\delta_{\mathrm{fl}}$, where
\[
\delta_{\mathrm{fl}}=\frac{515}{2048}E_8(\gamma)-d_{\mathrm{tr}}=\frac{20623674158457189646056701180070933}{3584000000000000000000000000000000000}
\approx0.0057544>\delta_{\mathrm{tr}}(6).
\]
This proves (1). For (2) with $\omega\ge257$, the sum is at least
$\frac5{16}\delta_{\mathrm{fl}}>d_{\mathrm{pair}}$. At $\omega=256$ it is at least
$\frac3{16}\delta_{\mathrm{fl}}+\frac18\cdot\frac{39886}{10^6}>0.0059>d_{\mathrm{pair}}$.
The inequalities in (3) and the identity for $\Omega$ are exact rational
comparisons between the numbers displayed above; the auxiliary rationals
are $d_{\mathrm{wh}}=\frac{6799086470348268519226291189141329}{1351270400000000000000000000000000000}$
and $\frac3{16}+\frac{6701}{26392}=\frac{23299}{52784}$.
\end{proof}

\begin{lemma}[Trace reserve]\label{lem_s02_rho_unit}\label{lem_s02_trace_term}
We have $0<\mathsf r_2<1$,
\begin{equation}\label{eq:trace}
0<290b_{\mathrm{tr}}-101\gamma-L_{\mathrm{tr}}<2^{-20},
\end{equation}
and
\[
e^{101\gamma}\mathsf r_2^{290}<e^{-L_{\mathrm{tr}}}\le\frac1{P_{\mathrm{tr}}}.
\]
Here $101=T_{\mathrm{tr}}$ and $290=r_{\mathrm{tr}}/2$.
\end{lemma}
\begin{proof}
All of $d_{\mathrm{tr}},\Omega,\mathsf r_2,b_{\mathrm{tr}}$ are explicit rationals, and $0<\mathsf r_2<1$
and \eqref{eq:trace} are exact rational comparisons. The middle quantity
in \eqref{eq:trace} is $2.954\ldots\cdot10^{-7}$. With $x=1-\mathsf r_2\in(0,1)$,
\[
-\log\mathsf r_2=\sum_{j\ge1}\frac{x^j}j>b_{\mathrm{tr}}.
\]
Hence $\mathsf r_2^{290}<e^{-290b_{\mathrm{tr}}}$, and \eqref{eq:trace} gives
$e^{101\gamma}\mathsf r_2^{290}<e^{-L_{\mathrm{tr}}}$. Finally,
Lemma~\ref{lem_s03_log3_bounds}(4) gives $0<P_{\mathrm{tr}}\le e^{L_{\mathrm{tr}}}$.
\end{proof}

\begin{definition}[Continuation constants]\label{def_s02_cont}%
\label{def_s02_lc}\label{def_s02_PW}
Put
\[
c_{\mathrm{cont}}=\frac{511}{10},\qquad
\ell_{\mathrm{cont}}=\frac{157263}{100000},\qquad
x_{\mathrm{wc}}=zT_{\mathrm{tr}}-A_*\ell_{\mathrm{cont}},\qquad
P_{\mathrm{wc}}=\sum_{j=0}^{20}\frac{x_{\mathrm{wc}}^j}{j!}.
\]
\end{definition}
In the proof of Case~3 (Lemma~\ref{lem_c3_case3}), a continuation outside $\mathrm{Bad}^p_m$ loses at
most the factor $(400/83)^{A_*}$, which $c_{\mathrm{cont}}$ bounds; one also outside $\mathrm{Low}^p$
loses at most $e^{-zT_{\mathrm{tr}}}(400/83)^{A_*}$, which $1/P_{\mathrm{wc}}$ bounds, with
$\ell_{\mathrm{cont}}>\log(400/83)$ and $x_{\mathrm{wc}}$ the exponent that remains.

\begin{lemma}[The budget closes]\label{lem_s02_budget}\label{lem_s02_cont}%
\label{lem_s02_white_term}
\leavevmode
\begin{enumerate}
\item $(400/83)^{A_*}<c_{\mathrm{cont}}$ and $c_{\mathrm{cont}}/3360<1/60$.
\item $e^{-zT_{\mathrm{tr}}}(400/83)^{A_*}<1/P_{\mathrm{wc}}$.
\item The error budget $\mathcal B$ of \eqref{eq:budget} satisfies
\[
\frac{14975371}{10^{18}}<1-\mathcal B<\frac{14975372}{10^{18}}.
\]
In particular $\mathcal B<1$.
\end{enumerate}
\end{lemma}
\begin{proof}
For (1), raising both sides to the power $4096$ turns the first claim into
the exact integer comparison
$400^{10241}\cdot10^{4096}<83^{10241}\cdot511^{4096}$. The second claim is
$511/33600<560/33600$.

For (2), Lemma~\ref{lem_s03_log3_bounds}(4) gives
$e^{\ell_{\mathrm{cont}}}\ge\sum_{j\le20}\ell_{\mathrm{cont}}^j/j!=4.819306\ldots>400/83=4.819277\ldots$.
Hence $(400/83)^{A_*}<e^{A_*\ell_{\mathrm{cont}}}$. Next
\[
x_{\mathrm{wc}}=\frac{126992817}{409600000}>0,
\]
so $0<P_{\mathrm{wc}}\le e^{x_{\mathrm{wc}}}$, again by Lemma~\ref{lem_s03_log3_bounds}(4).
Therefore $e^{-zT_{\mathrm{tr}}}(400/83)^{A_*}<e^{-x_{\mathrm{wc}}}\le1/P_{\mathrm{wc}}$.

For (3), $\mathcal B$ is an explicit rational, since $P_{\mathrm{tr}}$ and $P_{\mathrm{wc}}$ are
finite sums of rationals. Its six terms are $0.000125$,
$0.1881196438\ldots$, $0.0616718457\ldots$, $0.0166666666\ldots$,
$0.7334168436\ldots$ and $5.8\ldots\cdot10^{-12}$. The bounds on
$1-\mathcal B$ are exact rational comparisons.
\end{proof}

\subsection{Rational logarithms}
This subsection proves \eqref{eq:Clog}, the enclosure of $\log_2g$, the ceiling $j_0=6103914$ in \eqref{eq:thresholds}, and \eqref{eq:endpoint} by means of \eqref{eq:logcert}, which is finer than the decimal values of Lemma~\ref{lem_s03_log3_bounds}.

Write $(\log y)^{\mathrm{lo}},(\log y)^{\mathrm{hi}}$ for the lower and upper sides of \eqref{eq:logcert} with $N=80$, and put $\mathsf a=\floor{3200D_*/2^{34928}}$, the integer of \eqref{eq:bits}, so that $\mathsf a/2^{79}\le3200D_*/2^{35007}<(\mathsf a+1)/2^{79}$. By \eqref{eq:C}, $C+1=(477/20)(3200D_*)^{A_*}+\frac{169}{10}$, where $\log(477/20)=4\log2+\log(477/320)$ and the additive term changes the natural logarithm by less than $2^{-100}$. Hence the rationals
\begin{align*}
(\log(3200D_*))^{\mathrm{lo}}&=35007(\log 2)^{\mathrm{lo}}+(\log(\mathsf a/2^{79}))^{\mathrm{lo}},\\
(\log(3200D_*))^{\mathrm{hi}}&=35007(\log 2)^{\mathrm{hi}}+(\log((\mathsf a+1)/2^{79}))^{\mathrm{hi}},\\
(\log_2(C+1))^{\mathrm{lo}}&=\frac{4(\log 2)^{\mathrm{lo}}+(\log(477/320))^{\mathrm{lo}}+A_*(\log(3200D_*))^{\mathrm{lo}}}{(\log 2)^{\mathrm{hi}}},\\
(\log_2(C+1))^{\mathrm{hi}}&=\frac{4(\log 2)^{\mathrm{hi}}+(\log(477/320))^{\mathrm{hi}}+A_*(\log(3200D_*))^{\mathrm{hi}}+2^{-100}}{(\log 2)^{\mathrm{lo}}},\\
(\log_23)^{\mathrm{lo}}&=\frac{(\log 2)^{\mathrm{lo}}+(\log(3/2))^{\mathrm{lo}}}{(\log 2)^{\mathrm{hi}}},\qquad(\log_23)^{\mathrm{hi}}=\frac{(\log 2)^{\mathrm{hi}}+(\log(3/2))^{\mathrm{hi}}}{(\log 2)^{\mathrm{lo}}},\\
(\log_2g)^{\mathrm{lo}}&=\frac{(\log(101/100))^{\mathrm{lo}}}{(\log 2)^{\mathrm{hi}}},\qquad(\log_2g)^{\mathrm{hi}}=\frac{(\log(101/100))^{\mathrm{hi}}}{(\log 2)^{\mathrm{lo}}}
\end{align*}
satisfy $(\log_2(C+1))^{\mathrm{lo}}<\log_2 C<\log_2(C+1)<(\log_2(C+1))^{\mathrm{hi}}$, which gives \eqref{eq:Clog}, and $(\log_2g)^{\mathrm{lo}}\le\log_2g\le(\log_2g)^{\mathrm{hi}}$, which gives
\[
0.014355292977070041430<\log_2g<0.014355292977070041432.
\]

Hence the quantity $(23+\log_2(3(C+1))+535/8)/\log_2g$ in \eqref{eq:thresholds} lies between
\[
\frac{23+(\log_23)^{\mathrm{lo}}+(\log_2(C+1))^{\mathrm{lo}}+535/8}{(\log_2g)^{\mathrm{hi}}}\quad\text{and}\quad\frac{23+(\log_23)^{\mathrm{hi}}+(\log_2(C+1))^{\mathrm{hi}}+535/8}{(\log_2g)^{\mathrm{lo}}},
\]
and both have ceiling $6103914$. Finally, $16+j_0(\log_2g)^{\mathrm{lo}}$ and $16+j_0(\log_2g)^{\mathrm{hi}}$ give the enclosure \eqref{eq:endpoint}. All divisions in these formulas have positive denominators.

\subsection{Exact enumeration of the first five blocks}\label{app:enumeration}
This subsection computes the retained fraction \eqref{eq:five} of the five-block selector, used in Lemma~\ref{lem_s05_tuple_mass}. The masses $\mass(\FC(\delta,16))$ for $9\le\delta\le13$ are given by Lemma~\ref{lem_fc_dp_mass}; since the selector bounds rounded offsets, we refine the counts \eqref{eq:fc-counts} by the rounded offset. As $B_0(n)=\bl(3^n)$ for $n\ge1$, no logarithm is evaluated. For $\delta\le11$ the lower bounds $o_\delta(d)$ on the overshoot, in increasing order of $d=d^{\mathrm{lo}}_\delta+1,\ldots,d^{\mathrm{hi}}_\delta$, are
\begin{align*}
\delta=9 &: (1,1,2,1,1,0,0,1,0,1),\\
\delta=10&: (0,1,1,1,1,0,0,0,0,0,0,0),\\
\delta=11&: (0,1,0,1,0,0,0,0,0,0,0,0).
\end{align*}

Fix $\delta$ with $9\le\delta\le13$ and an integer precision $\mathfrak r$; the quantities below depend on $\delta$ and $\mathfrak r$, which the notation suppresses. The \emph{bin} of a word $v$ is $2^{\mathfrak r}\operatorname{rd}_{\mathfrak r}(v)=\sum_{i=1}^{|v|}I_{\mathfrak r}(i,\deg_i(v))$, where $I_{\mathfrak r}(i,l)=\ceil{2^{\mathfrak r}3^{i-1}2^{-l}}$ is the contribution to the bin. Put $\bar H_i=H_{\delta,r_\delta}(\max(i,d^{\mathrm{lo}}_\delta))$, and let $\mathcal N_i(l,y)$ count the words $v$ of length $i$, degree $l$ and bin $y$ with $\deg_{i'}(v)<\bar H_{i'}$ for $1\le i'\le i$. A word of length $i$ is a word of length $i-1$ followed by one positive letter, which raises the degree from some $l'<l$ to $l$ and the bin by $I_{\mathfrak r}(i,l)$. Hence $\mathcal N_0(l,y)=\mathbf 1[l=y=0]$ and
\begin{equation}\label{eq:enum-step}
\mathcal N_i(l,y)=\mathbf 1[l<\bar H_i]\sum_{l'<l}\mathcal N_{i-1}(l',y-I_{\mathfrak r}(i,l))\qquad(i\ge1),
\end{equation}
where cells of negative bin are zero. Since $\deg_{i'}(v)<\deg_{d^{\mathrm{lo}}_\delta}(v)$ for $i'<d^{\mathrm{lo}}_\delta$, for $d^{\mathrm{lo}}_\delta\le i<d^{\mathrm{hi}}_\delta$ the sum $\sum_y\mathcal N_i(l,y)$ is the count $\mathcal N_i(l)$ of \eqref{eq:fc-counts}.

As in the proof of Lemma~\ref{lem_fc_dp_mass}, a word of $\FC(\delta,16)$ of length $d$ is a word counted by $\mathcal N_{d-1}$ followed by the unique letter that gives degree $c$, where $d^{\mathrm{lo}}_\delta<d\le d^{\mathrm{hi}}_\delta$ and $H_{\delta,r_\delta}(d)+o_\delta(d)\le c\le H_{\delta,r_\delta}(d)+16$. For such $d$ and $c$, the number of words of $\FC(\delta,16)$ with length $d$, degree $c$ and bin $y$ is therefore
\begin{equation}\label{eq:enum-final}
\mathcal N^{\mathrm{done}}_d(c,y)=\sum_{l'<c}\mathcal N_{d-1}(l',y-I_{\mathfrak r}(d,c)),
\end{equation}
and we put $\mathcal N^{\mathrm{done}}_d(c,y)=0$ otherwise. For an integer \emph{bin bound} $y_{\max}\ge0$, the set $\FC^{\mathrm{bin}}(\delta,16)=\{v\in\FC(\delta,16):2^{\mathfrak r}\operatorname{rd}_{\mathfrak r}(v)\le y_{\max}\}$ has mass
\begin{equation}\label{eq:enum-mass}
\mass(\FC^{\mathrm{bin}}(\delta,16))=\sum_{d,c}\sum_{y=0}^{y_{\max}}2^{-c}\mathcal N^{\mathrm{done}}_d(c,y).
\end{equation}
Bins never decrease along prefixes, so only cells with $y\le y_{\max}$ are needed.

The following table lists, for the $(\mathfrak r,y_{\max})$ of row $\delta$, the numerators over $2^{\hat l_\delta}$, without cancellation, of $\mass(\FC(\delta,16))$ and $\mass(\FC^{\mathrm{bin}}(\delta,16))$; here $\hat l_\delta=H_{\delta,r_\delta}(d^{\mathrm{hi}}_\delta)+16$ bounds every degree. For $\delta\ge10$ the bin bound is the condition of the five-block selector on block $v_{\delta-9}$. For $\delta=9$, the bin bound $y\le1216$, that is $\operatorname{rd}_9(v_0)\le19/8$, is implied by the upper bound in the first condition of the selector; the lower bound $\frac38$ in that condition is imposed when the first two blocks are joined.
\begin{center}
\small
\begin{tabular}{rrrrrr}
$\delta$ & $\hat l_\delta$ & $\mathfrak r$ & $y_{\max}$ & numerator without bin bound & bin-bounded numerator\\
\hline
9 &42&9&1216&1121528720843&667669768647\\
10&46&6&1024&25912441038107&25532579216651\\
11&48&3&128&110172925800689&108161817817367\\
12&52&0&64&2156388303073866&2155020659894667\\
13&54&$-3$&32&8200094289238498&8199875792177575
\end{tabular}
\end{center}
Finally we join the first two blocks, which the first condition of the selector couples. Write $\mathcal N^{\mathrm{done}}_{9,d}$ and $\mathcal N^{\mathrm{done}}_{10,d}$ for the tables \eqref{eq:enum-final} with $(\delta,\mathfrak r)=(9,9)$ and $(10,6)$, and let $\mathfrak m(y_1)=\sum_{d,c}2^{-c}\mathcal N^{\mathrm{done}}_{10,d}(c,y_1)$ be the mass of the second blocks $v_1\in\FC(10,16)$ with bin $y_1=2^6\operatorname{rd}_6(v_1)$. Call the length, degree and bin $(d,l,y)$ of a first block $v_0$ a \emph{first-block triple}. With $y=2^9\operatorname{rd}_9(v_0)$ and $\wt(v_0)=3^d2^{-l}$, multiplying the first condition of the selector by $512\cdot2^l$ gives exactly
\[
192\cdot2^l\le y\cdot2^l+8\cdot3^dy_1\le1216\cdot2^l.
\]
With the condition $y_1\le1024$ on the second block, the allowed interval of $y_1$ is therefore $y_1^{\mathrm{lo}}(d,l,y)\le y_1\le y_1^{\mathrm{hi}}(d,l,y)$, where
\begin{align*}
y_1^{\mathrm{lo}}(d,l,y)&=\max\left(0,\ceil{\frac{(192-y)2^l}{8\cdot3^d}}\right),\\
y_1^{\mathrm{hi}}(d,l,y)&=\min\left(1024,\floor{\frac{(1216-y)2^l}{8\cdot3^d}}\right)
\end{align*}
(signed integer ceilings and floors). Hence the joint mass of the first two blocks is
\begin{equation}\label{eq:enum-joint}
P_{\mathrm{jt}}=\sum_{d,l,y}2^{-l}\mathcal N^{\mathrm{done}}_{9,d}(l,y)\sum_{y_1=y_1^{\mathrm{lo}}(d,l,y)}^{y_1^{\mathrm{hi}}(d,l,y)}\mathfrak m(y_1),
\end{equation}
an empty inner sum being zero. Exactly $109715$ first-block triples give a positive summand, and $2^{88}P_{\mathrm{jt}}$ is the integer $15856838138131582931885054$, so
\[
P_{\mathrm{jt}}=\frac{7928419069065791465942527}{2^{87}}.
\]
The last three conditions of the five-block selector concern separate blocks, so finite distributivity gives its retained fraction as
\[
\frac{P_{\mathrm{jt}}\,\mass(\FC^{\mathrm{bin}}(11,16))\,\mass(\FC^{\mathrm{bin}}(12,16))\,\mass(\FC^{\mathrm{bin}}(13,16))}{\prod_{\delta=9}^{13}\mass(\FC(\delta,16))}.
\]
Substitution yields exactly \eqref{eq:five}. Its strict comparison with $1/2$ is the integer comparison between twice the printed numerator and the printed denominator.

\section{Extracting tower bounds from Mazur's constants}\label{app:mazur}
This appendix proves \eqref{eq:mazur-bounds}: Mazur's theorem holds with
$(X_0,c)=(D_M,D_M^{-1})$, where $D_M=2^{2^{2^{2^{2^{2^{2^{173}}}}}}}$. Mazur proves it with the
density constant $c_{\mathrm M}$ and cutoff $X_{\mathrm M}$ defined in \cite[Section 8]{Mazur};
we show $1/c_{\mathrm M}<D_M$ and $X_{\mathrm M}<D_M$, using only their arithmetic
definitions, and neither modify Mazur's construction nor verify its analytic inputs.

Mazur's quantities carry the
subscript $\mathrm M$: his $E_*$, $T_*$, $R_*-1$, $g$, $P_*$, $D_{\mathrm{sc}}$, $D_{\mathrm{exp}}$,
$D_*$ and $C$ of \cite[\S8.1]{Mazur} are $a_{\mathrm M}$, $T_{\mathrm M}$, $r_{\mathrm M}$,
$\mathfrak g_{\mathrm M}$, $\mathfrak p_{\mathrm M}$, $D_{\mathrm{sc},\mathrm M}$,
$D_{\mathrm{exp},\mathrm M}$, $D_{*,\mathrm M}$ and $C_{\mathrm M}$, and his $b$, $F$, $N$, $N_t$,
$\beta(j)$, $q$, $M$, $m$, $J$, $E(j)$, $B$, $c$ and $X_0$ of \cite[(8.1)--(8.6)]{Mazur} are
$b_{0,\mathrm M}$, $F_{\mathrm M}$, $N_{*,\mathrm M}$, $N_{\mathrm t,\mathrm M}$, $b_{j,\mathrm M}$,
$q_{*,\mathrm M}$, $\calM_{\mathrm M}$, $m_{*,\mathrm M}$, $J_{*,\mathrm M}$, $\mathfrak l_{j,\mathrm M}$,
$B_{\mathrm{ht},\mathrm M}$, $c_{\mathrm M}$ and $X_{\mathrm M}$. His density constant and cutoff
play the roles of the internal density $\cint$ and internal cutoff $\Lint$. Symbols without the
subscript keep their meaning here.

Write $a_{\mathrm M}=2^{170}$, $T_{\mathrm M}=64090\,2^{3a_{\mathrm M}}$, and
\begin{gather*}
r_{\mathrm M}=64090\,2^{4a_{\mathrm M}}+24586\,2^{a_{\mathrm M}},\\
\mathfrak p_{\mathrm M}=\mathfrak g_{\mathrm M}^{\circ r_{\mathrm M}}(T_{\mathrm M}),\qquad
 \mathfrak g_{\mathrm M}(u)=u+10\cdot2^{98304}(u+1)^3+1+T_{\mathrm M}.
\end{gather*}
For $u\ge T_{\mathrm M}$, $u^3<\mathfrak g_{\mathrm M}(u)<2^{100000}u^3$. Iterating the logarithmic
inequality and summing the geometric series gives
\begin{equation}\label{eq:mazur-p}
3^{r_{\mathrm M}}<\log_2 \mathfrak p_{\mathrm M}<3^{r_{\mathrm M}}(\log_2 T_{\mathrm M}+50000)<4a_{\mathrm M}\,3^{r_{\mathrm M}}.
\end{equation}
The last inequality uses $\log_2 64090<16$ and $a_{\mathrm M}>50016$.

Let $C_{\mathrm M}=2(640\cdot6409\cdot D_{*,\mathrm M})^{6409}+2+2^{481}$ be Mazur's mixing
constant, where $D_{*,\mathrm M}$ is the maximum of his thresholds. By \eqref{eq:mazur-p},
$\log_2\mathfrak p_{\mathrm M}>3^{r_{\mathrm M}}>2^{4a_{\mathrm M}}$. His thresholds other than
$D_{\mathrm{sc},\mathrm M}$ and $D_{\mathrm{exp},\mathrm M}$ are less than
$2^{6a_{\mathrm M}+42}+2^{39}\mathfrak p_{\mathrm M}<\mathfrak p_{\mathrm M}^2$. The degree-nine term
of $D_{\mathrm{sc},\mathrm M}$ is at most $2^{294921}\mathfrak p_{\mathrm M}^9$, so
$\mathfrak p_{\mathrm M}^{10}\le D_{\mathrm{sc},\mathrm M}<\mathfrak p_{\mathrm M}^{10}+2^{294922}\mathfrak p_{\mathrm M}^9<\mathfrak p_{\mathrm M}^{11}$, and
\[
\log_2 D_{\mathrm{exp},\mathrm M}=8192\cdot6409\,2^{3a_{\mathrm M}}
 <2^{3a_{\mathrm M}+26}<2^{r_{\mathrm M}}<3^{r_{\mathrm M}}<\log_2 \mathfrak p_{\mathrm M}.
\]
Hence $\mathfrak p_{\mathrm M}^{10}\le D_{*,\mathrm M}<\mathfrak p_{\mathrm M}^{11}$. As
$640\cdot6409<2^{22}$ and $6409\cdot11<2^{17}$, it follows that
$64090\log_2\mathfrak p_{\mathrm M}<\log_2C_{\mathrm M}<2+6409(22+11\log_2\mathfrak p_{\mathrm M})
<2^{17}\log_2\mathfrak p_{\mathrm M}$, and \eqref{eq:mazur-p} gives
\begin{equation}\label{eq:mazur-L}
3^{r_{\mathrm M}}<\log_2C_{\mathrm M}<2^{20}a_{\mathrm M}\,3^{r_{\mathrm M}},\qquad \log_2C_{\mathrm M}>4\cdot2^{80}+1000.
\end{equation}

Put $b_{0,\mathrm M}=2^{80}$. The next parameters \cite[(8.1)--(8.2)]{Mazur} are
\[
F_{\mathrm M}=2^{467}b_{0,\mathrm M}\cdot16^{b_{0,\mathrm M}}(C_{\mathrm M}+1),\quad
N_{*,\mathrm M}=20000(\ceil{\log_2 F_{\mathrm M}}+64),\quad
N_{\mathrm t,\mathrm M}=20000(\ceil{\log_2(21F_{\mathrm M})}+96).
\]
Since $\log_2 F_{\mathrm M}=547+4b_{0,\mathrm M}+\log_2(C_{\mathrm M}+1)$,
\[
20000\log_2C_{\mathrm M}\le N_{*,\mathrm M}<40000\log_2C_{\mathrm M},\qquad N_{\mathrm t,\mathrm M}<40000\log_2C_{\mathrm M}.
\]
The recurrence
$b_{j+1,\mathrm M}=b_{j,\mathrm M}+\floor{b_{j,\mathrm M}/100}$ gives
$b_{j,\mathrm M}\le2^{80+j}$. Unfolding the definition \cite[(8.3)--(8.4)]{Mazur} of $q_{*,\mathrm M}$ yields
\[
q_{*,\mathrm M}=\floor{b_{N_{*,\mathrm M},\mathrm M}/4}
 +\sum_{i<N_{*,\mathrm M}}\left(b_{i,\mathrm M}+\floor{3b_{i,\mathrm M}/5}\right),
\qquad N_{*,\mathrm M}\le q_{*,\mathrm M}\le(2N_{*,\mathrm M}+1)b_{N_{*,\mathrm M},\mathrm M}.
\]
Consequently $\log_2 q_{*,\mathrm M}<2N_{*,\mathrm M}<100000\log_2C_{\mathrm M}$. Write
\[
\calM_{\mathrm M}=4^{2b_{0,\mathrm M}+1+3^{q_{*,\mathrm M}}},\qquad \log_2\calM_{\mathrm M}=4b_{0,\mathrm M}+2+2\cdot3^{q_{*,\mathrm M}},
\qquad m_{*,\mathrm M}=\floor{(88\calM_{\mathrm M}C_{\mathrm M})^{1/6}}+1.
\]
Then
\begin{equation}\label{eq:mazur-h}
1000\log_2C_{\mathrm M}<\log_2\calM_{\mathrm M}<3\cdot3^{q_{*,\mathrm M}},\qquad
m_{*,\mathrm M}\le100\calM_{\mathrm M}C_{\mathrm M},\qquad \log_2 m_{*,\mathrm M}<2\log_2\calM_{\mathrm M}.
\end{equation}

\subsection{Mazur's density constant}
We bound the iterated logarithms of $1/c_{\mathrm M}$. By \cite[(8.6)]{Mazur},
$c_{\mathrm M}=3/(256\calM_{\mathrm M}\cdot2^{m_{*,\mathrm M}})$. Thus
$\log_2(1/c_{\mathrm M})=m_{*,\mathrm M}+\log_2\calM_{\mathrm M}+8-\log_23<128\calM_{\mathrm M}C_{\mathrm M}$.
We obtain
\begin{gather*}
\log_2^{\circ2}(1/c_{\mathrm M})<2\log_2\calM_{\mathrm M},\qquad
\log_2^{\circ3}(1/c_{\mathrm M})<2q_{*,\mathrm M},\\
\log_2^{\circ4}(1/c_{\mathrm M})<1+\log_2 q_{*,\mathrm M}<100001\log_2C_{\mathrm M}.
\end{gather*}
Here $\log_2(2\log_2\calM_{\mathrm M})<\log_26+q_{*,\mathrm M}\log_23<2q_{*,\mathrm M}$, since $q_{*,\mathrm M}\ge N_{*,\mathrm M}>100$.
Taking another logarithm and using \eqref{eq:mazur-L} gives
\begin{equation}\label{eq:mazur-fifth}
\log_2^{\circ5}(1/c_{\mathrm M})<\log_2^{\circ2}(C_{\mathrm M})+17<r_{\mathrm M}\log_23+207.
\end{equation}

\subsection{Mazur's cutoff}
We bound the iterated logarithms of $X_{\mathrm M}$ in the same way. Mazur defines \cite[(8.4)--(8.6)]{Mazur}
\[
J_{*,\mathrm M}=\max\bigl(N_{*,\mathrm M}+N_{\mathrm t,\mathrm M},200(\ceil{\log_2 m_{*,\mathrm M}}-78)_+,20\cdot10^9\bigr),
\]
\begin{gather*}
\mathfrak l_{0,\mathrm M}=0,\quad \mathfrak l_{j+1,\mathrm M}=\mathfrak l_{j,\mathrm M}+17+\floor{j/100}
 +2b_{j,\mathrm M}-\floor{19b_{j,\mathrm M}/12},\\
B_{\mathrm{ht},\mathrm M}=\mathfrak l_{J_{*,\mathrm M},\mathrm M}+\floor{2b_{J_{*,\mathrm M},\mathrm M}/3},
\end{gather*}
and $X_{\mathrm M}=32(2^{B_{\mathrm{ht},\mathrm M}}\calM_{\mathrm M}+1)$. Equations
\eqref{eq:mazur-L}--\eqref{eq:mazur-h} imply $J_{*,\mathrm M}<1000\log_2\calM_{\mathrm M}$.
Using monotonicity of $j\mapsto b_{j,\mathrm M}$ and the upper bound $2b_{j,\mathrm M}$ for the last
two terms of the recurrence,
\[
B_{\mathrm{ht},\mathrm M}\le4(J_{*,\mathrm M}+17)^2b_{J_{*,\mathrm M},\mathrm M},
\]
\[
\log_2 B_{\mathrm{ht},\mathrm M}\le82+2\log_2(J_{*,\mathrm M}+17)+J_{*,\mathrm M}<4J_{*,\mathrm M}<4000\log_2\calM_{\mathrm M}.
\]
For $X_{\mathrm M}$ this gives
\[
\log_2(X_{\mathrm M})<16+B_{\mathrm{ht},\mathrm M}+\log_2\calM_{\mathrm M}<2^{4096\log_2\calM_{\mathrm M}},\qquad
\log_2^{\circ2}(X_{\mathrm M})<4096\log_2\calM_{\mathrm M},
\]
\[
\log_2^{\circ3}(X_{\mathrm M})<12+\log_2^{\circ2}(\calM_{\mathrm M})<2q_{*,\mathrm M}.
\]
The last inequality uses $\log_2\calM_{\mathrm M}<3\cdot3^{q_{*,\mathrm M}}$, $\log_23<8/5$, and
$q_{*,\mathrm M}>100$. Hence $\log_2^{\circ4}(X_{\mathrm M})<100001\log_2C_{\mathrm M}$, and the same fifth-logarithm
bound \eqref{eq:mazur-fifth} holds for $X_{\mathrm M}$.

\subsection{Counting the iterated logarithms}
It remains to count the logarithms. For either $Y=1/c_{\mathrm M}$ or $Y=X_{\mathrm M}$, the preceding bounds give
\[
\log_2^{\circ5}(Y)<2r_{\mathrm M}+207<2^{4a_{\mathrm M}+17}.
\]
Indeed $2r_{\mathrm M}=128180\,2^{4a_{\mathrm M}}+49172\,2^{a_{\mathrm M}}$, and
$49172\,2^{a_{\mathrm M}}+207<2892\,2^{4a_{\mathrm M}}$, where $128180+2892=2^{17}$.
Therefore
\[
\log_2^{\circ6}(Y)<4a_{\mathrm M}+17=2^{172}+17<2^{173},\qquad
\log_2^{\circ7}(Y)<173.
\]
Since $Y>\calM_{\mathrm M}$ and $q_{*,\mathrm M}>100$, the iterated logarithms
$\log_2^{\circ i}(Y)$ with $i\le6$ are positive, so exponentiating seven times gives
$1/c_{\mathrm M}<D_M$ and $X_{\mathrm M}<D_M$. A larger cutoff and a smaller density
constant preserve Mazur's theorem, so it holds with the cutoff $D_M>X_{\mathrm M}$ and
the density constant $1/D_M<c_{\mathrm M}$. This proves \eqref{eq:mazur-bounds}.

\section{AxiomProver Lean certificate}\label{app:lean}

This paper was written using human-AI collaboration with AxiomProver. AxiomProver is an AI system for mathematical research through formal proof, currently under development at Axiom Math.

The explicit positive lower density established in Theorem \ref{thm:main}, which is the main result in this paper, has been formalized with AxiomProver in Lean. The repository contains
the formal statements, proofs, and instructions for checking the certificate:
\begin{center}
\url{https://github.com/AxiomMath/CollatzPositiveDensity}
\end{center}
The human-written challenge file gives the definitions and hypotheses.
The solution proves the stated targets. The repository reports
successful verification with the Lean comparator.

\section{AI declaration}\label{app:ai}

This project was born out of our goal to formalize Mazur's paper, with the intent of improving the estimates. With this stated goal, AxiomProver autonomously produced the mathematics in this paper and formalized it in Lean. Although this paper was written with AI assistance, the human author is accountable for the statements and conclusions in the introduction, and summarized the main argument in Section \ref{sec:outline} to facilitate human understanding.




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\end{document}