Divisor Sum of Square-Free Integer/Examples/70
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Example of Divisor Sum of Square-Free Integer
- $\map {\sigma_1} {70} = 144$
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:
- $70 = 2 \times 5 \times 7$
Hence:
| \(\ds \map {\sigma_1} {70}\) | \(=\) | \(\ds \frac {2^2 - 1} {2 - 1} \times \frac {5^2 - 1} {5 - 1} \times \frac {7^2 - 1} {7 - 1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 3 1 \times \frac {24} 4 \times \frac {48} {6}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \times 6 \times 8\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \times \paren {3 \times 2} \times 2^3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3^2 \times 2^4\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {2^2 \times 3}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 12^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 144\) |
$\blacksquare$
Proof 2
We have that:
- $70 = 2 \times 5 \times 7$
Hence:
| \(\ds \map {\sigma_1} {70}\) | \(=\) | \(\ds \paren {2 + 1} \paren {5 + 1} \paren {7 + 1}\) | Divisor Sum of Square-Free Integer | |||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \times 6 \times 8\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3 \times \paren {3 \times 2} \times 2^3\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 3^2 \times 2^4\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {2^2 \times 3}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 12^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 144\) |
$\blacksquare$