I am interested in knowing the exact asymptotic formula (if it exists) of
$\displaystyle \sum_{n \leq X} \mu^2(n) 2^{-\omega(n)},$
where $\omega(n)$ denotes the number of distinct prime divisors of $n$. I've looked all over and it does not seem to be written down.
Since we are restricting to square-free numbers, this is equal to the sum
$\displaystyle \sum_{n \leq X} \frac{\mu^2(n)}{d(n)},$
where $d(n)$ is the number of divisors of $n$.