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$\large \sum_{n=1}^\infty \dfrac{\gcd(n,2016)}{n^2}= \dfrac{a}{b}\pi^2$

If the equation above holds true for positive integers $a$ and $b$, find $a+b$.

Clarification: $\gcd(m,n)$ denotes the greatest common divisor of $m$ and $n$.

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