% PREAMBLE \documentclass[11pt]{amsart} \usepackage[T1]{fontenc} \usepackage{lmodern} \usepackage{microtype} \usepackage{amsmath,amssymb} \usepackage{mathtools} \usepackage{graphicx} \usepackage{booktabs} \usepackage{tikz} \usepackage[colorlinks=true,linkcolor=bluedark,citecolor=bluedark,urlcolor=bluedark]{hyperref} \usepackage[capitalize]{cleveref} % COLORS \definecolor{black}{HTML}{000000} \definecolor{white}{HTML}{FFFFFF} \definecolor{red}{HTML}{FF3D40} \definecolor{redlight}{HTML}{FF9D95} \definecolor{reddark}{HTML}{A80016} \definecolor{orange}{HTML}{FF8F2C} \definecolor{orangelight}{HTML}{FFC093} \definecolor{orangedark}{HTML}{A25400} \definecolor{yellow}{HTML}{FFD100} \definecolor{yellowlight}{HTML}{FFE591} \definecolor{yellowdark}{HTML}{9E8100} \definecolor{green}{HTML}{32CC58} \definecolor{greenlight}{HTML}{5EEE79} \definecolor{greendark}{HTML}{007F2C} \definecolor{mint}{HTML}{00D1BB} \definecolor{mintlight}{HTML}{48EFD8} \definecolor{mintdark}{HTML}{008173} \definecolor{teal}{HTML}{00CAD8} \definecolor{teallight}{HTML}{48E9F7} \definecolor{tealdark}{HTML}{007C85} \definecolor{cyan}{HTML}{1EC9F3} \definecolor{cyanlight}{HTML}{86E2FF} \definecolor{cyandark}{HTML}{007C98} \definecolor{blue}{HTML}{008CFF} \definecolor{bluelight}{HTML}{84BDFF} \definecolor{bluedark}{HTML}{00559F} \definecolor{indigo}{HTML}{6768FA} \definecolor{indigolight}{HTML}{9EA9FF} \definecolor{indigodark}{HTML}{3C2ABC} \definecolor{purple}{HTML}{D332E9} \definecolor{purplelight}{HTML}{F08AFF} \definecolor{purpledark}{HTML}{870097} \definecolor{pink}{HTML}{FF325A} \definecolor{pinklight}{HTML}{FF9A9F} \definecolor{pinkdark}{HTML}{A50030} \definecolor{brown}{HTML}{B18462} \definecolor{brownlight}{HTML}{DFAF8C} \definecolor{browndark}{HTML}{754C2B} \definecolor{gray}{HTML}{8E8E93} \definecolor{graylight}{HTML}{BABABF} \definecolor{graydark}{HTML}{56565A} % COLORS END \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{fact}[theorem]{Fact} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \title{Menger Diagonal Slices: A Recurrence of Order $\lceil D/2 \rceil$} \author{Carlo Mitchener} \address{MrlyProd, Inc.} \email{carlo.mitchener@gmail.com} \date{First published 2026-08-23, revised 2026-09-08} % PAPER \begin{document} \begin{abstract} Slice 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. \end{abstract} % TITLE PAGE \makeatletter \global\let\titledate\@date \global\let\paperabstract\@setabstracta \global\let\@date\@empty \global\let\@setabstract\relax \makeatother \maketitle \begin{center} \normalfont\footnotesize MrlyProd, Inc.\\ \titledate \end{center} \vspace*{\stretch{1}} \begin{center} \includegraphics[width=0.8\textwidth]{figures/avatar-light.png} \end{center} \vspace*{\stretch{1.25}} \newpage \paperabstract % BODY \section{Introduction} Take 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}. This 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. \begin{figure}[ht] \centering \begin{tikzpicture}[scale=1.0] \foreach \d in {2,...,14}{ \pgfmathtruncatemacro{\free}{2*\d+1} \pgfmathtruncatemacro{\sharpo}{ceil(\d/2)} \fill[black!22] (0.46*\d-0.15,0) rectangle ++(0.13,0.052*\free); \fill[blue!45!black] (0.46*\d+0.01,0) rectangle ++(0.13,0.052*\sharpo); \node[font=\scriptsize] at (0.46*\d,-0.22) {\d}; } \draw[black!70] (0.70,0) -- (6.90,0); \fill[black!22] (0.72,1.72) rectangle ++(0.13,0.11); \node[font=\scriptsize,anchor=west] at (0.95,1.78) {order $2D+1$, free}; \fill[blue!45!black] (3.55,1.72) rectangle ++(0.13,0.11); \node[font=\scriptsize,anchor=west] at (3.78,1.78) {order $\lceil D/2 \rceil$, this paper}; \end{tikzpicture} \caption{How much memory the slice census needs, by dimension $D$.} \label{fig:order} \end{figure} The 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, \[ P_D(t) = (1+t^2)^{D-1}(1+Dt+t^2), \] which 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 $\lfloor (D-1)/2 \rfloor + 1 = \lceil D/2 \rceil$. \medskip \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}. \medskip \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}). \medskip \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. \section{Definitions} \begin{definition} \label{def:design} The \emph{$D$-dimensional Menger design} is \[ F_D = \bigl\{ v \in \{0,1,2\}^D : \#\{ i : v_i = 1 \} \le 1 \bigr\}, \] the digit vectors with at most one middle coordinate. Its \emph{fill} is $\lvert F_D \rvert$. \end{definition} \begin{definition} \label{def:level} For $L \ge 0$ 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} is \[ S_L^{(D)} = \bigl\{ x : v^{(j)} \in F_D \text{ for every } j \bigr\}, \] and the \emph{central diagonal census} is \[ a_D(L) = \#\Bigl\{ x \in S_L^{(D)} : \textstyle\sum_i x_i = D(3^L-1)/2 \Bigr\}, \qquad a_D(0) = 1. \] \end{definition} The 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. \begin{definition} \label{def:poly} The \emph{digit polynomial} is \[ P_D(t) = \sum_{v \in F_D} t^{v_1+\dots+v_D} = \sum_{s=0}^{2D} B_D(s)\, t^s . \] \end{definition} \begin{definition} \label{def:transfer} The \emph{transfer matrix} $M^{(D)}$ acts on carry states $c \in \mathbb{Z}$ by \[ M^{(D)}_{c'c} = B_D(c + D - 3c'), \] which 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. \end{definition} \section{Results} \begin{theorem} \label{thm:order} For 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 $\lceil D/2 \rceil$. \end{theorem} \begin{lemma} \label{lem:poly} For every $D \ge 1$, $P_D(t) = (1+t^2)^{D-1}(1+Dt+t^2)$, the fill is $\lvert F_D \rvert = 2^{D-1}(D+2)$, and \[ B_D(2k) = \binom{D}{k}, \qquad B_D(2k+1) = D\binom{D-1}{k}, \] with the convention that a binomial coefficient with an out-of-range lower index is zero. \end{lemma} \begin{lemma} \label{lem:carry} Every 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 $\lvert c' \rvert \le r_D$. \end{lemma} \begin{lemma} \label{lem:reflect} $P_D[s] = P_D[2D-s]$ 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 $\lceil D/2 \rceil$. \end{lemma} \begin{lemma} \label{lem:omega} Let $\omega = e^{2\pi i/3}$. Then $P_D(\omega) = (-1)^{D-1}\omega^D (D-1)$, so $\lvert P_D(\omega) \rvert = D-1$ for every $D \ge 1$. \end{lemma} \begin{lemma} \label{lem:classes} Write $\varepsilon_D = (D-1)(-1)^{D-1}/3$ and $f_D = 2^{D-1}(D+2)$. For $j \in \{0,1,2\}$, \[ \sum_{s \equiv j \ (\mathrm{mod}\ 3)} B_D(s) = \begin{cases} f_D/3 + 2\varepsilon_D, & j \equiv D \pmod 3, \\ f_D/3 - \varepsilon_D, & \text{otherwise}. \end{cases} \] \end{lemma} \begin{proposition} \label{prop:offcentre} Fix 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 $L$. \end{proposition} \begin{remark} \label{rem:sharp} The 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}. \end{remark} \begin{corollary} \label{cor:degree} $\limsup_L a_D(L)^{1/L}$ 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. \end{corollary} \begin{fact} \label{fact:exact} For 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}. \end{fact} \begin{proposition} \label{prop:reach} For 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. \end{proposition} \begin{fact} \label{fact:trace} For every $D$ with $2 \le D \le 24$, \[ \operatorname{tr} M_{\mathrm{even}}^{(D)} = \begin{cases} 3 \cdot 2^{D-2} - 1, & D \text{ even}, \\ 3D \cdot 2^{D-3}, & D \text{ odd}. \end{cases} \] The 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}. \end{fact} \begin{fact} \label{fact:anchor} At $D = 3$, $M_{\mathrm{even}}^{(3)} = \bigl(\begin{smallmatrix} 6 & 6 \\ 1 & 3 \end{smallmatrix}\bigr)$ 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 $\prod_{j 2$, $\Phi_D(\psi) < 0$ exactly when $\cos\psi < 0$. \end{lemma} \begin{theorem} \label{thm:product} For every $D \ge 2$ and $L \ge 0$, \[ b_D(L) = \sum_{c \in C_D} \bigl( M_C^L \bigr)_{c0} = \frac{1}{3^L} \sum_{m=0}^{3^L-1} \; \prod_{j=0}^{L-1} \Phi_D\!\Bigl( \frac{2\pi m 3^j}{3^L} \Bigr). \] In particular the right-hand side, a sum of $3^L$ products of cosines, is a positive integer. \end{theorem} \begin{corollary} \label{cor:step} For $k \ge 1$ let $U_k = \{ 1 \le u < 3^k : 3 \nmid u \}$ and \[ W_k = \sum_{u \in U_k} \prod_{j=0}^{k-1} \Phi_D\!\Bigl( \frac{2\pi u 3^j}{3^k} \Bigr). \] Then $3^L b_D(L) = \sum_{k=0}^{L} f_D^{\,L-k}\, W_k$ with $W_0 = 1$, equivalently \[ 3\, b_D(k) - f_D\, b_D(k-1) = 3^{1-k}\, W_k \qquad (k \ge 1). \] Moreover the innermost factor of every product in $W_k$ is $\Phi_D(2\pi u/3)$ with $3 \nmid u$, and \[ \Phi_D(2\pi/3) = \Phi_D(4\pi/3) = (-1)^{D-1}(D-1), \] so $W_k = (-1)^{D-1}(D-1)\, V_k$ with $V_k = \sum_{u \in U_k} \prod_{j=0}^{k-2} \Phi_D\bigl(2\pi u 3^j/3^k\bigr)$. \end{corollary} \begin{theorem} \label{thm:sign} Let $\rho_D$ 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 $D \ge 3$, \[ \rho_D \;\ge\; \frac{f_D}{3}, \] and if moreover $D \not\equiv 1 \pmod 3$, the inequality is strict: \[ \rho_D \;>\; \frac{f_D}{3} \qquad (D \text{ odd},\ D \equiv 0, 2 \pmod 3). \] This is the odd half of \cref{con:sign} for two of the three residue classes. \end{theorem} \begin{fact} \label{fact:signodd} For 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. \end{fact} \begin{theorem}[the pinning] \label{thm:pin} For every $D \ge 2$, \[ \Bigl\lvert \rho_D - \frac{f_D}{3} \Bigr\rvert \;\le\; \frac{2(D-1)}{3}, \] refined 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 $\rho_D = (f_D/3)\bigl(1 + O(D^2 2^{-D})\bigr)$, so \[ \log_3 \rho_D - \bigl( \log_3 f_D - 1 \bigr) \longrightarrow 0 : \] whatever 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 $D$. \end{theorem} \begin{proposition}[the mass identity] \label{prop:mass} Let $v$ 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, \begin{gather*} 3\rho_D \;=\; f_D + (-1)^{D-1}(D-1)\,\bigl(3p_D - 1\bigr), \\ \text{i.e.}\qquad \operatorname{sgn}\Bigl(\rho_D - \frac{f_D}{3}\Bigr) = (-1)^{D+1} \operatorname{sgn}\Bigl(p_D - \frac13\Bigr). \end{gather*} Consequently 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. \end{proposition} \begin{conjecture} \label{con:sign} For every even $D \ge 2$, and for every odd $D \equiv 1 \pmod 3$ beyond the verified range, \[ \operatorname{sgn}\bigl( \rho_D - f_D/3 \bigr) = (-1)^{D+1}, \qquad f_D = 2^{D-1}(D+2). \] The 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. Evidence: 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}. The 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 $p_D > 1/3$. Failure 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. \end{conjecture} \begin{remark}[note added, August 2026] \label{rem:noteadded} The 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. \end{remark} \begin{remark} \label{rem:rate} The size of the excess appears to be governed by the same tower of angles. Numerically, \[ \rho_D - \frac{f_D}{3} \;=\; (-1)^{D+1}\, \frac{2(D-1)}{3} \prod_{k \ge 2} \Bigl( \cos \frac{2\pi}{3^k} \Bigr)^{\!D-1} \frac{D + 2\cos(2\pi/3^k)}{D+2} \cdot \bigl(1 + o(1)\bigr), \] with 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. \end{remark} \begin{conjecture} \label{con:gap} There is no uniform spectral gap: writing $\lambda_2$ for a second-largest eigenvalue of $M_{\mathrm{even}}^{(D)}$ in modulus, \[ \frac{\rho_D}{\lvert \lambda_2 \rvert} = \frac{D+2}{D-2} + O(D^{-3}) \longrightarrow 1 . \] Evidence: 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 $6 \le D \le 20$. Failure 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. This 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. \end{conjecture} \section{Proofs} \label{sec:proofs} \begin{proof}[Proof of \cref{lem:poly}] Split $F_D$ 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 $F_D$, so \[ P_D(t) = (1+t^2)^D + Dt(1+t^2)^{D-1} = (1+t^2)^{D-1}\bigl(1+Dt+t^2\bigr). \] Setting $t = 1$ 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 $D\binom{D-1}{k}$. \end{proof} \begin{proof}[Proof of \cref{thm:order}] \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 $\sum_i x_i = \sum_{j 0$ 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 $M^{(D)}_{00} = B_D(D) > 0$. \end{proof} \begin{proof}[Proof of \cref{lem:reflect}] The 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 $s$. Now \begin{multline*} M^{(D)}_{-c',-c} = B_D(-c+D+3c') = B_D\bigl(2D - (c+D-3c')\bigr) \\ = B_D(c+D-3c') = M^{(D)}_{c'c}, \end{multline*} which 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. \end{proof} \begin{proof}[Proof of \cref{lem:omega}] Since $\omega^2 + \omega + 1 = 0$ we have $1 + \omega^2 = -\omega$ and $1 + D\omega + \omega^2 = (D-1)\omega$. Substituting into \cref{lem:poly}, \[ P_D(\omega) = (-\omega)^{D-1}(D-1)\omega = (-1)^{D-1}\omega^D (D-1), \] and $\lvert \omega \rvert = 1$ gives $\lvert P_D(\omega) \rvert = D-1$. \end{proof} \begin{proof}[Proof of \cref{lem:classes}] Roots of unity filter: for $j \in \{0,1,2\}$, \[ \sum_{s \equiv j} B_D(s) = \frac{1}{3}\sum_{m=0}^{2} \omega^{-jm} P_D(\omega^m) = \frac{f_D}{3} + \frac{2}{3}\,\mathrm{Re}\bigl( \omega^{-j} P_D(\omega) \bigr), \] using $P_D(1) = f_D$ 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. \end{proof} \begin{proof}[Proof of \cref{prop:offcentre}] By 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 $e_0$, so \[ a_D(L;-m) = e_0^{\top}(M^{(D)})^L R e_m = (Re_0)^{\top}(M^{(D)})^L e_m = a_D(L;m). \] Hence $a_D(L;m) = e_0^{\top}(M^{(D)})^L w$ 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. \end{proof} \begin{proof}[Proof of \cref{cor:degree}] By \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. \end{proof} \begin{remark} \label{rem:hankel} Exactness 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. \end{remark} \begin{proof}[Proof of \cref{lem:circle}] Write $1 + e^{2i\psi} = 2\cos\psi \, e^{i\psi}$ 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}, \[ P_D\bigl(e^{i\psi}\bigr) = \bigl(2\cos\psi\bigr)^{D-1} e^{i(D-1)\psi} \cdot e^{i\psi}\bigl(D + 2\cos\psi\bigr) = e^{iD\psi}\,\Phi_D(\psi). \] For 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 $\cos\psi$. \end{proof} \begin{proof}[Proof of \cref{thm:product}] \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}, \[ b_D(L) = \sum_{c' \in \mathbb{Z}} \bigl[t^{\,N_L - 3^L c'}\bigr] \prod_{j=0}^{L-1} P_D\bigl(t^{3^j}\bigr). \] Fix $c'$. 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 $T_p = \sum_{j f_D/3$. \end{proof} \begin{proof}[Proof of \cref{thm:pin}] By \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 $0$. \end{proof} \begin{proof}[Proof of \cref{prop:mass}] By \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 $\sum_c v_c$, \[ \rho_D = \frac{f_D}{3} + \varepsilon_D\,(3p_D - 1), \] which 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 $D$. \end{proof} \section{Reproducibility} Two scripts, plain Python~3, no dependencies, no arguments, every path relative to the lane; run them from the lane root. \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