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1% PREAMBLE2\documentclass[11pt]{amsart}3\usepackage[T1]{fontenc}4\usepackage{lmodern}5\usepackage{microtype}6\usepackage{amsmath,amssymb}7\usepackage{mathtools}8\usepackage{graphicx}9\usepackage{booktabs}10\usepackage{tikz}11\usepackage[colorlinks=true,linkcolor=bluedark,citecolor=bluedark,urlcolor=bluedark]{hyperref}12\usepackage[capitalize]{cleveref}1314% COLORS15\definecolor{black}{HTML}{000000}16\definecolor{white}{HTML}{FFFFFF}17\definecolor{red}{HTML}{FF3D40}18\definecolor{redlight}{HTML}{FF9D95}19\definecolor{reddark}{HTML}{A80016}20\definecolor{orange}{HTML}{FF8F2C}21\definecolor{orangelight}{HTML}{FFC093}22\definecolor{orangedark}{HTML}{A25400}23\definecolor{yellow}{HTML}{FFD100}24\definecolor{yellowlight}{HTML}{FFE591}25\definecolor{yellowdark}{HTML}{9E8100}26\definecolor{green}{HTML}{32CC58}27\definecolor{greenlight}{HTML}{5EEE79}28\definecolor{greendark}{HTML}{007F2C}29\definecolor{mint}{HTML}{00D1BB}30\definecolor{mintlight}{HTML}{48EFD8}31\definecolor{mintdark}{HTML}{008173}32\definecolor{teal}{HTML}{00CAD8}33\definecolor{teallight}{HTML}{48E9F7}34\definecolor{tealdark}{HTML}{007C85}35\definecolor{cyan}{HTML}{1EC9F3}36\definecolor{cyanlight}{HTML}{86E2FF}37\definecolor{cyandark}{HTML}{007C98}38\definecolor{blue}{HTML}{008CFF}39\definecolor{bluelight}{HTML}{84BDFF}40\definecolor{bluedark}{HTML}{00559F}41\definecolor{indigo}{HTML}{6768FA}42\definecolor{indigolight}{HTML}{9EA9FF}43\definecolor{indigodark}{HTML}{3C2ABC}44\definecolor{purple}{HTML}{D332E9}45\definecolor{purplelight}{HTML}{F08AFF}46\definecolor{purpledark}{HTML}{870097}47\definecolor{pink}{HTML}{FF325A}48\definecolor{pinklight}{HTML}{FF9A9F}49\definecolor{pinkdark}{HTML}{A50030}50\definecolor{brown}{HTML}{B18462}51\definecolor{brownlight}{HTML}{DFAF8C}52\definecolor{browndark}{HTML}{754C2B}53\definecolor{gray}{HTML}{8E8E93}54\definecolor{graylight}{HTML}{BABABF}55\definecolor{graydark}{HTML}{56565A}56% COLORS END5758\newtheorem{theorem}{Theorem}[section]59\newtheorem{proposition}[theorem]{Proposition}60\newtheorem{lemma}[theorem]{Lemma}61\newtheorem{corollary}[theorem]{Corollary}62\newtheorem{conjecture}[theorem]{Conjecture}63\theoremstyle{definition}64\newtheorem{definition}[theorem]{Definition}65\newtheorem{fact}[theorem]{Fact}66\theoremstyle{remark}67\newtheorem{remark}[theorem]{Remark}6869\title{Menger Diagonal Slices: A Recurrence of Order $\lceil D/2 \rceil$}70\author{Carlo Mitchener}71\address{MrlyProd, Inc.}72\email{carlo.mitchener@gmail.com}73\date{First published 2026-08-23, revised 2026-09-08}7475% PAPER76\begin{document}7778\begin{abstract}79Slice a Menger sponge along its main diagonal and count the cubes the cut meets; then do the same in every dimension. The count obeys a linear recurrence with constant coefficients of order at most $\lceil D/2 \rceil$ in dimension $D$, a fourfold sharpening of the order $2D+1$ that the standard carry automaton gives for free; exactness of that order is verified for $2 \le D \le 24$. A discrete Fourier extraction then turns a relaxed census into an exact product of cosines, and the formula decides half of the sign law this cut obeys: at every odd dimension $D$ with $D \not\equiv 1 \pmod 3$, the counting exponent sits strictly above the solid's dimension minus one, proved, with the parity carried by the single factor $(-1)^{D-1}(D-1)$ that every term of the formula ends in. Two exact identities frame what remains: the dominant root is pinned to within $2(D-1)/3$ of $f_D/3$ unconditionally, so the slice exponent converges to the generic Marstrand--Mattila value in every case, and the whole sign law is equivalent to the parity-free statement that the Perron carry vector puts more than a third of its mass on carries divisible by three. At $D = 3$ the machine outputs $\bigl(\begin{smallmatrix} 6 & 6 \\ 1 & 3\end{smallmatrix}\bigr)$, reproducing the published hexagon-triangle substitution spectrum and the integers $1, 6, 42, 306, 2250, 16578, 122202$ by a route that never mentions a hexagon.80\end{abstract}8182% TITLE PAGE83\makeatletter84\global\let\titledate\@date85\global\let\paperabstract\@setabstracta86\global\let\@date\@empty87\global\let\@setabstract\relax88\makeatother8990\maketitle9192\begin{center}93\normalfont\footnotesize94MrlyProd, Inc.\\95\titledate96\end{center}9798\vspace*{\stretch{1}}99100\begin{center}101\includegraphics[width=0.8\textwidth]{figures/avatar-light.png}102\end{center}103104\vspace*{\stretch{1.25}}105106\newpage107108\paperabstract109110% BODY111\section{Introduction}112113Take a cube. Cut it into $27$ equal subcubes and throw away the seven that touch the middle of a face or the middle of the cube; twenty survive. Repeat inside each survivor forever, and the limit is the Menger sponge. Now slice the sponge with the plane through its centre perpendicular to a main diagonal. The picture that falls out is famous: a hexagon full of six-pointed stars, first rendered by P\'erez-Duarte \cite{perezduarte}, popularised by Hart \cite{hart} and by Cook's slicing code \cite{cook}, and given its substitution rule and its dimension $1.8184$ by Abel \cite{abelstars}.114115This paper asks the same question one dimension at a time. In dimension $D$ the sponge becomes the set of cells with at most one coordinate digit equal to the middle digit $1$, and the diagonal cut becomes the hyperplane $\sum_i x_i = D(3^L-1)/2$ through the centre of the level-$L$ grid. Let $a_D(L)$ count the cells the hyperplane meets. The question is how much memory this sequence has: how many previous terms determine the next one.116117\begin{figure}[ht]118\centering119\begin{tikzpicture}[scale=1.0]120\foreach \d in {2,...,14}{121  \pgfmathtruncatemacro{\free}{2*\d+1}122  \pgfmathtruncatemacro{\sharpo}{ceil(\d/2)}123  \fill[black!22] (0.46*\d-0.15,0) rectangle ++(0.13,0.052*\free);124  \fill[blue!45!black] (0.46*\d+0.01,0) rectangle ++(0.13,0.052*\sharpo);125  \node[font=\scriptsize] at (0.46*\d,-0.22) {\d};126}127\draw[black!70] (0.70,0) -- (6.90,0);128\fill[black!22] (0.72,1.72) rectangle ++(0.13,0.11);129\node[font=\scriptsize,anchor=west] at (0.95,1.78) {order $2D+1$, free};130\fill[blue!45!black] (3.55,1.72) rectangle ++(0.13,0.11);131\node[font=\scriptsize,anchor=west] at (3.78,1.78) {order $\lceil D/2 \rceil$, this paper};132\end{tikzpicture}133\caption{How much memory the slice census needs, by dimension $D$.}134\label{fig:order}135\end{figure}136137The answer is \cref{thm:order}: at most $\lceil D/2 \rceil$ terms, in every dimension at once. \Cref{fig:order} is that half of the contribution in one picture; the other half is \cref{thm:sign}, a proved sign law at odd dimension. Three moves prove the order. First, the digit polynomial factors,138\[139P_D(t) = (1+t^2)^{D-1}(1+Dt+t^2),140\]141which is a two-line count of how a kept digit vector can sum to $s$. Second, the carry map $c \mapsto (c+D-s)/3$ contracts: its fixed point is $D/2$, integrality is a wall, and every carry reachable from zero obeys $\lvert c \rvert \le \lfloor (D-1)/2 \rfloor$. Third, the design is invariant under $v \mapsto (2,\dots,2)-v$, so $P_D[s] = P_D[2D-s]$, the carry reflection $c \mapsto -c$ commutes with the transfer matrix, and the orbit of the start state lives inside its $+1$ eigenspace, whose dimension is $\lfloor (D-1)/2 \rfloor + 1 = \lceil D/2 \rceil$.142143\medskip144\noindent\textbf{What is not new.} That some constant-coefficient recurrence exists is free and standard. A finite carry state space makes $a_D(L)$ a $3$-regular sequence in the sense of Allouche and Shallit \cite{allouche}, and the untrimmed automaton on states $-D,\dots,D$ already hands over order $2D+1$ with no argument at all. A paper that led with ``these slices obey a linear recurrence'' would be selling a textbook fact. The object is not new either: higher-dimensional Menger continua are classical, and Hocking \cite{hocking2023} cuts exactly the $D = 4$ diagonal for its geometry and its pictures. The $D = 3$ integers are catalogued as OEIS \href{https://oeis.org/A299916}{A299916} \cite{oeis}, whose Menger reading is a user comment counting the star-shaped holes rather than the tiles. The carry-automaton frame is classical too: with unrestricted digits it is the carries Markov chain of Holte's ``amazing matrix'' \cite{holte}, whose spectrum is known exactly; the design's joint digit restriction and pinned output digit take the matrix here outside that theory, and the precise relationship is spelled out in \cite{evenhalf}.145146\medskip147\noindent\textbf{What is new.} The exact factorisation of $P_D$, the sharp order $\lceil D/2 \rceil$ proved for every $D \ge 2$, the exactness of that order verified through $D = 24$, the closed form of the reduced matrix, and every census at $D \ge 4$. Roughly a factor of four, uniformly in the dimension. Then, from the same factorisation read on the unit circle: an exact trigonometric product formula for a relaxed census (\cref{thm:product}), and with it the odd half of the sign law -- $\rho_D > f_D/3$ for every odd $D \not\equiv 1 \pmod 3$, proved (\cref{thm:sign}), a statement that had resisted spectral routes because the relevant gap closes like $4/D$ (\cref{con:gap}).148149\medskip150\noindent\textbf{The cold anchor.} At $D = 3$ the construction, which never mentions a hexagon or a triangle, returns the $2 \times 2$ integer matrix $\bigl(\begin{smallmatrix} 6 & 6 \\ 1 & 3 \end{smallmatrix}\bigr)$, characteristic polynomial $\lambda^2 - 9\lambda + 12$, dominant root $(9+\sqrt{33})/2$ and census $1, 6, 42, 306, 2250, 16578, 122202$. Abel's tile substitution $\bigl(\begin{smallmatrix} 6 & 1 \\ 6 & 3 \end{smallmatrix}\bigr)$ \cite{abelstars} has the same trace $9$ and determinant $12$, A299916 has signature $(9,-12)$ and the same integers, and $\log_3\bigl((9+\sqrt{33})/2\bigr) = 1.818410$ is Abel's published dimension. Two different routes to one spectrum is the reason to believe the machine in the dimensions where nobody has looked.151152\section{Definitions}153154\begin{definition}155\label{def:design}156The \emph{$D$-dimensional Menger design} is157\[158F_D = \bigl\{ v \in \{0,1,2\}^D : \#\{ i : v_i = 1 \} \le 1 \bigr\},159\]160the digit vectors with at most one middle coordinate. Its \emph{fill} is $\lvert F_D \rvert$.161\end{definition}162163\begin{definition}164\label{def:level}165For $L \ge 0$ write each $x_i \in \{0,\dots,3^L-1\}$ in base $3$ as $x_i = \sum_{j=0}^{L-1} v^{(j)}_i 3^j$. The \emph{level-$L$ Menger analog} is166\[167S_L^{(D)} = \bigl\{ x : v^{(j)} \in F_D \text{ for every } j \bigr\},168\]169and the \emph{central diagonal census} is170\[171a_D(L) = \#\Bigl\{ x \in S_L^{(D)} : \textstyle\sum_i x_i = D(3^L-1)/2 \Bigr\}, \qquad a_D(0) = 1.172\]173\end{definition}174175The hyperplane is the one through the centre of the grid: the coordinate sum of the centre cell of $\{0,\dots,3^L-1\}^D$ is $D(3^L-1)/2$. At $D = 3$ this is the diagonal cut of the classical sponge.176177\begin{definition}178\label{def:poly}179The \emph{digit polynomial} is180\[181P_D(t) = \sum_{v \in F_D} t^{v_1+\dots+v_D} = \sum_{s=0}^{2D} B_D(s)\, t^s .182\]183\end{definition}184185\begin{definition}186\label{def:transfer}187The \emph{transfer matrix} $M^{(D)}$ acts on carry states $c \in \mathbb{Z}$ by188\[189M^{(D)}_{c'c} = B_D(c + D - 3c'),190\]191which is nonzero only when $0 \le c+D-3c' \le 2D$. The \emph{core} is $C_D = \{c \in \mathbb{Z} : \lvert c \rvert \le r_D\}$ with $r_D = \lfloor (D-1)/2 \rfloor$, and $M_{\mathrm{even}}^{(D)}$ is the matrix of $M^{(D)}$ restricted to the core and expressed in the reflection-symmetric basis $v_0 = e_0$, $v_k = e_k + e_{-k}$ for $1 \le k \le r_D$. Its size is $n_D = r_D + 1 = \lceil D/2 \rceil$, and $\chi_D$ denotes its characteristic polynomial.192\end{definition}193194\section{Results}195196\begin{theorem}197\label{thm:order}198For every $D \ge 2$ the sequence $a_D(L)$ satisfies the linear recurrence with constant integer coefficients given by $\chi_D$, the characteristic polynomial of $M_{\mathrm{even}}^{(D)}$. In particular $a_D$ satisfies a constant-coefficient linear recurrence of order at most $\lceil D/2 \rceil$.199\end{theorem}200201\begin{lemma}202\label{lem:poly}203For every $D \ge 1$, $P_D(t) = (1+t^2)^{D-1}(1+Dt+t^2)$, the fill is $\lvert F_D \rvert = 2^{D-1}(D+2)$, and204\[205B_D(2k) = \binom{D}{k}, \qquad B_D(2k+1) = D\binom{D-1}{k},206\]207with the convention that a binomial coefficient with an out-of-range lower index is zero.208\end{lemma}209210\begin{lemma}211\label{lem:carry}212Every carry state reachable from $c = 0$ lies in the core $C_D$, and the core is forward invariant: if $\lvert c \rvert \le r_D$ and $M^{(D)}_{c'c} \ne 0$ then $\lvert c' \rvert \le r_D$.213\end{lemma}214215\begin{lemma}216\label{lem:reflect}217$P_D[s] = P_D[2D-s]$ for all $s$, the reflection $R : e_c \mapsto e_{-c}$ satisfies $RM^{(D)} = M^{(D)}R$, and $Re_0 = e_0$. Consequently the orbit $\{ (M^{(D)})^L e_0 : L \ge 0 \}$ lies in the $+1$ eigenspace of $R$ inside the core, a space of dimension $\lceil D/2 \rceil$.218\end{lemma}219220\begin{lemma}221\label{lem:omega}222Let $\omega = e^{2\pi i/3}$. Then $P_D(\omega) = (-1)^{D-1}\omega^D (D-1)$, so $\lvert P_D(\omega) \rvert = D-1$ for every $D \ge 1$.223\end{lemma}224225\begin{lemma}226\label{lem:classes}227Write $\varepsilon_D = (D-1)(-1)^{D-1}/3$ and $f_D = 2^{D-1}(D+2)$. For $j \in \{0,1,2\}$,228\[229\sum_{s \equiv j \ (\mathrm{mod}\ 3)} B_D(s) =230\begin{cases}231f_D/3 + 2\varepsilon_D, & j \equiv D \pmod 3, \\232f_D/3 - \varepsilon_D, & \text{otherwise}.233\end{cases}234\]235\end{lemma}236237\begin{proposition}238\label{prop:offcentre}239Fix an integer offset $m$ with $\lvert m \rvert \le r_D$ and let $a_D(L;m)$ count the cells of $S_L^{(D)}$ on the parallel hyperplane $\sum_i x_i = D(3^L-1)/2 + m$. Then $a_D(\cdot\,;m)$ satisfies the same recurrence as $a_D$, the one given by $\chi_D$, and $a_D(L;m) = a_D(L;-m)$ for every $L$.240\end{proposition}241242\begin{remark}243\label{rem:sharp}244The hypothesis $\lvert m \rvert \le r_D$ is sharp at even $D$. At $D = 4$ the offset $m = 2 = D/2$ sits exactly on the fixed point of the carry map, never enters the core, and the residual of $\chi_4$ applied to $a_4(\cdot\,;2)$ is the constant $128$ rather than zero, for every window checked by \texttt{scripts/verify.py}.245\end{remark}246247\begin{corollary}248\label{cor:degree}249$\limsup_L a_D(L)^{1/L}$ is the modulus of a root of $\chi_D$, hence an algebraic number of degree at most $\lceil D/2 \rceil$. Whenever the counting exponent $\dim_{\mathrm{slice}}(D) = \lim_L \log_3 a_D(L) / L$ exists, it is $\log_3$ of that algebraic number.250\end{corollary}251252\begin{fact}253\label{fact:exact}254For every $D$ with $2 \le D \le 24$ the Hankel determinant $H_D = \det\bigl[ a_D(i+j+1) \bigr]_{0 \le i,j < n_D}$, computed in exact rational arithmetic, is nonzero; hence the minimal order of $a_D$ is exactly $\lceil D/2 \rceil$ for those $D$. The first values are $H_2 = 2$, $H_3 = 72$, $H_4 = -6336$, $H_5 = -1029600000$, $H_6 = -62272025640000$, and $H_{24}$ is a nonzero integer of $899$ digits. Checked by \texttt{scripts/verify.py}.255\end{fact}256257\begin{proposition}258\label{prop:reach}259For every $D \ge 2$ the set of carry states reachable from $c = 0$ is exactly the core $C_D$, of size $D-1$ at even $D$ and $D$ at odd $D$. Moreover $M_C$ is irreducible and aperiodic: every state reaches $0$ in one step, $0$ loops to itself, and $0$ reaches every state.260\end{proposition}261262\begin{fact}263\label{fact:trace}264For every $D$ with $2 \le D \le 24$,265\[266\operatorname{tr} M_{\mathrm{even}}^{(D)} =267\begin{cases}2683 \cdot 2^{D-2} - 1, & D \text{ even}, \\2693D \cdot 2^{D-3}, & D \text{ odd}.270\end{cases}271\]272The even values are $2, 11, 47, 191, 767, 3071, 12287, \dots$ and the odd values are $9, 60, 336, 1728, 8448, 39936, \dots$. Checked by \texttt{scripts/verify.py}.273\end{fact}274275\begin{fact}276\label{fact:anchor}277At $D = 3$, $M_{\mathrm{even}}^{(3)} = \bigl(\begin{smallmatrix} 6 & 6 \\ 1 & 3 \end{smallmatrix}\bigr)$ with $\chi_3(\lambda) = \lambda^2 - 9\lambda + 12$, dominant root $(9+\sqrt{33})/2$ and census $1, 6, 42, 306, 2250, 16578, 122202$, matching OEIS \href{https://oeis.org/A299916}{A299916} term for term and its signature $(9,-12)$, and $\log_3\bigl((9+\sqrt{33})/2\bigr) = 1.818410$ to six places. At $D = 2$ the census is $2^L$, of order $1$. At $D = 4$, $\chi_4(\lambda) = \lambda^2 - 11\lambda - 66$ with dominant root $(11+\sqrt{385})/2$ and census $1, 6, 132, 1848, 29040, 441408, 6772128$. At $D = 5$ and $D = 6$ the censuses begin $1, 30, 1000, 35700, 1321600$ and $1, 20, 4030, 242300, 24642700$. Checked by \texttt{scripts/verify.py}, which also confirms that three independent generators, brute-force digit enumeration, the substitution product $\prod_{j<L} P_D(t^{3^j})$, and the untrimmed carry automaton on states $-D,\dots,D$, agree term for term for $2 \le D \le 7$ and $0 \le L \le 4$.278\end{fact}279280\begin{definition}281\label{def:sheaf}282The \emph{diagonal sheaf census} is283\[284b_D(L) = \#\Bigl\{ x \in S_L^{(D)} : \textstyle\sum_i x_i \equiv D(3^L-1)/2 \ (\mathrm{mod}\ 3^L) \Bigr\}, \qquad b_D(0) = 1 :285\]286the total census of the central hyperplane together with its translates by multiples of $3^L$.287\end{definition}288289\begin{lemma}290\label{lem:circle}291For every real $\psi$,292\[293P_D\bigl(e^{i\psi}\bigr) = e^{iD\psi}\, \Phi_D(\psi), \qquad \Phi_D(\psi) = (2\cos\psi)^{D-1}\,\bigl(D + 2\cos\psi\bigr),294\]295and $\Phi_D$ is real-valued. For odd $D$, $\Phi_D(\psi) \ge 0$ for every $\psi$; for even $D > 2$, $\Phi_D(\psi) < 0$ exactly when $\cos\psi < 0$.296\end{lemma}297298\begin{theorem}299\label{thm:product}300For every $D \ge 2$ and $L \ge 0$,301\[302b_D(L) = \sum_{c \in C_D} \bigl( M_C^L \bigr)_{c0}303= \frac{1}{3^L} \sum_{m=0}^{3^L-1} \; \prod_{j=0}^{L-1} \Phi_D\!\Bigl( \frac{2\pi m 3^j}{3^L} \Bigr).304\]305In particular the right-hand side, a sum of $3^L$ products of cosines, is a positive integer.306\end{theorem}307308\begin{corollary}309\label{cor:step}310For $k \ge 1$ let $U_k = \{ 1 \le u < 3^k : 3 \nmid u \}$ and311\[312W_k = \sum_{u \in U_k} \prod_{j=0}^{k-1} \Phi_D\!\Bigl( \frac{2\pi u 3^j}{3^k} \Bigr).313\]314Then $3^L b_D(L) = \sum_{k=0}^{L} f_D^{\,L-k}\, W_k$ with $W_0 = 1$, equivalently315\[3163\, b_D(k) - f_D\, b_D(k-1) = 3^{1-k}\, W_k \qquad (k \ge 1).317\]318Moreover the innermost factor of every product in $W_k$ is $\Phi_D(2\pi u/3)$ with $3 \nmid u$, and319\[320\Phi_D(2\pi/3) = \Phi_D(4\pi/3) = (-1)^{D-1}(D-1),321\]322so $W_k = (-1)^{D-1}(D-1)\, V_k$ with $V_k = \sum_{u \in U_k} \prod_{j=0}^{k-2} \Phi_D\bigl(2\pi u 3^j/3^k\bigr)$.323\end{corollary}324325\begin{theorem}326\label{thm:sign}327Let $\rho_D$ be the spectral radius of $M_{\mathrm{even}}^{(D)}$; it is a root of $\chi_D$, and it is the dominant one. For every odd $D \ge 3$,328\[329\rho_D \;\ge\; \frac{f_D}{3},330\]331and if moreover $D \not\equiv 1 \pmod 3$, the inequality is strict:332\[333\rho_D \;>\; \frac{f_D}{3} \qquad (D \text{ odd},\ D \equiv 0, 2 \pmod 3).334\]335This is the odd half of \cref{con:sign} for two of the three residue classes.336\end{theorem}337338\begin{fact}339\label{fact:signodd}340For the remaining residue class $D \equiv 1 \pmod 3$, the integer $\det\bigl(f_D I - 3 M_{\mathrm{even}}^{(D)}\bigr)$ is nonzero for every $D \le 80$, checked in exact integer arithmetic by \texttt{scripts/verify.py}; with \cref{thm:sign} this gives $\rho_D > f_D/3$ for every odd $D \le 80$ with no exception. Two structural notes, both checked on the same sweep: for $D \equiv 1 \pmod 3$ the $3$-adic valuation of the determinant is at least $n_D$, so $\chi_D(f_D/3)$ is a nonzero \emph{integer} there; and its valuation is otherwise erratic ($2, 7, 6, 10, 17, 11, 11, 34, 16, 17, 33, 20, 20$ over $D \equiv 1 \pmod 3$, $4 \le D \le 40$, both parities), so no uniform $3$-adic argument is in sight for this class.341\end{fact}342343\begin{theorem}[the pinning]344\label{thm:pin}345For every $D \ge 2$,346\[347\Bigl\lvert \rho_D - \frac{f_D}{3} \Bigr\rvert \;\le\; \frac{2(D-1)}{3},348\]349refined by parity to $f_D/3 - 2(D-1)/3 \le \rho_D \le f_D/3 + (D-1)/3$ at even $D$ and $f_D/3 - (D-1)/3 \le \rho_D \le f_D/3 + 2(D-1)/3$ at odd $D$. In particular $\rho_D = (f_D/3)\bigl(1 + O(D^2 2^{-D})\bigr)$, so350\[351\log_3 \rho_D - \bigl( \log_3 f_D - 1 \bigr) \longrightarrow 0 :352\]353whatever the sign law does, the counting exponent of the recurrence converges to the generic Marstrand--Mattila value $\log_3 f_D - 1$, and the central diagonal plane becomes asymptotically generic. Only the \emph{side} of the approach remains in question, and only at even $D$.354\end{theorem}355356\begin{proposition}[the mass identity]357\label{prop:mass}358Let $v$ be the Perron vector of $M_C$, unique up to scale by \cref{prop:reach}, and let $p_D = \sum_{c \in C_D,\, 3 \mid c} v_c \,\big/ \sum_{c \in C_D} v_c$ be its mass on carries divisible by $3$. Then, exactly,359\begin{gather*}3603\rho_D \;=\; f_D + (-1)^{D-1}(D-1)\,\bigl(3p_D - 1\bigr), \\361\text{i.e.}\qquad362\operatorname{sgn}\Bigl(\rho_D - \frac{f_D}{3}\Bigr) = (-1)^{D+1} \operatorname{sgn}\Bigl(p_D - \frac13\Bigr).363\end{gather*}364Consequently the full sign law, both parities at once, is equivalent to the parity-free statement that $p_D > \tfrac13$ for every $D \ge 2$: the Perron carry vector always puts more than a third of its mass on the carries divisible by three.365\end{proposition}366367\begin{conjecture}368\label{con:sign}369For every even $D \ge 2$, and for every odd $D \equiv 1 \pmod 3$ beyond the verified range,370\[371\operatorname{sgn}\bigl( \rho_D - f_D/3 \bigr) = (-1)^{D+1}, \qquad f_D = 2^{D-1}(D+2).372\]373The comparison value $\log_3 f_D - 1$ is the generic one: for a set of Hausdorff dimension $s$ in $\mathbb{R}^D$, almost every hyperplane meets it in dimension at most $s-1$, with equality on a positive-measure set of translates once $s > 1$ \cite{marstrand,mattila}, and $\log_3 f_D$ is the dimension of the solid. The central diagonal hyperplane is one specific, highly symmetric slice, to which no such theorem applies; together with \cref{thm:sign} the conjecture says its counting exponent misses the generic value in every dimension, alternately above and below.374Evidence: the sign alternates without exception for $2 \le D \le 20$ computed from the roots of $\chi_D$, and the exact rational sign of $\chi_D(f_D/3)$ alternates as $(-1)^D$ for $2 \le D \le 30$, both in \texttt{scripts/verify.py}.375The sharpest reduced target on the even side is a two-step census contraction: for even $D$, exact computation finds $9\, b_D(2j+2) \le f_D^2\, b_D(2j)$ and $3\, b_D(2j+1) < f_D\, b_D(2j)$ with no violation through index $400$ for every even $D \le 50$, with the ratio $-W_{k+1}/(f_D W_k)$ at odd $k$ peaking near $(D-2)/(D+2)$, a uniform margin. By induction these two inequalities give $b_D(L) \le (f_D/3)^L$, which with \cref{prop:reach} (irreducibility) and the nonvanishing of \cref{fact:signodd} would give the even half; neither inequality is proved. Equivalently, by \cref{prop:mass}, everything left of the law -- the even half and the residue class $D \equiv 1 \pmod 3$ -- is the single parity-free inequality $p_D > 1/3$.376Failure mode, and why the even half is genuinely harder: for odd $D$ the integrand $\Phi_D$ of \cref{thm:product} is nonnegative, and \cref{thm:sign} follows; for even $D$ it is not, and the cancellation is real. Already $V_2 = 2\bigl[\Phi_D(2\pi/9) + \Phi_D(4\pi/9) + \Phi_D(8\pi/9)\bigr]$ is negative for every even $D \ge 6$, because $\Phi_D(8\pi/9) = -(2\cos(\pi/9))^{D-1}(D - 2\cos(\pi/9))$ outweighs the positive terms, $2\cos(\pi/9) > 2\cos(2\pi/9)$. And no finite sign table of the $W_k$ can decide the even half: since $b_D(L) \sim C \rho_D^L$ with $C > 0$, \cref{cor:step} forces $W_L \sim C\, 3^{L-1} \rho_D^{L-1} (3\rho_D - f_D)$, so the eventual sign of $W_k$ \emph{is} the assertion $\rho_D < f_D/3$ itself; the signs observed at small $k$ (they alternate transiently at base $3$) are no evidence either way.377\end{conjecture}378379\begin{remark}[note added, August 2026]380\label{rem:noteadded}381The even half of \cref{con:sign} has since been proved, in every even dimension at once, at base $3$ and at base $5$: see the companion paper \cite{evenhalf}, which builds exactly the second idea asked for above -- a Collatz--Wielandt certificate closed analytically at depth $\Theta(D^2)$, the transient identified in closed form. The statement above is kept as written, both for the record and because its remaining content, strictness at odd $D \equiv 1 \pmod 3$ beyond the verified range, is still open; \cite{evenhalf} reduces it to a single $2$-adic valuation bound.382\end{remark}383384\begin{remark}385\label{rem:rate}386The size of the excess appears to be governed by the same tower of angles. Numerically,387\[388\rho_D - \frac{f_D}{3} \;=\; (-1)^{D+1}\, \frac{2(D-1)}{3} \prod_{k \ge 2} \Bigl( \cos \frac{2\pi}{3^k} \Bigr)^{\!D-1} \frac{D + 2\cos(2\pi/3^k)}{D+2} \cdot \bigl(1 + o(1)\bigr),389\]390with the ratio of the two sides within $10^{-8}$ of $1$ at $D = 61$, where the left side is computed by exact rational bisection of $\chi_{61}$ against the bracket $\chi_{61}(f_{61}/3) < 0 < \chi_{61}(f_{61}/3 + 1)$ in \texttt{scripts/verify.py}. The per-dimension decay constant is then $\prod_{k\ge2} \cos(2\pi/3^k) = 0.7428747134\ldots$, and the linear factor $D-1$ is the parity factor of \cref{cor:step}. This asymptotic is a conjecture supported by computation; only the inequalities of \cref{thm:sign} carry theorem dress.391\end{remark}392393\begin{conjecture}394\label{con:gap}395There is no uniform spectral gap: writing $\lambda_2$ for a second-largest eigenvalue of $M_{\mathrm{even}}^{(D)}$ in modulus,396\[397\frac{\rho_D}{\lvert \lambda_2 \rvert} = \frac{D+2}{D-2} + O(D^{-3}) \longrightarrow 1 .398\]399Evidence: the computed ratios fall from $3.551785$ at $D = 4$ to $1.222222$ at $D = 20$, monotonically from $D = 6$ on; over $6 \le D \le 20$ they agree with $(D+2)/(D-2)$ to within $0.05$, and to six decimal places by $D = 18$. The two smallest dimensions in that range sit outside the agreement, $3.551785$ against $3$ at $D = 4$ and $2.253050$ against $2.333333$ at $D = 5$, which is where the $O(D^{-3})$ correction is largest. \texttt{scripts/verify.py} asserts the monotonicity, and the agreement over $6 \le D \le 20$.400Failure mode: the ratios are floating-point roots of exactly computed integer polynomials, so a genuine near-degeneracy at some larger $D$ could be misread; and the fitted shape $(D+2)/(D-2)$ is an observed match, not a derivation.401This is the lighthouse that explains why \cref{thm:sign} had to go around the spectrum. The classical route to the sign law reads it off $\operatorname{sgn} \chi_D(f_D/3) = \prod_i \operatorname{sgn}(f_D/3 - \lambda_i)$, which reports on $\rho_D$ alone only if every subdominant eigenvalue sits below $f_D/3$; any argument pinning that has to work inside a margin that shrinks like $4/D$, and so cannot be a fixed-size perturbation estimate. Four natural routes were tried and refuted by their own computations: $M_{\mathrm{even}}^{(D)}$ is not banded plus low rank, not totally nonnegative, not symmetrizable by a positive diagonal, and $f_D/3 \cdot I - M_{\mathrm{even}}^{(D)}$ is not an M-matrix. The census route of \cref{thm:sign} never mentions $\lambda_2$, which is how the odd half fell; the even half of \cref{con:sign} still awaits either this separation or a second idea of the same kind.402\end{conjecture}403404\section{Proofs}405\label{sec:proofs}406407\begin{proof}[Proof of \cref{lem:poly}]408Split $F_D$ by the number of middle coordinates. A vector with no coordinate equal to $1$ chooses $0$ or $2$ in each of the $D$ places, contributing $(1+t^2)^D$ to the generating function. A vector with exactly one coordinate equal to $1$ chooses which coordinate, in $D$ ways, and then $0$ or $2$ in the remaining $D-1$ places, contributing $Dt(1+t^2)^{D-1}$. No other vector lies in $F_D$, so409\[410P_D(t) = (1+t^2)^D + Dt(1+t^2)^{D-1} = (1+t^2)^{D-1}\bigl(1+Dt+t^2\bigr).411\]412Setting $t = 1$ gives $\lvert F_D \rvert = 2^{D-1}(D+2)$. For the coefficients, expand $(1+t^2)^{D-1}(1+t^2) = (1+t^2)^D$ for the even part and $Dt(1+t^2)^{D-1}$ for the odd part: the coefficient of $t^{2k}$ is $\binom{D}{k}$ and that of $t^{2k+1}$ is $D\binom{D-1}{k}$.413\end{proof}414415\begin{proof}[Proof of \cref{thm:order}]416\emph{Step 1: the census is a coefficient.} By \cref{def:level} a cell of $S_L^{(D)}$ is exactly a choice of $v^{(0)},\dots,v^{(L-1)} \in F_D$, and its coordinate sum is $\sum_i x_i = \sum_{j<L} 3^j \sigma_j$ with $\sigma_j = \sum_i v^{(j)}_i$. The number of vectors in $F_D$ with digit sum $\sigma$ is $B_D(\sigma)$, so for any target $T$,417\[418\#\Bigl\{ x \in S_L^{(D)} : \textstyle\sum_i x_i = T \Bigr\} = \bigl[t^T\bigr] \prod_{j=0}^{L-1} P_D\bigl(t^{3^j}\bigr).419\]420421\emph{Step 2: the carry recursion.} For $c \in \mathbb{Z}$ let $N_L(c) = \bigl[t^{T_L(c)}\bigr]\prod_{j<L} P_D(t^{3^j})$ with $T_L(c) = D(3^L-1)/2 + c$, so $a_D(L) = N_L(0)$ and $N_0(c) = \mathbf{1}[c = 0]$. Peel the least significant digit vector, of digit sum $s$. Since422\[423D(3^L-1)/2 = 3 \cdot D(3^{L-1}-1)/2 + D,424\]425the remaining target after dividing by $3$ is $T_{L-1}(c')$ with $c' = (c+D-s)/3$, and the peel is possible exactly when $3 \mid c+D-s$. Therefore426\[427N_L(c) = \sum_{\substack{0 \le s \le 2D \\ s \equiv c+D \ (\mathrm{mod}\ 3)}} B_D(s)\, N_{L-1}\Bigl(\frac{c+D-s}{3}\Bigr),428\]429which in the notation of \cref{def:transfer} reads $N_L = (M^{(D)})^L$ applied in the sense $a_D(L) = \bigl((M^{(D)})^L\bigr)_{00}$: the entry $M^{(D)}_{c'c} = B_D(c+D-3c')$ is the weight of the step from source $c$ to destination $c'$, and only finitely many entries are nonzero in each row and column because $0 \le c+D-3c' \le 2D$.430431\emph{Step 3: contraction.} This is \cref{lem:carry}, proved below. It gives an invariant finite state set, the core $C_D$, containing the start state $0$, so $a_D(L) = \bigl(M_C^L\bigr)_{00}$ where $M_C$ is the finite matrix $M^{(D)}$ restricted to $C_D$.432433\emph{Step 4: halving.} This is \cref{lem:reflect}, proved below. The vector $u_L = M_C^L e_0$ lies, for every $L$, in the $+1$ eigenspace $E$ of the reflection $R$ inside $\mathbb{R}^{C_D}$, and $E$ is $M_C$-invariant because $R$ and $M_C$ commute. A basis of $E$ is $v_0 = e_0$ and $v_k = e_k + e_{-k}$ for $1 \le k \le r_D$, so $\dim E = r_D + 1 = \lceil D/2 \rceil = n_D$. Let $M_{\mathrm{even}}^{(D)}$ be the matrix of $M_C|_E$ in that basis and $\chi_D$ its characteristic polynomial, of degree $n_D$. By Cayley--Hamilton, $\chi_D(M_C|_E) = 0$, hence for every $L \ge 0$434\[435\sum_{k=0}^{n_D} \chi_{D,k}\, u_{L+n_D-k} = \chi_D(M_C|_E)\, u_L = 0,436\]437where $\chi_D(\lambda) = \sum_k \chi_{D,k}\lambda^{n_D-k}$ with $\chi_{D,0} = 1$. Applying the linear functional $w \mapsto w_0$ to this vector identity gives438\[439\sum_{k=0}^{n_D} \chi_{D,k}\, a_D(L+n_D-k) = 0 ,440\]441a constant-coefficient linear recurrence of order $n_D = \lceil D/2 \rceil$ satisfied by $a_D$. The coefficients are integers because $M_{\mathrm{even}}^{(D)}$ has integer entries.442\end{proof}443444\begin{proof}[Proof of \cref{lem:carry}]445If $M^{(D)}_{c'c} \ne 0$ then $c' = (c+D-s)/3$ for some $0 \le s \le 2D$, so $c+D-s$ lies in $[c-D, c+D]$ and446\[447\lvert c' \rvert \le \frac{\lvert c \rvert + D}{3}.448\]449Suppose $\lvert c \rvert \le r_D = \lfloor (D-1)/2 \rfloor \le (D-1)/2$. Then450\[451\lvert c' \rvert \le \frac{(D-1)/2 + D}{3} = \frac{3D-1}{6} = \frac{D}{2} - \frac{1}{6} < \frac{D}{2}.452\]453Since $c'$ is an integer, $\lvert c' \rvert < D/2$ forces $\lvert c' \rvert \le \lfloor (D-1)/2 \rfloor = r_D$, whether $D$ is even or odd: at even $D$ an integer below $D/2$ is at most $D/2 - 1 = r_D$, and at odd $D$ it is at most $(D-1)/2 = r_D$. So the core is forward invariant. The start state $c = 0$ lies in the core because $r_D \ge 0$ for $D \ge 2$, and induction on the number of steps finishes the claim.454\end{proof}455456\begin{proof}[Proof of \cref{prop:reach}]457Containment in the core is \cref{lem:carry}. For the reverse, note first that every coefficient $B_D(s)$ with $0 \le s \le 2D$ is positive, by the closed forms of \cref{lem:poly}: $B_D(2k) = \binom{D}{k} > 0$ for $0 \le k \le D$ and $B_D(2k+1) = D\binom{D-1}{k} > 0$ for $0 \le k \le D-1$. Hence from carry $c$ the one-step reachable set is exactly the set of integers in $[(c-D)/3,\, (c+D)/3]$, an interval that always contains $0$ since $\lvert c \rvert \le r_D < D$. If the states reachable in at most $t$ steps include the interval $[-m, m]$, then one more step reaches every integer in $[-(m+D)/3,\, (m+D)/3]$, because the one-step intervals of consecutive sources overlap. So the reachable radius climbs as $m \mapsto \lfloor (m+D)/3 \rfloor$, starting from $\lfloor D/3 \rfloor$; while $m \le r_D - 1 \le (D-3)/2$ we have $m + D \ge 3m + 3$, so each step gains at least $1$, and by \cref{lem:carry} the climb stops exactly at $r_D$. Irreducibility follows since every state reaches $0$ in one step and $0$ reaches every state; aperiodicity from the loop $M^{(D)}_{00} = B_D(D) > 0$.458\end{proof}459460\begin{proof}[Proof of \cref{lem:reflect}]461The involution $v \mapsto (2,\dots,2)-v$ of $\{0,1,2\}^D$ preserves the number of coordinates equal to $1$, so it permutes $F_D$ and sends digit sum $s$ to $2D-s$. Hence $B_D(s) = B_D(2D-s)$ for all $s$. Now462\begin{multline*}463M^{(D)}_{-c',-c} = B_D(-c+D+3c') = B_D\bigl(2D - (c+D-3c')\bigr) \\464= B_D(c+D-3c') = M^{(D)}_{c'c},465\end{multline*}466which is exactly $RM^{(D)}R = M^{(D)}$, and $R^2 = I$ gives $RM^{(D)} = M^{(D)}R$. Clearly $Re_0 = e_0$. Therefore $R\,(M^{(D)})^L e_0 = (M^{(D)})^L R e_0 = (M^{(D)})^L e_0$ for every $L$, so the whole orbit lies in the $+1$ eigenspace of $R$. Restricted to the core, that eigenspace consists of the vectors with $w_c = w_{-c}$, spanned by $e_0$ and $e_k + e_{-k}$ for $1 \le k \le r_D$; these $r_D + 1 = \lceil D/2 \rceil$ vectors are visibly independent.467\end{proof}468469\begin{proof}[Proof of \cref{lem:omega}]470Since $\omega^2 + \omega + 1 = 0$ we have $1 + \omega^2 = -\omega$ and $1 + D\omega + \omega^2 = (D-1)\omega$. Substituting into \cref{lem:poly},471\[472P_D(\omega) = (-\omega)^{D-1}(D-1)\omega = (-1)^{D-1}\omega^D (D-1),473\]474and $\lvert \omega \rvert = 1$ gives $\lvert P_D(\omega) \rvert = D-1$.475\end{proof}476477\begin{proof}[Proof of \cref{lem:classes}]478Roots of unity filter: for $j \in \{0,1,2\}$,479\[480\sum_{s \equiv j} B_D(s) = \frac{1}{3}\sum_{m=0}^{2} \omega^{-jm} P_D(\omega^m) = \frac{f_D}{3} + \frac{2}{3}\,\mathrm{Re}\bigl( \omega^{-j} P_D(\omega) \bigr),481\]482using $P_D(1) = f_D$ and $P_D(\overline{\omega}) = \overline{P_D(\omega)}$ because $P_D$ has real coefficients. By \cref{lem:omega}, $\omega^{-j}P_D(\omega) = (-1)^{D-1}(D-1)\,\omega^{D-j}$, whose real part is $(-1)^{D-1}(D-1)$ when $j \equiv D \pmod 3$ and $-(-1)^{D-1}(D-1)/2$ otherwise, since $\mathrm{Re}\,\omega^k$ is $1$ for $k \equiv 0$ and $-1/2$ otherwise. Substituting $\varepsilon_D = (D-1)(-1)^{D-1}/3$ gives $f_D/3 + 2\varepsilon_D$ in the distinguished class and $f_D/3 - \varepsilon_D$ in each of the other two.483\end{proof}484485\begin{proof}[Proof of \cref{prop:offcentre}]486By Step 1 and Step 2 of the proof of \cref{thm:order} with target $T_L(m)$, the off-centre census is $a_D(L;m) = \bigl((M^{(D)})^L\bigr)_{0m} = e_0^{\top}(M^{(D)})^L e_m$: the source state is the offset $m$ and the accepted final state is $0$. By \cref{lem:reflect}, $R$ commutes with $M^{(D)}$, is symmetric, and fixes $e_0$, so487\[488a_D(L;-m) = e_0^{\top}(M^{(D)})^L R e_m = (Re_0)^{\top}(M^{(D)})^L e_m = a_D(L;m).489\]490Hence $a_D(L;m) = e_0^{\top}(M^{(D)})^L w$ with $w = \tfrac12 (e_m + e_{-m})$. Because $\lvert m \rvert \le r_D$, the vector $w$ lies in the $+1$ eigenspace $E$ of $R$ inside the core, which by \cref{lem:carry} and \cref{lem:reflect} is invariant under $M_C$ and has dimension $n_D$. So $(M_C)^L w \in E$ for all $L$, and Cayley--Hamilton applied to $M_C|_E$ annihilates the vector sequence exactly as in Step 4; taking the coordinate functional $w \mapsto w_0$ gives the same recurrence with the same coefficients.491\end{proof}492493\begin{proof}[Proof of \cref{cor:degree}]494By \cref{thm:order}, $a_D(L) = \sum_i p_i(L) \lambda_i^L$ where the $\lambda_i$ are the roots of $\chi_D$ and the $p_i$ are polynomials. Let $\Lambda$ be the largest modulus among the roots that actually occur with a nonzero polynomial; it is positive, because every entry of $M^{(D)}$ is nonnegative and so $a_D(L) = \bigl((M^{(D)})^L\bigr)_{00} \ge \bigl(M^{(D)}_{00}\bigr)^L = B_D(D)^L > 0$. Summing that expansion, the generating function $\sum_{L \ge 0} a_D(L) z^L$ is a rational function whose poles are exactly the points $1/\lambda_i$ over the occurring nonzero roots, so its radius of convergence is $1/\Lambda$, and Cauchy--Hadamard turns that into $\limsup_L a_D(L)^{1/L} = \Lambda$, the modulus of a root of $\chi_D$. And $\chi_D$ is a monic integer polynomial of degree $n_D = \lceil D/2 \rceil$, so every root is an algebraic number of degree at most $n_D$. If the limit defining $\dim_{\mathrm{slice}}(D)$ exists, then it equals $\log_3 \Lambda$ by taking logarithms.495\end{proof}496497\begin{remark}498\label{rem:hankel}499Exactness of the order is a Hankel condition, not an eigenvalue condition. Given the upper bound $n_D$ of \cref{thm:order}, nonvanishing of $H_D = \det\bigl[a_D(i+j+1)\bigr]_{0 \le i,j < n_D}$ forces the minimal order to be exactly $n_D$; that is \cref{fact:exact}. Indeed, if $a_D$ obeyed some recurrence $a_D(L+m) = \sum_{k=1}^{m} c_k a_D(L+m-k)$ of order $m < n_D$ for every $L \ge 0$, then for every $i \ge m$ row $i$ of that matrix would be the combination $\sum_{k=1}^{m} c_k \cdot (\text{row } i-k)$ of earlier rows, all $n_D$ rows would lie in the span of the first $m$, and $H_D$ would vanish. Only this direction holds: a vanishing $H_D$ proves nothing, as the sequence $5, 2, 4, 8, 16, \dots$ shows, whose minimal order is $2$ because $a(L+2) = 2a(L+1)$ for every $L \ge 0$ while no order-one recurrence fits, yet whose shifted Hankel determinant $\det\bigl(\begin{smallmatrix} 2 & 4 \\ 4 & 8 \end{smallmatrix}\bigr)$ is zero. Distinct eigenvalues would not suffice either: writing $a_D(L) = \sum_i \lambda_i^L\, (e_0^{\top} \Pi_i\, e_0)$ with spectral projections $\Pi_i$, exactness also needs every weight $e_0^{\top}\Pi_i e_0$ to be nonzero. The two-line counterexample is $M = \mathrm{diag}(1,2)$ with start vector $e_0 = (1,0)$: the eigenvalues are distinct and the matrix is non-derogatory, yet the sequence $e_0^{\top}M^L e_0 = 1$ has order $1$, not $2$. The Hankel determinant sees this where the eigenvalues do not, and it is what \texttt{scripts/verify.py} computes, in exact rational arithmetic.500\end{remark}501502\begin{proof}[Proof of \cref{lem:circle}]503Write $1 + e^{2i\psi} = 2\cos\psi \, e^{i\psi}$ and $1 + De^{i\psi} + e^{2i\psi} = e^{i\psi}\bigl(D + 2\cos\psi\bigr)$. Substituting into the factorisation of \cref{lem:poly},504\[505P_D\bigl(e^{i\psi}\bigr) = \bigl(2\cos\psi\bigr)^{D-1} e^{i(D-1)\psi} \cdot e^{i\psi}\bigl(D + 2\cos\psi\bigr) = e^{iD\psi}\,\Phi_D(\psi).506\]507For odd $D$ the power $D-1$ is even, so $(2\cos\psi)^{D-1} \ge 0$, and $D + 2\cos\psi \ge D - 2 > 0$ for $D \ge 3$. For even $D > 2$ the power is odd and $D + 2\cos\psi > 0$, so the sign of $\Phi_D$ is the sign of $\cos\psi$.508\end{proof}509510\begin{proof}[Proof of \cref{thm:product}]511\emph{Step 1: the sheaf census is the free-end census.} Set $N_p = D(3^p-1)/2$, an integer for every $D$ because $3^p - 1$ is even. The coordinate sums $T \equiv N_L \pmod{3^L}$ are exactly $T = N_L - 3^L c'$ over $c' \in \mathbb{Z}$, where terms with $T$ outside $[0, D(3^L-1)]$ vanish, so by Step 1 of the proof of \cref{thm:order},512\[513b_D(L) = \sum_{c' \in \mathbb{Z}} \bigl[t^{\,N_L - 3^L c'}\bigr] \prod_{j=0}^{L-1} P_D\bigl(t^{3^j}\bigr).514\]515Fix $c'$. A monomial choice from the product is a tuple $(s_0,\dots,s_{L-1})$ of digit sums with $\sum_j s_j 3^j = N_L - 3^L c'$, weighted by $\prod_j B_D(s_j)$. For $0 \le p \le L$ set $T_p = \sum_{j<p} s_j 3^j$ and $c_p = (N_p - T_p)/3^p \in \mathbb{Q}$; then $c_0 = 0$, the identity $3c_{p+1} = c_p + D - s_p$ holds over $\mathbb{Q}$, and $c_p \in \mathbb{Z}$ exactly when $T_p \equiv N_p \pmod{3^p}$. The full congruence $T \equiv N_L \pmod{3^L}$ implies every partial one: $T_p \equiv T \equiv N_L \equiv N_p \pmod{3^p}$, since $N_L - N_p = D(3^L - 3^p)/2 = 3^p N_{L-p}$. So each tuple in the class determines an integer carry path from $c_0 = 0$ with final carry $c_L = c'$, stepping with the weights of \cref{def:transfer}. Conversely, an integer carry path $0 = c_0 \to \dots \to c_L$ has $s_p = c_p + D - 3c_{p+1}$ and the same telescoping gives $T = N_L - 3^L c_L \equiv N_L \pmod{3^L}$. Hence the weighted count of tuples with final carry $c'$ is $\bigl((M^{(D)})^L\bigr)_{c'0}$, paths out of the start state $0$ (in \cref{def:transfer} the first index is the destination). By \cref{lem:carry} no path from $0$ leaves the core and $M_C$ is a submatrix of $M^{(D)}$ with equal entries, so $b_D(L) = \sum_{c' \in C_D} (M_C^L)_{c'0} = \mathbf{1}^{\top} M_C^L\, e_0$.516517\emph{Step 2: extraction.} With $\zeta = e^{2\pi i/3^L}$ and $N_L = D(3^L-1)/2$,518\[519b_D(L) = \sum_{T \equiv N_L (3^L)} \bigl[t^T\bigr] \prod_{j<L} P_D\bigl(t^{3^j}\bigr)520= \frac{1}{3^L}\sum_{m=0}^{3^L-1} \zeta^{-N_L m} \prod_{j=0}^{L-1} P_D\bigl(\zeta^{m 3^j}\bigr).521\]522523\emph{Step 3: the phases cancel.} Write $\psi_j = 2\pi m 3^j/3^L$. By \cref{lem:circle}, $P_D(\zeta^{m3^j}) = e^{iD\psi_j}\,\Phi_D(\psi_j)$, and524\[525\prod_{j=0}^{L-1} e^{iD\, 2\pi m 3^j/3^L} = \exp\Bigl( 2\pi i\, m D \tfrac{3^L-1}{2 \cdot 3^L} \Bigr) = \zeta^{\,m N_L},526\]527which cancels the extraction phase $\zeta^{-N_L m}$ exactly, leaving the stated real sum. It is a positive integer because it equals the census $b_D(L)$, and $b_D(L) \ge a_D(L) \ge B_D(D)^L \ge 1$.528\end{proof}529530\begin{proof}[Proof of \cref{cor:step}]531Sort the summation index $m$ of \cref{thm:product} by its $3$-adic valuation: $m = 0$ contributes $\Phi_D(0)^L = f_D^L$, and $m = 3^{L-k} u$ with $u \in U_k$, $1 \le k \le L$, has angles $2\pi u 3^j/3^k$ reduced mod $2\pi$: for $j \ge k$ the angle is an integer multiple of $2\pi$ and the factor is $\Phi_D(0) = f_D$; the remaining factors are $\prod_{j<k} \Phi_D(2\pi u 3^j/3^k)$. Hence $3^L b_D(L) = \sum_{k=0}^{L} f_D^{L-k} W_k$, and subtracting $f_D$ times the same identity at $L-1$ leaves $3^L b_D(L) - f_D\, 3^{L-1} b_D(L-1) = W_L$. For the innermost factor, $u 3^{k-1}/3^k = u/3$ with $3 \nmid u$, and $2\cos(2\pi/3) = 2\cos(4\pi/3) = -1$ gives $\Phi_D = (-1)^{D-1}(D-1)$ there.532\end{proof}533534\begin{proof}[Proof of \cref{thm:sign}]535\emph{The spectral radius is the dominant root.} $M_{\mathrm{even}}^{(D)}$ is a square matrix, so its spectral radius is the maximal modulus of the roots of $\chi_D$; that it is itself a root is Perron--Frobenius for nonnegative matrices (no irreducibility required): a nonnegative square matrix has its spectral radius as an eigenvalue, with an entrywise nonnegative eigenvector \cite[Theorem 8.3.1]{hornjohnson}. Nonnegativity of $M_{\mathrm{even}}^{(D)}$ holds because its entries are $B_D(D-3i)$ and $B_D(D+j-3i) + B_D(D-j-3i)$.536537\emph{Step 1: the census grows at least like $(f_D/3)^L$.} For odd $D$, $\Phi_D \ge 0$ everywhere by \cref{lem:circle}, so every $W_k \ge 0$, and \cref{cor:step} gives $3^L b_D(L) = \sum_k f_D^{L-k} W_k \ge f_D^L$, i.e. $b_D(L) \ge (f_D/3)^L$ for every $L$.538539\emph{Step 2: the growth is bounded by the spectral radius of the core.} $b_D(L)$ is the $0$-column sum of $M_C^L$, so $b_D(L) \le \lVert M_C^L \rVert_1$, the maximal column sum. By Gelfand's formula $\lVert M_C^L \rVert_1^{1/L} \to \rho(M_C)$, so $\rho(M_C) \ge \limsup_L b_D(L)^{1/L} \ge f_D/3$.540541\emph{Step 3: the core radius is $\rho_D$.} By Perron--Frobenius again there is $v \ge 0$, $v \ne 0$, with $M_C v = \rho(M_C)\, v$. Its even part $w = (v + Rv)/2$ is nonnegative, and nonzero: $w = 0$ would force $Rv = -v$ with both sides nonnegative, hence $v = 0$. Since $R$ commutes with $M_C$ (\cref{lem:reflect}), $w$ is an eigenvector of $M_C$ at $\rho(M_C)$ lying in the $+1$ eigenspace $E$, so $\rho(M_C)$ is an eigenvalue of $M_{\mathrm{even}}^{(D)} = M_C|_E$ and $\rho_D \ge \rho(M_C)$. Conversely every eigenvalue of $M_C|_E$ is an eigenvalue of $M_C$, so $\rho_D \le \rho(M_C)$. Hence $\rho_D = \rho(M_C) \ge f_D/3$.542543\emph{Step 4: strictness off the residue class $D \equiv 1$.} The matrix $J = f_D I - 3 M_{\mathrm{even}}^{(D)}$ has integer entries, and544\[545\chi_D(f_D/3) = \det\bigl( (f_D/3) I - M_{\mathrm{even}}^{(D)} \bigr) = 3^{-n_D} \det J .546\]547Mod $3$ the term $3 M_{\mathrm{even}}^{(D)}$ vanishes, so $\det J \equiv f_D^{\,n_D} \pmod 3$. Since $f_D = 2^{D-1}(D+2)$, $3 \mid f_D$ exactly when $D \equiv 1 \pmod 3$; for $D \not\equiv 1 \pmod 3$ we get $\det J \not\equiv 0 \pmod 3$, so $\chi_D(f_D/3) \ne 0$ and $f_D/3$ is not an eigenvalue. Since $\rho_D \ge f_D/3$ and $\rho_D$ is an eigenvalue, $\rho_D > f_D/3$.548\end{proof}549550\begin{proof}[Proof of \cref{thm:pin}]551By \cref{lem:carry} the core is forward invariant, so no column of $M_C$ loses mass under the restriction: the column of $M_C$ at source $c$ sums to $\sum_{s \equiv c + D \ (\mathrm{mod}\ 3)} B_D(s)$, which by \cref{lem:classes} equals $f_D/3 + 2\varepsilon_D$ when $c \equiv 0 \pmod 3$ and $f_D/3 - \varepsilon_D$ otherwise, with $\varepsilon_D = (D-1)(-1)^{D-1}/3$. The spectral radius of a nonnegative matrix lies between its smallest and largest column sums \cite[Theorem 8.1.22]{hornjohnson}, and $\rho(M_C) = \rho_D$ by Step 3 of the proof of \cref{thm:sign}. Reading off the two values by the parity of $\varepsilon_D$ gives the refined brackets, and both are contained in $\lvert \rho_D - f_D/3 \rvert \le 2(D-1)/3$. Finally $2(D-1)/3$ against $f_D/3 = 2^{D-1}(D+2)/3$ gives the relative error $O(D^2 2^{-D})$, and $\log_3$ of $1 + O(D^2 2^{-D})$ tends to $0$.552\end{proof}553554\begin{proof}[Proof of \cref{prop:mass}]555By \cref{prop:reach}, $M_C$ is irreducible, so its Perron vector $v \ge 0$ is unique up to scale and $M_C v = \rho_D v$ with $\rho(M_C) = \rho_D$ as above. Multiply by the all-ones row vector: $\rho_D \sum_c v_c = \mathbf{1}^{\top} M_C v = \sum_c \sigma(c)\, v_c$, where $\sigma(c)$ is the column sum at $c$, two-valued as in the proof of \cref{thm:pin}. Writing $\sigma(c) = f_D/3 + \varepsilon_D\,(3 \cdot \mathbf 1[3 \mid c] - 1)$ and dividing by $\sum_c v_c$,556\[557\rho_D = \frac{f_D}{3} + \varepsilon_D\,(3p_D - 1),558\]559which is the identity. Multiplying $\rho_D - f_D/3 = (-1)^{D-1}(D-1)(p_D - \tfrac13)$ by $(-1)^{D+1}$ gives the sign statement, and the equivalence with the sign law follows because $(-1)^{D+1}\operatorname{sgn}(\rho_D - f_D/3) = \operatorname{sgn}(p_D - \tfrac13)$ at every $D$.560\end{proof}561562\section{Reproducibility}563564Two scripts, plain Python~3, no dependencies, no arguments, every path relative to the lane; run them from the lane root.565566\texttt{scripts/verify.py} runs in about four seconds and dies loudly on the first mismatch, naming what it got and what it wanted. It covers, in order: the factorisation of \cref{lem:poly} and its closed entry form against brute-force enumeration of all $3^D$ digit tuples for $2 \le D \le 8$; agreement of three independent generators, brute enumeration, the substitution product $\prod_{j<L} P_D(t^{3^j})$, and the untrimmed carry automaton on states $-D,\dots,D$, for $2 \le D \le 7$ and $0 \le L \le 4$; the exact reachable carry set by breadth-first search for $2 \le D \le 24$, corroborating \cref{prop:reach}; nonvanishing of the exact rational Hankel determinant for $2 \le D \le 24$, which is \cref{fact:exact} and \cref{rem:hankel}; the two trace formulas of \cref{fact:trace} for $2 \le D \le 24$; the fill and the three residue-class sums of \cref{lem:classes} for $2 \le D \le 12$; \cref{prop:offcentre} for $3 \le D \le 7$ at every offset $1 \le m \le r_D$ over windows through $L = 2n_D+3$, together with the sharpness witness of \cref{rem:sharp} at $D = 4$, $m = 2$; the anchors of \cref{fact:anchor} at $D = 2,3,4,5,6$, including the matrix, its characteristic polynomial, the census, and $\log_3\bigl((9+\sqrt{33})/2\bigr)$ to six places; the spectral ratios of \cref{con:gap} for $4 \le D \le 20$, whose monotonicity and agreement with $(D+2)/(D-2)$ are asserted over $6 \le D \le 20$; and both sign checks of the sign law, floating-point roots for $2 \le D \le 20$ and the exact rational sign of $\chi_D(f_D/3)$ for $2 \le D \le 30$, now corroboration of \cref{thm:sign} on the odd side and evidence for \cref{con:sign} on the even side. The new results are covered by five further blocks: the sheaf census of \cref{def:sheaf} computed three independent ways (free-end transfer count, coefficient-class sums of the substitution product in exact integers, and the trigonometric sum of \cref{thm:product} in floating point) for $2 \le D \le 7$; the step identity and parity factor of \cref{cor:step} in exact integers; strict positivity of every $W_k$, $k \le 8$, at every odd $D \le 25$, the mechanism of \cref{thm:sign}; nonvanishing of the exact integer $\det(f_D I - 3M_{\mathrm{even}}^{(D)})$ for $2 \le D \le 80$ together with its congruence to $f_D^{n_D}$ mod $3$, which is \cref{fact:signodd}; and the sign of $W_2$ at even $6 \le D \le 24$, the obstruction quoted in \cref{con:sign}; and the mass identity of \cref{prop:mass} together with both parity-refined brackets of \cref{thm:pin}, by power iteration on the core matrix for $2 \le D \le 20$. Characteristic polynomials are computed by the Faddeev--LeVerrier recursion over exact rationals; only the spectral checks, the product-formula cross-check and the $D = 3$ dimension use floating point, and the spectral checks do so on integer polynomials rescaled by $f_D/3$ so that all roots have modulus near one.567568\texttt{scripts/figure.py} runs instantly and redraws \texttt{figures/order.svg}, the README's copy of \cref{fig:order}, from the same two formulas $2D+1$ and $\lceil D/2 \rceil$.569570The $D = 3$ census is OEIS \href{https://oeis.org/A299916}{A299916} \cite{oeis}. The ladders at $D = 4, 5, 6$ are not in the OEIS to the author's knowledge; no novelty search was run, and they are offered here only as computed values.571572\section*{Acknowledgments}573574This paper was developed and verified in collaboration with Claude (Anthropic). The author takes sole responsibility for every claim.575576\begin{thebibliography}{9}577578\bibitem{abelstars}579Z.~Abel, \emph{Seeing Stars}, 2012. The hexagon-triangle substitution and the dimension $1.8184$, stated there as a computation rather than a proof. \url{http://blog.zacharyabel.com/2012/02/seeing-stars/}580581\bibitem{allouche}582J.-P. Allouche and J.~Shallit, \emph{The ring of $k$-regular sequences}, Theoret. Comput. Sci. \textbf{98} (1992), 163--197. \url{https://doi.org/10.1016/0304-3975(92)90001-V}583584\bibitem{cook}585J.~D. Cook, \emph{Code to slice open a Menger sponge}, 2011. \url{https://www.johndcook.com/blog/2011/08/30/slice-a-menger-sponge/}586587\bibitem{evenhalf}588C.~Mitchener, \emph{The Even Half of the Slice Sign Law}, 2026. The companion paper: the even half of \cref{con:sign} proved at bases $3$ and $5$, and the strictness gap reduced to a $2$-adic valuation bound. \url{https://github.com/carlomitchener/carlomitchener/tree/main/research/slice-sign-even-half}589590\bibitem{hart}591G.~Hart, \emph{Mathematical Impressions: The Surprising Menger Sponge Slice}, Simons Foundation, 2012. \url{https://www.simonsfoundation.org/2012/12/10/mathematical-impressions-the-surprising-menger-sponge-slice/}592593\bibitem{hocking2023}594G.~Hocking, \emph{Three-Dimensional Diagonal Cross-Sections of Four-Dimensional Menger Sponges}, Bridges 2023 Conference Proceedings, 291--294. \url{https://archive.bridgesmathart.org/2023/bridges2023-291.pdf}595596\bibitem{holte}597J.~M. Holte, \emph{Carries, combinatorics, and an amazing matrix}, Amer. Math. Monthly \textbf{104} (1997), no.~2, 138--149. \url{https://doi.org/10.2307/2974981}598599\bibitem{hornjohnson}600R.~A. Horn and C.~R. Johnson, \emph{Matrix Analysis}, 2nd ed., Cambridge University Press, 2013. Theorem 8.3.1: a nonnegative square matrix has its spectral radius as an eigenvalue, with an entrywise nonnegative eigenvector; no irreducibility is assumed.601602\bibitem{marstrand}603J.~M. Marstrand, \emph{Some fundamental geometrical properties of plane sets of fractional dimensions}, Proc. London Math. Soc. \textbf{s3-4} (1954), 257--302. \url{https://doi.org/10.1112/plms/s3-4.1.257}604605\bibitem{mattila}606P.~Mattila, \emph{Hausdorff dimension, orthogonal projections and intersections with planes}, Ann. Acad. Sci. Fenn. Ser. A I Math. \textbf{1} (1975), 227--244. \url{https://doi.org/10.5186/aasfm.1975.0110}607608\bibitem{oeis}609OEIS Foundation Inc., Entry \emph{A299916}, The On-Line Encyclopedia of Integer Sequences. Terms $1, 6, 42, 306, 2250, 16578, 122202$ and signature $(9,-12)$; its Menger reading is a user comment counting star-shaped holes. \url{https://oeis.org/A299916}610611\bibitem{perezduarte}612S.~P\'erez-Duarte, \emph{Slice of Menger}, 2007. The first published render of the diagonal cut. \url{https://www.flickr.com/photos/sbprzd/1432723128/}613614\end{thebibliography}615616\end{document}