SOLVED
This has been resolved in some other way than a proof or disproof.
Suppose $a_1<a_2<\cdots$ is a sequence of integers such that for all integer sequences $t_n$ with $t_n\geq 1$ the sum\[\sum_{n=1}^\infty \frac{1}{t_na_n}\]is irrational. How slowly can $a_n$ grow?
One possible definition of an 'irrationality sequence' (see also
[263] and
[264]). An example of such a sequence is $a_n=2^{2^n}$ (proved by Erdős
[Er75c]), while a non-example is $a_n=n!$. It is known that if $a_n$ is such a sequence then $a_n^{1/n}\to\infty$.
This was essentially solved by Hančl
[Ha91], who proved that such a sequence needs to satisfy\[\limsup_{n\to \infty} \frac{\log_2\log_2 a_n}{n} \geq 1.\]More generally, if $a_n\ll 2^{2^{n-F(n)}}$ with $F(n)<n$ and $\sum 2^{-F(n)}<\infty$ then $a_n$ cannot be an irrationality sequence.
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This page was last edited 28 September 2025.
Additional thanks to: Vjekoslav Kovac and Terence Tao
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