Sum of Unitary Divisors of Integer
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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:
- $\ds n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
Then:
- $\ds \map {\sigma^*} n = \prod_{1 \mathop \le i \mathop \le r} \paren {1 + p_i^{k_i} }$
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:
- $\ds \map {\sigma^*} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
Hence the result.
$\blacksquare$