# Divisor function

The divisor function is denoted $\sigma_k(n)$ and is defined as the sum of the $k$th powers of the divisors of $n$. Thus $\sigma_k(n) = \sum_{d|n}d^k = d_1^k + d_2^k + \cdots + d_r^k$ where the $d_i$ are the positive divisors of $n$.

## Counting divisors

Note that $\sigma_0(n) = d_1^0 + d_2^0 + \ldots + d_r^0 = 1 + 1 + \ldots + 1 = r$, the number of divisors of $n$. Thus $\sigma_0(n) = d(n)$ is simply the number of divisors of $n$.

### Example Problems

#### Demonstration

Consider the task of counting the divisors of 72.

First, we find the prime factorization of 72: $72=2^{3} \cdot 3^{2}.$
Since each divisor of 72 can have a power of 2, and since this power can be 0, 1, 2, or 3, we have 4 possibilities. Likewise, since each divisor can have a power of 3, and since this power can be 0, 1, or 2, we have 3 possibilities. By an elementary counting principle, we have $3\cdot 4=12$ divisors.

We can now generalize. Let the prime factorization of $n$ be $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$. Any divisor of $n$ must be of the form $p_1^{f_1}p_2^{f_2} \cdots p_k^{e_k}$ where the $f_i$ are integers such that $0\le f_i \le e_i$ for $i = 1,2,\ldots, k$. Thus, the number of divisors of $n$ is $\sigma_0(n) = (e_1+1)(e_2+1)\cdots (e_k+1)$.

## Sum of divisors

The sum of the divisors, or $\sigma_1(n)$, is given by

$\sigma_1(n) = (1 + p_1 + p_1^2 +\cdots p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{e_k}).$

There will be $(e_1+1)(e_2+1)(e_3+1)\cdots (e_k+1)$ products formed by taking one number from each sum, which is the number of divisors of $n$. Clearly all possible products are divisors of $n$. Furthermore, all of those products are unique since each positive integer has a unique prime factorization.

Since all of these products are added together, we can conclude this gives us the sum of the divisors.

## Sum of kth Powers of Divisors

Inspired by the example of the sum of divisors, we can easily see that the sum of the $k^\text{th}$ powers of the divisors is given by \begin{align*} \sigma_k(n) &= (1+p_1^k+p_1^{2k}+\cdots +p_1^{e_1k})(1+p_2^k+p_2^{2k}+\cdots +p_2^{e_2k})\cdots (1+p_i^k+p_i^{2k}+\cdots +p_i^{e_ik}) \\ &= \prod_{a=1}^{i}\left(\sum_{b=0}^{e_a}p_a^{bk}\right) \end{align*} where $p_1,p_2,...,p_i$ are the distinct prime divisors of $n$.

This is proven in a very similar way to the $\sigma_1$ case.