That's a weird sum

\[\sum _{ n=1 }^{ \infty } \frac { { 0.5 }^{ \Omega (n) } }{ { n }^{ 2 } } \sum _{ k=1 }^{ \infty } \frac { \mu \left( k \right) { 0.5 }^{ \Omega (k) } }{ { k }^{ 2 } } \]

\(\mu(n)\) is the mobius function and \(\Omega (n)\) outputs the number of prime divisors of \(n\) counted with multiplicity.


\(\textbf{Bonus:}\) Find a closed form for the sum, (which I am unable to do)

\[\sum _{ n=1 }^{ \infty } \frac { { 2 }^{ \Omega (n) } }{ { n }^{ 2 } } \]

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