---

Statement as printed

Write $R(E)=\sum_{n\in E}1/n$, $E_d=\{n\in E:d\mid n\}$, and $\Omega(n)=\sum_p v_p(n)$. The printed lemma gives an absolute constant $C\ge1$ that works for every sufficiently large $N$ and every $\delta\in[0,1/2]$ under these hypotheses:

• $N^{0.99}\le M\le N/10$ and $A\subseteq[M,N]$;
• the prime power $q$ satisfies $q\le M\exp(- (\log N)^{1-\delta})$ and $qR(A_q)\ge\eta>0$;
• each $n\in A$ has $\Omega(n)\le5\log\log n$.

Define the scale

$$H=\exp\!\left( \frac{\eta(\log N)^{1-\delta}} {(\log\log N)^3\log(N/M)}\right).$$

The lemma then asserts that some positive integer $d$ and some subset $A^*_{qd}$ of $A_{qd}$ satisfy

$$\min A^*_{qd}\ge Hqd,\qquad qd\ge M\exp(- (\log N)^{1-\delta}),\qquad qdR(A^*_{qd})\ge\frac{\eta}{C(\log N)^\delta\log\log N}.$$

Source: Liu–Sawhney, arXiv:2404.07113v1, Lemma 5.1 and proof, printed/PDF p. 15. The PDF's condition literally reads $\max_{n\in A}\Omega(n)\le5\log\log n$; the pointwise formulation above removes its unbound $n$ without changing its apparent meaning.

Literal-scope limitation

The unrestricted $\eta>0$ statement above is false. For a sufficiently large prime $p$, take

$$A=\{2p\},\quad N=2p,\quad M=\lfloor N/10\rfloor,\quad q=2,\quad\delta=1/2,\quad\eta=1/p.$$

The stated hypotheses hold, while $\theta=M\exp(-\sqrt{\log N})>2$. Positive output mass forces $A^*_{qd}=\{2p\}$, hence $d\mid p$. For $d=1$ the lower bound $qd\ge\theta$ fails. For $d=p$, the lower bound $\min A^*_{qd}\ge Hqd$ fails since $H>1$. This concerns the literal v1 lemma, not Theorem 1.1.

Application form

The same conclusions hold with the following explicit changes to the hypotheses: require $H\ge2$, and replace the prime-factor condition by $\Omega(n)\le5\log\log N$ for every $n\in A$. All other hypotheses and the definition of $H$ stay as above. The proof below follows p. 15 with the divisor exponent and distinct-prime convention made explicit. The application in Proposition 5.2 satisfies $H\to\infty$. These are compilation corrections, not an erratum attributed to the authors or the unseen published version.

Rewritten proof of the application form

Put

$$L=\log N,\qquad\ell=\log\log N,\qquad w=\log(N/M), \qquad a=L^{1-\delta},\qquad y=e^{a/(10\ell)}.$$

The elementary harmonic upper bound gives $\eta\le qR(A_q)\le w+O(q/M)\le2w$ for large $N$. Consequently

$$\frac{\log H}{\log y}=\frac{10\eta}{\ell^2w} \le\frac{20}{\ell^2},$$

so $2\le H\le y$. For each $n\in A_q$, define

$$d_n=\prod_{\substack{p\mid(n/q)\\p>y}}p^{v_p(n/q)}.$$

Then $qd_n\mid n$. Its remaining quotient has all prime factors at most $y$, and at most $\Omega(n)\le5\ell$ of them counted with multiplicity. It follows that

$$qd_n\ge n/y^{\Omega(n)}\ge Me^{-a/2}\ge Me^{-a}.$$

Call $n\in A_q$ poor if it has fewer than two distinct prime factors in $[H,y]$, and write $A'_q$ for the poor elements. Then $n/q$ is poor whenever $n$ is poor. We estimate the poor integers $m$ in $[M/q,N/q]$ by full intervals $[X,2X)$ starting at $M/q$. Bound the final partial part by its containing full interval. There are $O(w)$ such intervals, since $w\ge\log10$.

By Lemma 2.4, the number in an interval with no prime factor in $[H,y]$ is $O(X\log H/\log y)$. To count those with exactly one distinct prime $p$ in that range, divide by $p$ and sieve out all the other primes in $[H,y]$. Leaving $p$ unsieved allows its higher powers and multiplies the sieve product by at most $(1-1/p)^{-1}\le2$. The resulting bound is $O((X/p)\log H/\log y)$. Summing over $p$ gives $O(X\ell\log H/\log y)$ by Theorem 2.1.

These sieve uses satisfy the cutoff uniformly. Indeed, $X\ge M/q\ge e^a$ and, for $p\le y$,

$$\log(X/p)\ge a-a/(10\ell),\qquad \log y=\frac a{10\ell} \le\frac{\log(X/p)}{\sqrt{\log\log(X/p)}}$$

for large $N$; the upper endpoint is at most $2N$, so the logarithm in the denominator is at most $\ell+o(1)$. Reciprocal summation over the intervals therefore gives

$$qR(A'_q) \le\sum_{\substack{M/q\le m\le N/q\\m\text{ poor}}}\frac1m \le C_0w\ell\frac{\log H}{\log y} =\frac{10C_0\eta}{\ell}\le\eta/2.$$

Here $C_0$ is an absolute sieve comparison constant, and the last inequality holds once $\ell\ge20C_0$.

Put $\widetilde A_q=A_q\setminus A'_q$. Its remaining mass satisfies $qR(\widetilde A_q)\ge\eta/2$. Each retained $n$ has two distinct primes in $[H,y]$. Division by the prime power $q$ can remove at most one of these primes. At least one still divides $n/(qd_n)$, because the prime factors placed into $d_n$ exceed $y$. Thus $n\ge Hqd_n$.

Partition the retained integers into the finite nonempty fibers $A^*_{qd}=\{n\in\widetilde A_q:d_n=d\}$. Every such fiber lies in $A_{qd}$ and has the required two size bounds. The possible values of $d$ have prime factors in $[y,N]$, so the convergent Euler product and Theorem 2.1 imply

$$\sum_{d:\,p\mid d\Rightarrow y\le p\le N}\frac1d =\prod_{y\le p\le N}(1-1/p)^{-1} \ll\frac L{\log y}=10L^\delta\ell.$$

As

$$\eta/2\le qR(\widetilde A_q) =\sum_d\frac1d\,qdR(A^*_{qd}),$$

one fiber has $qdR(A^*_{qd})\gg\eta/(L^\delta\ell)$. Taking a sufficiently large absolute $C$ proves all conclusions.

Source corrections and verification

The source defines $d_n$ with exponent $v_p(n)$ rather than $v_p(n/q)$. If $q=p^b$, $p>y$, and $v_p(n)>b$, that choice makes $qd_n\nmid n$. The quotient exponent used above restores the required divisibility. The two primes must be distinct to ensure that one survives division by $q$. The proof needs $H\ge2$ for the sieve cutoff; the unrestricted statement admits the counterexample above. The global bound $5\log\log N$ is precisely what the proof uses and what Proposition 5.2 supplies. The application form's proof and the counterexample passed independent blind review on 2026-09-18, retained as the fresh main-proof review with its distinct grade. The compilation's own reviews checked these bounded corrections and the counterexample separately before incorporation; see the preliminary review, source checks and the earlier main-proof review, which was ruled on 2026-09-18 a coordinated compilation check rather than an independent review.

Dependencies

• Lemma 2.4: bounds integers avoiding an interval of primes, including the single-prime case after division by that prime.
• Theorem 2.1: the reciprocal-prime and Euler-product estimates used in the proof.
• The source describes this as a simplification of Bloom [4, Lemma 5.1]. This is a provenance reference; the argument is reproduced above and does not substitute Bloom's statement for the application form.

Bears on

• Problem 298, through Theorem 1.1.
• Problem 299, through Theorem 1.1.
• Problem 300, through Proposition 5.2.
• Problem 310, through Proposition 5.2.