Divisor Sum of 616

From ProofWiki
Jump to navigation Jump to search

Example of Divisor Sum of Integer

$\map {\sigma_1} {616} = 1440$

where $\sigma_1$ denotes the divisor sum function.


Proof

From Divisor Sum of Integer:

$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

where $n = \ds \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i}$ denotes the prime decomposition of $n$.


We have that:

$616 = 2^3 \times 7 \times 11$


Hence:

\(\ds \map {\sigma_1} {616}\) \(=\) \(\ds \frac {2^4 - 1} {2 - 1} \times \paren {7 + 1} \times \paren {11 + 1}\)
\(\ds \) \(=\) \(\ds 15 \times 8 \times 12\)
\(\ds \) \(=\) \(\ds \paren {3 \times 5} \times 2^3 \times \paren {2^2 \times 3}\)
\(\ds \) \(=\) \(\ds 2^5 \times 3^2 \times 5\)
\(\ds \) \(=\) \(\ds 1440\)

$\blacksquare$