Sigma Function of Power of Prime

Theorem

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

Let $\sigma \left({n}\right)$ be the sigma function of $n$.

That is, let $\sigma \left({n}\right)$ be the sum of all positive divisors of $n$.

Then:

$\sigma \left({n}\right) = \dfrac {p^{k+1} - 1} {p - 1}$

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 Progression:

$\sigma \left({p^k}\right) = 1 + p + p^2 + \cdots + p^{k-1} + p^k = \dfrac {p^{k+1} - 1} {p - 1}$

$\blacksquare$

Examples

$\sigma$ of $81$

$\sigma \left({81}\right) = 121$

Sources

• 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes