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Statement

On p. 4 the paper says: "if one instead takes $A$ to be the set of squarefree numbers with exactly $k$ prime factors for any fixed $k\ge1$, a standard calculation again based on Mertens' theorem (see also Lemma 1 below) shows that the left-hand side of (1) now grows like $(1+o(1))\mathrm{Log}_2^{k-1}x/k!$, but the left-hand side of (2) stays bounded in the limit $x\to\infty$ (and the defect (9) decays like $(1+o(1))k^2/\mathrm{Log}_2x$). In particular, the answer to Problem 1 is negative; this result and construction was already implicitly observed in [1] (see the discussion after [1, Proposition 2.1]), although the authors seem to have been unaware of Problem 1." Here (1) is the hypothesis $(1/\mathrm{Log}_2x)\sum_{n\in A:n\le x}1/n\to\infty$ and (2) the conclusion of Problem 1 (the site's #442, p. 2), and [1] is Bergelson and Richter.

For $k=2$ the hypothesis holds, since $(1+o(1))\mathrm{Log}_2x/2\to\infty$, so the squarefree numbers with exactly two prime factors are the simplest counterexample the paper names. The paper also notes (pp. 2--4) that the primes, which it says likely motivated the $\log\log x$ threshold, give both quantities of order $1$.

Source. Terence Tao, Dense sets of natural numbers with unusually large least common multiples, arXiv:2407.04226v5 (11 November 2025); the remark on p. 4 = PDF p. 4, read in the text layer and on the page image.

Read depth. Claims checked: the remark was read clause by clause on the page image. It is asserted with a pointer to Lemma 1; the "standard calculation" is not written out at this point of the paper and was not reproduced here.

Proof pointer

Mertens' theorem, via the paper's Lemma 1, per the remark. Not read here.

Dependencies

Mertens' theorem; Bergelson and Richter (the paper's [1]) for the earlier implicit observation.

Bears on

• Problem 442: the elementary negative answer, weaker than Theorem 1 in growth rate but sufficient for the site's yes-or-no question; the external Lean file the problem page describes uses exactly the case $k=2$.