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Statement

For sufficiently large $N$, let $8\le4y<z\le(\log N)^{1/2}$. Let $Y\subseteq[1,N]\cap\mathbb N$ consist of the integers divisible by distinct primes $p_1,p_2\in[y,z]$ with $4p_1<p_2$. Then

$$|([1,N]\cap\mathbb N)\setminus Y| \ll N\left(\frac{\log y}{\log z}\right)^{1/2}.$$

Source. Bloom, arXiv:2112.03726v2, Lemma 2, pp. 7–8.

Rewritten proof

First assume that $\log z>16\log y$; otherwise the claimed right side is at least $N/4$, so the trivial bound $N$ proves the assertion after increasing the absolute constant. Put

$$w=\exp\sqrt{(\log y)(\log z)}.$$

In the nontrivial case $4y<w<z$. Integers with no prime divisor in $[w,z]$ number $\ll N\log w/\log z$ by the $[1,N]$ version of Lemma 1. Any remaining integer outside $Y$ has a prime divisor $p\in[w,z]$ but none in $[y,p/4)$. Write it as $pm$. Because $p$ is outside this latter interval, $m$ avoids every prime there. Lemma 1 bounds these $m\le N/p$ by

$$\ll\frac{N}{p}\frac{\log y}{\log p}.$$

Here $p/4\le z\le\sqrt{\log N}\le\log(N/p)$ for large $N$, so the sieve parameters are admissible. If $2\le y<3$, apply the lemma with lower endpoint $3$ and absorb the bounded ratio of logarithms. Open or closed upper endpoints affect only constants in the same product estimate.

Summing over $p$ (overcounting is harmless) gives

$$|[1,N]\setminus Y| \ll N\left(\frac{\log w}{\log z} +\log y\sum_{p\ge w}\frac1{p\log p}\right).$$

The elementary prime-counting bound $\pi(t)\ll t/\log t$ and partial summation give $\sum_{p\ge w}(p\log p)^{-1}\ll1/\log w$. Both remaining terms now equal $(\log y/\log z)^{1/2}$, proving the claim.

Dependencies and source detail

Lemma 1 and the external Chebyshev prime-counting estimate suffice. The paper directly chooses this $w$ inside $(4y,z)$; the trivial-case split above makes its endpoint requirement explicit when $z$ is close to $4y$.

Bears on

• Problem 298
• Problem 299