\section{The M\"obius side: the exact anchor and its square-root refinement}\label{ms:sec:intro}
Two facts frame this paper. The first is an identity, valid for every real $x\ge1$:
\begin{equation}\label{ms:eq:anchor}
\sum_{d\le x}\mu(d)\Bigl\lfloor\frac xd\Bigr\rfloor=\sum_{n\le x}\sum_{d\mid n}\mu(d)=1 .
\end{equation}
The summands have size $x/d$; the sum is always exactly $1$. This is the M\"obius orthogonality $\sum_{d\mid n}\mu(d)=[n=1]$ in its rawest form, and it is the engine of Legendre's sieve. The second fact is that the Riemann hypothesis is the square-root refinement of \eqref{ms:eq:anchor}. Writing $\{y\}=y-\lfloor y\rfloor$,
\begin{equation}\label{ms:eq:Phi}
\Phi(x)=\sum_{d\le x}\mu(d)\Bigl\{\frac xd\Bigr\}=x\sum_{d\le x}\frac{\mu(d)}d-1,
\end{equation}
and RH is equivalent to $\sum_{d\le x}\mu(d)/d\ll x^{-1/2+\varepsilon}$, that is, to
\begin{equation}\label{ms:eq:rh}
\Phi(x)\ll x^{1/2+\varepsilon}
\end{equation}
(see \cite[\S14.25--14.27]{T}; the equivalent statements for $M(x)=\sum_{d\le x}\mu(d)$ and for $\sum\mu(d)/d$ are classical, and Franel and Landau \cite{FL} recast them as the equidistribution of the Farey fractions). The prime number theorem gives only $\Phi(x)\ll x\exp(-c(\log x)^{3/5-\varepsilon})$.

So the question ``how far is RH?''\ has, in this language, a concrete form: \emph{how is the cancellation in \eqref{ms:eq:Phi} organized, which parts of it can be proved, and exactly which part cannot?} This paper answers the first two questions with theorems and measurements, and the third with a precise statement of the obstruction. The contributions, in order of appearance:
\begin{enumerate}
\item the $z$-smooth part of $\Phi$ is bounded, by an elementary complete-sieve argument, and the measurements locate the mass of the problem at the divisors $d\asymp x$ (Section~\ref{ms:sec:scales});
\item the long-run variance of the truncated sum obeys an exact, positive-definite identity, diagonalized by Jordan's totient \eqref{ms:eq:diag}; it identifies every constant observed in the data and shows that the gcd structure of $\mu$ cancels $46\%$ of the diagonal variance (Section~\ref{ms:sec:variance}); a decomposition of $\widetilde\Phi_D$ into orthogonal \emph{primitive sawtooths}, one per squarefree conductor, re-proves the identity and becomes the main tool of the paper (Proposition~\ref{ms:prop:primitive});
\item the distribution of $\widetilde\Phi_D$ is empirically Gaussian, out to $4.5$ standard deviations (Section~\ref{ms:sec:gauss});
\item the Gaussian defect of the fourth moment has an exact anatomy (Section~\ref{ms:sec:r4}): it splits into an intra-orbit part, computed exactly --- with the closed law $\varepsilon_4(p)=-(1-p^{-4})/120$ at prime conductors --- and a cross-orbit term $X(D)$. We prove that the conductor-$1$ orbit decouples completely (the classes carrying an integer frequency cancel identically in $X$) and that $X(p)=X(p-1)$ at every prime $p$, so the cross term moves only at composite squarefree conductors; the proofs are real-space identities needing no regularization, confirmed in exact rational arithmetic. Splitting $X$ into four sectors by conductor multiplicities, we prove the $(3,1)$ and $(2,2)$ sectors are $O(D\log^4D)$ unconditionally, prove the exact jump law of the $(2,2)$ sector at the conductors $2p$ (Theorem~\ref{ms:thm:2pjump}), prove a parity sieve that annihilates every sector term using an odd number of frequency slots of even conductor (Lemma~\ref{ms:lem:parity}), and prove $X_{211}\ll D^{3/2+o(1)}$ --- the first unconditional bound on a sector with three distinct conductors (Proposition~\ref{ms:prop:x211}). All four sectors are then computed to $D=800$ (Section~\ref{ms:sec:scale}): the conjectured object stays below $0.44$ in absolute value, and the growth of its absolute-value majorant indicates that, if the trend persists, the quadruple sector cannot be bounded by absolute values alone, locating exactly what kind of cancellation Conjecture~\ref{ms:conj:r4} still needs;
\item a precise statement of why none of this reaches the Riemann hypothesis (Section~\ref{ms:sec:wall}, placed with the other walls in Part~V).
\end{enumerate}
The data reach $x=2\cdot10^7$ for the full sum, $D=3\cdot10^4$ for distributions, $10^8$ samples for the fourth moment, and exact computations over the full period $\mathrm{lcm}(1,\dots,D)$ for $D\le18$. Everything is reproducible from the ancillary code. Part~I carries the complementary, one-sided form of RH that motivates this part: there, RH is the statement that a weighted mean of the prime-counting error stays below the explicit constant $-\bigl(1+\frac\gamma2-\frac12\log4\pi\bigr)$ for ever; here we study the M\"obius side of the same coin. We do not prove the Riemann hypothesis, and no statement below should be read as progress towards it beyond what it literally says.

Throughout, $B_1(y)=\{y\}-\frac12$ is the sawtooth, $\mu^2(k)=1$ marks squarefree $k$, $J_2(k)=k^2\prod_{p\mid k}(1-p^{-2})$ is Jordan's totient, and
\[
\widetilde\Phi_D(x)=\sum_{d\le D}\mu(d)B_1\Bigl(\frac xd\Bigr),\qquad
m_k(Y)=\sum_{\substack{m\le Y\\(m,k)=1}}\frac{\mu(m)}m
\]
are the truncated sawtooth sum and the restricted Mertens sums; $\Phi_D(x)=\sum_{d\le D}\mu(d)\{x/d\}$ is the plain truncation of \eqref{ms:eq:Phi}. ``Long-run'' averages, written $\langle\,\cdot\,\rangle=\lim_{T\to\infty}\frac1T\int_0^T(\cdot)\,dx$, exist because $\widetilde\Phi_D$ is periodic with period $\mathrm{lcm}(1,\dots,D)$. One convention is used silently throughout: every conditionally convergent sum over frequencies is a \emph{symmetric} (principal-value) limit, $\lim_{U\to\infty}\sum_{0<|n|\le U}$. This is the convergence in which $B_1(y)=\sum_{n\ne0}e(ny)/(-2\pi in)$ holds at every non-integer $y$, and all the moment computations below inherit it.

\section{The smooth part is tame, and the mass sits at the large divisors}\label{ms:sec:scales}

\begin{lemma}[the $z$-smooth part is bounded]\label{ms:lem:legendre}
Fix $z\ge2$ and let $P(z)=\prod_{p\le z}p$. The contribution to $\Phi(x)$ of the $d$ free of prime factors $>z$ is
\[
S_z(x)=\sum_{d\mid P(z)}\mu(d)\Bigl\{\frac xd\Bigr\}=x\prod_{p\le z}\Bigl(1-\frac1p\Bigr)-\#\{n\le x:(n,P(z))=1\}\qquad(x\ge P(z)),
\]
a periodic function of $x$ with period $P(z)$, mean $0$, and $|S_z(x)|\le2^{\pi(z)-1}$.
\end{lemma}

\begin{proof}
A squarefree $z$-smooth $d$ divides $P(z)$, and for $x\ge P(z)$ all divisors of $P(z)$ are $\le x$, so the sum is finite and complete; the displayed identity is \eqref{ms:eq:anchor} relativized to $P(z)$ (Legendre's formula). Periodicity is that of each $\{x/d\}$; the mean of $\{x/d\}$ over a period is $\frac12$ and $\sum_{d\mid P(z)}\mu(d)=0$ for $z\ge2$; the bound is $\frac12\cdot\#\{d\mid P(z)\}$.
\end{proof}

Numerically $\max|S_z|=1.37,\ 3.03,\ 5.46$ for $z=7,13,19$ on $x\le2\cdot10^7$ (Figure~\ref{ms:fig:scales}(d)), far below the trivial bound. Whatever is hard in \eqref{ms:eq:rh}, it is not the smooth divisors.

\begin{lemma}[the pairing $d\leftrightarrow pd$]\label{ms:lem:pairing}
For any function $g$ and any prime $p$,
\[
\sum_{d\le x}\mu(d)g(d)=\sum_{\substack{d\le x,\ p\nmid d}}\mu(d)\bigl[g(d)-g(pd)\,\mathbf1_{pd\le x}\bigr].
\]
\end{lemma}

\begin{proof}
A squarefree $d$ is either coprime to $p$ or equals $pd'$ with $p\nmid d'$, and $\mu(pd')=-\mu(d')$.
\end{proof}

Iterating over primes turns the M\"obius sum into difference operators $(\mathrm I-T_p)$ applied to the kernel: the Euler product in operator form. Its statistical signature is visible in the data. Splitting $\Phi(x)$ into octaves $d\in[2^j,2^{j+1})$, over $120$ samples of $x$ near $2\cdot10^7$: \emph{(i)} every octave shows square-root cancellation with the same constant, $\mathrm{sd}_j\approx0.22\sqrt{2^j}$ for all $j\le22$ (the constant is identified in Remark~\ref{ms:rem:constants} below as $1/\sqrt{2\pi^2}=0.2251$); \emph{(ii)} the octaves are not independent: $\mathrm{Var}(\Phi)=0.40\sum_j\mathrm{Var}(\text{octave }j)$, with neighbouring octaves anticorrelated by as much as $-0.56$ --- the pairing of Lemma~\ref{ms:lem:pairing} seen as statistics; \emph{(iii)} truncations $\Phi_D(x)$ stay negligible until $D\approx x/30$: the mass sits at $d\asymp x$, the type~II range, where $\{x/d\}=x/d-k$ on $(\frac x{k+1},\frac xk]$ is piecewise smooth and each block reduces exactly to short Mertens sums,
\[
\sum_{\frac x{k+1}<d\le\frac xk}\mu(d)\Bigl\{\frac xd\Bigr\}=x\!\!\sum_{\frac x{k+1}<d\le\frac xk}\!\frac{\mu(d)}d-k\Bigl[M\Bigl(\frac xk\Bigr)-M\Bigl(\frac x{k+1}\Bigr)\Bigr].
\]
Figure~\ref{ms:fig:scales} shows all four facts. The reduction localizes the problem --- RH is square-root cancellation in smooth-weighted short Mertens sums at the top octaves --- without solving any of it: unconditionally those blocks are only known to be $O(x\exp(-c(\log x)^{3/5-\varepsilon}))$.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/F1_scales.png}
\caption{The scales of $\Phi(x)$ to $2\cdot10^7$. (a) $\Phi/\sqrt x$ is bounded in the range. (b) Truncations: the mass sits at $d\asymp x$. (c) Square-root cancellation with the same constant at every octave. (d) The $z$-smooth part, bounded (Lemma~\ref{ms:lem:legendre}).}
\label{ms:fig:scales}
\end{figure}

\section{The variance: an exact, positive-definite identity}\label{ms:sec:variance}

\begin{lemma}[the Franel--Montgomery--Vaughan kernel; cf.\ \cite{FL,MV}]\label{ms:lem:kernel}
For $d,e\ge1$,
\[
\lim_{T\to\infty}\frac1T\int_0^TB_1\Bigl(\frac xd\Bigr)B_1\Bigl(\frac xe\Bigr)\,dx=\frac{(d,e)^2}{12\,de}.
\]
\end{lemma}

\begin{proof}
Both factors are periodic with common period $de/g$, $g=(d,e)$, so the long-run average is the integral over one period, and Parseval applies with absolute convergence: writing $B_1(y)=\sum_{n\ne0}e(ny)/(-2\pi in)$ (valid off the integers, a null set), the average keeps only the resonant pairs $n/d+m/e=0$, i.e.\ $n=kd/g$, $m=-ke/g$ with $k\ne0$; such a pair contributes $-1/(4\pi^2nm)=g^2/(4\pi^2k^2de)$, and $\sum_{k\ne0}k^{-2}=\pi^2/3$.
\end{proof}

\begin{theorem}[the variance identity]\label{ms:thm:variance}
For every $D\ge1$,
\begin{equation}\label{ms:eq:VD}
\lim_{T\to\infty}\frac1T\int_0^T\widetilde\Phi_D(x)^2\,dx=\frac{V(D)}{12},\qquad
V(D)=\sum_{d,e\le D}\frac{\mu(d)\mu(e)\,(d,e)^2}{de},
\end{equation}
and $V$ diagonalizes as a positive combination of squared restricted Mertens sums:
\begin{equation}\label{ms:eq:diag}
V(D)=\sum_{k\le D}\mu^2(k)\,\frac{J_2(k)}{k^2}\;m_k\!\Bigl(\frac Dk\Bigr)^{\!2}\;\ge\;0 .
\end{equation}
\end{theorem}

\begin{proof}
\eqref{ms:eq:VD} is Lemma~\ref{ms:lem:kernel} summed over $d,e\le D$ with weights $\mu(d)\mu(e)$. For \eqref{ms:eq:diag}, insert Jordan's identity $(d,e)^2=\sum_{k\mid(d,e)}J_2(k)$ and exchange sums:
\[
V(D)=\sum_{k\le D}J_2(k)\Bigl(\sum_{\substack{d\le D\\ k\mid d}}\frac{\mu(d)}d\Bigr)^{\!2},
\]
and non-squarefree $k$ contribute nothing, since no squarefree $d$ is divisible by such a $k$ and the inner sum vanishes. For squarefree $k$, writing $d=km$ with $(m,k)=1$ (any other $d$ has $\mu(d)=0$) gives $\sum_{k\mid d\le D}\mu(d)/d=(\mu(k)/k)\,m_k(D/k)$ by multiplicativity. Squaring removes the sign $\mu(k)^2=1$ and leaves \eqref{ms:eq:diag}.
\end{proof}

\begin{remark}[the constants of the data]\label{ms:rem:constants}
\begin{enumerate}
\item The diagonal $d=e$ alone gives variance $(6/\pi^2)D/12=D/(2\pi^2)$ as $D\to\infty$: per squarefree term, $1/12$; per integer, $1/(2\pi^2)$, i.e.\ standard deviation $0.22508$ per term --- the constant of Figure~\ref{ms:fig:scales}(c).
\item The measured $c_V(D)=V(D)/D$ settles quickly: $c_V(3000)=0.33043$, $c_V(10^4)=0.33138$, $c_V(10^5)=0.33111$, $c_V(2\cdot10^6)=0.33114$ --- three stable digits from $D=3000$ on. (Whether $\lim V(D)/D$ exists is not proved here; the identity \eqref{ms:eq:diag} is exact at every $D$.) Taking $c_V=0.33114$,
\[
\operatorname{sd}\widetilde\Phi_D=\sqrt{\frac{c_VD}{12}}=0.16612\,\sqrt D ,
\]
against $0.2251\sqrt D$ for independent terms: \emph{the gcd structure of $\mu$ cancels $46\%$ of the diagonal variance}, unconditionally. The moving-truncation measurement ($D=x$, Figure~\ref{ms:fig:scales}(a)) gave $0.17\sqrt x$.
\item In \eqref{ms:eq:diag}, the top octave $D/2<k\le D$, where $m_k(D/k)=1$ trivially, carries $81\%$ of $V(2\cdot10^6)$; the top two octaves carry $96\%$. The small $k$, which carry the restricted Mertens sums at large argument, are negligible \emph{unconditionally}, since $m_k(Y)\to0$ by the prime number theorem. The second moment at fixed $D$ is, in this sense, completely understood without any hypothesis.
\end{enumerate}
\end{remark}


The gcd kernel is the aggregate view. The sharper tool, used for the rest of the paper, is to group the Fourier frequencies of $\widetilde\Phi_D$ by their reduced denominator.

\begin{lemma}[class weights]\label{ms:lem:class}
Group the lifts of a reduced fraction $u/q$, $(u,q)=1$: for every $D\ge q$,
\[
\sum_{\substack{d\le D,\ n/d=u/q}}\frac{\mu(d)}{n}=\frac{\mu(q)}{u}\,m_q\!\Bigl(\frac Dq\Bigr).
\]
\end{lemma}

\begin{proof}
$d=qm$ with $(m,q)=1$, $n=um$, and $\mu(qm)/(um)$ summed over $m\le D/q$ is the statement, by multiplicativity.
\end{proof}

\begin{proposition}[primitive decomposition]\label{ms:prop:primitive}
Group the frequencies of $\widetilde\Phi_D$ by their reduced denominator. Then, exactly,
\[
\widetilde\Phi_D(x)=\sum_{\substack{q\le D\\ q\ \mathrm{squarefree}}}\mu(q)\,m_q\!\Bigl(\frac Dq\Bigr)\,\sigma_q\!\Bigl(\frac xq\Bigr),
\qquad
\sigma_q(y)=\sum_{t\mid q}\frac{\mu(t)}t\,B_1(ty),
\]
where $\sigma_q$ is the \emph{primitive sawtooth} of conductor $q$: its Fourier frequencies are exactly the $u$ with $(u,q)=1$. Distinct components have disjoint frequency sets, hence are mutually orthogonal, and
\[
\int_0^1\sigma_q^2=\frac{J_2(q)}{12\,q^2},
\]
which gives a second proof of \eqref{ms:eq:diag}: the variance identity is the orthogonality of the primitive components.
\end{proposition}

\begin{proof}
By Lemma~\ref{ms:lem:class} the aggregated coefficient of the reduced frequency $u/q$ in $\widetilde\Phi_D$ is $\mu(q)m_q(D/q)/(-2\pi iu)$; and $\sigma_q$ as defined has coefficient $1/(-2\pi iu)$ at $u$ coprime to $q$ and $0$ otherwise, by M\"obius inclusion--exclusion over $t\mid q$. The $L^2$ norm is $\sum_{(u,q)=1}(2\pi u)^{-2}=\frac1{2\pi^2}\zeta(2)\prod_{p\mid q}(1-p^{-2})$.
\end{proof}

The identity was verified against sampling: at $D=3000$, sampled variance $82.35$ against $V(D)/12=82.61$; the theoretical line fits from $D=300$ to $3\cdot10^4$ (Figure~\ref{ms:fig:variance}). Positivity of \eqref{ms:eq:diag} has a one-sided consequence worth noting: any single $k$ gives a free \emph{lower} bound for the variance; upper bounds for individual $m_k$ at large arguments are, of course, exactly the hard direction.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/F2_variance.png}
\caption{The variance identity. (a) Verified against sampling. (b) $c_V=0.33114$, between the diagonal $6/\pi^2$ and the moving measurement. (c) The spectrum of \eqref{ms:eq:diag} in $k$.}
\label{ms:fig:variance}
\end{figure}

\section{The distribution is empirically Gaussian}\label{ms:sec:gauss}
With the second moment exact, the next question is the shape; this section is measurement, not theorem. (For the distribution of the Mertens function $M(x)$ itself, see Ng \cite{Ng}.) Over $10^6$ samples at $D=300$, $3\cdot10^5$ at $D=3000$ and $6\cdot10^4$ at $D=3\cdot10^4$ (windows of $x$ near $10^8$), the normalized $\widetilde\Phi_D/\mathrm{sd}$ is Gaussian to remarkable accuracy: skewness $-0.003,\,+0.003,\,+0.006$; kurtosis $2.992,\,3.009,\,2.986$, each within $1.5$ standard errors of $3$; and the empirical tails follow the normal law out to $\lambda=4.5$, where the probability is $7\cdot10^{-6}$ (Figure~\ref{ms:fig:gauss}(a,b)). The symmetry is worth a remark: the one-sidedness that the Riemann hypothesis needs in its margin form (Part~I) is nowhere in the distribution of $\Phi$; it lives in the Fej\'er positivity of the explicit formula, not here.

\section{The fourth moment: anatomy of the Gaussian defect}\label{ms:sec:r4}

\subsection{The resonance expansion}
The method of Lemma~\ref{ms:lem:kernel} computes every moment. For the fourth, define the \emph{Gaussian defect}
\begin{equation}\label{ms:eq:res4}
e_4(D)=\lim_{T\to\infty}\frac1T\int_0^T\widetilde\Phi_D(x)^4\,dx
-3\Bigl(\frac{V(D)}{12}\Bigr)^{\!2}.
\end{equation}
The subtracted term collects the Wick pairings of resonances --- the count a Gaussian would give --- so $e_4(D)$ is what the genuine four-term resonances leave behind: tuples $(n_i,d_i)$, $n_i\ne0$, $d_i\le D$, with $\sum_{i\le4}n_i/d_i=0$ and no vanishing two-term subsum, weighted by $\prod\mu(d_i)/(-2\pi in_i)$, plus the lower-order corrections from coincident pairings. \emph{The excess kurtosis measures $e_4$.} By Lemma~\ref{ms:lem:class}, every four-term resonance class of reduced fractions $(u_i/q_i)$ with $\sum u_i/q_i=0$ enters $e_4$ with weight $\frac1{(2\pi i)^4}\prod_i\frac{\mu(q_i)}{u_i}m_{q_i}(D/q_i)$.

\emph{Regularization.} Two conventions make this expansion precise. Since $\widetilde\Phi_D$ is bounded and the symmetric partial sums of the Fourier series of $B_1$ are uniformly bounded and converge off the integers, bounded convergence computes every joint moment as the limit of truncated frequency sums, the truncations tending to infinity independently; at this stage the value is order-independent. The expansion then groups the surviving quadruples into the classes of Lemma~\ref{ms:lem:class}; within a class the numerator sum is only conditionally convergent, and it is always evaluated in the symmetric (principal-value) sense fixed in Section~\ref{ms:sec:intro}. The main theorems of this section turn out to need none of this: they are proved in real space, as identities between long-run averages of bounded periodic functions over a common period. The symmetric convention governs the class-by-class identities of Remark~\ref{ms:rem:freqview} and the heuristic discussion; we do not claim a general interchange theorem for the infinite sum over classes, and every assertion that ties class sums to the measurable quantities $e_4$, $I$, $X$ is backed by the exact full-period computations, which involve no limits at all.

Two structural consequences. \emph{(i) Classes containing an integer frequency cancel identically in $X$}: genuine such classes exist, with nonzero individual weights, at every fixed $D\ge6$ --- the conductor multiset $\{1,2,3,6\}$, with witness $\frac12+\frac13+\frac16-1=0$, is live from $D=6$ --- but the conductor-$1$ orbit decouples completely from the fourth moment (Theorem~\ref{ms:thm:decouple} below); the factor $m_1(D)\to0$ makes them small, the decoupling makes them irrelevant. \emph{(ii) The live classes need all $q_i\asymp D$}, where $m_{q_i}(D/q_i)=1+O(q_i/D)$ gives no damping; clearing denominators then forces the multiplicative matching of Proposition~\ref{ms:prop:valuation} below. The sign $\mu(q_1)\cdots\mu(q_4)$ is $+1$ whenever every prime divides an even number of the conductors, though not always (a prime may divide exactly three: the multiset $\{2,6,6,6\}$, with witness $-\frac12+\frac16+\frac16+\frac16=0$, carries sign $-1$); the real damping is that the numerators live on a rank-$3$ lattice of covolume $\asymp D$, so the weights $1/(u_1u_2u_3u_4)$ suppress every class. This is the structural reason to expect the non-pairing resonances to total far less than the trivial $O(D^2)$ scale of the expansion (two of the four conductors free, each $\asymp D$).

\subsection{The intra-orbit part is exact}
The decomposition of Proposition~\ref{ms:prop:primitive} splits the Gaussian defect into the \emph{intra-orbit} defects, which are exactly computable, and the \emph{cross-orbit} resonances:
\begin{equation}\label{ms:eq:IX}
e_4(D)
=\underbrace{\sum_{q\le D}\mu^2(q)\,m_q\!\Bigl(\frac Dq\Bigr)^{\!4}\varepsilon_4(q)}_{I(D)}\;+\;X(D),
\qquad \varepsilon_4(q)=\Bigl\langle\sigma_q^4\Bigr\rangle-3\Bigl\langle\sigma_q^2\Bigr\rangle^2 ,
\end{equation}
where $X(D)$ is \emph{defined} by this identity --- so it is as rigorous as $e_4$ and $I$ --- and, in the resonance expansion, collects the quadruples of frequencies from at least two different conductors. Each $\varepsilon_4(q)$ is a finite rational computation: $\sigma_q$ is piecewise linear with nodes on $\frac1q\mathbb Z$ and slope $\sum_{t\mid q}\mu(t)=[q=1]$, so its moments are finite sums. The first values are
\[
\varepsilon_4(1)=-\tfrac1{120},\quad \varepsilon_4(2)=-\tfrac1{128},\quad \varepsilon_4(3)=-\tfrac2{243},\quad
\varepsilon_4(5)=-\tfrac{26}{3125},\quad \varepsilon_4(6)=-\tfrac1{216},\quad\ldots
\]
all negative. For prime conductor there is a closed law with a pleasant proof.

\begin{lemma}[prime conductors]\label{ms:lem:primeorbit}
For $q=p$ prime, $\sigma_p$ is the \emph{discrete} uniform variable on the $p$ equally spaced values $\frac{2j+1-p}{2p}$, $0\le j<p$, and
\[
\varepsilon_4(p)=-\frac1{120}\bigl(1-p^{-4}\bigr).
\]
\end{lemma}

\begin{proof}
Write $y=(j+u)/p$ with $j$ uniform on $\{0,\dots,p-1\}$ and $u$ uniform on $[0,1)$, independent. Then $\sigma_p(y)=B_1(y)-\frac1pB_1(py)=\frac{j+u}p-\frac12-\frac{u-\frac12}p=\frac{2j+1-p}{2p}$: the continuous part cancels. The moments of the symmetric discrete uniform give $\langle\sigma_p^2\rangle=(1-p^{-2})/12$ and, with $\sum_{j}(2j+1-p)^4=p(p^2-1)(3p^2-7)/15$, the stated $\varepsilon_4(p)$.
\end{proof}

The law was also confirmed to $10^{-17}$ for all primes up to $1000$ by the finite algorithm.

\subsection{The measurement, split exactly}
With $I(D)$ computed exactly and the fourth moment known exactly for $D\le18$ (full-period computation) and to high precision at $D=100$ ($10^8$ samples) and $D=1000$:
\begin{center}
\begin{tabular}{lccccccc}
\toprule
$D$ & $6$ & $10$ & $12$ & $15$ & $18$ & $100$ & $1000$\\
\midrule
$e_4(D)$, eq.~\eqref{ms:eq:res4} & $-0.0181$ & $-0.0253$ & $-0.0336$ & $-0.0327$ & $-0.0411$ & $-0.2405(38)$ & $-1.60(74)$\\
$I(D)$ (exact) & $-0.0150$ & $-0.0210$ & $-0.0293$ & $-0.0429$ & $-0.0512$ & $-0.2751$ & $-2.679$\\
$X(D)$ & $-0.0031$ & $-0.0043$ & $-0.0043$ & $+0.0102$ & $+0.0102$ & $+0.035$ & $+1.1(7)$\\
\bottomrule
\end{tabular}
\end{center}
Three facts stand out. \emph{(i)} The intra part dominates and is exact; numerically $I(D)/D\to-0.0027$, and the measured law $e_4\approx-0.00228\,D$ of Figure~\ref{ms:fig:gauss}(c) splits as $-3.0=-3.5_{\rm intra}+0.5_{\rm cross}$ in the normalized units of the excess-kurtosis law $-3/D$. \emph{(ii)} $X$ is piecewise constant in $D$ and \emph{moves only at composite squarefree conductors}. This is now a theorem in full: nothing can move at non-squarefree $D$ (a new conductor, or a new term $m$ with $\mu(m)\ne0$ in a weight $m_q(D/q)$, requires $D=qm$ with $q,m$ squarefree and coprime, so $D$ squarefree), and $X(p)=X(p-1)$ at every prime $p$ (Theorems~\ref{ms:thm:decouple} and~\ref{ms:thm:primearrival} below). \emph{(iii)} $X$ changes sign, exactly: at $D=15$, where $X(15)=\frac{822653}{81033750}=+0.01015\ldots$; and it stays minute against the trivial $O(D^2)$ scale of the resonance expansion --- $X(1000)=+1.1(7)$.


\subsection{The prime-conductor cancellation}
The constancy of $X$ across prime arrivals has a complete and elementary proof, and --- unexpectedly --- it needs no regularization at all: everything below is an identity between long-run averages of bounded periodic functions over a common period. Three ingredients: the valuation lemma, which is the $p$-adic skeleton; a one-line \emph{unit-interval lemma}, which decouples the conductor-$1$ orbit completely; and the disjointness of Fourier supports. Call a zero-sum quadruple of reduced nonzero frequencies \emph{genuine} if no two-term subsum vanishes.

\begin{proposition}[valuation lemma]\label{ms:prop:valuation}
In any zero-sum quadruple of reduced frequencies $u_i/q_i$, every prime divides either none or at least two of the conductors $q_i$.
\end{proposition}

\begin{proof}
If $p\mid q_1$ and $p\nmid q_2q_3q_4$, the $p$-adic valuation of $u_1/q_1$ is negative and that of the other three terms is $\ge0$; the sum cannot be $0$.
\end{proof}

This is the multiplicative matching made precise: cross-orbit resonances require conductors that share their primes in pairs, which is why only composite conductors can couple to older ones in a new way.

\begin{lemma}[unit-interval lemma]\label{ms:lem:unit}
For squarefree $q>1$ the function $x\mapsto\sigma_q(x/q)$ is constant on every interval $(j,j+1)$, $j\in\mathbb Z$. Consequently, for any function $g$ of integer period that is constant on unit intervals --- in particular any polynomial in such $\sigma$'s, with any conductors $>1$ ---
\[
\widehat g(k)=0\quad\text{for every integer }k\ne0,
\qquad
\bigl\langle B_1(x)^m\,g(x)\bigr\rangle=\bigl\langle B_1^m\bigr\rangle\,\langle g\rangle\quad(m\ge0).
\]
\end{lemma}

\begin{proof}
$\sigma_q(x/q)=\sum_{t\mid q}(\mu(t)/t)B_1(tx/q)$ jumps only where $tx/q\in\mathbb Z$, i.e.\ at multiples of the integer $q/t$, and its slope is $\frac1q\sum_{t\mid q}\mu(t)=0$ for $q>1$. If $g$ has period $L$ and value $g_j$ on $(j,j+1)$, then $\widehat g(k)=\frac1L\sum_{j<L}g_j\int_j^{j+1}e(-kx)\,dx=0$ for $0\ne k\in\mathbb Z$, and $\langle B_1^mg\rangle=\frac1L\sum_{j<L}g_j\int_0^1B_1(t)^m\,dt=\langle g\rangle\langle B_1^m\rangle$.
\end{proof}

\begin{theorem}[the conductor-$1$ orbit decouples]\label{ms:thm:decouple}
Write $\widetilde\Phi_D=m_1(D)\,B_1+\Psi_D$ with $\Psi_D=\sum_{1<q\le D}\mu(q)\,m_q(D/q)\,\sigma_q(x/q)$. Then
\[
\bigl\langle\widetilde\Phi_D^4\bigr\rangle=m_1^4\,\langle B_1^4\rangle+6\,m_1^2\,\langle B_1^2\rangle\langle\Psi_D^2\rangle+\langle\Psi_D^4\rangle,
\qquad
\bigl\langle\widetilde\Phi_D^2\bigr\rangle=\frac{m_1^2}{12}+\langle\Psi_D^2\rangle,
\]
hence
\[
e_4(D)=m_1(D)^4\,\varepsilon_4(1)+\Bigl(\langle\Psi_D^4\rangle-3\langle\Psi_D^2\rangle^2\Bigr),
\]
and $X(D)$ does not depend on $m_1(D)$ at all: \emph{the resonance classes containing an integer frequency contribute nothing to $X$ in total}.
\end{theorem}

\begin{proof}
$\Psi_D$ and its powers are constant on unit intervals (Lemma~\ref{ms:lem:unit}), so every mixed term splits: $\langle B_1^a\Psi_D^b\rangle=\langle B_1^a\rangle\langle\Psi_D^b\rangle$, which vanishes for $a=1,3$ and equals $\frac1{12}\langle\Psi_D^b\rangle$ for $a=2$. In $e_4=\langle\widetilde\Phi_D^4\rangle-3\langle\widetilde\Phi_D^2\rangle^2$ the terms $6m_1^2\cdot\frac1{12}\langle\Psi_D^2\rangle$ cancel exactly, leaving $m_1^4(\frac1{80}-\frac1{48})=m_1^4\varepsilon_4(1)$ plus the pure $\Psi_D$ defect. Since the $q=1$ term of $I(D)$ is exactly $m_1^4\varepsilon_4(1)$, the weight $m_1$ drops out of $X=e_4-I$.
\end{proof}

\begin{remark}
The cancellation is also exact \emph{per conductor multiset}: a single multiset involves finitely many fixed functions, so the bounded-convergence expansion applies to it alone, and Lemma~\ref{ms:lem:unit} gives $\langle B_1^a\,\sigma_{q_1}\cdots\sigma_{q_b}\rangle=\langle B_1^a\rangle\langle\sigma_{q_1}\cdots\sigma_{q_b}\rangle$. For instance, the symmetric family sum of $\{1,2,3,6\}$ (witness $\frac12+\frac13+\frac16-1=0$) vanishes, as direct summation confirms.
\end{remark}

\begin{theorem}[prime arrivals leave $X$ unchanged]\label{ms:thm:primearrival}
$X(p)=X(p-1)$ for every prime $p$.
\end{theorem}

\begin{proof}
When $D$ moves from $p-1$ to $p$, the only coefficients of the primitive decomposition that change are $m_1(D)$, which jumps by $\mu(p)/p$, and the new $c_p=\mu(p)m_p(1)=\mu(p)$: a new term in $m_q(D/q)$ requires $qm=D$ with $\mu(m)\ne0$, which at $D=p$ forces $q\in\{1,p\}$. The change in $m_1$ is invisible to $X$ by Theorem~\ref{ms:thm:decouple}. Write $\Psi=\Psi_{p-1}$ for the unchanged part; every conductor in $\Psi$ is $\le p-1$, hence coprime to $p$. The Fourier support of $\sigma_p(x/p)$ is $\{u/p:p\nmid u\}$; the supports of $\Psi,\Psi^2,\Psi^3$ are contained in the fractions with denominator coprime to $p$ (sumsets of the $\Psi$-frequencies); the support of $\sigma_p^3$ lies in $\frac1p\mathbb Z$. All functions here are bounded and piecewise polynomial with a common period, so their Fourier coefficients are square-summable and Parseval applies. Then $\langle\sigma_p\Psi\rangle=\langle\sigma_p\Psi^3\rangle=0$ (a reduced denominator equal to $p$ never equals one coprime to $p$: Proposition~\ref{ms:prop:valuation} in $L^2$ form) and $\langle\sigma_p^3\Psi\rangle=0$ ($\frac1p\mathbb Z$ meets the support of $\Psi$ only in the integers, which $\Psi$ does not carry). Therefore adding $c_p\sigma_p$ changes $e_4$ by
\[
4c_p\langle\sigma_p\Psi^3\rangle
+6c_p^2\bigl(\langle\sigma_p^2\Psi^2\rangle-\langle\sigma_p^2\rangle\langle\Psi^2\rangle\bigr)
+4c_p^3\langle\sigma_p^3\Psi\rangle
+c_p^4\varepsilon_4(p),
\]
and $I$ by $c_p^4\varepsilon_4(p)$, so $\Delta X$ is the three mixed brackets. The first and third vanish as above. For the middle one, the supports of $\sigma_p^2$ ($\subset\frac1p\mathbb Z$) and $\Psi^2$ (denominators coprime to $p$) meet only in the integers --- and both functions are constant on unit intervals, so by Lemma~\ref{ms:lem:unit} neither carries any nonzero integer frequency: $\langle\sigma_p^2\Psi^2\rangle=\langle\sigma_p^2\rangle\langle\Psi^2\rangle$.
\end{proof}

\begin{remark}[the frequency-space view]\label{ms:rem:freqview}
Class by class, the same cancellations read as principal-value identities, which we found first and record for their own sake. A pair of conductor-$p$ frequencies summing to an integer $k\ne0$ carries the factor
\[
T_p(k)=\sum_{\substack{u\ne0,\ p\nmid u}}\frac1{u(pk-u)}
=\frac1{pk}\sum_{\substack{u\ne0,\ p\nmid u}}\Bigl(\frac1u+\frac1{pk-u}\Bigr)=0
\]
(two negation-symmetric harmonic sums), and a triple of them summing to $N=pk$ carries $U_p(N)=\sum_{u_1+u_2+u_3=N,\ p\nmid u_i}1/(u_1u_2u_3)$, which also vanishes: fixing $u_3$ and using partial fractions with the classical $\sum_{n\equiv a\,(p)}1/n=\frac\pi p\cot\frac{\pi a}p$ gives $\frac{2\pi}{p(N-u_3)}\cot\frac{\pi u_3}p$, and the sum over $u_3$ pairs the resulting $\sum_a\cot^2\frac{\pi a}p$ against its reflection with the opposite sign, by $\cot\frac{\pi(N-w)}p=-\cot\frac{\pi w}p$. (The restriction to genuine quadruples is vacuous in both: a vanishing two-term subsum would force $k=0$ or $p\mid u_i$.) These identities hold in the symmetric limit of Section~\ref{ms:sec:intro}; the real-space proof above is what makes that limit, and the bookkeeping it would require across infinitely many classes, unnecessary.
\end{remark}

The exact computations confirm the theorems and sharpen the picture. $X(D)=0$ for $D\le5$ (no composite conductor yet); the first nonzero value of the cross term is the exact rational $X(6)=-\frac1{324}$; and across every prime arrival in range $X$ indeed does not move:
\begin{gather*}
X(7)=X(6)=-\tfrac1{324},\qquad X(11)=X(12)=X(13)=-\tfrac{217}{50625}=-0.0042864198,\\
X(15)=X(18)=X(19)=+0.0101519799 .
\end{gather*}
The composite arrivals move it: $\Delta X=-\frac1{324}$ at $6$, $-\frac3{2500}$ at $10$, $-\frac3{4802}$ at $14$, $\frac{1830934}{121550625}$ at $15$, $-\frac{211124}{24310125}$ at $21$ and $-\frac{44110277}{22598383500}$ at $22$. The first three are \emph{exactly} $-(p^2-1)/(32p^4)$ for the conductors $2p$, $p=3,5,7$ --- and the out-of-sample test at $p=11$ refuted that law, $\Delta X(22)=-0.00195\ldots$ against a predicted $-0.00026$. When we first found this we recorded it as a caution against single-family formulas; the sector decomposition of the next subsection turns both halves of the experience into theorems: the law is exactly true, at \emph{every} prime $p$, for one of the four sectors of $X$, and the correction at $p=11$ is the other three sectors switching on, each by an exact, explained amount (Theorem~\ref{ms:thm:2pjump}). Only composite conductors drive $X$, exactly as Proposition~\ref{ms:prop:valuation} prescribes.

\subsection{The four sectors of the cross term}\label{ms:sec:sectors}
Grouping the four frequencies of a resonance by the multiplicities of their conductors splits $X$ into four \emph{sectors}, each an exact and finite object. Throughout this subsection $\sigma_q$ abbreviates $\sigma_q(x/q)$, all conductors are squarefree in $(1,D]$, $c_q=\mu(q)\,m_q(D/q)$, and $\operatorname{Cov}(f,g)=\langle fg\rangle-\langle f\rangle\langle g\rangle$.

\begin{proposition}[sector decomposition]\label{ms:prop:sectors}
For every $D$,
\[
X(D)=X_{31}(D)+X_{22}(D)+X_{211}(D)+X_{1111}(D),
\]
\begin{align*}
X_{31}&=4\sum_{q\ne r}c_q^3c_r\,\langle\sigma_q^3\sigma_r\rangle, &
X_{22}&=6\sum_{q<r}c_q^2c_r^2\,\operatorname{Cov}(\sigma_q^2,\sigma_r^2),\\
X_{211}&=12\sum_{q}\sum_{\substack{r<s\\ r,s\ne q}}c_q^2c_rc_s\,\langle\sigma_q^2\sigma_r\sigma_s\rangle, &
X_{1111}&=24\!\!\sum_{q<r<s<t}\!\!c_qc_rc_sc_t\,\langle\sigma_q\sigma_r\sigma_s\sigma_t\rangle,
\end{align*}
and each average is a finite exact computation over the common period of its conductors.
\end{proposition}

\begin{proof}
Expand $\langle\Psi_D^4\rangle$ multinomially over the orbits and subtract $3\langle\Psi_D^2\rangle^2=3(\sum_qc_q^2v_q)^2$, $v_q=\langle\sigma_q^2\rangle$; orthogonality kills nothing here but pairs the Wick term $6\sum_{q<r}c_q^2c_r^2v_qv_r$ with the two-pair moments, producing the covariances, while the all-equal terms give $\sum_qc_q^4(\langle\sigma_q^4\rangle-3v_q^2)=I(D)-m_1^4\varepsilon_4(1)$, which cancels in $X=e_4-I$ by Theorem~\ref{ms:thm:decouple}.
\end{proof}

\begin{lemma}[support and parity]\label{ms:lem:supp}
\emph{(i)} $\langle\sigma_q^3\sigma_r\rangle=0$ unless $r\mid q$. \emph{(ii)} $\sigma_2=\mp\frac14$ on unit intervals, so $\sigma_2^2\equiv\frac1{16}$ and $\operatorname{Cov}(\sigma_2^2,\,\cdot\,)=0$: conductor $2$ is inert in the $(2,2)$ sector. \emph{(iii)} For even squarefree $q$, $\sigma_q^2$ carries no half-integer frequency: its coefficient at $\xi$ with $2\xi$ odd sums $1/(uu')$ over $u+u'=q\xi$ with $u,u'$ odd units of $q$, and $q\xi=\frac q2(2\xi)$ is odd --- an odd number is not a sum of two odd ones. \emph{(iv)} Hence $\operatorname{Cov}(\sigma_q^2,\sigma_r^2)=0$ unless $q$ and $r$ share an \emph{odd} prime: the common frequencies lie in $\frac1{(q,r)}\mathbb Z$, the integers are blind by Lemma~\ref{ms:lem:unit}, and when $(q,r)=2$ the half-integers are blind by (iii).
\end{lemma}

\begin{lemma}[the parity sieve]\label{ms:lem:parity}
Let $q_1,\dots,q_k>1$ be distinct squarefree conductors with multiplicities $e_1,\dots,e_k\ge1$. If $\sum_{i:\,2\mid q_i}e_i$ is odd, then
\[
\bigl\langle\sigma_{q_1}^{e_1}\cdots\sigma_{q_k}^{e_k}\bigr\rangle=0 .
\]
\end{lemma}

\begin{proof}
If no $q_i$ is even the count is zero and there is nothing to prove; otherwise $L=\operatorname{lcm}(q_1,\dots,q_k)$ is squarefree and even, so $2\parallel L$. Split the product into two blocks, one factor against the rest. Both blocks are bounded periodic step functions, so Parseval applies with absolute convergence (Cauchy--Schwarz), and the coefficient of each block at a frequency $\xi$ is an absolutely convergent sum over the solutions of $\sum u_{ij}/q_i=\xi$ with every $u_{ij}$ coprime to its $q_i$ (Proposition~\ref{ms:prop:primitive}: each $\sigma_{q_i}$ is supported on the reduced fractions of denominator exactly $q_i$). The two blocks pair at opposite frequencies, so every nonzero contribution to the mean comes from a \emph{resonance}
\[
\sum_{i=1}^{k}\sum_{j=1}^{e_i}\frac{u_{ij}}{q_i}=0,\qquad(u_{ij},q_i)=1 .
\]
Multiply by $L$ and reduce modulo $2$. For even $q_i$ the cofactor $L/q_i$ is odd and so is $u_{ij}$: the term contributes $1$. For odd $q_i$ the cofactor $L/q_i$ is even: the term contributes $0$. The left side is therefore $\equiv\sum_{i:2\mid q_i}e_i\pmod2$, odd by hypothesis, while the right side is $0$. No resonance exists, and the mean vanishes.
\end{proof}

The sieve is the $2$-adic mechanism of parts (iii)--(iv) of Lemma~\ref{ms:lem:supp}, promoted from supports to full means: the prime $2$ is the only prime with a single unit, so a resonance must use an \emph{even} number of frequency slots of even conductor. Applied to the sectors it halves the problem:

\begin{corollary}[sector parities]\label{ms:cor:parity}
In Proposition~\ref{ms:prop:sectors}: every $(3,1)$ term with $q$ even and $r$ odd vanishes, so by Lemma~\ref{ms:lem:supp}(i) the $(3,1)$ sector is supported on $r\mid q$ with $2\mid r$ whenever $2\mid q$; every $(2,1,1)$ term with $r\not\equiv s\pmod 2$ vanishes; every $(1,1,1,1)$ term with an odd number of even conductors vanishes. The $(2,2)$ sector is untouched, its multiplicities being even.
\end{corollary}

\begin{proof}
Count the even slots: $3[2\mid q]+[2\mid r]$ is odd exactly when $q$ and $r$ have opposite parity --- the case $q$ odd, $r$ even is already void by Lemma~\ref{ms:lem:supp}(i), since then $r\nmid q$, leaving $2\mid q$, $2\nmid r$; $2[2\mid q]+[2\mid r]+[2\mid s]$ is odd exactly when $r\not\equiv s\pmod2$; $\sum_i[2\mid q_i]$ odd is the stated condition.
\end{proof}

The corollary is sharp in every clause: $\langle\sigma_6^3\sigma_3\rangle=0$ while $\langle\sigma_6^3\sigma_2\rangle=-\frac1{432}$; the means $\langle\sigma_q^2\sigma_r\sigma_s\rangle$ with $r\not\equiv s$ vanish identically while the all-odd $\langle\sigma_{15}^2\sigma_{21}\sigma_{35}\rangle=\frac{74}{55125}$ and $\langle\sigma_5^2\sigma_3\sigma_{15}\rangle=\frac2{1125}$ do not; and the \emph{first} admissible quadruple, $\{2,6,10,15\}$ at $D=15$, carries three even conductors and dies, $\langle\sigma_2\sigma_6\sigma_{10}\sigma_{15}\rangle=0$, while $\langle\sigma_2\sigma_6\sigma_{10}\sigma_{30}\rangle=-\frac1{1200}$ lives with four. Across the complete term lists to $D=800$ (ancillary \texttt{xsectors.py}), no multiset violating the parity condition carries a nonzero mean. The sieve also re-proves, in one line, the two vanishings of Proposition~\ref{ms:prop:ladder} --- $\langle\sigma_p\sigma_{2p}^3\rangle$ and $\langle\sigma_2\sigma_p\sigma_{2p}^2\rangle$ each use three even slots --- and several entries of the kill list in Theorem~\ref{ms:thm:2pjump}.

\begin{lemma}[the coefficient bound]\label{ms:lem:coef}
For squarefree $q>1$ and $0\ne\xi\in\frac1q\mathbb Z$, the Fourier coefficient of $\sigma_q^2$ at $\xi$ equals $-\widetilde T_q(q\xi)/(4\pi^2)$, where, with absolute convergence,
\[
\widetilde T_q(N)=\sum_{\substack{u+u'=N\\ (uu',q)=1}}\frac1{uu'}
=\frac{2\pi}{Nq}\sum_{a\in R_q(N)}\cot\frac{\pi a}q,
\qquad R_q(N)=\{a\bmod q:\ (a(N-a),q)=1\},
\]
and $|\widetilde T_q(N)|\le 4(1+\log q)/|N|$.
\end{lemma}

\begin{proof}
The coefficient is the convolution of the coefficients of $\sigma_q$, and $\sum_u1/(|u||N-u|)<\infty$. The involution $u\mapsto N-u$ preserves the index set and swaps the partial fractions $\frac1{uu'}=\frac1N(\frac1u+\frac1{u'})$, so $\widetilde T_q(N)=\frac2N\lim_U\sum_{|u|\le U}1/u$ over the set (the involution maps the window $|u|\le U$ to $|u-N|\le U$, at a cost $O(N/U)\to0$), which is $\frac2N\sum_{a\in R_q(N)}\frac\pi q\cot\frac{\pi a}q$ by $\sum_{u\equiv a\,(q)}\frac1u=\frac\pi q\cot\frac{\pi a}q$. The bound follows from $\cot x\le1/x$ on $(0,\pi/2)$: $\sum_{a=1}^{q-1}|\cot\frac{\pi a}q|\le\frac{2q}\pi(1+\log q)$.
\end{proof}

\begin{theorem}[two sectors are tame]\label{ms:thm:tame}
Unconditionally,
\[
|X_{31}(D)|\ll D(\log D)(\log\log D)^{6},
\qquad
|X_{22}(D)|\ll D(\log D)^{2}(\log\log D)^{4}.
\]
Hence $X(D)=X_{211}(D)+X_{1111}(D)+O(D\log^4D)$: the sectors with at most two distinct conductors obey the conjectured bound up to logarithmic factors, and Conjecture~\ref{ms:conj:r4} reduces, up to those factors, to the sectors with at least three distinct conductors.
\end{theorem}

\begin{proof}
For $X_{31}$: by Lemma~\ref{ms:lem:supp}(i) the sum has $\ll\sum_{q\le D}d(q)\ll D\log D$ terms, each at most $|c_q|^3|c_r|\cdot\|\sigma_q\|_\infty^2(\langle\sigma_q^2\rangle\langle\sigma_r^2\rangle)^{1/2}\ll(\log\log D)^{6}$ by $\|\sigma_q\|_\infty\le\frac12\prod_{p\mid q}(1+p^{-1})$ and $|c_q|\le q/\varphi(q)$. For $X_{22}$: by Lemmas~\ref{ms:lem:supp} and~\ref{ms:lem:coef}, for $g=(q,r)>2$,
\[
|\operatorname{Cov}(\sigma_q^2,\sigma_r^2)|\le\sum_{0\ne\xi\in\frac1g\mathbb Z}\frac{4(1+\log q)}{4\pi^2 q|\xi|}\cdot\frac{4(1+\log r)}{4\pi^2 r|\xi|}
=\frac{2\zeta(2)}{\pi^4}\,\frac{g^2(1+\log q)(1+\log r)}{qr},
\]
and summing $g^2/(qr)$ over $q=gq_1,r=gr_1\le D$ gives $\sum_{g\le D}\bigl(\sum_{n\le D/g}1/n\bigr)^2\ll D$, so the weighted total is $\ll D(\log D)^2(\log\log D)^4$.
\end{proof}

The same spectral method reaches into the $(2,1,1)$ sector. The key is the analogue of Lemma~\ref{ms:lem:coef} for a \emph{pair} of distinct conductors.

\begin{lemma}[the pair-coefficient bound]\label{ms:lem:paircoef}
Let $r\ne s$ be squarefree conductors $>1$, $g=(r,s)$, $r_1=r/g$, $s_1=s/g$. The Fourier coefficients of $\sigma_r(x/r)\,\sigma_s(x/s)$ are supported on the $\xi$ whose reduced denominator is divisible by $r_1s_1$ and divides $gr_1s_1$; at any $\xi\ne0$ with $rs\,\xi\in\mathbb Z$ the coefficient equals $-W_{r,s}(\xi)/(4\pi^2)$, where
\[
W_{r,s}(\xi)=\sum_{\substack{u/r+v/s=\xi\\ (u,r)=(v,s)=1}}\frac1{uv}
\qquad\text{(absolutely convergent),}\qquad
|W_{r,s}(\xi)|\le\frac{2(1+\log rs)(r+s)}{rs\,|\xi|}.
\]
\end{lemma}

\begin{proof}
The coefficient at $\xi$ is the convolution $\sum_{u/r+v/s=\xi}\frac1{2\pi iu}\cdot\frac1{2\pi iv}$, and the sum converges absolutely because $v=(rs\xi-us)/r$ grows linearly in $u$. The support: for a prime $p\mid r_1$ we have $p\nmid us$ and $p\,\|\,r$, so $p$ survives at full power in the reduced denominator of $u/r+v/s$, and symmetrically for $s_1$; all denominators divide $\operatorname{lcm}(r,s)=gr_1s_1$. For the bound, put $N=rs\,\xi=us+vr\in\mathbb Z\setminus\{0\}$ and use the partial fractions $\frac1{uv}=\frac1N\bigl(\frac ru+\frac sv\bigr)$. The solution set is a union of residue classes modulo $rs$ in $u$ (the congruence $us\equiv N\ (\mathrm{mod}\ r)$, the condition $(u,r)=1$, and the condition $(v,s)=1$ are all determined by $u\bmod rs$), and $v$ runs over a union of classes modulo $rs$ as well. Evaluating each half in symmetric windows --- the window $|u|\le U$ transports to a window in $v$ centred at $N/r$, at a recentring cost $O(N/(Us))\to0$ --- the classical $\sum_{u\equiv a\,(M)}\frac1u=\frac\pi M\cot\frac{\pi a}M$ and $\sum_{a=1}^{M-1}|\cot\frac{\pi a}M|\le\frac{2M}\pi(1+\log M)$ give $|\sum'1/u|,\ |\sum'1/v|\le2(1+\log rs)$, hence $|W|\le2(1+\log rs)(r+s)/|N|$.
\end{proof}

\begin{proposition}[the $(2,1,1)$ sector is $O(D^{3/2+o(1)})$]\label{ms:prop:x211}
Unconditionally,
\[
|X_{211}(D)|\ \ll\ D^{3/2}\,(\log D)^{7/2}\,(\log\log D)^{O(1)} .
\]
\end{proposition}

\begin{proof}
A term $(q;r<s)$ of $X_{211}$ is nonzero only if $r_1s_1\mid q$, where $g=(r,s)$, $r_1=r/g$, $s_1=s/g$: the support of $\sigma_r\sigma_s$ sits over denominators divisible by $r_1s_1$ (Lemma~\ref{ms:lem:paircoef}) and the support of $\sigma_q^2$ over denominators dividing $q$. Write $q=r_1s_1h$ and $E=(g,h)$; the parametrization $(g,r_1<s_1,h)\mapsto(q;r,s)$ is injective, and the constraints are $gs_1\le D$, $r_1s_1h\le D$. Each term carries two bounds. \emph{Trivial:} $|\langle\sigma_q^2\sigma_r\sigma_s\rangle|\le\|\sigma_q\|_\infty^2\sqrt{v_rv_s}\le\tau:=C(\log\log D)^2$, by $\|\sigma_q\|_\infty\le\frac12\prod_{p\mid q}(1+p^{-1})$ and $v_q\le\frac1{12}$. \emph{Spectral:} the common support of $\sigma_q^2$ and $\sigma_r\sigma_s$ lies in $\frac1{r_1s_1E}\mathbb Z$, so by Parseval, Lemma~\ref{ms:lem:coef} at $q\xi=hk/E$ and Lemma~\ref{ms:lem:paircoef} at $rs\,\xi=g^2k/E$ (writing $\xi=k/(r_1s_1E)$, $k\ne0$),
\[
|\langle\sigma_q^2\sigma_r\sigma_s\rangle|
\le\sum_{k\ne0}\frac{(1+\log q)E}{\pi^2h|k|}\cdot\frac{(1+\log rs)(r_1+s_1)E}{2\pi^2 g|k|}
\le C(\log D)^2\,\frac{E^2(r_1+s_1)}{hg},
\]
where $(r+s)/(rs)\cdot r_1s_1E=(r_1+s_1)E/g$ was used. Since $|c_q^2c_rc_s|\ll(\log\log D)^4$, it remains to sum $\min(\tau,S)$ over the parameters. Fix $(g,r_1,s_1)$ and sum over $h=E\tilde h$, $E\mid g$:
\[
\sum_{h\le D/(r_1s_1)}\min\Bigl(\tau,\ C\log^2D\,\frac{E(r_1+s_1)}{\tilde hg}\Bigr)
\le\sum_{E\mid g}\min\Bigl(\tau\,\frac D{r_1s_1E},\ C\log^3D\,\frac{E(r_1+s_1)}g\Bigr)
\le d(g)\sqrt{\tau C\log^3D}\,\sqrt{\frac{2D}{r_1g}},
\]
using $\min(x,y)\le\sqrt{xy}$, in which $E$ cancels, and $(r_1+s_1)/(r_1s_1)\le2/r_1$. The number of admissible $s_1$ is at most $\min(D/g,D/r_1)$, so, with the per-term logarithms absorbed into $\mathcal L=(\log D)^{3/4}(\log\log D)^{O(1)}$,
\[
|X_{211}(D)|\ \ll\ \mathcal L^2\sum_{g\le D}d(g)\sqrt{\frac Dg}\sum_{r_1\le D}\frac1{\sqrt{r_1}}\min\Bigl(\frac Dg,\frac D{r_1}\Bigr)
\ \ll\ \mathcal L^2\sum_{g\le D}d(g)\sqrt{\frac Dg}\cdot\frac{4D}{\sqrt g}
\ \ll\ D^{3/2}\mathcal L^2\sum_{g\le D}\frac{d(g)}g,
\]
and $\sum_{g\le D}d(g)/g\ll\log^2D$: collecting powers, $|X_{211}|\ll D^{3/2}(\log D)^{3/2+2}(\log\log D)^{O(1)}$. (The inner estimate: $\sum_{r_1\le g}r_1^{-1/2}(D/g)+\sum_{r_1>g}r_1^{-3/2}D\le4D/\sqrt g$.)
\end{proof}

This is far from the conjectured $O(D)$, but it is the first unconditional bound on a sector with three distinct conductors, and the proof shows exactly where the loss occurs: the absolute values surrender the sign of the cotangent sums in $W_{r,s}$. The data below suggest the truth is very much smaller.

The remaining structure is best seen in the data. In exact arithmetic (ancillary \texttt{sectors.py}):
\begin{center}
\begin{tabular}{lcccccc}
\toprule
$D$ & $6$ & $10$ & $14$ & $15$ & $21$ & $22$\\
\midrule
$X_{31}$ & $+0.00617$ & $+0.00712$ & $+0.00636$ & $+0.00810$ & $+0.00840$ & $+0.00727$\\
$X_{22}$ & $-0.00309$ & $-0.00429$ & $-0.00491$ & $-0.00045$ & $-0.00059$ & $-0.00084$\\
$X_{211}$ & $-0.00617$ & $-0.00712$ & $-0.00636$ & $+0.01077$ & $+0.01521$ & $+0.01465$\\
$X_{1111}$ & $0$ & $0$ & $0$ & $-0.00826$ & $-0.02155$ & $-0.02156$\\
\midrule
$X$ & $-0.00309$ & $-0.00429$ & $-0.00491$ & $+0.01015$ & $+0.00147$ & $-0.00048$\\
\bottomrule
\end{tabular}
\end{center}
Two striking facts: $X_{31}+X_{211}=0$ \emph{exactly} for all $D\le14$, so that below $D=15$ the cross term \emph{is} the $(2,2)$ sector; and the refuted $2p$ law lives entirely in $X_{22}$. Both have proofs.

\begin{proposition}[the $2p$ ladder]\label{ms:prop:ladder}
For every odd prime $p$,
\[
\operatorname{Cov}(\sigma_p^2,\sigma_{2p}^2)=-\frac{p^2-1}{48\,p^4},\qquad
\langle\sigma_2\sigma_p^2\sigma_{2p}\rangle=\frac{p^2-1}{96\,p^3},\qquad
\langle\sigma_2\sigma_{2p}^3\rangle=-\frac{p^2-1}{128\,p^3},
\]
\[
\langle\sigma_p\sigma_{2p}^3\rangle=\langle\sigma_2\sigma_p\sigma_{2p}^2\rangle=0,
\qquad\text{and in particular}\qquad
4\langle\sigma_2\sigma_{2p}^3\rangle+3\langle\sigma_2\sigma_p^2\sigma_{2p}\rangle=0 .
\]
Consequently the joint $(3,1)+(2,1,1)$ contribution of the cluster $\{2,p,2p\}$ equals
\[
12\,c_2c_{2p}\,\langle\sigma_2\sigma_p^2\sigma_{2p}\rangle\Bigl(c_p^2-\frac{c_{2p}^2}4\Bigr),
\]
which vanishes --- for \emph{any} value of $c_2$ --- whenever $m_p(D/p)=\frac12\,m_{2p}(D/2p)$.
\end{proposition}

\begin{proof}[Proof of Proposition~\ref{ms:prop:ladder}]
On $(j,j+1)$, $0\le j<2p$, the three functions take the explicit values
\[
\sigma_p=v_j=\frac{2(j\bmod p)+1-p}{2p},\qquad
\sigma_2=-\frac{(-1)^j}4,\qquad
\sigma_{2p}=w_j=\frac{2j+1-2p}{4p}+\frac{(-1)^j}{4p}-\frac{v_j}2
\]
(the last from $B_1(t\,\frac{2j+1}{4p})$, $t\mid2p$). Each identity is a sum over $j$ of a polynomial in $j,(-1)^j,v_j$, \emph{defined by these formulas for every odd $p$}; splitting $j$ by the Chinese remainder pair $(j\bmod2,\,j\bmod p)$ and applying Faulhaber's formulas, each side, cleared of its fixed power of $p$, becomes a polynomial in $p$ of degree at most $24$. The identities hold in exact arithmetic at the twenty-nine odd values $3\le p\le59$ (ancillary \texttt{ladder.py}, which also checks the value formulas against the true $\sigma_p,\sigma_{2p}$), hence identically in $p$; specializing to odd primes gives the statement.
\end{proof}

\begin{lemma}[double-ladder orthogonality]\label{ms:lem:dladder}
For distinct odd primes $r\ne p$,
\[
\langle\sigma_r\sigma_{2r}\sigma_p\sigma_{2p}\rangle=0 .
\]
\end{lemma}

\begin{proof}
On $(j,j+1)$ write $\varepsilon=(-1)^j$ and $\rho=j\bmod r$. From $j\bmod 2r=\rho+r\,\frac{1-\varepsilon(-1)^\rho}2$, the value formula of Proposition~\ref{ms:prop:ladder} splits as $\sigma_{2r}=A_r(\rho)+\varepsilon B_r(\rho)$ with $B_r(\rho)=\frac1{4r}-\frac{(-1)^\rho}4$, and $\sigma_r=v_r(\rho)$. As $j$ runs over $2rp$ consecutive integers the triple $(\varepsilon,\,j\bmod r,\,j\bmod p)$ is uniform on the product (Chinese remainders), so averaging the product $v_r(A_r+\varepsilon B_r)\,v_p(A_p+\varepsilon B_p)$ kills the $\varepsilon$-odd part and factorizes the rest over the two independent residues:
\[
\langle\sigma_r\sigma_{2r}\sigma_p\sigma_{2p}\rangle
=\overline{v_rA_r}\;\overline{v_pA_p}+\overline{v_rB_r}\;\overline{v_pB_p}.
\]
Both one-prime means vanish: $\overline{v_rA_r}=\langle\sigma_r\sigma_{2r}\rangle=0$ by orthogonality, and $\overline{v_rB_r}=\frac1{4r}\overline{v_r}-\frac1{4r}\sum_{\rho<r}v_r(\rho)(-1)^\rho=0$ since $\overline{v_r}=0$ and $\sum_{\rho=0}^{r-1}(-1)^\rho(2\rho+1-r)=2\cdot\frac{r-1}2+(1-r)=0$ for odd $r$. The vanishing is special to the double-ladder structure: for instance $\langle\sigma_3\sigma_6\sigma_{14}\sigma_{21}\rangle=\frac1{2646}\ne0$.
\end{proof}


\begin{theorem}[the $2p$ jumps, proved and explained]\label{ms:thm:2pjump}
At $D=2p$, $p$ an odd prime, the $(2,2)$ sector jumps by exactly
\[
6\,c_p^2c_{2p}^2\operatorname{Cov}(\sigma_p^2,\sigma_{2p}^2)=-\frac{p^2-1}{32\,p^4}
\qquad\text{for every }p
\]
($c_p=\mu(p)m_p(2)=-\frac12$ and $c_{2p}=1$ at arrival; no older conductor shares an odd prime with $2p$, and conductor $2$ is inert). For $p=3,5,7$ the \emph{joint} contribution of the other three sectors does not move --- the individual sectors do move, as the table above shows, but their changes cancel exactly: every new multiset involving $2p$ is annihilated by Proposition~\ref{ms:prop:valuation}, Lemma~\ref{ms:lem:supp}, Proposition~\ref{ms:prop:ladder} or Lemma~\ref{ms:lem:dladder}, except the cluster pair, whose total vanishes by the cluster identity with the relation $m_{p'}(D/p')=\frac12$, valid for every live odd prime throughout $D\le14$. Hence
\[
\Delta X(2p)=-\frac{p^2-1}{32p^4}\qquad\text{exactly, for }p=3,5,7 .
\]
From $D=15$ the weight relation breaks ($m_3(5)=\frac3{10}$, $m_5(3)=\frac16$), and at $D=22$ the other sectors move through the new term $-\frac1{11}$ in $m_2(11)$, by the exact amounts $-\frac{76512673}{67795150500}$ (sector $31$), $-\frac{37687127}{67795150500}$ ($211$) and $-\frac{16}{1414875}$ ($1111$):
\[
\Delta X(22)=-\frac{11^2-1}{32\cdot11^4}-\frac{7198}{4244625}=-\frac{44110277}{22598383500}=-0.0019519218\ldots
\]
The law, its three exact successes, and its failure at $p=11$ are thereby explained term by term.
\end{theorem}

\begin{proof}
The $(2,2)$ statement: the new pairs at $D=2p$ are $\{2p,q\}$, $q<2p$, and $\operatorname{Cov}(\sigma_{2p}^2,\sigma_q^2)=0$ unless $q$ shares an odd prime with $2p$ (Lemma~\ref{ms:lem:supp}), i.e.\ $q=p$; older pairs can move only through $c_2$ and $c_p$, but conductor $2$ is inert and no older pair involving $p$ has a partner sharing the prime $p$. The weights at arrival are $m_p(2)=1-\frac12$ and $m_{2p}(1)=1$, giving $6\cdot\frac14\cdot(-\frac{p^2-1}{48p^4})$. For the other sectors at $p\le7$: a new non-$(2,2)$ multiset uses $2p$ and, by Proposition~\ref{ms:prop:valuation}, must contain $p$ or a second even conductor. The complete list of multisets surviving Proposition~\ref{ms:prop:valuation}, and its fate: $\{p,2p,2p,2p\}$ and $\{2,p,2p,2p\}$, zero by the vanishings of Proposition~\ref{ms:prop:ladder}; $\{2,2,p,2p\}$, zero since $\sigma_2^2$ is constant and $\langle\sigma_p\sigma_{2p}\rangle=0$; the cluster pair $\{2,2p,2p,2p\}$ together with $\{2,p,p,2p\}$ --- jointly $4c_{2p}^3c_2\langle\sigma_2\sigma_{2p}^3\rangle+12c_p^2c_2c_{2p}\langle\sigma_2\sigma_p^2\sigma_{2p}\rangle$, which vanishes at $c_p^2=\frac14c_{2p}^2$; the multisets $\{q,q,p,2p\}$ and $\{q,q,r,2r\}$ with $q$ even (older, or $q=2p$ itself) and $r$ odd, which die because $\sigma_q^2$ carries no half-integer and no nonzero integer frequency while $\sigma_p\sigma_{2p}$ (resp.\ $\sigma_r\sigma_{2r}$) is supported on denominators $2p$ and $2$ (resp.\ $2r$ and $2$); and the double ladders $\{r,2r,p,2p\}$, $r<p$ odd prime, zero by Lemma~\ref{ms:lem:dladder}. Reweighting of older multisets at $D=2p$ goes only through $\Delta c_2$, $\Delta c_p$ and $\Delta m_1$, the last invisible by Theorem~\ref{ms:thm:decouple}; below $D=15$ the only nonzero older non-$(2,2)$ multisets are the clusters $\{2,p',2p'\}$ --- the older double ladders vanish by Lemma~\ref{ms:lem:dladder} --- and the clusters are linear in $c_2$ and zero for any $c_2$ by the weight relation, while no nonzero older multiset contains $p$. The values at $D=22$ are exact computations over the periods of the multisets (ancillary \texttt{sectors.py}).
\end{proof}

A prediction worth recording: the $(2,2)$ jump law holds at \emph{every} prime, so at $D=26$ the $(2,2)$ sector must move by exactly $-(13^2-1)/(32\cdot13^4)=-\frac{21}{114244}$, whatever the other sectors do.

\subsection{The four sectors at scale}\label{ms:sec:scale}
The sector means $\langle\,\cdot\,\rangle$ do not depend on $D$ --- only the coefficients $c_q=\mu(q)m_q(D/q)$ do --- so each admissible multiset need be integrated once, over its own period, and the four sectors can then be swept in $D$ through the coefficients alone. With the support prefilters ($r\mid q$ for $(3,1)$; an odd common prime for $(2,2)$; $r_1s_1\mid q$ for $(2,1,1)$; every prime dividing at least two conductors for $(1,1,1,1)$) this reaches $D=800$ in exact structure and double precision (ancillary \texttt{xsectors.py}; the values at $D\le22$ reproduce the exact rationals of the table above to $10^{-15}$). There are $78\,336$ multisets with nonzero mean in the $(2,1,1)$ sector and $146\,211$ in the $(1,1,1,1)$ sector up to $D=800$; the parity sieve of Corollary~\ref{ms:cor:parity} is confirmed across the complete term lists.
\begin{center}
\begin{tabular}{rrrrrrrr}
\toprule
$D$ & $X_{31}$ & $X_{22}$ & $X_{211}$ & $X_{1111}$ & $X_{211}{+}X_{1111}$ & $\Sigma|{\cdot}|_{211}$ & $\Sigma|{\cdot}|_{1111}$\\
\midrule
100 & $+0.0291$ & $-0.0138$ & $+0.0859$ & $-0.0673$ & $+0.0186$ & $0.25$ & $0.68$\\
200 & $+0.0656$ & $-0.0215$ & $+0.0990$ & $-0.0254$ & $+0.0736$ & $0.64$ & $2.15$\\
300 & $+0.0918$ & $-0.0288$ & $+0.0419$ & $-0.1149$ & $-0.0730$ & $0.99$ & $4.46$\\
400 & $+0.1203$ & $-0.0439$ & $+0.1694$ & $-0.4907$ & $-0.3212$ & $1.45$ & $7.33$\\
500 & $+0.1581$ & $-0.0582$ & $+0.1988$ & $-0.6003$ & $-0.4016$ & $1.86$ & $10.47$\\
600 & $+0.1864$ & $-0.0720$ & $+0.3129$ & $-0.6391$ & $-0.3261$ & $2.17$ & $13.82$\\
700 & $+0.2227$ & $-0.0859$ & $+0.3430$ & $-0.5301$ & $-0.1871$ & $2.66$ & $18.08$\\
800 & $+0.2537$ & $-0.0906$ & $+0.2983$ & $-0.3559$ & $-0.0576$ & $3.07$ & $22.66$\\
\bottomrule
\end{tabular}
\end{center}
The last two columns are the sums of \emph{absolute} term contributions $12\sum|c_q^2c_rc_s\langle\cdot\rangle|$ and $24\sum|c_qc_rc_sc_t\langle\cdot\rangle|$. Three facts stand out (Figure~\ref{ms:fig:sectors800}).

\emph{The conjecture has enormous room.} The running maximum of $|X_{211}+X_{1111}|$ on $2\le D\le800$ is $0.438$, attained at $D=506$; at $D=800$ the ratio $(X_{211}+X_{1111})/D$ is $-7\cdot10^{-5}$. The measured log--log slopes on $100\le D\le800$ (window fits, not asymptotics) are $\approx1.0$ for $X_{31}$ and $X_{22}$, $\approx1.2$ for the amplitude of $X_{211}$, $\approx1.5$ for $X_{1111}$ --- and $\approx0.26$ for the total $X(D)$, which at $D=800$ is $+0.105$: the four sectors cancel one another to a total that grows, in this window, like $D^{1/4}$.

\emph{The $(1,1,1,1)$ sector cannot be tamed by absolute values.} The absolute sums grow like $D^{1.2}$ ($(2,1,1)$) and $D^{1.65}$ ($(1,1,1,1)$) in the window. Both are still far below $D$ at $D=800$, but the trend of the second crosses every linear bound: if it persists, no estimate that surrenders the signs of the quadruple means --- including the method of Proposition~\ref{ms:prop:x211} --- can prove $X_{1111}=O(D)$. The conjecture, if true, is true because of sign cancellation \emph{among} the quadruples, not within them.

\emph{The individual means are small in a structured way.} Across all $78\,336$ nonzero $(2,1,1)$ means, the normalized size $|\langle\sigma_q^2\sigma_r\sigma_s\rangle|\cdot gr_1s_1$ has median $10^{-4}$ and $99$th percentile $\approx1.3$, so the typical mean behaves like $(gr_1s_1)^{-1}$ --- summing that model over the admissible parameters gives $O(D\log D)$, which is why we read the $D^{3/2}$ of Proposition~\ref{ms:prop:x211} as an artefact of the cotangent absolute values --- but a thin tail (maximum $21.7$, at $(q;r,s)=(253;374,782)$) keeps the uniform model false. In the $(1,1,1,1)$ sector the largest mean in absolute value is $3.3\cdot10^{-3}$ and does not grow across the range, while $|\langle\sigma_q\sigma_r\sigma_s\sigma_t\rangle|\cdot\operatorname{lcm}(q,r,s,t)$ has $99$th percentile $\approx19$: the quadruple means live at the scale of one over their own period.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/F4_sectors800.png}
\caption{The four sectors to $D=800$. Left: signed values; the black curve is $X_{211}+X_{1111}$, the object of Conjecture~\ref{ms:conj:r4} after Theorem~\ref{ms:thm:tame}. Right: $|X_{211}+X_{1111}|$ against the absolute-value sums of the two sectors, log--log, with the conjectured slope $D$ and the proved $D^{3/2}$ for scale.}
\label{ms:fig:sectors800}
\end{figure}

\subsection{What remains}
\begin{conjecture}\label{ms:conj:r4}
$X(D)=O(D)$.
\end{conjecture}

The intra part needs no conjecture: $|m_q(Y)|\le q/\varphi(q)$ --- from $m_q(Y)=\sum_{s}s^{-1}\,u(Y/s)$, the sum running over the $q$-smooth integers $s$ (every prime factor of $s$ divides $q$), where $u(Z)=\sum_{m\le Z}\mu(m)/m$ satisfies the elementary $|u|\le1$ and $\sum_s1/s=\prod_{p\mid q}(1-p^{-1})^{-1}=q/\varphi(q)$ --- and $|\varepsilon_4(q)|\ll(\log\log q)^2$, so $I(D)\ll D(\log\log D)^6$ unconditionally, with the data saying $I(D)\approx-0.0027\,D$. Conjecture~\ref{ms:conj:r4} would therefore give $e_4(D)\ll D(\log\log D)^6$, and since $V(D)\ge\sum_{D/2<k\le D}\mu^2(k)J_2(k)/k^2\gg D$ by positivity, excess kurtosis $O\bigl(D^{-1}(\log\log D)^6\bigr)$ --- the measured law is the cleaner $-3/D$. By Theorem~\ref{ms:thm:tame} the two sectors with at most two distinct conductors are $O(D\log^4D)$ --- the conjectured bound up to logarithmic factors --- so what remains is $X_{211}(D)+X_{1111}(D)=O(D)$, which would give $X(D)=O(D\log^4D)$. Proposition~\ref{ms:prop:x211} settles a first piece unconditionally, $X_{211}\ll D^{3/2+o(1)}$, so the open problem now has two named parts: improve the $(2,1,1)$ bound from $D^{3/2}$ to $D$ --- a problem about the true size of the pair sums $W_{r,s}$, which the data of Section~\ref{ms:sec:scale} put at $(gr_1s_1)^{-1}$ for the typical mean, a model that sums to $O(D\log D)$ --- and bound $X_{1111}$, where the same data say absolute values cannot suffice and the cancellation \emph{among} quadruple resonances must enter. Unlike everything else in this circle, Conjecture~\ref{ms:conj:r4} is \emph{not} the Riemann hypothesis in disguise: it concerns fixed-$D$ averages of periodic functions --- finitely many conductor multisets at each $D$, each a finite exact computation --- and the intra-orbit part of the problem is already settled exactly by \eqref{ms:eq:IX} and Lemma~\ref{ms:lem:primeorbit}. Its proof would make the $-3/D$ law unconditional. It would not touch the moving diagonal $D=x$.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/F3_gauss.png}
\caption{Gaussian statistics. (a) Histogram at $D=300$ against the normal density, log scale. (b) Tails against the Gaussian law, to $\lambda=4.5$. (c) The Gaussian defect of the fourth moment: exact values for $D\le18$, high-precision Monte Carlo at $D=100,1000$, and the law $-3/D$, which splits as intra $-3.5$ plus cross $+0.5$ by \eqref{ms:eq:IX}.}
\label{ms:fig:gauss}
\end{figure}