## Sum of squares of divisors

### Problem 401

Published on Saturday, 10th November 2012, 04:00 pm; Solved by 793

The divisors of 6 are 1,2,3 and 6.
The sum of the squares of these numbers is 1+4+9+36=50.

Let sigma2(n) represent the sum of the squares of the divisors of n. Thus sigma2(6)=50.

Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=sigma2(i) for i=1 to n.
The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.

Find SIGMA2(1015) modulo 109.