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Source. Wouter van Doorn and GPT-6 Astra Pro (the author line as printed), Practical numbers and Egyptian fractions, Lemma 3.1, statement on physical p. 2 and proof on p. 3 of the seven-page PDF held by van Doorn (2026). Read in the text extracted from the PDF and checked against the page images. Consumed by the Corollary 3.4 reconstruction and by the base case of the Proposition 4.1 reconstruction.

Standing. Author-recorded reconstruction of a claimed result: the note is a proof claim on the erdosproblems.com proof-claims tab, mostly AI-generated by its own account, not refereed, with an author-side Lean file that was not built here. This page is not an independent review; it changes no status and assigns no tier.

Definitions

A positive integer $n$ is practical if every positive integer $m\le n$ is a sum of distinct positive divisors of $n$. For practical $n$, $h(n)$ is the least integer such that every positive integer $m\le n$ is a sum of at most $h(n)$ distinct divisors of $n$. A residue class $c$ modulo $A$ is represented by a sum of divisors $s$ when $s\equiv c\pmod A$; the empty sum, with value $0$, is allowed.

Statement

Let $A\ge2$ and let $n$ be practical. Suppose that every residue modulo $A$ is represented by a sum of at most $L$ distinct divisors of $n$, with total at most $n$ and no summand divisible by $A$. Then $An$ is practical and

$$h(An)\le h(n)+L .$$

Proof

Every divisor of $n$ divides $An$. Let $1\le m\le An$.

If $m\le n$, a representation of $m$ by at most $h(n)$ distinct divisors of $n$ is a representation by divisors of $An$.

If $m=An$, the single divisor $An$ represents it.

If $n<m<An$, choose one of the prescribed sums $s\equiv m\pmod A$. Then $0\le s\le n<m$, so $m-s$ is a positive multiple of $A$, and $m-s<An$ because $s\ge0$; hence $0<(m-s)/A<n$. Since $n$ is practical, write $(m-s)/A=e_1+\cdots+e_r$ with distinct divisors $e_i$ of $n$ and $r\le h(n)$. Then

$$m=s+Ae_1+\cdots+Ae_r .$$

Each $Ae_i$ divides $An$ and is divisible by $A$, and the $Ae_i$ are distinct; each summand of $s$ divides $n$, hence $An$, and is not divisible by $A$, and the summands of $s$ are distinct. So the two groups do not overlap, and $m$ is a sum of at most $L+h(n)$ distinct divisors of $An$.

Every $1\le m\le An$ is thus a sum of distinct divisors of $An$, so $An$ is practical, and every such $m$ uses at most $\max(h(n),1,L+h(n))=h(n)+L$ divisors, so $h(An)\le h(n)+L$.