# Problem 401: Sum of squares of divisors
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.