# How is this possible?

Let $$f(n)$$ be the smallest divisor of $$n$$ which is greater than or equal to $$\sqrt{n}$$.

$\large \lim_{m \to \infty} \frac{\ln(m)}{m^2} \sum\limits_{n=1}^{m}f(n)$

If the value of the limit above is equal to $$\dfrac{\pi^A}{B}$$ for integers $$A$$ and $$B$$, find $$A+B$$.

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