Divisor Sum of Integer

Theorem

Let $n$ be an integer such that $n \ge 2$.

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 $n$.

Let the prime decomposition of $n$ be:

$\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

Then:

$\ds \map {\sigma_1} n = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

Corollary

$\ds \map {\sigma_1} n = \prod_{\substack {1 \mathop \le i \mathop \le r \\ k_i \mathop > 1} } \frac {p_i^{k_i + 1} - 1} {p_i - 1} \prod_{\substack {1 \mathop \le i \mathop \le r \\ k_i \mathop = 1} } \paren {p_i + 1}$

Proof

We have that the Divisor Sum Function is Multiplicative.

From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have:

$\map f n = \map f {p_1^{k_1} } \map f {p_2^{k_2} } \ldots \map f {p_r^{k_r} }$

From Divisor Sum of Power of Prime, we have:

$\ds \map {\sigma_1} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

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 $14 \, 288$

$\map {\sigma_1} {14 \, 288} = 29 \, 760$

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): sigma function (sum function)