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.

Answer 1

For my mathhomework

Answer 2

Pls help

Answer 3

So sigma2(i) = 1?

Answer 4

Pls respond

Answer 5

Could you draw out ∑sigma2(i)

Answer 6

I copy paste the entire problem from my online course.

Answer 7

1015=5*7*29
then Sigma2(1015)=sigma2(5)+sigma2(7)+sigma2(29)
Please try from there

Answer 8

This is an interesting question...

Answer 9

sorry I was wrong :)

Answer 10

Would you paste the entire problem?

Answer 11

I did.

Answer 12

I misread. I thought you said you could copy and paste the entire problem.

Answer 13

Could you draw ∑sigma2(i)?

Answer 14

∑sigma2(i) for i=1 to n.

Answer 15

Here is the table of sigma2(n) until n=1000. If you just want to know the answer you can continue until you compute sigma2(1015) :D :D

Answer 16

sorry here is the link : http://oeis.org/A001157/b001157.txt

Answer 17

Is this suppose to be SIGMA2(1015) = 109?