Fan: Strongly complete sets and a conjecture of Erdős

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remark_4_2: If every set with at least two elements in each dyadic interval and divergent sums of distances to integers were strongly complete, the nonzero floors of the doubling multiples of two reals whose ratio is not a power of two, one of them not a dyadic rational, would be strongly complete; the proved threshold is five; context for Problem 354.

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Steve Fan, Strongly complete sets and a conjecture of Erdős, arXiv:2607.14071 [math.NT; math.CO], MSC 11B13, 11B75, 11J71; five versions: v1 15 July 2026, v2 23 July, v3 25 July, v4 9 September 2026 (22:13 UTC), v5 16 September 2026 (22:53 UTC; "35 pages; This version fixed several typos and expanded Remark 4.1"). No journal reference or DOI on the arXiv record on 2026-09-28; a preprint, not refereed; license CC BY-NC-ND 4.0.

Two versions were read for this card. The copy the result pages cite is v5, the current version, 36 PDF pages (the references end on p. 36): downloaded from https://arxiv.org/pdf/2607.14071v5; 530,449 bytes. The earlier v4, the version the bounty site's review of the Problem 354 record cites ("Fan v4"), 35 pages: downloaded from https://arxiv.org/pdf/2607.14071v4; 526,506 bytes. Label map: the remark on Problem 354 is the second paragraph of Remark 4.1 of v4 (p. 19) and Remark 4.2 of v5 (p. 20); v5's Remark 4.1 (pp. 19--20) expands the first paragraph of v4's Remark 4.1, which gives $2\le M_2^*\le5$, to $M_\rho^*\ge u_\rho$ for $\rho\ge2$; Corollary 1.2 (p. 4) and the introduction's (1.8)--(1.9) (p. 4) are unchanged between the two. Result pages cite v5. The arXiv record (https://arxiv.org/abs/2607.14071, read 2026-10-02) names the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 license for both v5 and v4.

Read status. Claims checked for Corollary 1.2 (p. 4), the definitions (1.8)--(1.9) (p. 4) and Remark 4.2 (p. 20), read clause by clause in the text layer of v5 and compared with v4; the remark's half-page argument was read through and not independently reviewed; Theorem 1.1 (p. 3) and the rest of the paper were read at statement level only. The paper's AI disclosure (p. 35) states that "ChatGPT 5.6 was used for proofreading the manuscript" and "suggested a core idea underlying the current shorter proof of Lemma 3.2", the author taking "full responsibility"; recorded as the source's own disclosure. Consumed here as context for Problem 354 only; the paper's main results concern Erdős's 1961 conjecture (the site's Problem 254) and the Burr--Erdős--Graham--Li problem on mixed power sets (the site's Problem 124), and are not triaged here.

Overview

Definitions (pp. 2--4): $A\subseteq\mathbb N$ is complete if every sufficiently large integer is a sum of distinct elements of $A$, and strongly complete if $A\setminus B$ is complete for every finite $B\subseteq A$; condition (1.5) is $\sum_{a\in A}\|a\theta\|=\infty$ for every $\theta\in\mathbb T\setminus\{0\}$, where $\|x\|$ is the distance to the nearest integer, a condition every strongly complete set satisfies (p. 3). Theorem 1.1 (p. 3): for $\rho>1$ put $u_\rho=\lceil\rho(\rho-1)\rceil$, $v_\rho=\lceil\rho^3/(\rho+1)\rceil$ and $M_\rho=\min\{2u_\rho+1,2v_\rho\}$; if $A$ satisfies (1.5) and $|A\cap(\rho^k,\rho^{k+1}]|\ge M\ge M_\rho$ for all large $k$, then the number of representations of $n$ as a sum of distinct elements of $A\setminus F$ grows faster than $n^{(M-M_\rho)\log_\rho2}$ for every finite $F$, so $A$ is strongly complete. Corollary 1.2 (p. 4), the case $\rho=2$: every $A$ with (1.5) and at least five elements in every $(2^k,2^{k+1}]$ for large $k$ is strongly complete, which the paper presents as confirming Erdős's 1961 conjecture in a strong form. $M_\rho^*$ (1.8) is the least positive integer such that every $A$ with (1.5) and at least $M_\rho^*$ elements in every $(\rho^k,\rho^{k+1}]$ for large $k$ is strongly complete, so $M_2^*\le5$; v5's expanded Remark 4.1 (pp. 19--20) shows $M_\rho^*\ge u_\rho$ for $\rho\ge2$ (at $\rho=2$ through $A=\{2^k+1\}$, which satisfies (1.5) and is incomplete), so $2\le M_2^*\le5$, and reports that random sets with $M$ elements per interval are strongly complete almost surely when $M\ge u_\rho$ and incomplete almost surely when $M<u_\rho$.

The connection with Problem 354 (p. 4): with $A_{\alpha,\beta}=\{\lfloor2^k\alpha\rfloor,\lfloor2^k\beta\rfloor:k\ge0\}\setminus\{0\}$ (1.9), $\alpha\sim\beta$ when $\alpha/\beta=2^n$ for some $n\in\mathbb Z$ and "dyadic rational" meaning $\alpha\sim n$ for a nonzero integer $n$, the paper recalls what Hegyvári's argument gives with small modifications, that for pairwise nonequivalent $\alpha,\beta,\gamma>0$ the three-ray set $A_{\alpha,\beta,\gamma}$ is strongly complete if and only if one of them is not a dyadic rational, Hegyvári's conjecture that $A_{\alpha,\beta}$ is complete when $\alpha\not\sim\beta$ and one of $\alpha,\beta$ is not a dyadic rational, and Hegyvári's proof of the case where exactly one of them is a dyadic rational; Remark 4.2 (p. 20) shows that $M_2^*=2$ would imply the full conjecture. The rest of the paper: Theorem 1.3 (p. 5), a partition of a set with Erdős's conditions into countably many strongly complete sets with prescribed local growth; Theorem 1.4 (p. 6), strong completeness of polynomially perturbed ray sets $\{\lfloor t\alpha^n\rfloor,\lfloor t\alpha^n\rfloor+P(n)\}$; Theorem 1.5, mixed power sets; all at statement level only here.

Bears on. #354: context only. Remark 4.2 (p. 20; Remark 4.1 in v4, p. 19) shows that $M_2^*=2$ would imply that $A_{\alpha,\beta}$ is strongly complete, hence Hegyvári's conjecture that it is complete, when $\alpha/\beta$ is not a power of $2$ and one of $\alpha,\beta$ is not a dyadic rational; the paper proves only $2\le M_2^*\le5$, so it resolves neither question of the problem, as the bounty site's review of the accepted part (i) proof also notes ("leaves the relevant two-ray case unresolved"). Its Corollary 1.2 is the criterion Geneson's preprint cites for strong completeness.

Results.

• Remark 4.2 (p. 20; Remark 4.1 in v4): $M_2^*=2$ would imply Hegyvári's conjecture; with Corollary 1.2 and v5's Remark 4.1, $2\le M_2^*\le5$.

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