Divisor Sum of 105

Example of Divisor Sum of Square-Free Integer

$\map {\sigma_1} {105} = 192$

where $\sigma_1$ denotes the divisor sum function.

Proof

We have that:

$105 = 3 \times 5 \times 7$

Hence:

\(\ds \map {\sigma_1} {105}\)

\(=\)

\(\ds \paren {3 + 1} \paren {5 + 1} \paren {7 + 1}\)

Divisor Sum of Square-Free Integer

\(\ds \)

\(=\)

\(\ds 4 \times 6 \times 8\)

\(\ds \)

\(=\)

\(\ds 2^2 \times \paren {2 \times 3} \times 2^3\)

\(\ds \)

\(=\)

\(\ds 2^6 \times 3\)

\(\ds \)

\(=\)

\(\ds 192\)

$\blacksquare$