SOLVED
This has been resolved in some other way than a proof or disproof.
Let $A=\{a_1<a_2<\cdots\}\subseteq \mathbb{N}$ be infinite and let $A(x)$ count the number of indices for which $\mathrm{lcm}(a_i,a_{i+1})\leq x$. Is it true that $A(x) \ll x^{1/2}$?
How large can\[\liminf \frac{A(x)}{x^{1/2}}\]be?
Taking $A=\mathbb{N}$ shows that\[\liminf \frac{A(x)}{x^{1/2}}=1\]is possible. Erdős and Szemerédi
[ErSz80] proved that it is always $\leq 1$.
Tao in the comments has given a simple proof that $A(x) \ll x^{1/2}$. van Doorn
has proved that $A(x) \leq (c+o(1))x^{1/2}$ where\[c=\sum_{n\geq 1}\frac{1}{n^{1/2}(n+1)}\approx 1.86.\]This was already proved by Erdős and Szemerédi
[ErSz80], who showed that this constant is the best possible.
There are more related results (particularly for the more general case of $\mathrm{lcm}(a_i,a_{i+1},\ldots,a_{i+k})$) in
[ErSz80].
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This page was last edited 27 December 2025. View history
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Additional thanks to: Zachary Chase, Terence Tao, and Wouter van Doorn
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #440, https://www.erdosproblems.com/440, accessed 2026-04-11
