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For a positive integer $n$, let $\omega(n)$ be the number of distinct prime divisors of $n$. For non-zero integers $m,n$ let $\left(\frac{m}{n}\right)$ be the Jacobi symbol.

For positive $X$, interpreted as a quantity that tends to infinity, how does one evaluate asymptotically the sum

$$\displaystyle \sideset{}{^\ast}\sum_{mn \leq X} 2^{\omega(m) - \omega(n)} \left(\frac{m}{n}\right)?$$

Here the summation $\sideset{}{^\ast}\sum$ denotes that the sum is over square-free $mn$. In particular, $m,n$ are square-free and co-prime.

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    $\begingroup$ Where does it appear, what is it useful for ? $\endgroup$
    – reuns
    Commented Aug 3, 2019 at 23:44
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    $\begingroup$ It might be helpful to write the sum as $\sum_{k\le x} \mu^2(k) 2^{\omega(k)}f(k)$ where $f(k) = \sum_{d\mid k} 4^{-\omega(d)} \big( \frac{k/d}d \big)$. For example, the trivial upper bound on $f(k)$ is $\sum_{d\mid k} 4^{-\omega(d)} = (5/4)^{\omega(k)}$, so that your sum is bounded above by $\sum_{k\le x} \mu^2(k) (5/2)^{\omega(k)} \sim c x(\log x)^{3/2}$ for some constant $c$. Numerical explorations of the (not quite multiplicative) function $f(k)$ might yield some insight. $\endgroup$
    – Greg Martin
    Commented Aug 4, 2019 at 6:02
  • $\begingroup$ @GregMartin Thank you, I'll try to follow this idea. $\endgroup$
    – Stanley Yao Xiao
    Commented Aug 5, 2019 at 21:30

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