The partition exponent in the graded pieces of the Claude 3-SAT algorithm is optimal for pieces of that shape

In Appendix B of "A $2^{o(n)}$ time algorithm for 3-SAT" (Claude, Anthropic, hexagon:2610.00577) the gain parameter $\rho = 1/\sqrt{2}$ comes from the Hardy–Ramanujan growth $\log p(E) \sim c\sqrt{E}$ of the partition numbers, and Appendix H observes that growth of order $E^\alpha$ would give $\rho = 2^{\alpha-1}$ by the same calculation. We point out that the rank floor stated in the same Appendix H already excludes $\alpha < 1/2$ for every graded family of the same shape: the pieces at weight $h$ have total rank at most $(4h+1)\dim V_{\le 4h}$ in one of the two directions, and the crossing they must add up to has rank $(\dim V_h)^2$, so the construction needs $\log \dim V_{\le 4h} \ge 2\log\dim V_h - O(\log h)$, that is $\alpha \ge 1/2$. Restoring the rank with a larger weight costs $\rho \ge 2^{-\alpha}$ in the weight count, so the count never goes below $\max(2^{\alpha-1}, 2^{-\alpha}) \ge 1/\sqrt{2}$, with equality only at $\alpha = 1/2$; when the identity holds on the whole weight space, as Lemma 9 of the paper gives it, the bound sharpens to $2^\alpha/(2(4^\alpha-1))$, which exceeds 1 for $\alpha < 0.357$. Theorem B of the paper is rank-tight up to a polynomial factor. The inequality is exact and we tabulate it: for the partition numbers it holds for every $h$, for partitions into squares ($\alpha = 1/3$) it fails for all $h \ge 5891$. We also record that the Heisenberg Fock spaces on a sparse set of modes, the natural candidates for slower growth, do not carry Lemma 9, and that no choice of coefficients repairs it. The floor $\rho \ge 1/2$ remains open.