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Notation and statement

Write $R(A)=\sum_{n\in A}1/n$, $A_q=\{n\in A:q\mid n,\ \gcd(q,n/q)=1\}$, $\mathcal Q_A=\{q=p^a:A_q\ne\varnothing\}$ and $R(A;q)=\sum_{n\in A_q}q/n$. Thus $A_q$ uses the exact prime power in $n$, not every prime power divisor.

For sufficiently large $N$ and $A\subseteq[1,N]\cap\mathbb N$, there is $B\subseteq A$ such that

$$R(B)\ge R(A)-(\log N)^{-1/200},\qquad R(B;q)\ge2(\log N)^{-1/100}\quad(q\in\mathcal Q_B).$$

Source. Bloom, arXiv:2112.03726v2, Lemma 6, p. 17.

Rewritten proof

Start with $A_0=A$. If $q_i\in\mathcal Q_{A_i}$ has $R(A_i;q_i)<2(\log N)^{-1/100}$, delete the entire fiber: $A_{i+1}=A_i\setminus(A_i)_{q_i}$. Otherwise stop. A deletion removes at least one element, so the process stops at some $B$ satisfying the required fiber inequalities.

The loss at step $i$ is $R(A_i;q_i)/q_i$, less than $2/[q_i(\log N)^{1/100}]$. No $q_i$ can recur: after deletion no remaining integer has that exact prime power, and subsequent steps only remove integers. Every such $q_i$ is at most $N$. Consequently

$$R(A)-R(B) \le2(\log N)^{-1/100}\sum_{q\le N}\frac1q \ll(\log N)^{-1/100}\log\log N \le(\log N)^{-1/200}$$

for all sufficiently large $N$.

Dependencies

The external Mertens prime-power estimate $\sum_{q\le x}1/q=\log\log x+c+O(1/\log x)$, equation (1), p. 3. The proof refines the pruning approach of Croot's Proposition 2.

Bears on

• Problem 298
• Problem 299