SOLVED

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 $a_n=2^{2^n}$, 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.