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GCD of Divisors

 Published on Sunday, 18th October 2015, 01:00 am and solved by 669
Difficulty: Level 23 [60%]

Problem 530

Every divisor $d$ of a number $n$ has a complementary divisor $n/d$.

Let $f(n)$ be the sum of the greatest common divisor of $d$ and $n/d$ over all positive divisors $d$ of $n$, that is $f(n)=\displaystyle\sum_{d\mid n}\gcd(d,\frac n d)$.

Let $F$ be the summatory function of $f$, that is $F(k)=\displaystyle\sum_{n=1}^k f(n)$.

You are given that $F(10)=32$ and $F(1000)=12776$.

Find $F(10^{15})$.



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