Sum of Unitary Divisors of Integer

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

Let $\map {\sigma^*} n$ be the sum of all positive unitary 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:
 * $\displaystyle \map {\sigma^*} n = \prod_{1 \mathop \le i \mathop \le r} \paren {1 + p_i^{k_i}}$

Proof
We have that the Sum of Unitary Divisors is Multiplicative.

From Value of Multiplicative Function is Product of Values of Prime Power Factors, we have:
 * $\map {\sigma^*} n = \map {\sigma^*} {p_1^{k_1} } \map {\sigma^*} {p_2^{k_2} } \ldots \map {\sigma^*} {p_r^{k_r} }$

From Sum of Unitary Divisors 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.