{"app":{"locale":"en-GB"},"problem":{"content":"

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.

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