Divisor Sum of 1260
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Example of Divisor Sum of Integer
- $\map {\sigma_1} {1260} = 4368$
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:
- $1260 = 2^2 \times 3^2 \times 5 \times 7$
Hence:
\(\ds \map {\sigma_1} {1260}\) | \(=\) | \(\ds \frac {2^3 - 1} {2 - 1} \times \frac {3^3 - 1} {3 - 1} \times \paren {5 + 1} \times \paren {7 + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 7 1 \times \frac {26} 2 \times 6 \times 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 7 \times 13 \times \times \paren {2 \times 3} \times 2^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^4 \times 3 \times 7 \times 13\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4368\) |
$\blacksquare$