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
- $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
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:
\(\ds \map {\sigma_1} {2620}\) | \(=\) | \(\ds \frac {2^3 - 1} {2 - 1} \times \paren {5 + 1} \times \paren {131 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 7 1 \times 6 \times 132\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 7 \times \paren {2 \times 3} \times \paren {2^2 \times 3 \times 11}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^3 \times 3^2 \times 7 \times 11\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5544\) |
$\blacksquare$