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PROJECT EULER · #0530

GCD of Divisors

Statement only · SolvedOriginal problem ↗

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)=dngcd(d,nd).

Let F be the summatory function of f, that is F(k)=n=1kf(n).

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

Find F(1015).

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