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Source. Scott D. Hughes, Sums of distinct divisors of factorials, arXiv:2609.10902v1, Remark 6, physical pp. 4–5 of the five-page PDF held by Hughes (2026); the library records it on its result page. Read in the canonical conversion beside the PDF and checked against the page images.

Standing. Author-recorded reconstruction; not an independent review; it changes no status and assigns no tier. The only imported input is Chebyshev's bound $\pi(x)\ll x/\log x$.

Definitions

$h(N)$ is defined on the Theorem 1 page. $\tau(N)$ is the number of positive divisors of $N$, $v_p$ the $p$-adic valuation, and $\pi(x)$ the number of primes up to $x$. Chebyshev's bound is used in the form $\pi(x)\le Cx/\log x$ for $x\ge2$ with an absolute $C$.

Statement

$h(n!)\gg(\log n)^2$: there is an absolute constant $c>0$ with $h(n!)\ge c(\log n)^2$ for all sufficiently large $n$.

Proof

Write $T=\tau(n!)$ and $k=h(n!)$, and let $n\ge4$.

Counting. Every integer $1\le m\le n!$ is the sum of some set of at most $k$ distinct divisors of $n!$, and distinct $m$ need distinct sets. Hence

$$n!\le\sum_{i=0}^{k}\binom Ti .$$

The binomial tail. If $1\le k\le T$ then $\sum_{i\le k}\binom Ti\le(eT/k)^k$: for $0<x\le1$,

$$\sum_{i\le k}\binom Ti\le x^{-k}\sum_{i=0}^{T}\binom Tix^i =x^{-k}(1+x)^T\le x^{-k}e^{xT},$$

and $x=k/T$ gives $(T/k)^ke^k$. Taking logarithms in the counting inequality, for $1\le k\le T$,

$$\log n!\le k\log\frac{eT}k\le k\,(1+\log T).$$

The divisor count. $\log T=\sum_{p\le n}\log\bigl(v_p(n!)+1\bigr)$. The primes $p\le\sqrt n$ are at most $\sqrt n$ in number and each has $v_p(n!)\le n/(p-1)\le n$, so they contribute $O(\sqrt n\log n)$. A prime $p>\sqrt n$ has $p^2>n$, hence $v_p(n!)=\lfloor n/p\rfloor$. Such a prime lies in $(n/2^{r+1},n/2^r]$ for exactly one integer $r\ge0$ with $n/2^r>\sqrt n$; there $\lfloor n/p\rfloor<2^{r+1}$, so $\log(v_p(n!)+1)\le(r+1)\log2$, and the number of primes in the range is at most $\pi(n/2^r)\le Cn/(2^r\log(n/2^r))\le2Cn/(2^r\log n)$, using $\log(n/2^r)>\tfrac12\log n$. The primes above $\sqrt n$ therefore contribute at most

$$\frac{2Cn\log2}{\log n}\sum_{r\ge0}\frac{r+1}{2^r}\ll\frac n{\log n},$$

and altogether $\log T\ll n/\log n+\sqrt n\log n\ll n/\log n$.

Conclusion. Since $\log n!\ge\sum_{n/2<j\le n}\log j\ge\tfrac n2\log\tfrac n2\gg n\log n$, the case $k\le T/2$ (so $k\le T$) gives

$$n\log n\ll\log n!\le k\,(1+\log T)\ll k\,\frac n{\log n},$$

that is, $k\gg(\log n)^2$. In the case $k>T/2$, every integer $1,\dots,n$ divides $n!$, so $T\ge n$ and $k>n/2\gg(\log n)^2$. In both cases $h(n!)\gg(\log n)^2$.

Qualifications

The source splits at $k\le T/2$; the binomial-tail bound holds for all $k\le T$, so the split is only for convenience. The source also restates the two questions of Erdős and Graham (pp. 37–38) on $h(n!)$, which are questions (b) and (c) of Problem 18; the remark proves that any answer to (c) has exponent at least $2$.