Sigma Function of Integer

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Theorem

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

Let $\map \sigma n$ be the sigma function of $n$.

That is, let $\map \sigma n$ be the sum of all positive divisors of $n$.

Let the prime decomposition of $n$ be:

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


Then:

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


Corollary

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


Proof

We have that the Sigma 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 Sigma Function of Power of Prime, we have:

$\displaystyle \map \sigma {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

Hence the result.

$\blacksquare$


Examples

$\sigma$ of $1$

$\map \sigma 1 = 1$


$\sigma$ of $3$

$\map \sigma 3 = 4$


$\sigma$ of $12$

$\sigma \left({12}\right) = 28$


$\sigma$ of $20$

$\sigma \left({20}\right) = 42$


$\sigma$ of $24$

$\sigma \left({24}\right) = 60$


$\sigma$ of $40$

$\map \sigma {40} = 90$


$\sigma$ of $44$

$\map \sigma {44} = 84$


$\sigma$ of $48$

$\sigma \left({48}\right) = 124$


$\sigma$ of $207$

$\sigma \left({207}\right) = 312$


$\sigma$ of $12 \, 496$

$\sigma \left({12 \, 496}\right) = 26 \, 784$


$\sigma$ of $14 \, 288$

$\sigma \left({14 \, 288}\right) = 29 \, 760$