Sum of all the divisors of a number

Imagine you wish to work out the sum of divisors of the number 72. It would not take long to list the divisors, and then find their sum: 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72 = 195.

However, this method would become both tedious and difficult for large numbers like 6483658785. There is a simple and elegant method.

Let σ(n) be the sum of divisors of the natural number, n.

For any prime, p: σ(p) = p + 1, as the only divisors would be 1 and p

.

Consider p^a: σ(p^a) = 1 + p + p² + ... + p^a (1).

Multiplying by p: pσ(p^a) = p + p² + p³ + ... + p^{a + 1} (2).

Subtracting (1) from (2): pσ(p^a)−σ(p^a) = (p−1)σ(p^a) = p^a+1 − 1.

Hence σ(p^a) = (p^a+1 − 1)/(p − 1).

For example, σ(3⁴)=(3⁵−1)/(3−1) = 242/2 = 121,
and checking: 1 + 3 + 9 + 27 + 81 = 121.

Although no proof is supplied here, the usefulness of the function, σ(n), is its multiplicitivity, which means that σ(a×b×...)=σ(a)×σ(b)×..., where a, b, ..., are relatively prime.

Returning to example, we use the fact that σ(72) = σ(2³×3²). As 2³ and 3² are relatively prime, we can separately work out σ(2³) = 2⁴ − 1 = 15 and σ(3²) = (3³ − 1)/2 = 13. Therefore, σ(72) = 15×13 = 195.