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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. For every infinite A={a1<a2<⋯ }⊆NA=\{a_1<a_2<\cdots\}\subseteq\mathbb N, with A(x)A(x) the number of indices ii with lcm⁡(ai,ai+1)≤x\operatorname{lcm}(a_i,a_{i+1})\le x: A(x)≤(c+o(1))x1/2A(x)\le(c+o(1))x^{1/2} with c=∑k≥1(k1/2−(k−1)1/2)/k=1.8600…c=\sum_{k\ge1}(k^{1/2}-(k-1)^{1/2})/k=1.8600\ldots, so the answer to the first question is yes; and lim inf⁡A(x)/x1/2≤1\liminf A(x)/x^{1/2}\le1, with A=NA=\mathbb N attaining 11, so the largest possible value of the liminf is exactly 11.

The result. P. Erdős and E. Szemerédi, Megjegyzések az American Mathematical Monthly egy problémájához (Remarks on a problem of the American Mathematical Monthly), Mat. Lapok 28 (1980), no. 1--3, 121--124, in Hungarian (MR 82c:10066, Zbl 476.10045); the year is the only date the volume gives, so the page name uses its first day. Library home: erdos_1980_megjegyzesek_az_american_mathematical_monthly_egy. The paper writes the count as F(A,X,2)F(A,X,2) inside the family F(A,X,i)F(A,X,i) counting blocks of ii consecutive terms whose least common multiple is at most XX; F(A,X,2)=A(X)F(A,X,2)=A(X). Theorem I (printed p. 121): lim sup⁡X→∞F(A,X,2)/X1/2≤c\limsup_{X\to\infty}F(A,X,2)/X^{1/2}\le c, and if equality holds then lim inf⁡F(A,X,2)/X1/2=0\liminf F(A,X,2)/X^{1/2}=0. Theorem II (printed p. 121): lim inf⁡X→∞F(A,X,2)/X1/2≤1\liminf_{X\to\infty}F(A,X,2)/X^{1/2}\le1 for every AA. The example A=NA=\mathbb N is the site's, not the paper's: A(x)=⌊(4x+1−1)/2⌋A(x)=\lfloor(\sqrt{4x+1}-1)/2\rfloor, so A(x)/x1/2→1A(x)/x^{1/2}\to1 (an elementary check made in this corpus). Theorem I's proof (pp. 122--123) splits the terms into the ranges ((k−1)x,kx ](\sqrt{(k-1)x},\sqrt{kx}\,], where a pair with least common multiple at most xx forces k−1k-1 consecutive integers out of AA, so each range holds at most (kx−(k−1)x)/k(\sqrt{kx}-\sqrt{(k-1)x})/k good pairs; Theorem II's proof (p. 123) runs the same count on [xi,xi2][x_i,x_i^2] along values xix_i where the counting function of AA is near its lower density. That proof has a numerical error: its last step asserts that γ=∑j≥2(j−j−1)/(j−1)\gamma=\sum_{j\ge2}(\sqrt j-\sqrt{j-1})/(j-1) is less than 11, but γ=1.1840…\gamma=1.1840\ldots (the j=2j=2 term alone is 0.4140.414, and the partial sums pass 11 at j=30j=30), so the displayed bound F(A,xi2,2)≤αxi+(1−α)γxi+o(xi)F(A,x_i^2,2)\le\alpha x_i+(1-\alpha)\gamma x_i+o(x_i) exceeds xix_i by a constant factor for every α<1\alpha<1, and the argument as printed does not give lim inf⁡≤1\liminf\le1. The theorem is true: the problem page records an authored averaging proof, a note of this corpus and not acceptance evidence.

What the paper does not settle. The site's commentary says the constant cc is optimal, attributing this to [ErSz80]; the paper exhibits no sequence attaining it, and Theorem I says only that equality would force the liminf to zero. This point is outside the problem's two questions and is left as recorded on the problem page. The paper's Theorem III and its i=3i=3 bounds concern longer blocks and are not part of the problem.

Acceptance. Reviewed: a thread comment of 26 October 2025 asked whether Theorem II holds for every AA, since A=NA=\mathbb N would then settle the problem, and Thomas Bloom, the site's curator and independent of the authors, answered on 27 December 2025 that, going by a machine translation of the Hungarian, it does, and relabeled the problem SOLVED (page last edited the same day); the commentary credits [ErSz80] for both the liminf bound and the constant (site page and thread accessed 2026-09-05 and 2026-09-18). The commentary also credits two later proofs of the first answer: Tao's thread comment of 26 October 2025 (post) gives a short argument for A(x)≪x1/2A(x)\ll x^{1/2}, and van Doorn's note (last changed 12 August 2025) proves the finite version with the constant cc, whose proof, its author writes in the thread, carries over to the problem as stated. Refereed: Matematikai Lapok is the refereed journal of the Bolyai Society. No refereed English account is known. The statement collection added a file for the problem on 20 September 2026, linked above as a record at that commit: three statements, the first question, the second question and the limsup constant, each tagged research solved with a sorry body and a formal_proof attribute naming the Lean file below, so it is a statement file and not a formalization. Read depth: the three theorem statements (p. 121) are checked clause by clause; the proof of Theorem I was read for its structure and not checked step by step, and the proof of Theorem II to its final display, where the error above was found; nothing is independently reviewed by this project.

Formalization. The repository plby/lean-proofs (Boris Alexeev) holds src/latest/ErdosProblems/Erdos440.lean, with four supporting files under Erdos440/, linked above at the repository head of 15 September 2026; the file was added on 17 August 2026 and last changed on 31 August 2026 (its commit history). Its header calls it a formalization of a solution to the problem and names Erdős and Szemerédi as informal authors and Codex and GPT-5.6 Sol as formal authors, so it is a formalization link on this page and not a claim of its own. Its closing theorem erdos_440 asserts five conjuncts: for every infinite AA the count A(x)A(x) is O(x1/2)O(x^{1/2}); the series cc is the universal limsup coefficient; some sequence attains cc; every normalized liminf is at most 11; and 11 is attained by the positive integers. The first, fourth and fifth answer the problem's two questions; the third asserts what the paper does not state. The file contains no sorry and no axiom; nothing was built, replayed or audited in this corpus, no statement-fidelity review exists, the site does not label the problem Lean, and the file gives no formalized evidence.

Depends on. No page of this wiki. The proofs are self-contained in the paper.