Divisor Sum of 651
Jump to navigation
Jump to search
Example of Divisor Sum of Integer
- $\map {\sigma_1} {651} = 1024$
where $\sigma_1$ denotes the divisor sum function.
Proof
We have that:
- $651 = 3 \times 7 \times 31$
Hence:
\(\ds \map {\sigma_1} {651}\) | \(=\) | \(\ds \paren {3 + 1} \paren {7 + 1} \paren {31 + 1}\) | Divisor Sum of Square-Free Integer | |||||||||||
\(\ds \) | \(=\) | \(\ds 4 \times 8 \times 32\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^2 \times 2^3 \times 2^5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^{10}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^5}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1024\) |
$\blacksquare$