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Source. Scott D. Hughes, Sums of distinct divisors of factorials, arXiv:2609.10902v1, Theorem 1 (physical p. 1) and its proof in Section 3 (pp. 2–4), in the five-page PDF held by Hughes (2026); the library records the theorem on its result page. Read in the canonical conversion beside the PDF and checked against the page images (pp. 1–4). The two ingredients are reconstructed on the Lemma 4 page and the Corollary 3 page.

Standing. Author-recorded reconstruction; not an independent review; it changes no status and assigns no tier. The argument imports the Berend–Harmse gap estimate through Corollary 3 (second-hand; see that page) and the elementary asymptotic $\sum_{j\le n}1/\log j\sim n/\log n$. As a bound on $h(n!)$ the theorem is superseded by the site-accepted $h(n!)<n^{o(1)}$ recorded on the Problem 18 page; its interest is the explicit constant of the greedy route.

Definitions

An integer $N\ge1$ is practical if every integer $1\le m\le N$ is a sum of distinct divisors of $N$; for practical $N$, $h(N)$ is the least $k$ such that every $1\le m\le N$ is a sum of at most $k$ distinct divisors of $N$ (a fresh set of divisors for each $m$). Greedy expansion, bracketing divisors, chosen divisors and nonterminal steps are defined on the Lemma 4 page; $\varepsilon_j$ and the windows $[\sqrt{(j-1)!},\sqrt{j!}]$ on the Corollary 3 page. Throughout, $N=n!$,

$$j_0=2^{16},\qquad T_0=2\sqrt{(j_0+1)!},\qquad \ell(R)=\log R ,$$

and for integers $j\ge j_0+1$

$$\delta_j=6\,\varepsilon_{j-1},\qquad s_j=\log\frac1{\delta_j}=\log\frac1{\varepsilon_{j-1}}-\log6 .$$

By display (1) of the Corollary 3 page applied at $j-1$, and since $\log(j-1)=\log j+O(1/j)$,

$$s_j=\frac{(\log j)^2}{2\log2} \Bigl(1+O\Bigl(\frac{\log\log j}{\log j}\Bigr)\Bigr). \tag{2}$$

The sequence $s_j$ is increasing in $j$, because $\varepsilon_{j-1}$ is decreasing, and $s_j\ge64\log2-\log6>0$ for every $j\ge j_0+1$.

For real $1\le R\le\sqrt N$ let $j(R)$ be the least integer $j\in[j_0+1,n]$ with $R\le\sqrt{j!}$; it exists because $\sqrt{n!}=\sqrt N$, and it is nondecreasing in $R$. If $R\ge T_0$ then $R>\sqrt{(j_0+1)!}$, so $j(R)\ge j_0+2$, and by minimality $\sqrt{(j(R)-1)!}<R\le\sqrt{j(R)!}$.

Statement

$$h(n!)\le(2\log2+o(1))\,\frac n{\log n}\qquad(n\to\infty).$$

Proof

Since the statement is asymptotic, assume $n\ge j_0^2$; then $T_0<\sqrt N$. Fix $1\le m\le N$ and run the greedy expansion $R_0=m>R_1>\cdots$ of Lemma 4. Consecutive divisors of $N=n!$ have ratio at most $2$ (proved on the Lemma 4 page), so Lemma 4 applies at every nonterminal step: the divisors used are distinct and $m$ is their sum. The number of divisors used is the number of nonterminal steps plus one. The nonterminal steps are counted by the size of the remainder $R_i$ at which they start, in four ranges: $N/T_0<R\le N$, $\sqrt N<R\le N/T_0$, $T_0\le R\le\sqrt N$ and $1\le R<T_0$. Since the remainders decrease, the steps of each range form a consecutive run.

Locating the window of a step

Let $d<R<b$ be the bracketing divisors of a nonterminal remainder $R$. Two symmetric facts hold.

(i) If $R\le\sqrt N$ then $db\le N$. Otherwise $N/d<b$; but $N/d$ is a divisor of $N$ with $N/d\ge N/R\ge\sqrt N\ge R>d$, so $N/d$ would be a divisor of $N$ strictly between the consecutive divisors $d$ and $b$.

(ii) If $R>\sqrt N$ then $db\ge N$. Otherwise $N/b>d$; but $N/b<N/R<\sqrt N<R<b$, so $N/b$ would be a divisor of $N$ strictly between $d$ and $b$.

Also $d\ge b/2>R/2$ and $b\le2d<2R$, so $R^2/2<db<2R^2$ and $R/2<\sqrt{db}<2R$.

The window bound

Let $T_0\le R\le\sqrt N$ be a nonterminal remainder with bracketing divisors $d<R<b$, and put $j=j(R)$, so that $\sqrt{(j-1)!}<R\le\sqrt{j!}$ and $j\ge j_0+2$. Then

$$\sqrt{db}>\frac R2>\frac{\sqrt{(j-1)!}}2\ge\sqrt{(j-2)!}, \qquad \sqrt{db}<2R\le2\sqrt{j!}\le\sqrt{(j+1)!},$$

using $\sqrt{j-1}\ge2$ and $\sqrt{j+1}\ge2$; and $\sqrt{db}\le\sqrt N$ by (i). Hence $\sqrt{db}$ lies in the window $[\sqrt{(j'-1)!},\sqrt{j'!}]$ of some index $j'\in\{j-1,j,j+1\}$ with $j'\le n$ (the value $\sqrt N$ itself belongs to the window of index $n$). As $j'\ge j-1\ge j_0+1>2^{16}$ and $n\ge j'$, Corollary 3 gives $\log(b/d)\le3\varepsilon_{j'}\le3\varepsilon_{j-1}$ by the monotonicity of $\varepsilon$. That is,

$$\log\frac bd\le\frac12\,\delta_{j(R)} . \tag{3}$$

Lower range $T_0\le R\le\sqrt N$

For a nonterminal step starting at $R_i$ in this range, Lemma 4 and (3) give $R_{i+1}\le2R_i\log(b/d)\le R_i\,\delta_{j(R_i)}$, that is,

$$\ell(R_{i+1})\le\ell(R_i)-s_{j(R_i)} . \tag{4}$$

For $\ell$ in the interval $[\ell(R_{i+1}),\ell(R_i)]$ one has $e^\ell\le R_i$, hence $j(e^\ell)\le j(R_i)$ and $s_{j(e^\ell)}\le s_{j(R_i)}$. Therefore, by (4),

$$\int_{\ell(R_{i+1})}^{\ell(R_i)}\frac{d\ell}{s_{j(e^\ell)}} \ge\frac{\ell(R_i)-\ell(R_{i+1})}{s_{j(R_i)}}\ge1 .$$

The intervals $[\ell(R_{i+1}),\ell(R_i)]$ of the steps starting in the lower range have disjoint interiors and lie inside $[\log T_0,\tfrac12\log N]$, except that the last of them may extend below $\log T_0$. Hence the number of such steps is at most

$$1+\int_{\log T_0}^{\frac12\log N}\frac{d\ell}{s_{j(e^\ell)}} .$$

The set of $\ell$ with $j(e^\ell)=j$ is contained in $(\tfrac12\log(j-1)!,\tfrac12\log j!]$, an interval of length $\tfrac12\log j$, and there the integrand is $1/s_j$. So

$$\#\{\text{steps with }T_0\le R\le\sqrt N\} \le1+\sum_{j=j_0+1}^{n}\frac{\tfrac12\log j}{s_j} =1+\sum_{j=j_0+1}^{n}\Bigl(\frac{\log2}{\log j} +O\Bigl(\frac{\log\log j}{(\log j)^2}\Bigr)\Bigr),$$

by (2), since $\tfrac12\log j\cdot2\log2/(\log j)^2=\log2/\log j$ and $(1/\log j)\cdot\log\log j/\log j=\log\log j/(\log j)^2$. The elementary asymptotic

$$\sum_{2\le j\le n}\frac1{\log j} =\frac n{\log n}\Bigl(1+O\Bigl(\frac1{\log n}\Bigr)\Bigr),$$

obtained by comparing the sum with $\int_2^n dt/\log t$ and integrating by parts once, and the crude bound $\sum_{j\le n}\log\log j/(\log j)^2\ll n\log\log n/(\log n)^2$ (split the sum at $\sqrt n$), give

$$\#\{\text{steps with }T_0\le R\le\sqrt N\} \le\Bigl(\log2+O\Bigl(\frac{\log\log n}{\log n}\Bigr)\Bigr)\frac n{\log n}.$$

Upper range $\sqrt N<R\le N/T_0$

Put $U=N/R$, so $T_0\le U<\sqrt N$. If $d<R<b$ are the bracketing divisors of $R$, then $N/b<U<N/d$, and $N/b$, $N/d$ are consecutive divisors of $N$, because $x\mapsto N/x$ is an order-reversing bijection of the divisor set; their ratio is again $b/d$, and their geometric mean $N/\sqrt{db}$ satisfies $U/2<N/\sqrt{db}<2U$ by the bounds on $\sqrt{db}$ above and $N/\sqrt{db}\le\sqrt N$ by (ii). So the argument that proved (3) applies word for word to the pair $N/b<U<N/d$ in place of $d<R<b$: with $j=j(U)\ge j_0+2$, the geometric mean lies in a window of index $j'\in\{j-1,j,j+1\}$, $j'\le n$, and Corollary 3 gives $\log(b/d)\le\tfrac12\delta_{j(U)}$. Lemma 4, applied to the actual remainder $R$ with its bracketing divisors $d<R<b$, then gives $R_{i+1}\le\delta_{j(U_i)}R_i$; with $U_i=N/R_i$ this reads

$$\log U_{i+1}-\log U_i\ge s_{j(U_i)} . \tag{5}$$

Along the expansion $U_i$ increases, so $j(U_i)$ and $s_{j(U_i)}$ are nondecreasing in $i$. The charging integral of the lower range cannot be mirrored: there the descent of a step was bounded below by the value of $s$ at the upper end of the step's interval, which dominated the integrand on the whole interval, whereas (5) bounds the ascent of a step by the value of $s$ at the lower end, which does not. The steps are counted in dyadic blocks of window indices instead.

Let

$$R_n=\Bigl\lfloor\log_2\frac n{2j_0}\Bigr\rfloor,\qquad J_r=\frac n{2^{r+1}}\quad(0\le r\le R_n),\qquad \mathcal B_r=\{j\in\mathbb Z: J_r<j\le2J_r\}.$$

Then $2^{R_n}\le n/(2j_0)<2^{R_n+1}$ gives $j_0\le J_{R_n}<2j_0$, and the blocks $\mathcal B_0,\dots,\mathcal B_{R_n}$ partition the integers in $(J_{R_n},n]$.

Steps starting in a block. Fix $r$ and consider the nonterminal steps of the upper range whose start satisfies $j(U_i)\in\mathcal B_r$; they form a consecutive run of the expansion because $j(U_i)$ is nondecreasing in $i$. For each of them $\sqrt{(j(U_i)-1)!}<U_i\le\sqrt{j(U_i)!}$, so their values $\log U_i$ lie in an interval of length at most

$$L_r=\tfrac12\sum_{J_r<j\le2J_r}\log j=\tfrac12J_r\log J_r+O(J_r),$$

the last by comparing the sum with $\int_{J_r}^{2J_r}\log t\,dt=2J_r\log(2J_r)-J_r\log J_r-J_r$. By (5) and the monotonicity of $s$, consecutive starts of the run are separated in $\log U$ by at least $s_{\lfloor J_r\rfloor+1}$, because $j(U_i)>J_r$ forces $j(U_i)\ge\lfloor J_r\rfloor+1$. A run of $M$ starts therefore has $(M-1)\,s_{\lfloor J_r\rfloor+1}\le L_r$, so the run has at most

$$1+\frac{L_r}{s_{\lfloor J_r\rfloor+1}} =\Bigl(\log2+O\Bigl(\frac{\log\log J_r}{\log J_r}\Bigr)\Bigr) \frac{J_r}{\log J_r}+O(1)$$

steps, by (2) at $\lfloor J_r\rfloor+1$, where $\log(\lfloor J_r\rfloor+1)=\log J_r+O(1/J_r)$, and $\tfrac12J_r\log J_r\cdot2\log2/(\log J_r)^2=\log2\cdot J_r/\log J_r$; the term $O(J_r)/s_{\lfloor J_r\rfloor+1}$ is $O(J_r/(\log J_r)^2)$ and is absorbed.

Steps below the blocks. The remaining upper-range steps have $j(U_i)\le J_{R_n}<2j_0$, hence $T_0\le U_i\le\sqrt{(2j_0)!}$. By (5) each of them raises $\log U$ by at least $s_{j_0+2}>0$, an absolute constant, inside the fixed interval $[\log T_0,\tfrac12\log(2j_0)!]$; so there are $O(1)$ of them.

Summing the blocks. The main terms satisfy $\sum_{r=0}^{R_n}J_r/\log J_r\le(1+o(1))\,n/\log n$: for fixed $0<\eta<1$, the blocks with $J_r\ge n^{1-\eta}$ have $\log J_r\ge(1-\eta)\log n$ and $\sum_rJ_r\le n$, so they contribute at most $n/((1-\eta)\log n)$, while the blocks with $J_r<n^{1-\eta}$ contribute at most $\sum_{J_r<n^{1-\eta}}J_r\le2n^{1-\eta}=o(n/\log n)$; letting $n\to\infty$ and then $\eta\to0$ gives the claim. The relative-error terms satisfy

$$\sum_{r=0}^{R_n}\frac{J_r\log\log J_r}{(\log J_r)^2} =O\Bigl(\frac{n\log\log n}{(\log n)^2}\Bigr):$$

the function $g(t)=\log\log t/(\log t)^2$ is decreasing for $t\ge j_0$ (its derivative has the sign of $1-2\log\log t$), so the blocks with $J_r\ge\sqrt n$ have $g(J_r)\le g(\sqrt n)\le4\log\log n/(\log n)^2$ and $\sum_rJ_r\le n$, and the blocks with $J_r<\sqrt n$ have $g(J_r)\le g(j_0)$ and $\sum J_r\le2\sqrt n$. The $O(1)$ terms of the $R_n+1=O(\log n)$ blocks total $O(\log n)$. Altogether

$$\#\{\text{steps with }\sqrt N<R\le N/T_0\}\le(\log2+o(1))\,\frac n{\log n}.$$

Endgames

For a nonterminal step starting at $R_i>N/T_0$: since $d_i>R_i/2$, $R_{i+1}=R_i-d_i<R_i/2$. If the first $M$ steps all start above $N/T_0$, then $N/T_0<R_{M-1}<R_0/2^{M-1}\le N/2^{M-1}$, so $M<\log_2T_0+1$: at most $O(1)$ steps. The same halving bounds the number of steps starting at $1\le R<T_0$ by $\log_2T_0+1$.

Conclusion

Adding the four ranges and the terminal divisor,

$$h(n!)\le\Bigl(\log2+O\Bigl(\frac{\log\log n}{\log n}\Bigr)\Bigr)\frac n{\log n} +(\log2+o(1))\frac n{\log n}+O(1) =(2\log2+o(1))\frac n{\log n},$$

uniformly in $1\le m\le N$, which is the theorem. The source's Remark 5 (p. 4) records that the $o(1)$ is $O(\log\log n/\log n)$, inherited from the $\lg(\lg n)$ term of the gap estimate.

Gaps and qualifications

• The gap estimate is imported second-hand through Corollary 3; the 1993 paper is not held.
• The source writes $j(R)-1\ge j_0$ "by the choice of $T_0$"; in fact $R\ge T_0$ forces $j(R)\ge j_0+2$, which is what the window bound uses.
• The asymptotic $\sum_{j\le n}1/\log j\sim n/\log n$ and the block sum $\sum_rJ_r/\log J_r\le(1+o(1))n/\log n$ are stated by the source without proof; the justifications above are supplied by the compilation.
• The source counts the steps of the lower range from $j=j_0+1$; the window of index $j_0+1$ lies below $\log T_0$ and contributes nothing, so this is only an upper bound, as used.