Cubefree

$\large \sum_{n \text{ cube-free}} \dfrac{(-2)^{2\omega(n)-\Omega(n)}}{n^2}$

If the sum above can be expressed in the form of $$\dfrac a{b \pi^m}$$, where $$a,b$$ and $$m$$ are positive integers with $$a,b$$ coprime, find $$a+b+m$$.

Clarifications:

• $$n$$ runs through all positive integers.

• $$\omega(n)$$ denotes the number of distinct prime divisors of $$n$$.

• $$\Omega(n)$$ counts the prime divisors of $$n$$ with multiplicity.

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