Divisor Sum of Integer/Corollary
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Corollary to Divisor Sum of Integer
Let $n$ be an integer such that $n \ge 2$.
Let the prime decomposition of $n$ be:
- $\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Let $\map {\sigma_1} n$ be the divisor sum of $n$.
That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.
Then:
- $\ds \map {\sigma_1} n = \prod_{\substack {1 \mathop \le i \mathop \le r \\ k_i \mathop > 1} } \frac {p_i^{k_i + 1} - 1} {p_i - 1} \prod_{\substack {1 \mathop \le i \mathop \le r \\ k_i \mathop = 1} } \paren {p_i + 1}$
Proof
From Divisor Sum of Integer:
- $\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
Suppose $k_i = 1$.
Then we have:
\(\ds \frac {p_i^{k_i + 1} - 1} {p_i - 1}\) | \(=\) | \(\ds \frac {p_i^2 - 1} {p_i - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {p_i + 1} \paren {p_i - 1} } {p_i - 1}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds p_i + 1\) |
Thus the contribution from a prime factor which is square-free can be expressed in the simpler form $p_i + 1$ instead of the more unwieldy form $\dfrac {p_i^2 - 1} {p_i - 1}$.
Hence the result.
$\blacksquare$
Examples
$\sigma_1$ of $1$
- $\map {\sigma_1} 1 = 1$
$\sigma_1$ of $3$
- $\map {\sigma_1} 3 = 4$
$\sigma_1$ of $12$
- $\map {\sigma_1} {12} = 28$
$\sigma_1$ of $20$
- $\map {\sigma_1} {20} = 42$
$\sigma_1$ of $24$
- $\map {\sigma_1} {24} = 60$
$\sigma_1$ of $40$
- $\map {\sigma_1} {40} = 90$
$\sigma_1$ of $44$
- $\map {\sigma_1} {44} = 84$
$\sigma_1$ of $48$
- $\map {\sigma_1} {48} = 124$
$\sigma_1$ of $207$
- $\map {\sigma_1} {207} = 312$
$\sigma_1$ of $12 \, 496$
- $\map {\sigma_1} {12 \, 496} = 26 \, 784$
$\sigma_1$ of $14 \, 288$
- $\map {\sigma_1} {14 \, 288} = 29 \, 760$