Divisor Sum of Non-Square Semiprime

Theorem

Let $n \in \Z_{>0}$ be a semiprime with distinct prime factors $p$ and $q$.

Then:

$\map {\sigma_1} n = \paren {p + 1} \paren {q + 1}$

where $\map {\sigma_1} n$ denotes the divisor sum function.

Proof 1

As $p$ and $q$ are distinct prime numbers, it follows that $p$ and $q$ are coprime.

Thus by Divisor Sum Function is Multiplicative:

$\map {\sigma_1} n = \map {\sigma_1} p \map {\sigma_1} q$

From Divisor Sum of Prime Number:

$\map {\sigma_1} p = \paren {p + 1}$

$\map {\sigma_1} q = \paren {q + 1}$

Hence the result.

$\blacksquare$

Proof 2

A semiprime with distinct prime factors is a square-free integer.

By Divisor Sum of Square-Free Integer:

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

Hence the result.

$\blacksquare$

Examples

$\sigma_1$ of $14$

$\map {\sigma_1} {14} = 24$

$\sigma_1$ of $15$

$\map {\sigma_1} {15} = 24$

$\sigma_1$ of $22$

$\map {\sigma_1} {22} = 36$

$\sigma_1$ of $26$

$\map {\sigma_1} {26} = 42$

$\sigma_1$ of $33$

$\map {\sigma_1} {33} = 48$

$\sigma_1$ of $35$

$\map {\sigma_1} {35} = 48$

$\sigma_1$ of $38$

$\map {\sigma_1} {38} = 60$

$\sigma_1$ of $58$

$\map {\sigma_1} {58} = 90$

$\sigma_1$ of $62$

$\map {\sigma_1} {62} = 96$

$\sigma_1$ of $65$

$\map {\sigma_1} {65} = 84$

$\sigma_1$ of $87$

$\map {\sigma_1} {87} = 120$

$\sigma_1$ of $94$

$\map {\sigma_1} {94} = 144$

$\sigma_1$ of $115$

$\map {\sigma_1} {115} = 144$

$\sigma_1$ of $206$

$\map {\sigma_1} {206} = 312$

$\sigma_1$ of $362$

$\map {\sigma_1} {362} = 546$

$\sigma_1$ of $1257$

$\map {\sigma_1} {1257} = 1680$