Problem 1101

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 $u_n\leq e^{o(n)}$ ?

Source: erdosproblems.com/1101

Number theory

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Open

No claim settles this problem.

Open. The site labels the problem OPEN (page last edited 19 October 2025). Its two questions are the problem's two parts; Erdős expected the first to have a negative answer and the second a positive one. The second, whether some good sequence has $u_n\leq e^{o(n)}$ , has one pending partial claim, Li's subexponential good sequence of 2026, which constructs a good sequence of primes with $u_n=\exp((1+o(1))\sqrt n)$ . The first, whether some good sequence has $u_n<n^{O(1)}$ , has no claim.

Problem 1101

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Proof claims

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