Divisor Sum of 2620

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Example of Divisor Sum of Integer

$\map {\sigma_1} {2620} = 5544$

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:

$2620 = 2^2 \times 5 \times 131$

Hence:

\(\ds \map {\sigma_1} {2620}\) \(=\) \(\ds \frac {2^3 - 1} {2 - 1} \times \paren {5 + 1} \times \paren {131 + 1}\)
\(\ds \) \(=\) \(\ds \frac 7 1 \times 6 \times 132\)
\(\ds \) \(=\) \(\ds 7 \times \paren {2 \times 3} \times \paren {2^2 \times 3 \times 11}\)
\(\ds \) \(=\) \(\ds 2^3 \times 3^2 \times 7 \times 11\)
\(\ds \) \(=\) \(\ds 5544\)

$\blacksquare$