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.

Question

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.

shubhamsrg · Answer

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anonymous · Answer

well can you help me? i am writing a program to do this btw it's 10^15 not 1015, also i can calculate SIGMA2 for small values like 10^6 but it get's impossible for even 10^7

anonymous · Answer

and also it's modulo 10^9 the notation got off

shubhamsrg · Answer

hmm, I am trying here, not too sure if I'll actually be able to help, but I am giving it a shot.
@mukushla @phi @robtobey

anonymous · Answer

the small sigma2 can easily be calculated. But the sumation of 10^15 sigma is the problem. Also there must be some trick to the fact i need it modulo 10^9. But I can't seem to use it. Maybe some multiplicative properties of the functions should be used, but I can't find them.