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Let $N$ be a sufficiently large integer, $1\le M\le N/10$, and $N^{.9999}\le S\le N/2$. Write $L=\log N$ and $\ell=\log\log N$. Let $A$ be all integers in $[M,N]$ such that every prime-power divisor is at most $S$ and

$$\widetilde\Omega(n)\le5\ell,\qquad \Omega(n)\le10\ell.$$

Here $\Omega$ counts prime factors with multiplicity and $\widetilde\Omega$ is the largest prime exponent. Then

$$|A|\ge .89N.$$

Source. Liu–Sawhney, arXiv:2404.07113v1, Lemma 3.3, p. 10. The proof below supplies the omitted deduction using the sufficient proved form of Lemma 2.2. The definition uses the member-wise $\Omega(n)$ restriction needed by the argument; the source's preceding display prints $\Omega(N)$. The source states the lemma for $A$ as in Proposition 3.2, whose hypotheses also impose $N^{.9999}\le S\le K\le M$ and $N(\log N)^{-10}\le K\le10^{-7}N(\log N)^{-1}$; the ranges $1\le M\le N/10$ and $N^{.9999}\le S\le N/2$ above are wider, and the proof below covers them.

Bears on. Problem 297, through the major arc in the counting proof.

Proof

The union of the two exponent-exception sets is contained in $\{n\le N:\Omega(n)>5\ell\}$. The sufficient reciprocal estimate in Lemma 2.2 gives, for $\beta=5\log(3/2)-3/2>0$,

$$\#\{n\le N:\Omega(n)>5\ell\} \le N\sum_{\substack{n\le N\\\Omega(n)>5\ell}}\frac1n \ll NL^{-\beta}=o(N).$$

This uses the proved reciprocal deduction, not the false printed $O(N/L^3)$ count.

Put $t=N/S$. The hypotheses give $2\le t\le N^{.0001}$, so Lemma 2.3 applies. At most

$$\frac{2N\log(N/S)}{\log N}\le .0002N$$

integers at most $N$ have a prime-power divisor exceeding $S$. There are at least $.9N-1$ integers in $[M,N]$. Removing both exceptional sets therefore leaves

$$|A|\ge .8998N-o(N)\ge .89N$$

for sufficiently large $N$.

Scope

This proves the density input to the restricted Proposition 3.2. The external prime-number estimates enter through the two linked lemmas. No published-version equivalence or stronger literal Lemma 2.2 count is asserted.