Boundary unit factor: owner 5 cell certificate

This is the per-cell part of the numerical certificate for the unit-scale boundary dimension factor of the released global candidate, for owner $o=5$.

Fix the owner $o=5$. For a level-3 shape $s=(s_0,s_1,s_2)$ (one of the $45$ triples of naturals with $s_0+s_1+s_2=8$) and a mode $m\in\{0,1,2\}$, let $h_{s,m}(w)=\mathrm{wordCounts}(o,m,s,w)$ be the released unit-scale histogram of the joint rows of the cell $s$ in mode $m$, over the $81$ words $w\in\{0,1,2\}^4$; it equals $\alpha_{o,s}$ times an integer histogram of total mass $10^{48}$, where $\alpha_{o,s}$ is the released global weight of the cell. Put $N=\sum_w h_{s,m}(w)$, let $\mathrm{ones}(w)$ be the number of letters of $w$ equal to $1$, and write

for the mass entropy (with $0\log 0=0$). Let $\pi_o$ be the owner's hash orientation (hashMode o).

Claim. For every shape index $s$ and every mode $j$ such that the coordinate $\pi_o(j)$ of $s$ is zero, with $m=\pi_o(j+1)$,

where $g_{o,s}$ is the explicit integer listed in the formal statement (entry $s$ of the list; entries of shapes without a zero coordinate are $0$ and unused).

Numerical evidence. The integers $g_{o,s}$ were obtained from rigorous rational upper bounds for $\log(h(w)/N)$ (the series $\log x=2\,\mathrm{artanh}\frac{x-1}{x+1}$ with $14$ terms after scaling by a power of $2$, rounded outward at $10^{-20}$), the rational bound $\log 5\ge 1.6094379124340\ldots$, and the crude bound $81\log(6(N+1))\le 81\cdot 60\log 10\le 11340$. Recomputing the right side with 80-digit arithmetic, every cell exceeds its $g_{o,s}$, the smallest excess (over all owners) being about $1.29\cdot10^{3}$. The corner cells $s\in\{(0,0,8),(0,8,0),(8,0,0)\}$ carry a point-mass histogram and get $g=-11340$.

Role. Summing these six owner certificates over the $144$ boundary cells gives $\sum g_{o,s}=6\cdot10^{60}\cdot0.450512205+4.94\cdot10^{51}$, which proves the refentropy certificate for the boundary unit factor.

Formalization Note. The histogram is wordCounts o (hashMode o (j + 1)) (shape s) w, which is exactly zCountAt 1 c w for the boundary cell $c=(o,\mathrm{shape}\,s)$ when $j$ is its chosen zero mode; for shapes with two zero coordinates every choice gives a point mass. $\mathcal H$ is MME.RegionRate.massEntropy. The bound $g_{o,s}$ is written as (L : List ℤ).getD s.val 0.