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Claim. With $H(x,z)$ the least number of integers up to $x$ divisible by no element of a set $\mathcal S$ of integers greater than $1$ with $\sum_{s\in\mathcal S}1/s\le z$ (the paper's condition (3)), Theorem 5 states that for every fixed $Z>1$, uniformly for $1\le z\le Z$ and $x\ge2$,

$$H(x,z)\asymp\frac{x^{\exp(1-z)}}{\log x}.$$

Taking $z=C$: for $C=1$ the count has exact order $x/\log x$, which gives the asked lower bound with $c=1$, and for a fixed $C>1$ the count is at most a constant times $x^{e^{1-C}}/\log x$, below $x/(\log x)^c$ for every $c>0$ and all large $x$, so the asked lower bound fails. With the union bound for $0<C<1$, the question is answered yes for $0<C\le1$ and no for $C>1$, the same answer, and the same disproof of the bound read for every $C>0$, as Ruzsa's claim, whose Theorem I the paper names as the result it improves. The first part of Theorem 5, together with Theorem 4 and Corollary 2, gives $H(x,1)\le(ae^{-\delta}+o(1))x/\log x$ with $ae^{-\delta}\approx0.878$, and Question 2 asks whether this is the asymptotic. The source is A. Weingartner, The Schinzel-Szekeres function, Res. Number Theory 11 (2025), no. 3, Paper No. 63, 32 pp., DOI 10.1007/s40993-025-00643-9 (published 17 June 2025); arXiv:2310.13038, v1 of 19 October 2023, v2 of 13 June 2025. Library home: weingartner_2025_schinzel_szekeres_function.

Formulation. The paper's $H(x,z)$ excludes $1$ from the sifting set, as the problem page's corrected Statement does; see the formulation note on Ruzsa's claim page.

Acceptance. Refereed: the paper appeared in Research in Number Theory. Reviewed: the site's curator, Thomas Bloom, who is independent of the author, credits the paper in the curator's commentary (page last edited 8 April 2026, accessed 2026-09-05 and 2026-10-07) with the exact order $H_C(x)\asymp x^{e^{1-C}}/\log x$ for fixed $C>1$ and with the finer estimates at $C=1$, and a thread comment of 18 December 2025 derives the negative answer for $C>1$ from Theorem 5.

Depends on. No page of this wiki.