Number of divisors / sum of divisors

In this article we discuss how to compute the number of divisors $d(n)$ and the sum of divisors $\sigma(n)$ of a given number $n$.

Number of divisors

It should be obvious that the prime factorization of a divisor $d$ has to be a subset of the prime factorization of $n$, e.g. $6 = 2 \cdot 3$ is a divisor of $60 = 2^2 \cdot 3 \cdot 5$. So we only need to find all different subsets of the prime factorization of $n$.

Usually the number of subsets is $2^x$ for a set with $x$ elements. However this is no longer true, if there are repeated elements in the set. In our case some prime factors may appear multiple times in the prime factorization of $n$.

If a prime factor $p$ appears $e$ times in the prime factorization of $n$, then we can use the factor $p$ up to $e$ times in the subset. Which means we have $e+1$ choices.

Therefore if the prime factorization of $n$ is $p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$, where $p_i$ are distinct prime numbers, then the number of divisors is: $$d(n) = (e_1 + 1) \cdot (e_2 + 1) \cdots (e_k + 1)$$

A way of thinking about it is the following:

Sum of divisors

We can use the same argument of the previous section.

Multiplicative functions

A multiplicative functions is a function $f(x)$ which satisfies $$f(a \cdot b) = f(a) \cdot f(b)$$ if $a$ and $b$ are coprime.

Both $d(n)$ and $\sigma(n)$ are multiplicative functions.

Multiplicative functions have a huge variety of interesting properties, which can be very useful in number theory problems. For instance the Dirichlet convolution of two multiplicative functions is also multiplicative.

Practice Problems