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 ${sigma}_{2} (n)$ represent the sum of the squares of the divisors of $n$ . Thus ${sigma}_{2} (6) = 50$ .

Let

{SIGMA}_{2}

represent the summatory function of

{sigma}_{2}

, that is

{SIGMA}_{2} (n) = \sum {sigma}_{2} (i)

for

i = 1

n

.
The first

6

values of

{SIGMA}_{2}

are:

1, 6, 16, 37, 63

and

113

Find ${SIGMA}_{2} (10^{15})$ modulo $10^{9}$ .

Write-up coming later

The complete problem is available here. An approach, code, and answer will be added later.