\[S=\sum_{n=1}^{\infty}\frac{(-1)^{\Omega(n)}}{n^4}\]

Let \(\Omega(n)\) denote the number of prime factors of \(n\), counted with their multiplicities. For example, \(\Omega(2016)=\Omega(2^5\times 3^2\times 7)=5+2+1=8\).

Submit \(\dfrac{\pi^4}{S}\) as your answer.

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