Lower bound for the log of the unit-scale boundary factor

This is a numerical lower bound for the logarithm of the unit-scale boundary dimension factor of the released global candidate.

Let $D=10^{12}$ be the released denominator, so that $\mathrm{blocks}(1)=D^5=10^{60}$. For each of the $144$ boundary cells $c$ (an owner $o\in\mathrm{Fin}\,6$ together with a level-3 shape having a zero coordinate), let $z(c,s)=z_1(c,s)$, for words $s$, be the released unit-scale histogram of the cell in the mode following its zero mode, and let $N_c=\sum_s z(c,s)$. For a word $s$ let $\mathrm{ones}(s)$ be the number of letters of $s$ equal to $1$. The unit-scale boundary dimension factor is

Claim.

Role. $B$ is the dimension contributed by the level-3 zero cells, the boundary blocks closed by the exact-profile boundary end. It enters the refhashed construction at a chosen reference only through $k^2\log B$. This lemma is the level-3 boundary part of the dimension certificate (WP7 in the mission plan): a single fixed numerical fact. A computation with the exact released integers and 80-digit log-gamma gives $\log B/(6D^5)=0.4505122058230850\ldots$, so the stated constant leaves a relative margin of about $1.8\cdot10^{-9}$, which is $4.9\cdot10^{51}$ in absolute terms. Stirling-type error terms are of order $10^{4}$.

Formalization Note. $B$ is the natural number ∏ c : Fin zCells, ((∑ s, zCountAt 1 c s)! / ∏ s, (zCountAt 1 c s)!) * 5 ^ (∑ s, zCountAt 1 c s * Boundary.ones s), cast to ℝ. The division is exact. zCountAt 1 c s is wordCounts, the global weight times the joint row counts, so every entry is an explicit integer of size at most $10^{60}$. A cell with two zero grades contributes a factor $1$ whichever zero mode is chosen, and for a cell with one zero grade both nonzero modes give the same factor, so the value does not depend on the choose in zMode.