---

Let $N$ be sufficiently large, $L=\log N$, $\ell=\log\log N$, and $S\ge N^{.9999}$. Let $1\le M\le N/2$, and let $A$ consist of all integers in $[M,N]$ whose prime-power divisors are at most $S$ and for which $\widetilde\Omega(n)\le5\ell$, $\Omega(n)\le10\ell$. Suppose a positive integer $b$ satisfies

$$\omega(b)\le5,\quad \widetilde\Omega(b)\le5\ell,\quad \Omega(b)\le5\ell+4, \quad 2000\le U:=N/b\le L^9/40,$$

and all prime-power divisors of $b$ are at most $S$. Then there is $n\in A\cap[N/2,N]$ divisible by $b$. Here $\omega$ counts distinct prime factors; $\Omega$ counts them with multiplicity, and $\widetilde\Omega$ is the largest exponent.

Source. This expands the implicit multiple-selection step in Liu–Sawhney, arXiv:2404.07113v1, Proposition 3.2, p. 12. The argument is a compilation-supplied justification of that step. The precise prime reciprocal estimate is external at Theorem 2.1.

Bears on. Problem 297, through the nonempty fibers in the counting minor-arc proof.

Proof

First find an integer $u\in[U/2,U]$ coprime to $b$ and satisfying $\Omega(u)\le\ell$. For every positive integer $d$, the number of its multiples in this closed real interval differs from $U/(2d)$ by at most one. Inclusion–exclusion over the at most five prime divisors of $b$ therefore gives at least

$$\frac U2\prod_{p\mid b}(1-1/p)-32 \ge\frac U2\frac{16}{77}-32 >\frac U{10}-32\ge\frac U{12}$$

integers coprime to $b$. The product $16/77$ is that for the five smallest primes, and the last inequality holds for $U\ge2000$.

The identity

$$\sum_{u\le U}\Omega(u) =\sum_{p^j\le U}\left\lfloor\frac U{p^j}\right\rfloor$$

counts each prime-power divisor once. The prime reciprocal estimate and convergence of $\sum_p\sum_{j\ge2}p^{-j}=\sum_p1/(p(p-1))$ give an absolute constant $C_1$ with

$$\sum_{u\le U}\Omega(u)\le U(\log\log U+C_1) \qquad(U\ge2000).$$

Thus the number of integers at most $U$ with $\Omega(u)>\ell$ is at most $U(\log\log U+C_1)/\ell$. Uniformly for $2000\le U\le L^9/40$, this is $o(U)$, since

$$\frac{\log\log U+C_1}{\ell} \le\frac{\log(9\ell)+C_1}{\ell}\longrightarrow0.$$

For large $N$ it is less than $U/12$. Consequently one of the coprime integers in the interval has $\Omega(u)\le\ell$.

Set $n=bu$. Then $n\in[N/2,N]\subseteq[M,N]$. Coprimality prevents combining prime exponents across the two factors, so

$$\widetilde\Omega(n) =\max\{\widetilde\Omega(b),\widetilde\Omega(u)\}\le5\ell, \qquad \Omega(n)\le6\ell+4\le10\ell$$

for large $N$. All prime-power divisors of $b$ are at most $S$ by hypothesis; those of $u$ are at most $u\le L^9/40<S$. Therefore $n\in A$, as required.

Scope

The restricted Proposition 3.2 constructs $b=qp'rp$ and verifies every hypothesis above. This lemma does not assume a prime in an interval whose lower endpoint may stay bounded.