Divisor Sum of Integer/Corollary

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

Let the prime decomposition of $n$ be:
 * $\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$

Let $\sigma \left({n}\right)$ be the sigma function of $n$.

That is, let $\sigma \left({n}\right)$ be the sum of all positive divisors of $n$.

Then:
 * $\displaystyle \sigma \left({n}\right) = \prod_{\substack {1 \mathop \le i \mathop \le r \\ k_i \mathop > 1} } \frac {p_i^{k_i + 1} - 1} {p_i - 1} \prod_{\substack {1 \mathop \le i \mathop \le r \\ k_i \mathop = 1} } \left({p_i + 1}\right)$

Proof
From Sigma Function of Integer:
 * $\displaystyle \sigma \left({n}\right) = \prod_{1 \mathop \le i \mathop \le r} \frac {p_i^{k_i + 1} - 1} {p_i - 1}$

Suppose $k_i = 1$.

Then we have:

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.