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\title[Two frequency extrema over $\{1,3\}$]{Two exact frequency extrema\\for smooth sequences over $\{1,3\}$}
\author{Will Howe}
\email{howew@purdue.edu}
\author{Grant Huffman}
\email{huffma97@purdue.edu}
\author{Jensen Kohlmeyer}
\email{jenwin@purdue.edu}
\author{Liam Kruer}
\email{kruerl@purdue.edu}
\thanks{Prepared with AI assistance from OpenAI Codex in mathematical exploration,
  checking, and manuscript preparation.}
\date{November 7, 2026}

\begin{document}
\begin{abstract}
Using the frequency coding of Jamet, Marcovici, Poirier, and de la Rue,
we determine the largest frequency of $1$ in a smooth sequence over
$\{1,3\}$ and the largest such frequency below $1/2$. They are
$1/2+1/(3+\sqrt5)$ and $1/2-1/(13-\sqrt5)$, respectively.
The unique attaining type directives are $1^\infty$ and $0001^\infty$.
The proof is an order comparison for two planar homographies. The first
value corrects the numerical expression in the stated extremum question.
\end{abstract}
\maketitle

\section{Statement and frequency coding}

For a one-sided sequence $w$ over $\{1,3\}$, its derivative $\Delta w$
is the sequence of lengths of its successive maximal constant blocks.
A sequence is \emph{smooth} if $\Delta^n w$ is defined and belongs to
$\{1,3\}^{\N}$ for every $n\geq0$. Its frequency of $1$, when it exists,
is
\[
 \operatorname{freq}_1(w)=
 \lim_{N\to\infty}\frac{\#\{0\leq j<N:w_j=1\}}{N}.
\]
Write $F$ for the set of these frequencies over all smooth sequences.
The bi-infinite convention gives the same set, as justified at the end
of Section~\ref{sec:reduction}.
For such a sequence put $c_n=(\Delta^n w)_0$ and define its
\emph{type directive} $\tau\in\{0,1\}^{\N}$ by
\[
 \tau_n=0\quad\Longleftrightarrow\quad c_n=c_{n+1}.
\]
This is the one-sided form of the type convention in
\cite[Proposition 3.11]{JMPdR}; in particular, $0$ and $1$ are not
interchanged in the assertions below.

\begin{theorem}\label{thm:main}
Every smooth sequence over $\{1,3\}$ has a frequency of $1$, and
\begin{align}
 \max F&=\frac12+\frac1{3+\sqrt5},\label{eq:max}\\
 \max\bigl(F\cap[0,1/2]\bigr)&=
   \frac12-\frac1{13-\sqrt5}.\label{eq:lower}
\end{align}
Equality in \eqref{eq:max} holds exactly for type directive $1^\infty$;
equality in \eqref{eq:lower} holds exactly for type directive
$0001^\infty$.
\end{theorem}

The existence of frequencies and their coding are established results
of Jamet, Marcovici, Poirier, and de la Rue. We state the needed
letter-frequency version precisely and give its proof in
Section~\ref{sec:reduction}, before optimizing its image.

\begin{theorem}[Frequency coding; cf.\ \cite{JMPdR}]
\label{thm:coding}
Let
\[
 E=\{(a,b):0\leq a,b\leq1/2,\ a+b\leq3/4\},
\]
and define
\begin{equation}\label{eq:h}
 h_0(a,b)=\frac{(1-a,\,1/2-a)}{3-2(a+b)},\qquad
 h_1(a,b)=\frac{(1/2-a,\,1-a)}{3-2(a+b)}.
\end{equation}
For every $\tau\in\{0,1\}^{\N}$ the intersection
\[
 \bigcap_{n\geq1}h_{\tau_0}\circ\cdots\circ h_{\tau_{n-1}}(E)
   =\{(a(\tau),b(\tau))\}
\]
is a singleton. Every directive is realized by a smooth sequence;
every smooth sequence with directive $\tau$ has frequency
\begin{equation}\label{eq:sourcefreq}
 f_\tau=\frac{3/2-2a(\tau)}{3-2(a(\tau)+b(\tau))}.
\end{equation}
Consequently $F=\{f_\tau:\tau\in\{0,1\}^{\N}\}$.
\end{theorem}

The source ingredients are \cite[Proposition 2.14, Lemmas 3.3--3.5,
Corollary 3.6, Sections 3.2--3.3, and equation (25)]{JMPdR}.
The extremum question appears in \cite[Section 5.1, Problem 1]{JMPdR}
and \cite[Section 5, Problem 1]{JMPdRshort}.
The present result concerns its two numerical extrema only.
It does not determine the Hausdorff dimension of $F$ and does not
address the frequency problem for the classical alphabet $\{1,2\}$.

\section{A contracting planar model}

Introduce coordinates
\begin{equation}\label{eq:coords}
 u=2(a+b)-1,\qquad v=2(b-a),\qquad
 a=\frac{1+u-v}{4},\quad b=\frac{1+u+v}{4}.
\end{equation}
The image of $E$ is the convex compact set
\begin{equation}\label{eq:D}
 D=\{(u,v): |u|+|v|\leq1,\ u\leq1/2\}.
\end{equation}
With the sign convention $0\leftrightarrow-$ and
$1\leftrightarrow+$, direct substitution in \eqref{eq:h} gives
\begin{equation}\label{eq:H}
 H_\sigma(u,v)=\left(\frac{v}{2-u},\frac{\sigma}{2-u}\right),
 \qquad \sigma\in\{-1,1\}.
\end{equation}
We write $H_-=H_{-1}$ and $H_+=H_1$. The frequency function becomes
\begin{equation}\label{eq:f}
 f(u,v)=\frac12+\frac{v}{2(2-u)}.
\end{equation}
All denominators in these formulas are positive on $D$.

For completeness, we record the contraction and the precise coding
properties used in the comparison argument.

\begin{lemma}\label{lem:contract}
The maps $H_-$ and $H_+$ preserve $D$ and are contractions with
constant $2/3$ for the norm $\|(u,v)\|_1=|u|+|v|$.
For $\sigma=(\sigma_0,\sigma_1,\ldots)\in\{-1,1\}^{\N}$ there is
a unique coded point $p(\sigma)\in D$. It satisfies
\begin{equation}\label{eq:recurrence}
 p(\sigma)=H_{\sigma_0}\bigl(p(S\sigma)\bigr)
 =\lim_{n\to\infty}
 H_{\sigma_0}\circ\cdots\circ H_{\sigma_{n-1}}(0,0),
\end{equation}
where $S$ deletes the first sign. The coding is continuous.
\end{lemma}

\begin{proof}
On the square $0\leq a,b\leq1/2$, let $d=3-2(a+b)\geq1$.
For $h_0$, the two numerators are nonnegative, and
\[
 2(1-a)\leq d,\qquad 2(1/2-a)\leq d.
\]
The first inequality is equivalent to $b\leq1/2$, and the second
follows from $b\leq1$. Moreover,
\[
 4(3/2-2a)\leq3d
 \quad\Longleftrightarrow\quad 6b-2a\leq3,
\]
which also holds. Thus $h_0$ maps the square into $E$;
$h_1$ does so by exchanging its two output coordinates.
In particular, $E$ is invariant.

On $E$ the Jacobian of $h_0$ is
\[
 J_{h_0}(a,b)=\frac1{d^2}
 \begin{pmatrix}2b-1&2-2a\\2b-2&1-2a\end{pmatrix}.
\]
Each absolute row sum equals $d/d^2=1/d\leq2/3$.
The same bound holds for $h_1$. Integrating the derivative along a
line segment in the convex set $E$ proves the contraction in the
supremum norm. For coordinate differences, \eqref{eq:coords} gives
\[
 |\delta u|+|\delta v|
   =2\bigl(|\delta a+\delta b|+|\delta b-\delta a|\bigr)
   =4\max\{|\delta a|,|\delta b|\}.
\]
The conjugate maps therefore have the stated contraction constant
in $\|\cdot\|_1$ and preserve $D$.

For a fixed sign sequence the nonempty compact prefix images of $D$
are nested and have diameters at most $(2/3)^n\operatorname{diam}_1D$.
Their intersection is consequently one point, and the finite
compositions applied to $(0,0)\in D$ converge to it. Continuity of
$H_{\sigma_0}$ gives the recurrence. If two sequences agree for their
first $n$ signs, their coded points lie in the same prefix image,
which proves continuity of the coding.
\end{proof}

Let $\mathcal T=\{p(\sigma):\sigma\in\{-1,1\}^{\N}\}$.
Theorem~\ref{thm:coding} and the change of coordinates identify
$F=f(\mathcal T)$ with the indicated sign convention.

\section{Derivation of the frequency coding}\label{sec:reduction}

Here we supply the letter-frequency argument underlying
Theorem~\ref{thm:coding}. This also fixes the realization and limiting
issues needed to pass from homographies to smooth sequences.
Use four symbols to encode runs:
\[
 A=1,\qquad B=3,\qquad C=111,\qquad D=333.
\]
Call a sequence over $\{A,B,C,D\}$ \emph{alternating} if its symbols
alternate between $\{A,C\}$ and $\{B,D\}$. Reading the displayed
blocks decodes an alternating sequence into a unique sequence over
$\{1,3\}$ with run lengths in $\{1,3\}$.
Define non-erasing substitutions by
\begin{equation}\label{eq:substitutions}
\begin{array}{c|cccc}
 &A&B&C&D\\ \hline
 \varphi_0&A&D&ABA&DCD\\
 \varphi_1&B&C&BAB&CDC
\end{array}
\end{equation}
Both substitutions preserve alternation.

\begin{lemma}\label{lem:towers}
If $w$ is smooth and $y^j$ encodes the runs of $\Delta^j w$, then
\begin{equation}\label{eq:tower}
 y^j=\varphi_{\tau_j}(y^{j+1})\qquad(j\geq0).
\end{equation}
Conversely, any infinite family of alternating sequences satisfying
\eqref{eq:tower} decodes to a smooth sequence with directive $\tau$.
Such a family exists for every $\tau\in\{0,1\}^{\N}$.
\end{lemma}

\begin{proof}
A run of $\Delta^j w$ has odd length, so consecutive runs start at
positions of opposite parity. A run in $\Delta^{j+1}w$ therefore
groups either one or three consecutive runs of $\Delta^j w$.
If the first symbols of these two derivative levels agree, their
four possible run encodings give the first row of
\eqref{eq:substitutions}; if they disagree, they give the second row.
The same row applies to every subsequent run: after a group of odd
size the starting symbol flips at both levels, preserving agreement
or disagreement. This proves \eqref{eq:tower}.

For the converse, inspection of each of the eight entries of
\eqref{eq:substitutions} shows that differentiating its decoded word
gives the decoded input symbol. Alternation ensures that adjacent
decoded runs do not merge. Thus the words $w^j$ decoded from $y^j$
satisfy $\Delta w^j=w^{j+1}$. They also have the prescribed equality
or inequality of first symbols. Hence $w^0$ is smooth with directive
$\tau$.

For existence, let $\mathcal A$ be the nonempty compact space of
alternating one-sided sequences over $\{A,B,C,D\}$. It is closed in
the product of a finite discrete alphabet. Non-erasing substitutions
are continuous in the product topology, since any finite output
prefix is determined by a finite input prefix. In the compact space
$\mathcal A^{\N}$, each equation in \eqref{eq:tower} is a closed
condition. Any finite collection of these conditions can be met:
choose an arbitrary alternating sequence at a level beyond all of
them, substitute downwards, and fill unused higher levels
arbitrarily. Compactness gives a family satisfying all the equations.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:coding}]
Lemma~\ref{lem:contract} gives the singleton intersections by conjugacy.
We prove their frequency interpretation.

Fix a smooth sequence and its tower from Lemma~\ref{lem:towers}.
For a prefix of length $m$ of any $y^j$, denote its letter counts by
$A_m,B_m,C_m,D_m$. Alternation gives
\begin{equation}\label{eq:alternatingcounts}
 A_m+C_m=m/2+O(1),\qquad B_m+D_m=m/2+O(1),
\end{equation}
where each error has absolute value at most $1/2$.
If this prefix is substituted once, its output length is exactly
$m+2(C_m+D_m)$. For $\varphi_0$, the output counts of $A$ and $B$
are $A_m+2C_m$ and $C_m$; for $\varphi_1$ these two counts are
exchanged. Therefore, if the normalized input counts of $A,B$
converge to $(a,b)$, the normalized output counts converge to
$h_0(a,b)$ or $h_1(a,b)$, respectively. This follows by inserting
\eqref{eq:alternatingcounts}; the limiting denominator is
$3-2(a+b)\geq1$.

The same conclusion holds for arbitrary output prefixes. Indeed,
such a prefix differs from the substitution of a complete input
prefix by at most two symbols, and its input-prefix length tends to
infinity. Now let $(a_0,b_0)$ be any accumulation point of normalized
$A,B$ counts in prefixes of $y^0$. Fix $L\geq1$ and follow the
corresponding complete prefixes through levels $1,\ldots,L$.
Their lengths tend to infinity, because each substitution expands
length by a factor between $1$ and $3$. By successive subsequence
extraction, their normalized counts converge to points
$(a_j,b_j)$ in the square $K=[0,1/2]^2$ for every $j\leq L$.
The preceding count calculation gives
\[
 (a_j,b_j)=h_{\tau_j}(a_{j+1},b_{j+1})
 \quad(0\leq j<L).
\]
Thus $(a_0,b_0)$ belongs to every finite prefix image of $K$.
Since $h_i(K)\subset E$, it belongs to every finite prefix image of
$E$. Their intersection is the singleton $(a(\tau),b(\tau))$.
All accumulation points coincide, so the normalized $A,B$ counts
converge to this point. Equation~\eqref{eq:alternatingcounts} gives
the remaining limiting counts $c=1/2-a$ and $d=1/2-b$.

A prefix ending at a run boundary in the decoded word contains
$A_m+3C_m$ symbols equal to $1$ and has total length
$A_m+B_m+3(C_m+D_m)$. Its frequency therefore converges to
\[
 \frac{a+3c}{a+b+3(c+d)}
   =\frac{3/2-2a}{3-2(a+b)}.
\]
An arbitrary prefix differs from one ending at a run boundary by at
most two symbols, so it has the same limit. Finally,
Lemma~\ref{lem:towers} realizes every directive by a smooth sequence.
This proves all the assertions for one-sided sequences.

For bi-infinite sequences, index each derivative so that its zeroth
run contains the origin. The same odd-length argument gives the
substitution tower up to a shift of at most two symbols at each level.
At every fixed level these shifts change prefix counts by a bounded
amount, so the preceding limit argument applies unchanged. Conversely,
the compactness proof of Lemma~\ref{lem:towers}, now for alternating
sequences indexed by $\Z$, produces every directive with origins
aligned at run boundaries. Decoding gives bi-infinite smooth sequences.
Thus the bi-infinite frequency set is also exactly
$\{f_\tau:\tau\in\{0,1\}^{\N}\}$.
\end{proof}

\section{The order comparison}

Set
\begin{equation}\label{eq:t}
 t=\frac{\sqrt5-1}{2},\qquad P=(t^2,t).
\end{equation}
We will repeatedly use
\begin{equation}\label{eq:tid}
 0<t<1,\quad t^2=1-t,\quad t>1/2,\quad
 2-t^2=1+t=1/t.
\end{equation}
In particular $P\in D$, since $t^2+t=1$ and $t^2<1/2$,
and \eqref{eq:H} gives $H_+(P)=P$.
The all-plus point is therefore $p(+^\infty)=P$.

For points in $\R^2$ we use coordinatewise order.
\begin{lemma}\label{lem:dominate}
Every $q\in\mathcal T$ satisfies $q\leq P$.
If $q=p(\sigma)$ and $\sigma\neq+^\infty$, then at least one of
these coordinate inequalities is strict.
\end{lemma}

\begin{proof}
Suppose $q=(u,v)\in D$ satisfies $q\leq P$. If $v<0$, the first
coordinate of $H_\sigma(q)$ is negative and hence less than $t^2$.
If $v\geq0$, then
\[
 \frac{v}{2-u}\leq\frac{t}{2-t^2}=t^2.
\]
For the second coordinate, the minus sign gives a negative number,
while the plus sign gives
\[
 \frac1{2-u}\leq\frac1{2-t^2}=t.
\]
Thus both sign maps carry $\{q\in D:q\leq P\}$ into itself.
Because $(0,0)\leq P$, the finite compositions in
\eqref{eq:recurrence} are bounded above by $P$, and taking limits
proves the first assertion.

We also need a strictness observation. If $q\leq P$ and $u<t^2$,
then the second coordinate of $H_+(q)$ is strictly less than $t$;
the same is true of $H_-(q)$ because it is negative. If instead
$u=t^2$ and $v<t$, the first coordinate of either image is strictly
less than $t^2$. Consequently a strict inequality in at least one
coordinate persists through every outer sign map.

If $\sigma_j=-1$ for some $j$, write
$r=p(S^{j+1}\sigma)$. The second coordinate of $H_-(r)$ is negative,
hence strictly less than $t$. Applying the first $j$ maps and the
strictness observation proves the second assertion.
\end{proof}

\begin{corollary}\label{cor:box}
Every $(u,v)\in\mathcal T$ satisfies $|u|\leq t^2$ and $|v|\leq t$.
\end{corollary}

\begin{proof}
Write $(u,v)=H_\sigma(r)$. Lemma~\ref{lem:dominate} gives $r_u\leq t^2$,
and hence $|v|=1/(2-r_u)\leq t$. Apply this bound also to the tail $r$.
Then $|u|=|r_v|/(2-r_u)\leq t/(2-t^2)=t^2$.
\end{proof}

The linear functional that controls the frequency after one sign is
\begin{equation}\label{eq:X}
 X(u,v)=2u+v.
\end{equation}
Substitution into \eqref{eq:f} yields the useful identity
\begin{equation}\label{eq:branch}
 f\bigl(H_\sigma(u,v)\bigr)
   =\frac12+\frac{\sigma}{2(4-X(u,v))}.
\end{equation}
Here $4-X(u,v)>0$ on $D$, for instance from $u\leq1/2$ and
$v\leq1$. Frequencies below $1/2$ correspond exactly to a first
minus sign; none equals $1/2$.

\begin{lemma}\label{lem:Xextrema}
On $\mathcal T$ the function $X$ has extrema
\[
 \max X=2-t,\qquad \min X=t-2.
\]
The maximizing directive is uniquely $+^\infty$ and the minimizing
directive is uniquely $--+^\infty$.
\end{lemma}

\begin{proof}
The coefficients of $X$ are positive, so
Lemma~\ref{lem:dominate} gives
\[
 X(q)\leq X(P)=2t^2+t=2-t,
\]
with equality exactly for the all-plus directive.

Corollary~\ref{cor:box} also gives
\[
 X(q)\geq-2t^2-t=t-2.
\]
Equality forces $q=-P$. To identify its directive, write
$q=H_\sigma(r)$. The second coordinate $q_v=-t$ forces
$\sigma=-1$ and $r_u=t^2$. The first coordinate $q_u=-t^2$ then
forces $r_v=-t$, because $t^2(2-t^2)=t$. Next write $r=H_\eta(z)$.
Its second coordinate again forces $\eta=-1$ and $z_u=t^2$;
its first coordinate $r_u=t^2$ forces $z_v=t$. Thus $z=P$.
By Lemma~\ref{lem:dominate} its directive is $+^\infty$.
Conversely, direct substitution gives $H_-H_-(P)=-P$.
The minimum is therefore attained exactly by $--+^\infty$.
\end{proof}

\section{Evaluation of the frequency extrema}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
The existence of every frequency and the realization of every
directive follow from Theorem~\ref{thm:coding}. It remains to optimize
\eqref{eq:branch} over its two branches.

On the plus branch, the right-hand side of \eqref{eq:branch} is a
strictly increasing function of $X(u,v)$. Lemma~\ref{lem:Xextrema}
therefore gives
\[
 \max_{q\in\mathcal T} f(H_+(q))
  =\frac12+\frac1{2(4-(2-t))}
  =\frac12+\frac1{2(2+t)}
  =\frac12+\frac1{3+\sqrt5}.
\]
Every minus-branch value is less than $1/2$, so this is the global
maximum. Equality forces a first plus sign and the all-plus tail,
which gives the unique directive $1^\infty$.

On the minus branch, \eqref{eq:branch} is a strictly decreasing
function of $X(u,v)$. Thus
\[
 \max_{q\in\mathcal T} f(H_-(q))
  =\frac12-\frac1{2(4-(t-2))}
  =\frac12-\frac1{2(6-t)}
  =\frac12-\frac1{13-\sqrt5}.
\]
This branch is exactly $F\cap[0,1/2]$. Equality forces the tail
$--+^\infty$, so the full directive is $---+^\infty$, or
$0001^\infty$ in the binary convention. This proves both assertions.
\end{proof}

\begin{corollary}\label{cor:gap}
The open interval
\[
 \left(\frac12-\frac1{13-\sqrt5},\,
       \frac12+\frac1{13-\sqrt5}\right)
\]
contains no point of $F$. Both endpoints belong to $F$.
The upper endpoint has unique type directive $1001^\infty$.
\end{corollary}

\begin{proof}
At a fixed tail, \eqref{eq:branch} gives
$f(H_+(q))=1-f(H_-(q))$. Thus $F$ is invariant under
$x\mapsto1-x$, with this reflection obtained by changing only the
first sign. Apply Theorem~\ref{thm:main} to the lower half of $F$.
The uniqueness assertion follows from the same change of first sign.
\end{proof}

\begin{remark}
The value printed for the global supremum in
\cite[Problem 1]{JMPdR} and \cite[Problem 1]{JMPdRshort} is
$1/2+2/(3+\sqrt5)$. The maximizing directive stated there is correct,
but its value is $1/2+1/(3+\sqrt5)$, as follows already by evaluating
\eqref{eq:f} at the fixed point $P$. The lower-half value in those
questions agrees with \eqref{eq:lower}.
\end{remark}

\begin{thebibliography}{2}
\bibitem{JMPdR}
D.~Jamet, I.~Marcovici, L.~Poirier, and T.~de~la~Rue,
\emph{Frequency of patterns in smooth sequences over the alphabet
$\{1,3\}$}, arXiv:2604.11387v3, June 29, 2026,
\url{https://arxiv.org/abs/2604.11387v3}.

\bibitem{JMPdRshort}
D.~Jamet, I.~Marcovici, L.~Poirier, and T.~de~la~Rue,
\emph{Frequency of Patterns in Smooth Sequences over the Alphabet
$\{1,3\}$}, arXiv:2606.13846v1, June 11, 2026,
\url{https://arxiv.org/abs/2606.13846v1}.
\end{thebibliography}
\end{document}