Divisor Sum of Square-Free Integer/Examples/66
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Example of Divisor Sum of Square-Free Integer
- $\map {\sigma_1} {66} = 144$
where $\sigma_1$ denotes the divisor sum.
Proof 1
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:
- $66 = 2 \times 3 \times 11$
Hence:
\(\ds \map {\sigma_1} {66}\) | \(=\) | \(\ds \frac {2^2 - 1} {2 - 1} \times \frac {3^2 - 1} {3 - 1} \times \frac {11^2 - 1} {11 - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 3 1 \times \frac 8 2 \times \frac {120} {10}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 4 \times 12\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 2^2 \times \paren {2^2 \times 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^4 \times 3^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^2 \times 3}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 144\) |
$\blacksquare$
Proof 2
We have that:
- $66 = 2 \times 3 \times 11$
Hence:
\(\ds \map {\sigma_1} {66}\) | \(=\) | \(\ds \paren {2 + 1} \paren {3 + 1} \paren {11 + 1}\) | Divisor Sum of Square-Free Integer | |||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 4 \times 12\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \times 2^2 \times \paren {2^2 \times 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^4 \times 3^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {2^2 \times 3}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 12^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 144\) |
$\blacksquare$