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We prove L66.

Pair the exponents

Write $N=\prod_p p^{e_p}$. Partition the exponent choices $0,1,\ldots,e_p$ for each $p\in P$ into consecutive pairs $\{0,1\},\{2,3\},\ldots$, allowing a final singleton. Fix the chosen pair for every selected prime and fix all exponents at the other primes. Within this cell a divisor has the form

$$b\prod_{p\in P}p^{\delta_p},\qquad \delta_p\in\{0,1\},$$

with some choices possibly forced by the singleton cells. The factor $b$ is fixed and invertible modulo $q$. The products of selected primes are distinct positive integers, by unique factorization, and all are smaller than $q$. Thus this cell contains at most one divisor in any specified residue class modulo $q$.

There are $\lceil(e_p+1)/2\rceil$ cells at prime $p$. Dividing the total number of cells by the number $\tau(N)=\prod_p(e_p+1)$ of divisors gives

$$\frac{\#\{d\mid N:d\equiv a\pmod q\}}{\tau(N)} \le\prod_{p\in P} \frac{\lceil(e_p+1)/2\rceil}{e_p+1} \le(2/3)^{|P|}. \tag{1}$$

The last inequality holds for every integer $e_p\ge1$; the largest ratio is $2/3$, at $e_p=2$. The empty choice of $P$ gives the trivial bound one.

All dyadic levels through K

Let $N$ be the odd part of $K!$. We use the standard Chebyshev bounds in the explicit forms $\sum_{p\le y}\log p\le C_1y$ for $y\ge2$ and $\pi(y)\ge c_1y/\log y$ for all sufficiently large $y$. Choose a fixed $\alpha>0$ smaller than both $1$ and $(\log2)/(2C_1)$. For sufficiently large $j\le K$, take $P$ to be the odd primes at most $\alpha j$. They all divide $N$, and

$$\prod_{p\in P}p\le\exp(C_1\alpha j)<2^j, \qquad |P|\gg\frac{j}{\log j}\ge c_2\frac{j}{\log(K+1)}.$$

Equation (1) gives the claimed residue bound for these $j$. There are only finitely many smaller integers $j\ge2$ to cover. For each of them use $P=\{3\}$, since $3<2^j$ and $3\mid K!$ once $K\ge3$. Reducing the absolute constant $c$ covers all these levels simultaneously. This proves that, for every sufficiently large $K$, the uniform law $\mu_K$ on the odd divisors of $K!$ satisfies

$$\max_a\mu_K(a\bmod 2^j) \le2^{-g_Kj}\quad(2\le j\le K),\qquad g_K=\frac{c}{\log_2(K+1)}. \tag{2}$$

Reduce $c$ further if necessary so that $g_K\le1/2$.

A necessary number of multiplicative factors

Fix $0<g\le1/2$ and suppose $W(g;r,t,J_0)$ holds. For large $J$, set $b_J=\lceil gJ\rceil+1$ and let $X_J$ be all positive odd integers smaller than $2^{b_J}$. At levels $2\le l\le b_J$ the largest residue mass is $2^{1-l}\le2^{-gl}$. At levels $b_J<l\le J$ it is $2^{1-b_J}\le2^{-gJ}\le2^{-gl}$. Thus $X_J$ meets every required nonconcentration bound.

If $rg<1$, every product $z$ of $r$ members is smaller than $2^{rb_J}$, and $2^{rb_J-J}\to0$. Hence $e^{2\pi i z/2^J}\to1$ uniformly over these products. The primitive Fourier coefficient of their product law has modulus tending to one. This contradicts $W$'s bound $2^{-tJ}\to0$. Therefore

$$rg\ge1. \tag{3}$$

This is a restriction on the universal mixing assertion, not on one particular factorial-divisor law.

For the separated-block construction, L64 gives an additional consequence. If its factorial index is linear up to logarithms and its block count is polylogarithmic, an admissible exponent $g_K$ must tend to zero, because the top-level atom bound requires $g_KK\le\log_2|X_K|=o(K)$. Hence (3) forces $r_K\to\infty$. The source's displayed term bound $s_K((\operatorname{binaryLength}(K)+1)^{r_K}+1)$, with $s_K\ge1$, then exceeds every fixed power of $\log K$. Thus changing the parameters alone cannot turn that displayed construction into L63's required bound. This does not rule out a new representation argument using the same numbers more efficiently.

Use quantitative mixing without expanding digit sums

Assume the last clause's quantitative hypothesis, and choose its witnesses $r,t,J_0$ at $g=g_K$. They obey

$$r,J_0,1/t=O((\log K)^A).$$

Let $\sigma$ be the law of a product of $r$ independent odd divisors of $K!$, viewed modulo $2^K$. For every character of conductor $2^j$ with $J_0\le j\le K$ and $j\ge2$, (2) and $W$ give

$$|\widehat\sigma(a/2^j)|\le2^{-tj}\qquad(a\text{ odd}). \tag{4}$$

The same integer product law may be reduced modulo each $2^j$, so the application uses the same $r,t$ at every conductor.

Put $b=\max(1,J_0)$ and choose a positive integer $q$ with $tq\ge2$ and $q=O(1+1/t)$. For large $K$, $b<K$. Let $\nu_b$ be the uniform law on $\{0,1,\ldots,2^b-1\}$, embedded modulo $2^K$. It kills exactly the needed low conductors: its Fourier coefficient is zero at every nontrivial character of conductor $2^j$ with $j\le b$. Therefore the sum of the nontrivial Fourier magnitudes of the additive convolution $\sigma^{*q}*\nu_b$ is at most

$$\sum_{j=b+1}^K2^{j-1}2^{-tqj} \le\sum_{j=b+1}^{\infty}2^{-j-1}<1. \tag{5}$$

Fourier inversion makes this convolution positive on every residue. Accordingly every residue modulo $2^K$ has an integer representative

$$S=z_1+\cdots+z_q+B,\qquad 0\le B<2^b,$$

where each $z_i$ is a product of $r$ divisors of $K!$. Such a product is a single divisor of $(rK)!$, since $(K!)^r\mid(rK)!$. Write $B$ in binary, with at most $b$ summands among $1,2,\ldots,2^{b-1}$. These divide $(2b)!$. Thus $S$ is a sum of at most $q+b$ divisors of $F!$, repetitions allowed, for $F=rK+2b$. It satisfies

$$\begin{gathered} F=O(K(\log K)^A),\qquad q+b=O((\log K)^A),\\ 0\le S\le q(K!)^r+2^b,\qquad \log_2(q(K!)^r+2^b)=O(K(\log K)^{A+1}). \end{gathered} \tag{6}$$

These are L63's hypotheses with exponents $a=A$, $b=A$, $d=A+1$ (the exponent called $b$ there is separate from the binary cutoff here). Rounding the height exponent to an integer and enlarging its fixed constant does not change the bounds. L63 gives $h(n!)=O((\log n)^{3A+4})$.

The polynomial quantitative hypothesis has not been proved. The argument shows precisely why it would suffice; it does not infer such bounds from a mixing theorem that supplies only existential constants for each fixed $g$.

The loss in the displayed source witnesses

In the held source's exists_dyadic_unit_product_fourier_decay, $R$ flattening steps give a power $N=2^{R+2}$ and decay $t=(g-\min(g/2,1/2))/N$. The final weighted_dyadic_mixing takes $r=R+2$. Thus for $0<g\le1/2$ the displayed witnesses obey

$$t=\frac{g}{2^{R+3}},\qquad R\ge\frac1g-2,\qquad \frac1t\ge\frac2g\,2^{1/g}, \tag{7}$$

where the middle inequality follows from (3), since these witnesses satisfy the universal property $W$. This lower bound is not polynomial in $1/g$. At the value $g_K$ in (2) it is at least $2(K+1)^{1/c}/g_K$, so the additive-copy count certified by this decay bound is not polylogarithmic in $K$.

This stops the extraction through these displayed witnesses. It does not refute the polynomial quantitative hypothesis for different witnesses, stronger concentration of the particular full-divisor law, or a different way of obtaining coverage. In particular, reducing the number of flattening steps to $O(\log(1/g))$ while retaining $r=R+2$ cannot repair the universal argument: (3) forbids that. The exponential loss in the Fourier conversion, or the use of a universal small-$g$ mixing statement, must be replaced.