OPEN
This is open, and cannot be resolved with a finite computation.
If $u=\{u_1<u_2<\cdots\}$ is a sequence of integers such that $(u_i,u_j)=1$ for all $i\neq j$ and $\sum \frac{1}{u_i}<\infty$ then let $\{a_1<a_2<\cdots\}$ be the sequence of integers which are not divisible by any of the $u_i$. For any $x$ define $t_x$ by\[u_1\cdots u_{t_x}\leq x< u_1\cdots u_{t_x}u_{t_x+1}.\]We call such a sequence $u_i$ good if, for all $\epsilon>0$, if $x$ is sufficiently large then\[\max_{a_k<x} (a_{k+1}-a_k) < (1+\epsilon)t_x \prod_{i}\left(1-\frac{1}{u_i}\right)^{-1}.\]Is there a good sequence such that $u_n< n^{O(1)}$? Is there a good sequence such that $u_n\leq e^{o(n)}$?
Erdős
[Er81h] believed the answer to the first question is no and the second question is yes. He proved the existence of some good sequence (in which all the $u_i$ are primes).
An easy sieve argument proves that we always have, for any sequence $u$ with those properties,\[\max_{a_k<x} (a_{k+1}-a_k)> (1+o(1))t_x \prod_{i}\left(1-\frac{1}{u_i}\right)^{-1}.\]The strong form of
[208] is asking whether if $u_i=p_i^2$, the sequence of prime squares, is good.
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This page was last edited 19 October 2025.
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