Divisor Sum of 5,391,411,025

Example of Divisor Sum of Integer

$\map {\sigma_1} {5 \, 391 \, 411 \, 025} = 10 \, 799 \, 308 \, 800$

where $\sigma_1$ denotes the divisor sum function.

Proof

We have that:

$5 \, 391 \, 411 \, 025 = 5^2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29$

Hence:

\(\ds \map {\sigma_1} {5 \, 391 \, 411 \, 025}\)

\(=\)

\(\ds \dfrac {5^3 - 1} {5 - 1} \times \paren {7 + 1} \times \paren {11 + 1} \times \paren {13 + 1} \times \paren {17 + 1} \times \paren {19 + 1} \times \paren {23 + 1} \times \paren {29 + 1}\)

Divisor Sum of Integer

\(\ds \)

\(=\)

\(\ds \dfrac {124} 4 \times 8 \times 12 \times 14 \times 18 \times 20 \times 24 \times 30\)

\(\ds \)

\(=\)

\(\ds 31 \times 8 \times 12 \times 14 \times 18 \times 20 \times 24 \times 30\)

\(\ds \)

\(=\)

\(\ds 31 \times 2^3 \times \paren {2^2 \times 3} \times \paren {2 \times 7} \times \paren {2 \times 3^2} \times \paren {2^2 \times 5} \times \paren {2^3 \times 3} \times \paren {2 \times 3 \times 5}\)

\(\ds \)

\(=\)

\(\ds 2^{13} \times 3^5 \times 5^2 \times 7 \times 31\)

\(\ds \)

\(=\)

\(\ds 10 \, 799 \, 308 \, 800\)

$\blacksquare$