Divisors calculator

How it works

Once $n = p_1^{e_1} \cdots p_k^{e_k}$ is known, its divisors are exactly the products $p_1^{j_1} \cdots p_k^{j_k}$ with $0 \le j_i \le e_i$. Counting the choices gives the number of divisors $\tau(n) = (e_1 + 1) \cdots (e_k + 1)$, and multiplying out $\prod_i (1 + p_i + \dots + p_i^{e_i})$ gives their sum $\sigma(n) = \prod_i \frac{p_i^{e_i + 1} - 1}{p_i - 1}$.

A number is perfect when it is the sum of its proper divisors, $\sigma(n) = 2n$, like $6 = 1 + 2 + 3$ and $28$. Euclid showed that $2^{p-1}(2^p - 1)$ is perfect whenever $2^p - 1$ is prime, and Euler that every even perfect number is of this form. Whether an odd perfect number exists is still open. When the proper divisors add up to more than $n$, it is abundant ($12$); when less, deficient (every prime).

Divisors come in pairs $d$ and $n/d$, one of them at most $\sqrt n$, so $\tau(n)$ is odd exactly when $n$ is a perfect square.