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Problem #69: Is\[\sum_{n\geq 2}\frac{\omega(n)}{2^n}\]irrational?

Is\[\sum_{n\geq 2}\frac{\omega(n)}{2^n}\]irrational? (Here $\omega(n)$ counts the number of distinct prime divisors of $n$.)

Problem Statement

Is\[\sum_{n\geq 2}\frac{\omega(n)}{2^n}\]irrational? (Here $\omega(n)$ counts the number of distinct prime divisors of $n$.)
Categories: Number Theory Irrationality

Progress

Erdős [Er48] proved that $\sum_n \frac{d(n)}{2^n}$ is irrational, where $d(n)$ is the divisor function.

Pratt [Pr24] has proved this is irrational, conditional on a uniform version of the prime $k$-tuples conjecture.

Tao has observed that this is a special case of [257], since\[\sum_{n\geq 2}\frac{\omega(n)}{2^n}=\sum_p \frac{1}{2^p-1}.\]This sum was proved to be irrational unconditionally by Tao and Teräväinen [TaTe25].

Source: erdosproblems.com/69 | Last verified: January 13, 2026

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