Euler in Babylon

GCD of Divisors

October 18, 2015

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 $f(n)=\displaystyle\sum\limits_{d|n}\, \text{gcd}(d,\frac n d)$.

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

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

Find F(1015).


gamwe6

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