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Statement

There is an absolute constant $C\ge1$ such that, for every $\varepsilon>0$ and all sufficiently large $N$ depending on $\varepsilon$, if $A\subseteq[1,N]$, $|A|\ge\alpha N$, and

$$(\log N)^{-1/7+\varepsilon}\le\alpha\le1/2,$$

then there are integers $1\le s\le t\le\exp(C/\alpha)$ and $B\subseteq A$ with $\sum_{n\in B}1/n=s/t$. The paper does not require $s/t$ to be in lowest terms.

Source. Liu–Sawhney, arXiv:2404.07113v1, Proposition 1.4, p. 2; remarks p. 3; proof p. 21. Read depth: claims checked (the statement on p. 2 and the remarks on p. 3 were re-read clause by clause in the text layer on 2026-09-18 and agree with the statement above); the proof is not verified, and the coverage gap below stands. The published version (Int. Math. Res. Not. 2026, no. 2, rnaf382) was not compared.

Proof pointer and sketch

Localize to a dyadic interval retaining reciprocal mass comparable to $\alpha$. Delete non-smooth integers and those with too many prime factors, and prune thin prime-power fibers. The paper applies Proposition 5.2 with $\delta=1/7$ and its second term in $\Gamma$. A sieve argument locates a prime divisor $p\in[10/\alpha^2,\exp(C/\alpha)]$ of a surviving denominator. Since $p$ divides the common denominator $Q$, the proof chooses a rational with denominator $p$ as its target, subject to the checks below.

The exponential dependence is sharp up to constants: consider integers in $[N/2,N]$ with no prime factor below $L=\exp(c/\alpha)$, where $c>0$ is sufficiently small. The sieve gives at least $\alpha N$ such integers for sufficiently large $N$. Their total reciprocal mass is below one. Every nonempty subsum, after reduction, has a denominator greater than one with no prime divisor below $L$; its denominator is therefore at least $L$.

Coverage gap. Full proof and independent verification are required for Problem 310. The printed choice $M=N/2$ violates Proposition 5.2's stated $M\le N/10^4$. The fixed choice $\epsilon=1/1000$ in the proof's parameter check does not establish the claimed inequality for arbitrarily small $\varepsilon_0>0$. The displayed numerator of $\Gamma$ also has $1-\delta$ where the proposition states $1-2\delta$. Finally the chosen target $\lfloor R(A)p/5\rfloor/p$ must be checked against the earlier announced interval $[\alpha/128,\alpha/32]$. These discrepancies are recorded without inventing replacements; compare the published version first.

Dependencies and historical relation

Same-paper Lemmas 2.2, 2.3, 2.4, 6.2 and Proposition 5.2. The remarks on p. 3 explain that Bloom's Proposition 1 already yields a denominator $O_\alpha(1)$ for fixed positive $\alpha$, solving the original qualitative question. Liu–Sawhney supply the stated dependence on $\alpha$ and the logarithmically small density range.

Bears on

• Problem 310