Sum of Divisors of a Prime Power

Add up every divisor of $2^{10}$ — that is $1 + 2 + 4 + 8 + \cdots + 1024$ — and you get 2047. The formula gets there without the addition because the divisors of a prime power are a geometric series: each is the previous one times p. The closed form for such a series is $\sigma(p^k) = (p^{k+1} - 1)/(p - 1)$, which for base 2 collapses to the pleasant $2^{k+1} - 1$. Those are the Mersenne numbers, and when one of them is prime it generates an even perfect number, which is how this formula ended up at the centre of a 2,000-year-old open problem.

The σ function matters beyond curiosities because it is multiplicative: for numbers with no common factor, $\sigma(mn) = \sigma(m)\sigma(n)$. So the prime-power case is not a special case, it is the whole thing. To get σ(360) you compute σ(2³) = 15, σ(3²) = 13 and σ(5) = 6, then multiply: 1170. Every prime-power piece comes from this formula.

Two cautions. First, σ counts the number itself among its divisors — σ(6) = 12, not 6. The "aliquot sum" used in the definition of perfect and amicable numbers is σ(n) − n, so subtract before comparing. Second, p really must be prime. Feed this 6³ and it returns $(6^4-1)/5 = 259$, whereas the divisors of 216 actually sum to 600. Nothing in the arithmetic complains; the answer is simply about a series, not about a divisor set.