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Subject and independence

Role: independent reviewer in a fresh context, commissioned for refutation. The reviewer took no part in writing the page, received only the assignment text, and read nothing beyond the material listed below.

Subject: path wiki/research/erdos_18/doorn_lemma_3_1_reconstruction.md as it stood at 2026-09-28T05:03:27Z, read in full as of that time. The review worktree was checked out at that state, and every allowed file read from the tree was checked to be identical to the committed copy.

Artifact: the seven-page PDF held by the library card (author line as printed on p. 1: Wouter van Doorn and GPT-6 Astra Pro, Practical numbers and Egyptian fractions). Physical pages 1 to 3 were read in the extracted layout text: p. 1 for the definitions of a practical number and of $h(n)$ and for the author line, p. 2 for the Lemma 3.1 statement (last paragraph of the page) and the notation paragraph, p. 3 for the proof of Lemma 3.1 (first paragraph of the page). The statement and proof were read line by line; the rest of pp. 1 to 3 was skimmed for conventions only. Page images: pp. 1 to 3 rendered at 110 dpi and read; zoomed crops of the Lemma 3.1 statement (p. 2) and proof (p. 3) rendered at 220 dpi and read, confirming every inequality sign and the label. The PDF metadata reports seven pages, and the printed page numbers 2 and 3 sit on the physical pages 2 and 3. No canonical conversion sits beside the PDF.

Allowed material actually read: the provenance paragraphs of the library card _index.md (author line, posting, registration, page count, retained Lean file, AI-usage summary); the Statement paragraph of Problem 18; in docs/verification.md the shared section "Audit checklist" and the Erdos-specific subsections "Whole-claim report" and "Audit checklist"; in docs/evidence.md the section "Source fidelity"; docs/math_authoring.md in full. The page cites no reconstruction page as an input, so none was read. The two consumer pages the page links (the Corollary 3.4 and Proposition 4.1 reconstructions) were confirmed to exist in the tree listing as of that time; their content was not read.

Exposures: two fragments of excluded text reached the reviewer through the heading scans used to locate allowed paragraphs: the first line of the library card's "Read status" paragraph together with the first line of its "Bears on" paragraph, and the first line of the Problem 18 "Status" paragraph; the heading names "Current assessment", "Known results" and "Linked library material" of the problem page were also seen. None of these was used.

Restatement

Convention. Divisors are positive divisors. A sum of distinct divisors may be empty, with value $0$ (the page's stated convention; see F1). A positive integer $n$ is practical when every integer $m$ with $1\le m\le n$ is a sum of distinct divisors of $n$; for practical $n$, $h(n)$ is the least integer $k$ such that every such $m$ is a sum of at most $k$ distinct divisors of $n$.

Statement. Let $A\ge2$ be an integer, let $n$ be a practical number, and let $L$ be a nonnegative integer. Suppose that for every residue class $c$ modulo $A$ there is a set $S_c$ of distinct divisors of $n$ with $|S_c|\le L$, with no member divisible by $A$, and with

$$\sum_{d\in S_c} d\le n,\qquad \sum_{d\in S_c} d\equiv c\pmod A .$$

Then $An$ is practical and $h(An)\le h(n)+L$. The conclusion is exact, with no implied constant, and $L$ is the one bound given in the hypothesis, the same for every residue.

This is the source's Lemma 3.1 (p. 2) clause for clause: the source says "with total sum at most $n$ and none of the summands divisible by $A$"; the page says "with total at most $n$ and no summand divisible by $A$".

Checklist

• Quantifiers and scope. Pass. The hypothesis is universal over residues and existential over the representing sum, with $L$ fixed before the residues, and the page keeps that order. The conclusion ranges over all $1\le m\le An$; the page's three cases $m\le n$, $n<m<An$ and $m=An$ are exhaustive because $An>n$ for $A\ge2$ and $n\ge1$. The boundary case $m=An$ and an empty prescribed sum ($s=0$) are handled explicitly. No "almost all" or eventual quantifier appears.
• Circularity. Pass. The proof uses only that $n$ is practical, the definition of $h(n)$, and the residue hypothesis; neither "$An$ is practical" nor any bound on $h(An)$ is assumed.
• Model and convention changes. Pass. The objects are the source's (divisor sums of $n$ and of $An$, residues modulo $A$); the one convention the page adds, the empty sum, is explicit in its Definitions (F1). No relaxed or averaged system replaces the actual objects.
• Finite and statistical overreach. Inapplicable. The argument is a direct construction for every $m$; no finite check or heuristic is cited.
• Uniformity. Pass. The bound $h(n)+L$ carries no hidden constant; $L$ is the hypothesis's uniform bound and enters the count additively. No limits or sums are exchanged.
• Extremal conclusions. Pass. $h(An)$ is the least admissible count; the admissible counts form a nonempty set once $An$ is practical, and the page exhibits the admissible count $h(n)+L$, so the minimum is at most it. The page claims no sharpness.
• Consequences and composition. Pass. Each "hence" was re-derived (see Weakest steps). The page consumes no local claim. The page names two consumers; that relationship is those pages' claim and was not checked here.
• Computation. Inapplicable. No computation is used or claimed.
• Reproduction. Inapplicable. The page states no rerun command and no coverage claim.
• Source and verdict fidelity. Pass. The statement, the labels (Section 3, Lemma 3.1), the physical pages (statement p. 2, proof p. 3), the page count and the author line were checked against the PDF and its images and agree. The Standing paragraph claims only an author-recorded reconstruction of a claimed result and assigns no tier; F3 records a one-word tightening.

Weakest steps

W1. The range of $(m-s)/A$. Fix $m$ with $n<m<An$ and let $s$ be the prescribed sum for the residue of $m$. By hypothesis $0\le s\le n$, where $s=0$ occurs only for the empty sum. Then $m-s\ge m-n>0$, and $A\mid m-s$ because $s\equiv m\pmod A$, so $m-s=Aq$ with $q$ a positive integer. Also $m-s\le m<An$, so $q<n$. Hence $1\le q\le n-1$, and since $n$ is practical, $q$ is a sum of $r\le h(n)$ distinct divisors $e_1,\dots,e_r$ of $n$, with $r\ge1$. The source writes the weaker $0\le(m-s)/A<n$; the page's strict $0<(m-s)/A$ is a valid sharpening, and nothing later depends on which is used. This step feeds W2 by supplying the $e_i$.

W2. Distinctness of the assembled representation. The summands are the members of $S_c$ (distinct divisors of $n$, none divisible by $A$) and $Ae_1,\dots,Ae_r$. Each $Ae_i$ divides $An$ because $e_i\mid n$, and the $Ae_i$ are pairwise distinct because the $e_i$ are. Each member of $S_c$ divides $n$ and hence $An$. A member $d$ of $S_c$ cannot equal any $Ae_j$, since $A\mid Ae_j$ and $A\nmid d$. So the combined family is a set of distinct divisors of $An$ with sum $s+Aq=m$ and size $|S_c|+r\le L+h(n)$. This is the only place where "no summand divisible by $A$" is used, and dropping it breaks exactly this step (a divisor $d$ of $n$ with $A\mid d$ could coincide with some $Ae_j$). The step composes with W1, which supplies the $e_i$, and W3, which counts.

W3. From the three cases to $h(An)\le h(n)+L$. For $m\le n$ the definition of $h(n)$ gives at most $h(n)$ distinct divisors of $n$, which are distinct divisors of $An$; for $m=An$ one divisor; for $n<m<An$ at most $L+h(n)$ by W2. So every $1\le m\le An$ is a sum of distinct divisors of $An$, whence $An$ is practical and $h(An)$ is defined, and every $m$ admits a representation with at most $\max(h(n),1,L+h(n))$ summands. That maximum equals $h(n)+L$ because $L\ge0$ and $h(n)\ge1$ (the integer $1\le n$ needs at least one summand); in fact $L\ge1$, since the residue $1$ modulo $A\ge2$ is not the empty sum. By minimality of $h(An)$, $h(An)\le h(n)+L$. The page states the equality without the two trivial inequalities (F2).

Strongest attack

The strongest attempt aimed at W2, the disjointness of the two summand groups, which is the only nontrivial content of the lemma. The attack: find a practical $n$, a modulus $A\ge2$, a prescribed sum $s$ and a representation $q=e_1+\dots+e_r$ such that some summand of $s$ coincides with some $Ae_j$, making the assembled family a multiset rather than a set of distinct divisors, or such that $Ae_j=Ae_k$ for some $j\ne k$. Both fail under the hypotheses: the second because the $e_i$ are distinct and multiplication by $A$ is injective, the first because a coincidence $d=Ae_j$ forces $A\mid d$, which the hypothesis forbids for every summand of $s$. The attempt does show that the hypothesis is not decorative. Take $n=6$ (practical, divisors $1,2,3,6$) and $A=2$, and allow the forbidden summand $2$ in $s$, so that the residue $0$ is represented by $s=2$. For $m=8$ this gives $q=(8-2)/2=3$, which $n$ represents as $1+2$, and then

$$8=2+2\cdot1+2\cdot2$$

repeats the divisor $2$. The page's argument would break exactly where it invokes the hypothesis, so the reconstruction uses it in the right place.

A second attempt targeted the empty-sum convention: if the source meant only nonempty sums, the page's statement has the weaker hypothesis and is formally the stronger result. The proof was re-run with $s=0$: then $m=Aq$ with $1\le q\le n-1$, and $m=Ae_1+\dots+Ae_r$ is a sum of $r\le h(n)\le h(n)+L$ distinct divisors of $An$. The same argument therefore proves the stronger form, and the source's own proof, which allows $(m-s)/A=0$ and then writes that value as a sum of distinct divisors, is consistent with the convention. The attempt reduces to a labeling point (F1), not a defect.

A third attempt tried the boundary $m=An$ and the degenerate ranges. For $n=1$ (practical, with $h(1)=1$) and $A=2$, the range $n<m<An$ is empty, and the hypothesis asks that the residue $1$ modulo $2$ be a sum of divisors of $1$ not divisible by $2$ with total at most $1$: $s=1$ works, so $L\ge1$, and the lemma gives that $2$ is practical with $h(2)\le2$, which is true ($h(2)=1$). No boundary case escapes the case split.

Premises

• Imported theorems. None. The page imports no result beyond the source's definitions, and it names no imported standing because there is nothing to import.
• Source interface. Lemma 3.1 of the held note, statement on p. 2 and proof on p. 3, read line by line in text and in image (220 dpi crops). The page's Statement is the source's statement with "total sum" shortened to "total".
• Definitions. Practical number and $h(n)$ as on p. 1 of the source, which the page's Definitions reproduce, and the source's p. 2 convention that $\mathcal D(n)$ is the set of positive divisors.
• Explicit assumptions in the reconstruction. $A\ge2$ and $n$ are integers, with $n\ge1$ by the definition of practical; $L$ is a nonnegative integer, a count of summands; the empty sum is an admissible sum of divisors, with value $0$ (page convention, F1). Under these the hypothesis forces $L\ge1$.
• Consumed local claims. None; the page depends on no other reconstruction page. Its Standing paragraph records the source as a claimed result, not refereed, with an author-side Lean file not built in this repository, and holds the page itself to author-recorded standing.
• Batch acceptance order. Not applicable; this is a single focused review.

Findings

F1. Severity: suggested. Location: Definitions, "the empty sum, with value $0$, is allowed." Defect: the convention is stated as a definition and not marked as a reading supplied by the page. The source's Lemma 3.1 (p. 2) says only "a sum of at most $L$ distinct divisors of $n$" and never says whether zero summands are allowed; the page's form has the weaker hypothesis and so is formally the stronger statement. Witness: source p. 2, the lemma's second sentence; source p. 3, the proof's bound "$0\le(m-s)/A<n$", which is consistent with the convention but does not state it. The reconstructed argument proves the stronger form (Strongest attack, second attempt), so nothing mathematical changes. Proposed replacement text: "the empty sum, with value $0$, is allowed (a reading supplied here: the source does not say, and its proof's bound $0\le(m-s)/A$ is consistent with it; the argument below covers both readings)."

F2. Severity: note. Location: Proof, last paragraph, "$\max(h(n),1,L+h(n))=h(n)+L$". Defect: the equality is asserted without its two trivial supports, $L\ge0$ and $h(n)\ge1$. Witness: both hold ($L$ counts summands; the integer $1\le n$ needs one summand of $n$), and the hypothesis even forces $L\ge1$ because the residue $1$ modulo $A\ge2$ is not the empty sum. Proposed replacement text: "$\max(h(n),1,L+h(n))=h(n)+L$ (as $L\ge0$ and $h(n)\ge1$)".

F3. Severity: note. Location: Standing, "the note is a proof claim on the erdosproblems.com proof-claims tab". Defect: the library card's provenance paragraph records the registration as a partial proof claim, and Problem 18 asks three questions of which the note bears on the first; dropping "partial" can be read as a full-solution claim. Witness: the card's provenance paragraph ("registered ... as a partial proof claim"); the three questions of the Problem 18 Statement. Proposed replacement text: "the note is a partial proof claim on the erdosproblems.com proof-claims tab".

F4. Severity: note. Location: frontmatter desc, "a short sum of divisors ... h(An) grows by at most the length of those sums". Defect: the summary omits the hypothesis that each sum has total at most $n$, which the proof uses (W1), and "the length" means the common bound $L$. Acceptable as a compressed description, since the Statement is exact. Proposed replacement text: "if every residue modulo A is a sum of at most L distinct divisors of a practical n, of total at most n and avoiding multiples of A, then An is practical and h(An) is at most h(n) + L."

Verdict

Source fidelity: faithful. The page's Statement is the source's Lemma 3.1 clause for clause; the locators (Lemma 3.1 in Section 3; statement on physical p. 2, proof on p. 3; seven pages; author line) are correct; the one convention the source leaves unsaid is made explicit on the page (F1, a suggested labeling only).

The argument as reconstructed: sound. Every deduction was re-derived (W1 to W3); the hypotheses are used exactly where needed, and the only nontrivial step, the disjointness of the two summand groups, rests on the "no summand divisible by $A$" hypothesis and fails without it (Strongest attack).

Limitations: the review covers the lemma's statement and proof and the page's locators and standing text; it does not check the consumer pages the Source paragraph names, and it judges none of the source's other results. There are no required corrections. This focused review assigns no tier and changes no status.