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Statement

The paper's definitions (p. 71): "Let AxA_x be a set of positive integers with the least common multiple of each pair of terms not exceeding xx and ∣Ax∣|A_x| being the largest", and "let BxB_x be the union of the set of positive integers not exceeding x/2\sqrt{x/2} and the set of even integers between x/2\sqrt{x/2} and 2x\sqrt{2x}". Theorem (p. 71).

∣Ax∖Bx∣=o(x).|A_x\setminus B_x|=o(\sqrt x).

In particular ∣Ax∣=∣Bx∣+o(x)=98x+o(x)|A_x|=|B_x|+o(\sqrt x)=\sqrt{\tfrac98x}+o(\sqrt x). The Note after the theorem says the o(x)o(\sqrt x) can be given explicitly from the proof, and that ∣Ax∩Bx∣=9x/8+o(x)|A_x\cap B_x|=\sqrt{9x/8}+o(\sqrt x) and ∣Bx∖Ax∣=o(x)|B_x\setminus A_x|=o(\sqrt x) (pp. 71--72).

The introduction records (p. 71) that Erdős proposed the problem in 1951 (the paper's [3], the Mat. Lapok problem), that 9x/8+O(1)≤∣Ax∣≤4x+O(1)\sqrt{9x/8}+O(1)\le|A_x|\le\sqrt{4x}+O(1) with a proof in the paper's [4] (Erdős 1965), that Choi improved the upper bound to 1.638x1.638\sqrt x and then to 1.43x1.43\sqrt x, and that the problem is E2 and part of B26 in Guy's book.

Source. Yong-Gao Chen, Sequences with bounded l.c.m. of each pair of terms, Acta Arith. 84 (1998), no. 1, 71--95, DOI 10.4064/aa-84-1-71-95 (the Crossref record was read); the retained file is the publisher's 25-page PDF; the Theorem is on printed p. 71 = PDF p. 1 and the Note runs over pp. 71--72, read in the text layer and on the page images.

Read depth. Claims checked: the definitions, the Theorem and the Note were read clause by clause on the page images of pp. 71--72. The proof (Sections 1--2, pp. 72--95) was not read beyond the statement of Lemma 1 (p. 72).

Proof pointer

Section 1 (pp. 72--76) proves sieve lemmas; Lemma 1 uses the Eratosthenes--Legendre sieve to find kk in (c1x1/2+c2, c1(x1/2+x1/4)+c2)(c_1x^{1/2}+c_2,\,c_1(x^{1/2}+x^{1/4})+c_2) such that every prime factor of ∏i(aik+bi)\prod_i(a_ik+b_i) exceeds log⁡x/(6log⁡M)\log x/(6\log M). Section 2 (pp. 76--95) proves the Theorem by a case analysis of the elements of AxA_x against BxB_x. Not read here.

Dependencies

Standard sieve results (Halberstam--Richert, the paper's [6]).

Bears on

  • Problem 441: answers the first question asymptotically: g(N)=∣AN∣∼(9N/8)1/2g(N)=|A_N|\sim(9N/8)^{1/2}, the value of Erdős's construction, which is the site's "Chen established the asymptotic".