SOLVED
This has been resolved in some other way than a proof or disproof.
Is it true that if $A\subseteq\{1,\ldots,n\}$ is a set such that $[a,b]>n$ for all $a\neq b$, where $[a,b]$ is the least common multiple, then\[\sum_{a\in A}\frac{1}{a}\leq \frac{31}{30}?\]Is it true that there must be $\gg n$ many $m\leq n$ which do not divide any $a\in A$?
The first bound is best possible as $A=\{2,3,5\}$ demonstrates.
Resolved by Schinzel and Szekeres
[ScSz59] who proved the answer to the first question is yes and the answer to the second is no, and in fact there are examples with at most $n/(\log n)^c$ many such $m$, for some constant $c>0$. They further proved that, for any $\epsilon>0$, there is such a sequence for which the sum is $>1-\epsilon$.
Chen
[Ch96] has proved that if $n>172509$ then\[\sum_{a\in A}\frac{1}{a}< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}.\]In
[Er73] Erdős further speculates that in fact\[\sum_{a\in A}\frac{1}{a}\leq 1+o(1),\]where the $o(1)$ term $\to 0$ as $n\to \infty$.
In
[Er98] Erdős mentions that he, Schinzel, and Szekeres conjectured that $2,3,5$ and $3,4,5,7,11$ are the only two sequences for which the sum is $>1$.
See also
[784].
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This page was last edited 17 October 2025.
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