Divisor Sum of Power of Prime

Theorem

Let $n = p^k$ be the power of a prime number $p$.

Let $\map {\sigma_1} n$ be the divisor sum of $n$.

That is, let $\map {\sigma_1} n$ be the sum of all positive divisors of $n$.


Then:

$\map {\sigma_1} n = \dfrac {p^{k + 1} - 1} {p - 1}$


Proof

From Divisors of Power of Prime, the divisors of $n = p^k$ are $1, p, p^2, \ldots, p^{k - 1}, p^k$.

Hence from Sum of Geometric Sequence:

$\map {\sigma_1} {p^k} = 1 + p + p^2 + \cdots + p^{k - 1} + p^k = \dfrac {p^{k + 1} - 1} {p - 1}$

$\blacksquare$


Examples

$\sigma_1$ of $81$

$\map {\sigma_1} {81} = 121$


Sources

  • 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): sigma function or $\sigma$ function: 1.
  • 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
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