Divisor Sum of 5002

Example of Divisor Sum of Square-Free Integer

$\map {\sigma_1} {5002} = 7812$

where $\sigma_1$ denotes the divisor sum function.

Proof

We have that:

$5002 = 2 \times 41 \times 61$

Hence:

\(\ds \map {\sigma_1} {5002}\)

\(=\)

\(\ds \paren {2 + 1} \paren {41 + 1} \paren {61 + 1}\)

Divisor Sum of Square-Free Integer

\(\ds \)

\(=\)

\(\ds 3 \times 42 \times 62\)

\(\ds \)

\(=\)

\(\ds 3 \times \paren {2 \times 3 \times 7} \times \paren {2 \times 31}\)

\(\ds \)

\(=\)

\(\ds 2^2 \times 3^2 \times 7 \times 31\)

\(\ds \)

\(=\)

\(\ds 7812\)

$\blacksquare$