\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}). The spectral analysis then turns bounds into evaluations: the pair sums collapse to $O(r+s)$ cotangent sums with partial-$L(1)$ coefficients and a closed Euler-product law in the ladder family (Theorem~\ref{ms:thm:collapse}, Corollary~\ref{ms:cor:euler}); the profile sums obey a proved limit law with closed constants and conjectured supremum $6-2\pi/\sqrt3$ (Theorem~\ref{ms:thm:profile}); and on every structure with $E\le2$ or $E$ an odd prime the sector means are computed exactly as finite bilinear forms in $L(1)$-values with an $L(2)$ kernel (Theorems~\ref{ms:thm:lformula}, \ref{ms:thm:hdecor} and~\ref{ms:thm:Estrat}: the decorated formula computes exactly the stratum $E\mid k$, and the complementary generic stratum is a second bilinear form one level up, at conductor $r_1s_1E$), so the conjecture $X_{211}+X_{1111}=O(D)$ becomes, there, a cancellation statement about $L$-values; the $(2,1,1)$ structure sum itself then factorizes into bilinear sums of twisted truncated Euler sums against the $L(2)$ kernel (Lemma~\ref{ms:lem:hsum}), verified at two scales to $10^{-11}$ and computed in double precision to $D=4000$, where it grows like $D^{0.93}$ --- consistent with the conjectured $O(D)$, with $X_{211}/D=4.1\cdot10^{-4}$ at $D=4000$. 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 $-\bigl(1+\frac{\gamma_E}2-\frac12\log4\pi\bigr)$ for ever; here we study the M\"obius side of the same coin.

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}

Over a full period, $\sup|S_z|=1.514$ $(=\frac{53}{35})$, $3.547$, $8.659$ for $z=7,13,19$, far below the trivial bounds $8$, $32$, $128$ (Figure~\ref{ms:fig:scales}(d) shows a sampled range, on which the maxima seen are $1.37$, $3.03$, $5.46$). 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. On the diagonal $D=x$ the fixed-$D$ law does not apply: $\Phi(x)/\sqrt x$ has root mean square $\approx0.17$ (Figure~\ref{ms:fig:scales}(a)), which is the zero-sum constant $\bigl(2\sum_{\gamma>0}|\rho\zeta'(\rho)|^{-2}\bigr)^{1/2}$ of Ng \cite{Ng} (under RH and simple zeros), and $\widetilde\Phi_x(x)/\sqrt x=(\Phi(x)-\frac12M(x))/\sqrt x$ has root mean square $\approx0.086$, against $0.166$ at fixed $D$. The agreement of $0.17$ with $0.166$ is a coincidence of size, not an identification.
\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 in the normalized units $D\times{}$(excess kurtosis) $=144\,e_4/(c_V^2D)$ the intra part alone is $-3.5$; the measured total is $-2.61$ at $D=18$, $-3.07$ at $D=100$ and $-3.34$ at $D=800$, so the cross part, about $+0.5$ at $D=100$, shrinks in these units and the total drifts towards $-3.5$, not towards a constant $-3$ (Figure~\ref{ms:fig:gauss}(c)). \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 spectral law of the pair coefficients}\label{ms:sec:spectral}
The absolute values surrendered in Proposition~\ref{ms:prop:x211} hide an exact structure, and it is worth recording, because it identifies \emph{what} the pair sums $W_{r,s}$ are: finite combinations of Dirichlet $L$-values at $1$. Everything in this subsection is elementary and exact; the quantitative question it opens is stated at the end.

\begin{theorem}[collapse of the pair sums]\label{ms:thm:collapse}
Let $r\ne s$ be squarefree conductors $>1$, $\xi\ne0$ with $N=rs\,\xi\in\mathbb Z$, and $W_{r,s}(\xi)$ as in Lemma~\ref{ms:lem:paircoef}. Then
\[
W_{r,s}(\xi)\;=\;\frac\pi N\Big[\sum_{a\in A}\cot\frac{\pi a}{r}\;+\;\sum_{b\in B}\cot\frac{\pi b}{s}\Big],
\]
where
\[
A=\Big\{a\ \mathrm{mod}\ r:\ (a,r)=1,\ as\equiv N\ (\mathrm{mod}\ r),\ \big(v(a),s\big)=1\Big\},
\qquad v(a)=\frac{N-as}{r},
\]
and $B$ is the mirror set mod $s$. In particular $N\,W_{r,s}(\xi)$ is an odd function of $N$ mod $rs$; on each stratum of fixed $(N,rs)$ it expands in odd Dirichlet characters of the corresponding quotients (Corollary~\ref{ms:cor:euler} is the cleanest instance); and $W_{r,s}$ at any frequency costs $O(r+s)$ cotangent evaluations instead of the $O(rs)$ class sum implicit in the proof of Lemma~\ref{ms:lem:paircoef}.
\end{theorem}

\begin{proof}
In the proof of Lemma~\ref{ms:lem:paircoef} the solution set in $u$ is a union of residue classes mod $rs$. But all three defining conditions depend on $u$ only mod $r$: the congruence $us\equiv N$ and the coprimality $(u,r)=1$ visibly so, and the third because $v(u+r)=v(u)-s$, so the class of $v$ mod $s$ is unchanged under $u\mapsto u+r$. Each class $a$ mod $r$ is therefore either wholly in or wholly out, and contains the $s$ classes $a+jr$ mod $rs$, $0\le j<s$. The cotangent multiplication formula $\sum_{j=0}^{n-1}\cot\frac{\pi(x+j)}n=n\cot(\pi x)$, applied with $n=s$ and $x=a/r$, collapses each class:
\[
\sum_{j=0}^{s-1}\cot\frac{\pi(a+jr)}{rs}=s\,\cot\frac{\pi a}{r},
\]
so $r\cdot\frac\pi{rs}\sum_{a'\in A\ \mathrm{mod}\ rs}\cot\frac{\pi a'}{rs}=\pi\sum_{a\in A}\cot\frac{\pi a}r$, and symmetrically for the $v$-side. The solution set negates under $(u,v,N)\mapsto(-u,-v,-N)$ and $\cot$ is odd, so $NW$ is odd in $N$; it depends on $N$ mod $rs$ because $A$ and $B$ do, and the stratified character expansion follows by Fourier analysis on the unit groups of the quotients. The identity is verified in exact structure against the residue formula to $10^{-13}$ across the structures of the ancillary \texttt{wcollapse.py}.
\end{proof}

\begin{lemma}[{cotangent sums are $L$-values; classical, cf.\ \cite[\S9]{Dav}}]\label{ms:lem:cotL}
Let $\chi$ be an odd character mod $f$ (not necessarily primitive). Then
\[
\sum_{\substack{a=1\\(a,f)=1}}^{f-1}\chi(a)\cot\frac{\pi a}{f}\;=\;\frac{2f}{\pi}\,L^{(f)}(1,\chi),
\]
where $L^{(f)}(1,\chi)=\sum_{n\ge1}\chi(n)n^{-1}$ is the $L$-value of the induced character, i.e.\ $L(1,\chi^\ast)$ with the Euler factors at the primes dividing $f$ removed.
\end{lemma}

\begin{proof}
By the classical $\sum_{u\equiv a\,(f)}\frac1u=\frac\pi f\cot\frac{\pi a}f$ (symmetric limit),
\[
\frac\pi f\sum_{(a,f)=1}\chi(a)\cot\frac{\pi a}f=\sum_{\substack{u\in\mathbb Z\setminus\{0\}\\(u,f)=1}}\frac{\chi(u)}u=\big(1-\chi(-1)\big)\sum_{u\ge1}\frac{\chi(u)}u=2\,L^{(f)}(1,\chi)
\]
for odd $\chi$, the negative $u$ contributing $\chi(-1)$ times the positive.
\end{proof}

So every coefficient in the character expansion of $N\,W_{r,s}$ is an explicit partial $L(1)$-value. In the $2p$-ladder family the whole expansion collapses to a single character and the law closes:

\begin{corollary}[exact law in the ladder family]\label{ms:cor:euler}
Let $g$ be squarefree with $(g,6)=1$, and $(k,6)=1$. Then
\[
W_{2g,3g}\Big(\frac k6\Big)\;=\;-\,\chi_3(gk)\,\frac{\pi}{\sqrt3}\,\frac1{gk}\prod_{p\mid g}\Big(1-\frac{\chi_3(p)}p\Big),
\]
where $\chi_3$ is the odd character mod $3$; that is, $3L(1,\chi_3)=\pi/\sqrt3$ times a mobile Euler product. The identity holds to $10^{-15}$ across the family (ancillary \texttt{euler\_law.py}).
\end{corollary}

\begin{proof}
Here $r=2g$, $s=3g$, $N=rs\xi=g^2k$. \emph{The $u$-side vanishes.} The congruence $as\equiv N$ mod $r$ reads $3a\equiv gk$ mod $2$, which holds for every odd $a$; and for $p\mid g$, $v(a)=(gk-3a)/2\equiv-3a\cdot2^{-1}$ mod $p$ (since $p\mid gk$), a unit exactly when $a$ is; and $(v(a),3)=1$ since $v(a)\equiv gk\cdot2^{-1}\not\equiv0$ mod $3$ by $(gk,3)=1$. So $A$ is the full set of units mod $2g$, which is symmetric under $a\mapsto-a$, and the odd cotangent sums to zero. \emph{The $v$-side is a character.} The congruence $br\equiv N$ mod $s$ reads $2b\equiv gk$ mod $3$, i.e.\ $b\equiv 2gk$ mod $3$, one of the two unit classes mod $3$; the condition $(u(b),r)=1$ is automatic by the same computation as above. So $B=\{b:\ (b,3g)=1,\ \chi_3(b)=\chi_3(2gk)\}$, whose indicator on units is $\frac12\big(1+\chi_3(2gk)\chi_3(b)\big)$. The principal half dies by oddness; the character half gives, by Lemma~\ref{ms:lem:cotL},
\[
W=\frac{\pi}{g^2k}\cdot\frac{\chi_3(2gk)}2\cdot\frac{2\cdot3g}{\pi}L^{(3g)}(1,\chi_3)
=\frac{3\,\chi_3(2gk)}{gk}\,L(1,\chi_3)\prod_{p\mid g}\Big(1-\frac{\chi_3(p)}p\Big),
\]
and $\chi_3(2)=-1$, $3L(1,\chi_3)=\pi/\sqrt3$ give the statement.
\end{proof}

Two structural checks. For $s=2r$ the collapse reproduces $W\equiv0$ at the minimal denominators (both $A$ and $B$ are then full unit sets, symmetric under negation) --- the analogue, in pair form, of the identity $T_p(k)=0$ of Remark~\ref{ms:rem:freqview}. And the sign and Euler structure of Corollary~\ref{ms:cor:euler} are visible in the data of Section~\ref{ms:sec:scale}: the means $\langle\sigma_q^2\sigma_r\sigma_s\rangle$ in the ladder families inherit the character twist.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/F5_spectral.png}
\caption{The spectral law. Left: the exact $L$-value law of Corollary~\ref{ms:cor:euler} across the family $r=2g$, $s=3g$ (measured dots inside predicted circles; maximal difference $1.6\cdot10^{-15}$). Centre: $k\,W_{3,5}(k/15)$, odd and periodic mod $15$ --- the odd-character expansion made visible. Right: the profile sums $\Sigma^\ast$ (blue) against the proved cotangent bound (gray) over $61$ structures.}
\label{ms:fig:spectral}
\end{figure}

\emph{The profile sums, and where their supremum lives.} For the application to $X_{211}$ the relevant quantity is the profile sum
\[
\Sigma^\ast(r,s,E)=\sum_{k\ge1}\frac{\big|W_{r,s}\big(k/(r_1s_1E)\big)\big|}{k},\qquad E\mid(g,h),
\]
through which $|\langle\sigma_q^2\sigma_r\sigma_s\rangle|\le\frac{C\log D}{\pi^4}\,\frac Eh\,\Sigma^\ast$. The cotangent bound of Lemma~\ref{ms:lem:paircoef} gives only $\Sigma^\ast\ll(r_1+s_1)(\log rs)E/g$. A first scan of $61$ structures stayed below $2.2$ with the shape $\min(1,\,s_1E/g)\cdot O(1)$ (Figure~\ref{ms:fig:spectral}, right); a wider scan shows that $2.2$ is \emph{not} a uniform bound. The structures that break it are the divisor pairs $r\mid s$ taken at full depth $E=g$: since the terms of $\Sigma^\ast$ are nonnegative, every partial sum is a rigorous lower bound, and
\[
\Sigma^\ast(13,39,13)\ge2.323,\qquad \Sigma^\ast(211,633,211)\ge2.3718 .
\]
These are not accidents; they sit on a convergent family whose limit the collapse computes exactly.

\begin{theorem}[the limit profile]\label{ms:thm:profile}
Fix a squarefree $m\ge2$ and let $p\to\infty$ through primes $p\nmid m$. Take $(r,s,E)=(p,mp,p)$, so $r_1=1$, $s_1=m$, $g=E=p$, and the sampled frequencies are $\xi=k/(mp)$ with $N=pk$. Then for every fixed $k$ with $(k,m)=1$,
\[
W_{p,mp}\Bigl(\frac k{mp}\Bigr)\;\longrightarrow\;W_\infty(k):=\frac{\pi\cot(\pi k/m)}{k}-\frac m{k^2},
\]
while $W\equiv0$ when $(k,m)>1$, and the profile sums converge to the limit profile
\[
\Sigma^\ast(p,mp,p)\;\longrightarrow\;\Sigma^\ast_\infty(m):=\sum_{(k,m)=1}\frac{|W_\infty(k)|}{k}.
\]
For $m=3$ and $m=6$ --- the squarefree moduli with $\varphi(m)=2$ --- the limit is closed and elementary:
\[
\Sigma^\ast_\infty(3)=6-\frac{2\pi}{\sqrt3}=2.3724012\ldots,\qquad
\Sigma^\ast_\infty(6)=12-2\sqrt3\,\pi=1.1172038\ldots,
\]
and for $m=2$, where the cotangent vanishes at every admissible $k$, $\Sigma^\ast_\infty(2)=\tfrac74\zeta(3)=2.1035995\ldots$
\end{theorem}

\begin{proof}
With $r_1=1$ the solutions of $us+vr=N$ are $mu+v=k$ with $(u,p)=1$, $(v,mp)=1$; for $(k,m)=1$ the congruence $v\equiv k\ (\mathrm{mod}\ m)$ makes $(v,m)=1$ automatic, so the conditions at $p$ delete only the solutions with $p\mid u$ or $p\mid v$ (when $p\mid k$ only the first can occur), and the deleted subprogressions contribute $O_m((1+\log2k)/p)$ for fixed $k$ by the trivial estimate. The limit sum is evaluated exactly by symmetric truncation: from $mu+v=k$,
\[
\frac1{u(k-mu)}=\frac1k\Bigl(\frac1u+\frac m{k-mu}\Bigr),
\]
the odd part $\sum_{0<|u|\le U}1/u$ vanishes identically, and $\sum_{|u|\le U}(k-mu)^{-1}\to(\pi/m)\cot(\pi k/m)$ by the partial-fraction expansion of the cotangent along the progression $k+m\mathbb Z$; removing the excluded term $u=0$, which contributes $m/k$, gives $W_\infty(k)=\pi\cot(\pi k/m)/k-m/k^2$. For the sum over $k$, the trivial absolute bound $|W_{p,mp}(k/(mp))|\le\sum_{\mathrm{sols}}1/|uv|\ll_m(1+\log2k)/k$ holds uniformly in $p$, so $(1+\log2k)/k^2$ is a summable dominant free of $p$, and splitting at $k=p$ gives the convergence of the profile sums.

For $\varphi(m)=2$ write $c_m=\cot(\pi/m)$; the cotangent takes only the values $\pm c_m$, and $\cot(\pi k/m)=\chi_m(k)\,c_m$ with $\chi_m$ the odd quadratic character mod $m$. The sign of $W_\infty(k)$ agrees with $\chi_m(k)$ for every $k$ except $k=1$, where $\pi c_m-m<0$ because $m/(\pi c_m)<2$ for $m=3,6$ (and $4$); hence $|W_\infty(k)|=\chi_m(k)W_\infty(k)$ away from $k=1$ and
\[
\Sigma^\ast_\infty(m)=\pi c_m\!\!\sum_{(k,m)=1}\!\frac1{k^2}\;-\;m\,L(3,\chi_m)\;+\;2\bigl(m-\pi c_m\bigr).
\]
The classical evaluations $L(3,\chi_3)=4\pi^3/(81\sqrt3)$ and $L(3,\chi_6)=\bigl(1+\tfrac18\bigr)L(3,\chi_3)=\pi^3/(18\sqrt3)$ make the first two terms cancel \emph{exactly},
\[
\pi c_m\,\zeta(2)\prod_{p\mid m}\bigl(1-p^{-2}\bigr)=m\,L(3,\chi_m),
\]
leaving $\Sigma^\ast_\infty(m)=2m-2\pi\cot(\pi/m)$. For $m=2$ the cotangent vanishes at every odd $k$, so $W_\infty(k)=-2/k^2$ and $\Sigma^\ast_\infty(2)=2\sum_{k\ \mathrm{odd}}k^{-3}=\tfrac74\zeta(3)$.
\end{proof}

The same cancellation evaluates the non-squarefree analogue formally, $\sum_{k\ \mathrm{odd}}|\pi\cot(\pi k/4)/k-4/k^2|/k=8-2\pi$, and the identical computation for a general base pair shows that $(gr_1,gs_1,g)$, $g\to\infty$ through primes, has limit profile $\sum_k|W_{r_1,s_1}(k/(r_1s_1))|/k$ --- the base pair sampled at its own minimal denominators; for the ladder base $(2,3)$ this is $(\pi/\sqrt3)\sum_{(k,6)=1}k^{-2}=\pi^3/(9\sqrt3)=1.9890535\ldots$ by Corollary~\ref{ms:cor:euler}. The measured landscape organizes itself around these constants (Figure~\ref{ms:fig:profile}): $\Sigma^\ast_\infty(m)$ is maximal at $m=3$ and decays in $m$ (checked for $m\le40$); the family values increase toward their limits ($2.1235$, $2.3162$, $2.3681$, $2.3717$, $2.3722$, $2.3724$ at $g=5,11,41,101,211,499$ against $2.3724013$) while staying strictly below them; and composite $g$ only lowers the limit through its Euler factors (e.g.\ $\Sigma^\ast(385,1155,385)=1.91$). We therefore record, as measured evidence organized by Theorem~\ref{ms:thm:profile}:

\begin{conjecture}\label{ms:conj:sup}
$\displaystyle\sup_{(r,s,E)}\Sigma^\ast(r,s,E)=6-\frac{2\pi}{\sqrt3}$, a supremum approached along $(g,3g,g)$, $g$ prime, and not attained.
\end{conjecture}

A proof of any uniform bound $\Sigma^\ast\le C$ would clean the logarithms of Proposition~\ref{ms:prop:x211} but, by the accounting of its proof, not lower the exponent $3/2$; the exponent is guarded by the bilinear sums $\sum_{k\ge1}F(k)G(k)/k^2$ in which both factors are odd periodic functions with partial-$L(1)$ coefficients --- one from $\sigma_q^2$ through Lemma~\ref{ms:lem:coef}, one from $\sigma_r\sigma_s$ through Theorem~\ref{ms:thm:collapse}. The same bilinear forms govern the quadruple sector through the pairings $\langle(\sigma_q\sigma_r)(\sigma_s\sigma_t)\rangle$. The two remaining problems of this section are, in this precise sense, one problem.

\begin{figure}[htbp]\centering\includegraphics[width=\textwidth]{fig/F6_profile.png}
\caption{The limit profile. Left: $\Sigma^\ast_\infty(m)$ for $2\le m\le30$, with the four closed forms marked; the maximum is $\Sigma^\ast_\infty(3)=6-2\pi/\sqrt3$. Right: the family $\Sigma^\ast(g,3g,g)$ climbing monotonically toward $6-2\pi/\sqrt3$ (dashed), past the old measured bound $2.2$ (dotted); composite $g$ (open triangles) stays below through its Euler factors.}
\label{ms:fig:profile}
\end{figure}

\subsection{The bilinear $L$-formula for the sector means}\label{ms:sec:lformula}
The spectral analysis so far produced bounds. This subsection turns it into \emph{evaluations}: an exact Parseval identity writes every $(2,1,1)$ mean as a single sum over frequencies of two cotangent closed forms, a Jacobi-type factorization diagonalizes the $\sigma_q^2$ side in Dirichlet characters, and the collapse of Theorem~\ref{ms:thm:collapse} does the same for the pair side --- with the striking simplification that the pair coefficients separate by conductor. The result, for the structures $q=r_1s_1$, is a closed formula: the mean is a finite bilinear form in $L(1,\chi)$-values with an $L(2)$ kernel. No absolute values are surrendered anywhere.

\begin{lemma}[the Parseval identity]\label{ms:lem:parseval}
Let $q,r,s$ be squarefree, $r\ne s$ both $>1$, $r_1s_1\mid q$, and write $q=r_1s_1h$, $E=(g,h)$. Then, with absolute convergence,
\[
\langle\sigma_q^2\sigma_r\sigma_s\rangle=\frac1{8\pi^4}\sum_{\substack{k\ge1\\ E\mid hk}}\widetilde T_q\Bigl(\frac{hk}E\Bigr)\,W_{r,s}\Bigl(\frac k{r_1s_1E}\Bigr).
\]
\end{lemma}

\begin{proof}
The $\sigma$'s are bounded, so Parseval applies to the product of $\sigma_q^2$ and $\sigma_r\sigma_s$ over the common period, and by Lemmas~\ref{ms:lem:coef} and~\ref{ms:lem:paircoef} the two coefficient functions are $-\widetilde T_q(q\xi)/(4\pi^2)$ and $-W_{r,s}(\xi)/(4\pi^2)$ on the common support $\xi\in\frac1{r_1s_1E}\mathbb Z$; each product coefficient is an absolutely convergent convolution of square-summable sequences. Three structural facts assemble the sum. \emph{Both coefficient functions are even}: $(u,u')\mapsto(-u,-u')$ and $(u,v)\mapsto(-u,-v)$ are bijections of the defining solution sets, so $\widetilde T_q(-N)=\widetilde T_q(N)$ and $W_{r,s}(-\xi)=W_{r,s}(\xi)$ (in the closed forms, the oddness of the cotangent sums cancels against the prefactor $1/N$). Hence the two half-lines double and the sign is $(-1)^2\cdot2/(16\pi^4)=1/(8\pi^4)$, with no convention left free. \emph{The $\xi=0$ term vanishes}: $W_{r,s}(0)=0$, because a solution of $us+vr=0$ has $u=-r_1t$, $v=s_1t$, and any prime of $r_1$ divides $(u,r)$ --- while if $r_1=1$ then $s_1>1$ (as $r\ne s$ are squarefree) and any prime of $s_1$ divides $(v,s)$; so the solution set is empty. \emph{Absolute convergence}: both closed forms decay like $1/N$ times a constant depending only on the conductor, by Lemmas~\ref{ms:lem:coef} and~\ref{ms:lem:paircoef}, so the terms are $O(1/k^2)$. The identity is verified against the exact rationals of this section to $10^{-11}$ and, by independent quadrature, to $10^{-18}$.
\end{proof}

\begin{lemma}[the Jacobi factorization]\label{ms:lem:jacobi}
Let $q$ be squarefree and write $C_q(N)=\sum_{a\in R_q(N)}\cot(\pi a/q)$, as in Lemma~\ref{ms:lem:coef}. For every Dirichlet character $\chi$ mod $q$,
\[
\sum_{N\in(\mathbb Z/q)^\times}\overline\chi(N)\,C_q(N)=S_q(\overline\chi)\,J_q(\overline\chi),\qquad
S_q(\overline\chi)=\!\!\sum_{(a,q)=1}\!\overline\chi(a)\cot\frac{\pi a}q,\qquad
J_q(\overline\chi)=\!\!\prod_{\substack{p\mid q\\ \chi_p\ne\chi_0}}\!(-1)\prod_{\substack{p\mid q\\ \chi_p=\chi_0}}\!(p-2),
\]
where $\chi=\prod_{p\mid q}\chi_p$ is the prime factorization of $\chi$. Moreover $S_q(\overline\chi)=0$ for even $\chi$ and $S_q(\overline\chi)=\frac{2q}\pi L^{(q)}(1,\overline\chi)$ for odd $\chi$; consequently, for $(N,q)=1$,
\[
C_q(N)=\frac{2q}{\pi\,\varphi(q)}\sum_{\chi\ \mathrm{odd}}J_q(\overline\chi)\,L^{(q)}(1,\overline\chi)\,\chi(N).
\]
\end{lemma}

\begin{proof}
Since $C_q(N)=\sum_{a,c}\cot(\pi a/q)$ over pairs of units with $a+c\equiv N$, and $\overline\chi$ vanishes on non-units,
\[
\sum_N\overline\chi(N)C_q(N)=\sum_{a,c\ \mathrm{units}}\overline\chi(a+c)\cot\frac{\pi a}q
=\sum_{a}\overline\chi(a)\cot\frac{\pi a}q\sum_{t\ \mathrm{units}}\overline\chi(1+t),
\]
by the substitution $c=ta$ (the extension of $\overline\chi$ by zero makes the restriction to unit $N$ automatic). The $t$-sum factors over the primes of $q$ by the Chinese remainder theorem, and mod $p$ it is $\sum_{u\ne1}\overline\chi_p(u)$, which is $-1$ for $\chi_p$ nontrivial (orthogonality) and $p-2$ for $\chi_p$ trivial. The evaluation of $S_q$ is Lemma~\ref{ms:lem:cotL}; its proof uses only periodicity mod $q$ and the support $(a,q)=1$, never primitivity, so it applies verbatim to imprimitive (induced) characters --- a fact the next lemma needs. The factorization is verified for every character mod $15$, $21$, $35$ and $105$ to $8\cdot10^{-13}$.
\end{proof}

Note the factor $p-2$ at $p=2$: it is $0$. For even $q$ every character mod $q$ is trivial at $2$, so $J_q\equiv0$ and all the coefficients below vanish --- the parity sieve of Corollary~\ref{ms:cor:parity}, rediscovered by the $L$-formula as the vanishing of a Jacobi factor.

\begin{lemma}[the pair coefficients separate by conductor]\label{ms:lem:wside}
Let $r\ne s>1$ be squarefree, $g=(r,s)$, and sample at the minimal depth: $N=g^2k$ with $(k,r_1s_1)=1$. Then the $u$-side solution set of Theorem~\ref{ms:thm:collapse} is exactly
\[
A(k)=\{u\bmod r:\ (u,g)=1,\ u\equiv gk\overline{s_1}\ (\mathrm{mod}\ r_1)\},
\]
i.e.\ the condition $(v(u),s)=1$ is automatic: $v\equiv gk\overline{r_1}$ mod $s_1$ is a fixed unit class, and $vr_1\equiv-us_1$ mod $g$ makes $(v,g)=1$ equivalent to $(u,g)=1$. Consequently $B(k)=k\,W_{r,s}(k/(r_1s_1))$ is odd, periodic mod $r_1s_1$, supported on $(k,r_1s_1)=1$, and its character coefficients $B(k)=\sum_\psi b_\psi\psi(k)$ vanish unless $\psi$ comes from a single conductor:
\[
b_{\psi_{r_1}\times1}=\frac{2r_1}{g\,\varphi(r_1)}\,\psi_{r_1}(g\overline{s_1})\,L^{(r)}(1,\overline\psi_{r_1}),\qquad
b_{1\times\psi_{s_1}}=\frac{2s_1}{g\,\varphi(s_1)}\,\psi_{s_1}(g\overline{r_1})\,L^{(s)}(1,\overline\psi_{s_1}),
\]
for odd $\psi_{r_1}$, $\psi_{s_1}$, and $b_\psi=0$ whenever both components of $\psi$ are nontrivial.
\end{lemma}

\begin{proof}
From $us+vr=g^2k$: dividing by $g$, $us_1\equiv gk$ mod $r_1$ fixes the unit class $u\equiv gk\overline{s_1}$ mod $r_1$; $v=(gk-us_1)/r_1$ gives $vr_1\equiv gk$ mod $s_1$ --- a unit class, since $(gk,s_1)=1$ --- and $vr_1\equiv-us_1$ mod $g$, so $(v,s)=1$ reduces to $(u,g)=1$, already imposed. For the coefficients: $C_A(k)$ depends on $k$ only mod $r_1$, so in $b_\psi=\frac1{\varphi(r_1s_1)}\sum_k\overline\psi(k)(C_A+C_B)(k)$ the sum over $k$ mod $s_1$ annihilates $C_A$ unless $\psi_{s_1}$ is trivial (orthogonality), and symmetrically for $C_B$ --- whence the vanishing for mixed $\psi$. On the surviving characters, substituting $u\equiv gk\overline{s_1}$ and summing over the class,
\[
\sum_{k\ (r_1)}\overline\psi_{r_1}(k)\,C_A(k)=\psi_{r_1}(g\overline{s_1})\sum_{(u,r)=1}\overline\psi_{r_1}(u)\cot\frac{\pi u}r
=\psi_{r_1}(g\overline{s_1})\,\frac{2r}\pi L^{(r)}(1,\overline\psi_{r_1}),
\]
by Lemma~\ref{ms:lem:cotL} for the induced character mod $r$, and the normalizations $B=(\pi/g^2)(C_A+C_B)$, $\varphi(r_1s_1)=\varphi(r_1)\varphi(s_1)$ give the statement. All four coefficients at $(r,s)=(21,35)$, and the mixed-character vanishing at $g=7$ and $g=2$, are verified to $10^{-15}$ against the discrete Fourier transform.
\end{proof}

\begin{theorem}[the bilinear $L$-formula]\label{ms:thm:lformula}
Let $r\ne s$ both $>1$ be squarefree and $q=r_1s_1$. Then
\[
\langle\sigma_q^2\sigma_r\sigma_s\rangle=\frac1{8\pi^4}\sum_{\chi,\psi\ \mathrm{odd}\ \mathrm{mod}\ q}a_\chi\,b_\psi\,L^{(q)}(2,\chi\psi),\qquad
a_\chi=\frac4{\varphi(q)}\,J_q(\overline\chi)\,L^{(q)}(1,\overline\chi),
\]
with $b_\psi$ as in Lemma~\ref{ms:lem:wside}: a finite bilinear form in $L(1)$-values with an $L(2)$ kernel. (The hypotheses are only these: $q$ squarefree and $(q,g)=1$ follow automatically from the squarefreeness of $r$ and $s$.)
\end{theorem}

\begin{proof}
In Lemma~\ref{ms:lem:parseval} we have $h=E=1$: the sum is $\frac1{8\pi^4}\sum_{k\ge1}A(k)B(k)/k^2$ with $A(k)=k\widetilde T_q(k)=(2\pi/q)C_q(k)$ and $B(k)=kW_{r,s}(k/q)$. The frequency argument of the pair side is $N=rs\cdot k/q=g^2k$, so $\gcd(N,r_1s_1)=\gcd(k,q)$ and $B$ vanishes off the unit stratum --- the non-unit strata of $A$, which do not vanish, are killed by the pair factor at no cost. On the units, expand both factors by Lemmas~\ref{ms:lem:jacobi} and~\ref{ms:lem:wside} and sum: $\sum_{(k,q)=1}\chi\psi(k)/k^2=L^{(q)}(2,\chi\psi)$.
\end{proof}

The formula is verified at machine precision ($10^{-18}$, independent exact quadrature) against the exact rationals
\[
\langle\sigma_{15}^2\sigma_{21}\sigma_{35}\rangle=\tfrac{74}{55125},\quad
\langle\sigma_{15}^2\sigma_3\sigma_5\rangle=\tfrac2{3375},\quad
\langle\sigma_5^2\sigma_3\sigma_{15}\rangle=\tfrac2{1125},\quad
\langle\sigma_{15}^2\sigma_6\sigma_{10}\rangle=\tfrac7{6750},\quad
\langle\sigma_{35}^2\sigma_{10}\sigma_{14}\rangle=\tfrac{81}{42875},
\]
the last two obtained from the formula and then identified; and at $q$ even it correctly returns $0$ through $J_q\equiv0$ (e.g.\ $\langle\sigma_6^2\sigma_{10}\sigma_{15}\rangle=0$). For $h>1$ the argument $N=hk$ of the coefficient sums shares the primes of $h\mid q$ with $q$ at every $k$, so Lemma~\ref{ms:lem:jacobi} no longer applies on the nose --- but its stratified form does, and the outcome is simpler than one could hope: the entire $h$-dependence is a local decoration.

\begin{lemma}[the stratified closed form]\label{ms:lem:strat}
For squarefree $q$ and every integer $N$,
\[
C_q(N)=\sum_{\substack{d,e\mid q\\ p\mid(d,e)\Rightarrow p\mid N}}\mu(d)\mu(e)\,\frac q{[d,e]}\,\cot\frac{\pi c_{d,e}(N)}{[d,e]},
\qquad c_{d,e}(N)\equiv N\ (d),\quad c_{d,e}(N)\equiv0\ (e),
\]
with the convention $\cot(\pi c/m):=0$ when $m\mid c$ --- a load-bearing convention, since such singular terms do occur on the strata below. Each term is a single cotangent whose argument has denominator $d'=d/(d,e)$: $c/[d,e]=\langle N\bar e\rangle_{d'}/d'$, $\bar e=e^{-1}$ mod $d'$.
\end{lemma}

\begin{proof}
Expand both coprimality conditions of $R_q(N)$ by M\"obius, $[(N-a,q)=1]=\sum_{d\mid(N-a,q)}\mu(d)$ and $[(a,q)=1]=\sum_{e\mid(a,q)}\mu(e)$; a pair $(d,e)$ is compatible exactly when every prime of $(d,e)$ divides $N$, and then $a$ runs over the full residue class $c_{d,e}$ mod $[d,e]$. The multiplication formula $\sum_{j\bmod n}\cot(\pi(x+j)/n)=n\cot(\pi x)$, with $n=q/[d,e]$, collapses each class to one cotangent; a class through the integers contributes $0$ by the odd symmetry of its fiber. Verified for $q=15,21,30,35,105,210$ at every residue $N$ to $10^{-12}$.
\end{proof}

\begin{theorem}[the $h$-decoration]\label{ms:thm:hdecor}
Let $r\ne s>1$ be squarefree, $P=r_1s_1$, and $q=Ph$ squarefree with $E=(g,h)\le2$. Then the bilinear $L$-formula of Theorem~\ref{ms:thm:lformula} holds verbatim with the single change
\[
a_\chi\ \longmapsto\ \hat a_\chi(h)\;=\;\chi(h)\,\frac1h\prod_{p\mid h}\Bigl(1-\frac{\overline\chi(p)}p\Bigr)\cdot a_\chi,
\qquad a_\chi=\frac4{\varphi(P)}J_P(\overline\chi)\,L^{(P)}(1,\overline\chi):
\]
\[
\langle\sigma_q^2\sigma_r\sigma_s\rangle=\frac1{8\pi^4}\sum_{\chi,\psi\ \mathrm{odd}\ \mathrm{mod}\ P}\hat a_\chi(h)\,b_\psi\,L^{(P)}(2,\chi\psi).
\]
The entire dependence on $h$ is a character value, a factor $1/h$, and Euler factors at the primes of $h$.
\end{theorem}

\begin{proof}
Take $E=1$ first. $A_h(k)=k\,\widetilde T_q(hk)=(2\pi/(hq))\,C_q(hk)$ is periodic mod $P$ (as $hk$ mod $Ph$ is determined by $k$ mod $P$), and the pair factor still kills every $k$ with $(k,P)>1$, so only $N=hj$ with $(j,P)=1$ enters. In Lemma~\ref{ms:lem:strat} the compatibility forces $(d,e)\mid h$. Split $d'=d/(d,e)=d'_h\,d'_P$ into its $h$- and $P$-parts: then $\gcd(N\bar e,d')=d'_h$ exactly, the cotangent reduces to conductor $d'_P$ with argument $b\equiv(h/d'_h)\,j\,\bar e$ mod $d'_P$, and the inverse of Lemma~\ref{ms:lem:cotL} expands it over odd characters $\chi'$ mod $d'_P$ with coefficients $\frac{2d'_P}{\pi\varphi(d'_P)}L^{(d'_P)}(1,\overline{\chi'})$. Parametrizing $\delta=(d,e)\mid h$, $d=\delta d'_hd'_P$, $e=\delta e_1$ and summing the M\"obius weights, each prime $p\mid h$ offers four options --- absent, in $\delta$, in $d'_h$, in $e_1$ --- whose weights $1+\overline{\chi'}(p)/p-\overline{\chi'}(p)/p-\overline{\chi'}(p)/p$ collapse to $1-\overline{\chi'}(p)/p$, while $\chi'(h/d'_h)=\chi'(h)\overline{\chi'}(d'_h)$ supplies the prefactor $\chi'(h)$; the primes $p\mid P/d'_P$ contribute $(1-\overline{\chi'}(p)/p)$. (Here and throughout, the Euler-factor values are those of the \emph{primitive} character: the induced character mod $P$ would vanish at $p\mid P/d'_P$.) These factors assemble with $L^{(d'_P)}$ into $L^{(P)}(1,\overline{\chi'})\prod_{p\mid h}(1-\overline{\chi'}(p)/p)$, independently of the conductor $t=d'_P$; and the sum over the admissible $t$ with $f_\chi\mid t\mid P$ --- on all of which $\chi_t$ is odd, since $\chi_t(-1)=\chi_{f_\chi}(-1)$ --- collapses by
\[
\sum_{f\mid t\mid P}\frac{\mu(t)}{\varphi(t)}=\frac{\mu(f)}{\varphi(f)}\prod_{p\mid P/f}\Bigl(1-\frac1{p-1}\Bigr)=\frac{J_P(\overline\chi)}{\varphi(P)},
\]
recovering the Jacobi factor of Lemma~\ref{ms:lem:jacobi}. The $W$-side coefficients $b_\psi$ involve only $(r,s)$ and never see $h$. For $E=2$: then $2\mid q$, and $C_q$ of an odd argument vanishes identically (all units mod an even modulus are odd, so $a+c$ is even), so every frequency outside the $E$-lattice drops out of the Parseval identity term by term and the $E=1$ computation applies verbatim after reindexing; for odd $E\ge3$ the extra strata are genuinely present, and the formula provably fails by exactly their sum.
\end{proof}

The theorem is verified at machine precision ($10^{-17}$, independent exact quadrature) against six exact rationals spanning prime and composite $h$ and both parities of $E$:
\[
\langle\sigma_{30}^2\sigma_{21}\sigma_{35}\rangle=-\tfrac{353}{330750},\quad
\langle\sigma_{10}^2\sigma_3\sigma_{15}\rangle=\tfrac1{2250},\quad
\langle\sigma_{30}^2\sigma_7\sigma_{35}\rangle=-\tfrac1{55125},\quad
\langle\sigma_{15}^2\sigma_2\sigma_{10}\rangle=\tfrac1{2250},\quad
\langle\sigma_{42}^2\sigma_{15}\sigma_{35}\rangle=-\tfrac{499}{463050},\quad
\langle\sigma_{70}^2\sigma_{10}\sigma_{14}\rangle=-\tfrac{33}{171500},
\]
with every character coefficient checked against the discrete Fourier transform ($\le2\cdot10^{-15}$). (Ancillary: \texttt{bilinear\_L.py}.)

\subsubsection*{The $E$-stratification and the factorization of the structure sum}
For odd $E\ge3$ the frequencies $k$ with $\gcd(k,E)>1$ survive on both sides of the Parseval identity, and the stratification closes there as well. Throughout this subsection $E=p$ is an odd prime --- squarefree $q$ forces $E$ squarefree, and a composite squarefree $E$, two or more shared primes, iterates the same surgery once per prime and is left open --- and we write $g=pg'$, $h=ph'$; the pair-side frequency is $N=(g^2/p)k$.

\begin{lemma}[the generic pair side]\label{ms:lem:wgen}
Let $(k,PE)=1$. Then
\[
k\,W_{r,s}\Bigl(\frac k{r_1s_1E}\Bigr)\;=\;\frac{\pi p}{g^2}\Bigl[\,\mathrm{Cl}_r(r_1;\,g'k\bar s_1)-\mathrm{Cl}_r(pr_1;\,g'k\bar s_1)+\mathrm{Cl}_s(s_1;\,g'k\bar r_1)-\mathrm{Cl}_s(ps_1;\,g'k\bar r_1)\,\Bigr],
\]
where $\mathrm{Cl}_t(m;\beta)=\sum_{a\equiv\beta\,(m),\ (a,t)=1}\cot(\pi a/t)$ and $\bar s_1$ denotes the inverse of $s_1$ to the modulus of its class ($r_1$, resp.\ $pr_1$; mirror for $\bar r_1$). Each class expands over odd characters $Z$ mod $m$ as $\mathrm{Cl}_t(m;\beta)=\frac1{\varphi(m)}\sum_{Z\ \mathrm{odd}}\overline Z(\beta)\,\frac{2t}\pi L^{(t)}(1,Z)$: the pair coefficients of the generic stratum live at the conductors $r_1,s_1$ --- the class terms, the analogue of Lemma~\ref{ms:lem:wside} --- and at $pr_1,ps_1$ --- single-class corrections, the genuinely new $E$-terms.
\end{lemma}

\begin{proof}
In the collapsed form of the pair coefficient, the $r$-run is over $(a,r)=1$ with $as\equiv N$ mod $r$ and $(v,g)=1$, $v=(N-as)/r$. With $N=pg'^2k$: $r\mid as-N\iff r_1\mid as_1-g'k\iff a\equiv g'k\bar s_1$ $(r_1)$, and $v=(g'k-as_1)/r_1$. Modulo $g'$, $v\equiv-as_1\bar r_1$ is a unit multiple of $a$, so $(v,g')=1\iff(a,g')=1$, which already holds: the $g'$-condition is free. Modulo $p$, $v\equiv0\iff a\equiv g'k\bar s_1$ $(pr_1)$: the $p$-condition deletes exactly the subclass mod $pr_1$ from the class mod $r_1$. (Here, as in Lemma~\ref{ms:lem:wside}, the residual condition $(v,s)=1$ reduces to $(v,g)=1$: $v\equiv N\bar r$ is a fixed unit class mod $s_1$.) The prefactor is $k\cdot\pi/N=\pi p/g^2$; the $s$-run mirrors. The character expansion is Lemma~\ref{ms:lem:cotL} applied classwise, with $S_t(Z)=\sum_{(a,t)=1}Z(a)\cot(\pi a/t)=(2t/\pi)L^{(t)}(1,Z)$ for the induced character. The law is verified at every residue $k$ mod $PE$ in the six structures below, to $1.8\cdot10^{-15}$.
\end{proof}

\begin{lemma}[the decorated $T$-side stands alone]\label{ms:lem:tside}
Let $\tilde q=\tilde P\tilde h$ be squarefree and let $j$ be any integer with $(j,\tilde P)=1$. Then
\[
\frac{2\pi}{\tilde h\tilde q}\,C_{\tilde q}(\tilde hj)\;=\;\sum_{\chi\ \mathrm{odd}\ \mathrm{mod}\ \tilde P}\hat a_\chi(\tilde h)\,\chi(j),
\qquad
\hat a_\chi(\tilde h)=\chi(\tilde h)\,\frac1{\tilde h}\prod_{p'\mid\tilde h}\Bigl(1-\frac{\overline\chi(p')}{p'}\Bigr)\cdot\frac4{\varphi(\tilde P)}\,J_{\tilde P}(\overline\chi)\,L^{(\tilde P)}(1,\overline\chi).
\]
The only hypotheses are the two displayed; in particular $j$ may share primes with $\tilde h$, and no pair structure enters: at $\tilde h=1$ this is Lemma~\ref{ms:lem:jacobi}.
\end{lemma}

\begin{proof}
This is the $T$-side computation inside the proof of Theorem~\ref{ms:thm:hdecor}, which uses nothing beyond the two hypotheses; we rewrite its bookkeeping once, self-contained. In Lemma~\ref{ms:lem:strat} at $N=\tilde hj$ the primes of $\tilde P$ never divide $N$ --- $(\tilde h,\tilde P)=1$ and $(j,\tilde P)=1$ --- so the compatibility forces $\delta:=(d,e)\mid\tilde h$, while every prime of $\tilde h$ divides $N$ whether or not it divides $j$. Parametrize $d=\delta d'_ht$ and $e=\delta e_1$ with $d'_h\mid\tilde h/\delta$, $t\mid\tilde P$, $e_1\mid\tilde q/(\delta d'_ht)$; then $\mu(d)\mu(e)=\mu(d'_h)\mu(t)\mu(e_1)$, $q/[d,e]=\tilde q/(\delta d'_hte_1)$, and the term's cotangent has denominator $t$ and argument $m\equiv(\tilde h/d'_h)\,j\,\bar e\ (\mathrm{mod}\ t)$ --- a unit class, each factor being a unit mod $t$; the singular terms of Lemma~\ref{ms:lem:strat} are exactly $t=1$, zero on both sides (the empty odd-character sum mod $1$). Expanding each class over odd characters $\chi'$ mod $t$ (Lemma~\ref{ms:lem:cotL} for the induced character), the weight of $\overline{\chi'}(j)L^{(t)}(1,\overline{\chi'})$ factors over the primes of $\tilde q$. At $p'\mid\tilde h$ the four options --- absent, in $e_1$, in $\delta$, in $d'_h$ --- weigh
\[
1-\frac{\overline{\chi'}(p')}{p'}+\frac{\overline{\chi'}(p')}{p'}-\frac{\overline{\chi'}(p')}{p'}=1-\frac{\overline{\chi'}(p')}{p'},
\]
and $\overline{\chi'}(\tilde h/d'_h)\overline{\chi'}(d'_h)=\overline{\chi'}(\tilde h)$ supplies the prefactor; at the primes of $\tilde P$ the Euler factors assemble $L^{(t)}$ into $L^{(\tilde P)}$ independently of $t$ and the conductor sum collapses by $\sum_{f\mid t\mid\tilde P}\mu(t)/\varphi(t)=J_{\tilde P}(\overline{\chi'})/\varphi(\tilde P)$, exactly as in the proof of Theorem~\ref{ms:thm:hdecor}. At no point do $g$, $E$ or the conductors $r,s$ appear.
\end{proof}

\begin{theorem}[the $E$-stratification]\label{ms:thm:Estrat}
Let $r\ne s>1$ and $q=Ph$ be squarefree with $E=(g,h)=p$ an odd prime. Then the mean splits exactly along $\gcd(k,p)$,
\[
\langle\sigma_q^2\sigma_r\sigma_s\rangle\;=\;S_{\mathrm{gen}}+S_{\mathrm{div}},
\]
and both strata are finite bilinear forms in $L$-values:
\[
S_{\mathrm{div}}=\frac1{8\pi^4}\sum_{\chi,\psi\ \mathrm{odd}\ \mathrm{mod}\ P}\hat a_\chi(h)\,b_\psi\,L^{(P)}(2,\chi\psi)
\]
--- the bilinear $L$-formula of Theorem~\ref{ms:thm:hdecor}, \emph{verbatim}: the decorated formula, which at $E\le2$ computes the whole mean, computes at odd prime $E$ exactly the stratum $p\mid k$ --- and
\[
S_{\mathrm{gen}}=\frac1{8\pi^4}\sum_{X,Y\ \mathrm{odd}\ \mathrm{mod}\ PE}\hat a_X(h')\,b^{\mathrm{gen}}_Y\,L^{(PE)}(2,XY),
\]
with $\hat a_X(h')$ the $h'$-decorated coefficients one level up, at modulus $PE$ (plain $a_X=\frac4{\varphi(PE)}J_{PE}(\overline X)L^{(PE)}(1,\overline X)$ when $h=p$), and $b^{\mathrm{gen}}_Y$ the coefficients of Lemma~\ref{ms:lem:wgen}: each $Z$ odd of modulus $m\in\{r_1,pr_1\}$ contributes
\[
\pm\,\frac{2rp}{g^2\varphi(m)}\,\overline Z(g'\bar s_1)\,L^{(r)}(1,Z)
\qquad(+\ \text{at}\ m=r_1,\ -\ \text{at}\ m=pr_1)
\]
to the coefficient of the lifted character $Y=\overline Z$, the mirror families at $s$ contribute likewise, and $b^{\mathrm{gen}}_Y$ is the \emph{sum} of the contributions from all four families landing on $Y$ (a character mod $r_1$ and its lift to $pr_1$ land on the same $Y$); every $Y$ outside the four lifted families receives zero.
\end{theorem}

\begin{proof}
The split itself is the partition of the $k$-sum of Lemma~\ref{ms:lem:parseval}. On the divisible stratum, $k=pj$ gives $A(pj)=(2\pi E/(hq))C_q(hj)=p\,A_h(j)$ --- the $T$-side object of Theorem~\ref{ms:thm:hdecor}, whose $T$-side collapse never uses $(g,h)\le2$ --- and $pj\,W=p\,B(j)$ with $B$ the base pair function of the structure $(P;r,s)$, so the stratum equals $p\cdot p/p^2=1$ times the decorated sum, with kernel $L^{(P)}$ because $j$ runs over all $(j,P)=1$. On the generic stratum $(k,PE)=1$ the $T$-side argument $h'k$ shares with $q$ only the primes of $h'$, so Lemma~\ref{ms:lem:jacobi} (when $h=p$), or the $T$-side of Theorem~\ref{ms:thm:hdecor} (when $h'>1$), expands $A$ over odd characters mod $PE$; Lemma~\ref{ms:lem:wgen} expands the pair side; and $\sum_{(k,PE)=1}XY(k)/k^2=L^{(PE)}(2,XY)$. Both reuses of the $h$-decoration are instances of Lemma~\ref{ms:lem:tside}: the divisible stratum takes $(\tilde P,\tilde h)=(P,h)$ --- the prime $p$ sits in $\tilde h$, its four options collapse like any other prime of $h$, and the $T$-side never sees $g$ --- and the generic stratum takes $(\tilde P,\tilde h)=(PE,h')$, where $p$ has moved to the $\tilde P$-side and enters through the conductor sum and the Jacobi factor. The theorem is proved; the verification below is confirmation.
\end{proof}

The formula closes the target flagged above and five companions across $E\in\{3,5,7\}$, $g'\in\{1,2,5\}$, $h'\in\{1,2\}$; the exact rationals are computed by rational integration over the common period, every character coefficient is checked against the discrete Fourier transform ($\le8\cdot10^{-15}$), each stratum separately against its exact rational --- obtained by averaging the pair factor over the $p$ translates $x\mapsto x+tP$, which projects the Parseval sum onto $p\mid k$ --- to $\le3\cdot10^{-17}$, and the assembled formula reproduces
\[
\langle\sigma_{105}^2\sigma_{21}\sigma_{35}\rangle=\tfrac{236}{2701125},\quad
\langle\sigma_{105}^2\sigma_{15}\sigma_{35}\rangle=-\tfrac{92}{643125},\quad
\langle\sigma_{165}^2\sigma_{15}\sigma_{33}\rangle=\tfrac{914}{1497375},
\]
\[
\langle\sigma_{330}^2\sigma_{15}\sigma_{33}\rangle=-\tfrac{1489}{8984250},\quad
\langle\sigma_{105}^2\sigma_{30}\sigma_{42}\rangle=\tfrac{3083}{2315250},\quad
\langle\sigma_{231}^2\sigma_{105}\sigma_{165}\rangle=\tfrac{16372}{102719925}
\]
to $10^{-17}$. With Theorems~\ref{ms:thm:lformula}, \ref{ms:thm:hdecor} and~\ref{ms:thm:Estrat}, the sector means are finite bilinear forms in $L$-values on \emph{every} squarefree structure whose $E$ is $1$, $2$ or an odd prime. What survives at the level of single structures is composite squarefree $E$ alone, and the surgery is built to iterate: for $E=p_1p_2$ the Parseval sum stratifies over the divisors of $E$, the four strata carrying kernels at the levels $P$, $Pp_1$, $Pp_2$, $Pp_1p_2$, and the pair side one M\"obius deletion per shared prime and per side --- class sums at the levels $m\in\{r_1,\,p_1r_1,\,p_2r_1,\,p_1p_2r_1\}$ and their mirrors; the generalized pair law is already verified at $E=15$ on the pair $(105,165)$, at all $480$ unit residues, to $5.3\cdot10^{-15}$, with $\langle\sigma_{1155}^2\sigma_{105}\sigma_{165}\rangle$ as the exact target. And the structure sum is reshaped a third time: the odd-$E$ strata enter through the same decorated coefficients $\hat a_\chi(h)$ --- the divisible stratum --- and through Jacobi data one level up, at conductor $PE$ --- the generic stratum --- so the $h$- and $E$-sums remain multiplicative, $L$-type sums. Sector by sector, variable by variable, the conjecture $X_{211}+X_{1111}=O(D)$ continues its rewriting in the language of $L$-functions. (Ancillary: \texttt{estrata\_E.py}.)


With the sector means in closed form on every structure with $E\in\{1,2\}\cup\{\text{odd primes}\}$, the structure sum itself can be reorganized. Write $P=r_1s_1$ (odd: on every even $P$ block Lemma~\ref{ms:lem:parity} kills the means and $J_P=0$ kills the coefficients), and for the odd characters $\chi$ mod $P$ and the one-sided odd characters $\psi$ of Lemma~\ref{ms:lem:wside} set
\[
\delta_\chi(h)=\frac{\chi(h)}h\prod_{p\mid h}\Bigl(1-\frac{\overline\chi(p)}p\Bigr),
\qquad
\delta_\psi(g)=\frac{\psi(g)}g\prod_{p\mid g}\Bigl(1-\frac{\overline\psi(p)}p\Bigr):
\]
the $h$- and $g$-dependences of Theorems~\ref{ms:thm:lformula} and~\ref{ms:thm:hdecor} are the \emph{same} multiplicative decoration, acting on the two independent variables of the structure. Define the twisted truncated Euler sums
\[
H_\chi(D;P)=\sum_{\substack{h\ \mathrm{sqfree},\,(h,P)=1\\ Ph\le D}}m_{Ph}\Bigl(\frac D{Ph}\Bigr)^{\!2}\,\delta_\chi(h),
\qquad
G_\psi(D;r_1,s_1)=\sum_{\substack{g\ \mathrm{sqfree},\,(g,P)=1\\ gr_1>1,\ gs_1\le D}}m_{gr_1}\Bigl(\frac D{gr_1}\Bigr)m_{gs_1}\Bigl(\frac D{gs_1}\Bigr)\,\delta_\psi(g),
\]
where the constraint $gr_1>1$ excludes, in the blocks with $r_1=1$ only, the terms with $g=1$ --- these have $r=gr_1=1$, which is not a conductor, hence not a structure of the sector.

\begin{lemma}[factorization of the structure sum]\label{ms:lem:hsum}
With $a_\chi=\frac4{\varphi(P)}J_P(\overline\chi)\,L^{(P)}(1,\overline\chi)$ as in Theorem~\ref{ms:thm:hdecor}, with $b_\psi=\beta_\psi\,\delta_\psi(g)$ by Lemma~\ref{ms:lem:wside}, where $\beta_\psi$ is the $g$-free part of that lemma's coefficients,
\[
\beta_{\psi_{r_1}\times\mathbf1}=\frac{2r_1}{\varphi(r_1)}\,\psi_{r_1}(\overline{s_1})\,L^{(r_1)}(1,\overline{\psi_{r_1}})
\]
(inverse taken mod $r_1$; mirror formula at $s_1$), and with the kernel $K_{\chi\psi}=L^{(P)}(2,\chi\psi)$,
\[
X_{211}(D)\;=\;\frac{12}{8\pi^4}\sum_{\substack{r_1<s_1\ \mathrm{odd\ sqfree}\\ (r_1,s_1)=1,\ r_1s_1\le D}}\mu(r_1)\mu(s_1)\Bigl[\sum_{\chi,\psi}a_\chi H_\chi(D;P)\,K_{\chi\psi}\,\beta_\psi G_\psi(D;r_1,s_1)\;-\;\Delta_{r_1=1}\Bigr]\;+\;\mathcal C(D),
\]
where $\Delta_{r_1=1}$ removes, in the blocks with $r_1=1$ only, the diagonal $g=h\ge2$ (the excluded coincidence $q=s$). Two exclusions meet the support of the unsifted double sum, both in the $r_1=1$ blocks: $r=1$ (iff $g=1$, removed by the constraint $gr_1>1$ in $G_\psi$) and $q=s$ (iff $g=h$, removed by $\Delta_{r_1=1}$); the coincidence $q=r$ is impossible. The correction term is carried entirely by the genuine structures of the sector ($gr_1>1$, $q\ne s$) whose $(g,h)$ share an odd prime:
\[
\mathcal C(D)=12\!\!\sum_{\substack{(g,r_1,s_1,h)\ \mathrm{as\ above}\\ (g,h)\ \mathrm{shares\ an\ odd\ prime}}}\!\! c_q^2c_rc_s\,\bigl(\langle\sigma_q^2\sigma_r\sigma_s\rangle-\mathrm{naive}\bigr),
\]
with $\mathrm{naive}$ the bilinear form of Theorem~\ref{ms:thm:hdecor} evaluated blindly at $(g,h)$; the tuples with even $P$ contribute $0$ identically to $\mathcal C$ (the mean by Lemma~\ref{ms:lem:parity}, the naive term by $J_P=0$), so $\mathcal C$ is carried by the odd-$P$ structures alone. The identity is exact: on $E\le2$ the naive term \emph{is} the mean (Theorem~\ref{ms:thm:hdecor}), and at $E$ an odd prime the naive term is exactly the divisible stratum, so $\mathcal C$ collects precisely the generic strata of the $E$-stratification --- plus the rare composite-$E$ residuals. No coprimality sieve couples $H$ and $G$: the two exclusions above are support exclusions, not a sieve between $g$ and $h$, and the unsifted double sum is legitimate because the naive form evaluates something \emph{identified} at every $(g,h)$ with $E\le2$ or $E$ an odd prime --- the full mean, resp.\ the divisible stratum --- while the remaining (composite-$E$) structures are quarantined in $\mathcal C$ by construction.
\end{lemma}

The proof is bookkeeping on top of the three theorems cited, and the identity was verified in full at two scales: factorized assembly against the class-form truth agrees to $5.2\cdot10^{-12}$ at $D=100$ and $2.5\cdot10^{-11}$ at $D=200$, with the naive part $+0.1064385$ and the correction layer $-0.0205507$ summing to the exact $+0.0858878$ at $D=100$. At $D=100$ the correction is carried by the $20$ parity-live odd-sharing structures --- of $66$ structures whose $(g,h)$ share an odd prime ($51$ distinct pairs $(g,h)$), the $46$ with even $P$ contribute zero identically. Three facts of measurement then frame the conjecture $X_{211}=O(D)$:

\emph{(i) The factors are bounded.} Every $|m_q|$ is $\le1$: $\bigl|\sum_{n\le x,\,(n,q)=1}\mu(n)/n\bigr|\le1$ for all $q$ and $x$ is a theorem (Granville and Soundararajan; Tao \cite{Tao}; the measured maximum over $q\le4000$ is $1.0000$), so $|H_\chi|\ll\log D$ and $|G_\psi|\ll\log D$ absolutely --- and the measured $H_\chi(D)$ are bounded and in fact decay across the range ($|H_\chi(D;3)|$ falls from $0.168$ at $D=100$ to $0.023$ at $D=4000$): the twisted Euler sums carry no growth.

\emph{(ii) The correction layer is flat.} $\mathcal C(100)=-0.0206$, $\mathcal C(200)=+0.0184$, $\mathcal C(300)=-0.0165$: bounded and oscillating, with the naive part $3$--$5$ times larger in the computed window (ratios $5.2$, $4.4$, $3.5$). Whatever growth $X_{211}$ has, it lives in the naive bilinear part --- that is, in $L$-value language.

\emph{(iii) The sector grows like $D^{0.93}$ over the computed range --- consistent with the conjectured $O(D)$, and clearly below the $D^{1.2}$ termwise majorant.} Computing $X_{211}(D)$ exactly (class form, validated to double precision against the exact-arithmetic sector table of Section~\ref{ms:sec:sectors} at $D\le22$) on a grid to $D=4000$ --- $2\,008\,584$ structures, $1\,018\,496$ of them parity-live --- gives $+0.086$ at $D=100$, $+0.298$ at $800$, $+0.496$ at $1500$, $+1.09$ at $2100$, $+1.49$ at $3000$ and $+1.62$ at $4000$: against the absolute-value majorant $D^{3/2+o(1)}$ of Proposition~\ref{ms:prop:x211} and the measured $D^{1.2}$ growth of its termwise version, the sector itself crawls. The conjectured $O(D)$ holds in this range with margin, and the log--log regression slope of the (oscillating) record is $0.93$ over $100\le D\le4000$ ($0.77$ when fit only to $D\le1500$: the exponent drifts upward as the range extends, and the data cannot yet separate $\Theta(D/\log^A D)$ from $\Theta(D)$; what it does separate, decisively, is the sector from its absolute majorants). At $D=4000$, $X_{211}/D=4.1\cdot10^{-4}$.

We label precisely: the Lemma is proved given Theorems~\ref{ms:thm:lformula} and~\ref{ms:thm:hdecor} and the $E$-stratification; the boundedness of the $m$'s is a theorem \cite{Tao}; items (i)--(iii) are measurements. (Ancillary: \texttt{x211\_structure.py}.)

\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 values are $-3.07/D$ at $D=100$ and $-3.34/D$ at $D=800$. 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 give excess kurtosis $O(1/D)$; the constant would be $144\lim I(D)/(c_V^2D)\approx-3.5$ if moreover $X=o(D)$, as the data to $D=800$ suggest. 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 reference curve $-3/D$; the intra part alone is $-3.5/D$ and the measured total drifts from $-3.07/D$ at $D=100$ to $-3.34/D$ at $D=800$ \eqref{ms:eq:IX}.}
\label{ms:fig:gauss}
\end{figure}