Problem #257: Let $A\subseteq \mathbb{N}$ be an infinite set

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Progress

If $A=\mathbb{N}$ then this series is $\sum_{n}\frac{d(n)}{2^n}$, where $d(n)$ is the number of divisors of $n$, which Erdős [Er48] proved is irrational. In general, if $f_A(n)$ counts the number of divisors of $n$ which are elements of $A$ then\[\sum_{n\in A}\frac{1}{2^n-1}=\sum_n \frac{f_A(n)}{2^n}.\]The case when $A$ is the set of primes is [69]. This case (and when $A$ is the set of prime powers) was settled in the affirmative by Tao and Teräväinen [TaTe25].

Erdős [Er68d] proved this sum is irrational whenever $(a,b)=1$ for all $a\neq b\in A$ and $\sum_{n\in A}\frac{1}{n}<\infty$ (and thought that the condition $(a,b)=1$ could be dropped by complicating his proof).

There is nothing special about $2$ here, and this sum is likely irrational with $2$ replaced by any integer $t\geq 2$.

In [Er88c] Erdős goes further and speculates that $\sum_{n\in A}\frac{1}{2^n-t_n}$ is irrational for every infinite set $A$ and bounded sequence $t_n$ (presumably of integers, and presumably excluding the case when $t_n=0$ for all $n$). This was disproved by Kovač and Tao [KoTa24], and in the comments Kovač has sketched a proof that there exists some choice of $t_n$ with $1\leq t_n\leq 6$ such that this sum is rational.