../

---

Source. Scott D. Hughes, Sums of distinct divisors of factorials, arXiv:2609.10902v1, Lemma 4 (greedy step) with its proof, physical p. 2 of the five-page PDF held by its library source card, Hughes (2026), whose result page is Lemma 4. The statement and proof were read in the canonical conversion beside the PDF and checked against the page image. The lemma is consumed by the Theorem 1 reconstruction.

Standing. Author-recorded reconstruction; not an independent review; it changes no status and assigns no tier. The ratio property of the divisors of $n!$ is cited by the source from Tenenbaum–Yokota and Yokota and not proved there; the proof given at the end of this page is supplied by the compilation and labeled as such, as is the consequence on termination and the representation of $m$, which extends the source's lemma.

Definitions

For an integer $N\ge1$ and an integer $1\le R\le N$ that does not divide $N$, the bracketing divisors of $R$ are the consecutive divisors $d<R<b$ of $N$: $d$ is the largest divisor of $N$ below $R$ and $b$ the smallest above it. Both exist because $1\mid N$ and $N\mid N$.

The greedy expansion of an integer $1\le m\le N$ is the sequence $R_0=m>R_1>R_2>\cdots$ defined as follows: if $R_i$ divides $N$ the expansion terminates and $R_i$ is its last term; otherwise $R_{i+1}=R_i-d_i$, where $d_i<R_i<b_i$ are the bracketing divisors of $R_i$, and $d_i$ is the divisor chosen at step $i$. A step at which $R_i$ does not divide $N$ is nonterminal.

Statement

Let $N$ be an integer whose consecutive divisors have ratio at most $2$, and let $1\le R\le N$. If $R$ divides $N$ (in particular if $R=N$), the greedy expansion terminates at this step. Otherwise let $d<R<b$ be the bracketing divisors of $R$. Here $\log$ is the natural logarithm, as the source fixes on p. 1. Then

$$R-d<d,\qquad R-d\le2R\log\frac bd .$$

Consequently the divisors chosen at successive nonterminal steps of the greedy expansion strictly decrease, hence are distinct.

Consequence (compilation-supplied). For $1\le m\le N$ the greedy expansion of $m$ terminates after finitely many steps at some $R_k$ dividing $N$, and $m=d_0+\cdots+d_{k-1}+R_k$ is a sum of distinct divisors of $N$. The source's lemma ends at "strictly decreasing, hence distinct"; the termination rule and the counting of the final divisor are stated in the opening of its Section 3 (p. 2), and the representation of $m$ is used there without being stated.

Proof

Since $d<b$ are consecutive divisors of $N$, the hypothesis gives $b\le2d$, and $R<b$. Hence

$$R-d<b-d\le2d-d=d .$$

For the second inequality, $R-d<b-d=d\,(b/d-1)\le R\,(b/d-1)$, because $d<R$ and $b/d-1\ge0$. On the interval $[1,2]$ the function $x\mapsto x-1-2\log x$ vanishes at $x=1$ and has derivative $1-2/x\le0$, so $x-1\le2\log x$ there. As $1<b/d\le2$, this gives $R\,(b/d-1)\le2R\log(b/d)$.

For the consequence (the source's proof gives only that the next chosen divisor is below $d$; the terminal-divisor case and the rest are supplied here): at a nonterminal step $i$ the chosen divisor is $d_i$, and the next remainder satisfies $R_{i+1}=R_i-d_i<d_i$ by the first inequality. The divisor used at the next step, whether the chosen divisor $d_{i+1}<R_{i+1}$ or the terminal divisor $R_{i+1}$ itself, is therefore below $d_i$. So the divisors used form a strictly decreasing sequence of positive integers. The remainders are positive integers that strictly decrease, so some $R_i$ divides $N$ (at the latest $R_i=1$), and then $m=d_0+\cdots+d_{i-1}+R_i$ is a sum of distinct divisors of $N$.

The ratio property for factorials (compilation-supplied proof)

The source recalls from Tenenbaum–Yokota (J. Number Theory 35 (1990), 150–156, proof of Lemma 4) and Yokota (Res. Bull. Hiroshima Inst. Tech. 29 (1995), 25–28, Lemma 2) that consecutive divisors of $n!$ have ratio at most $2$. Neither paper is held here. The following proof is supplied by the compilation so that the reconstruction does not rest on an unread citation.

Let $n\ge2$, let $d$ be a divisor of $n!$ with $d<n!$, and put $q=n!/d>1$. It suffices to find a divisor $d'$ of $n!$ with $d<d'\le2d$. If $q$ is even, $d'=2d$ divides $n!$. If $q$ is odd, let $p$ be a prime factor of $q$; then $p$ is odd, $p\le n$, and $v_p(d)<v_p(n!)$. Let $2^c$ be the largest power of $2$ below $p$, so that $2^c<p<2^{c+1}$. Since $2^c<p\le n$, $2^c$ divides $n!$; and since $q$ is odd, $v_2(d)=v_2(n!)\ge c$. Put $d'=dp/2^c$. It is an integer, $v_p(d')=v_p(d)+1\le v_p(n!)$, $v_2(d')=v_2(d)-c\ge0$, and every other prime has the same exponent in $d'$ as in $d$; so $d'$ divides $n!$. Finally $d<d'<2d$ because $1<p/2^c<2$.