Divisor Sum of Non-Square Semiprime
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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 $1257$
- $\map {\sigma_1} {1257} = 1680$