\[\sum _{n=1}^{\infty}\frac{d\left(2048n\right)}{n^2}=\frac{A\pi^B}{C}\]

Where \(d(n)\) counts the number of divisors of \(n\). Find \(A+B+C\)

Hint: 2048 is 2 to the power of 11. 2 is a prime.

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