# Divisor function

(Redirected from Sum-of-divisors function)
Divisor function σ0(n) up to n = 250
Sigma function σ1(n) up to n = 250
Sum of the squares of divisors, σ2(n), up to n = 250
Sum of cubes of divisors, σ3(n) up to n = 250

In mathematics, and specifically in number theory, a divisor function is an arithmetic function related to the divisors of an integer. When referred to as the divisor function, it counts the number of divisors of an integer. It appears in a number of remarkable identities, including relationships on the Riemann zeta function and the Eisenstein series of modular forms. Divisor functions were studied by Ramanujan, who gave a number of important congruences and identities.

A related function is the divisor summatory function, which, as the name implies, is a sum over the divisor function.

## Definition

The sum of positive divisors function σx(n), for a real or complex number x, is defined as the sum of the xth powers of the positive divisors of n. It can be expressed in sigma notation as

$\sigma_{x}(n)=\sum_{d|n} d^x\,\! ,$

where ${d|n}$ is shorthand for "d divides n". The notations d(n), ν(n) and τ(n) (for the German Teiler = divisors) are also used to denote σ0(n), or the number-of-divisors function[1][2] (sequence A000005 in OEIS). When x is 1, the function is called the sigma function or sum-of-divisors function,[1][3] and the subscript is often omitted, so σ(n) is equivalent to σ1(n) ().

The aliquot sum s(n) of n is the sum of the proper divisors (that is, the divisors excluding n itself, ), and equals σ1(n) − n; the aliquot sequence of n is formed by repeatedly applying the aliquot sum function.

## Example

For example, σ0(12) is the number of the divisors of 12:

\begin{align} \sigma_{0}(12) & = 1^0 + 2^0 + 3^0 + 4^0 + 6^0 + 12^0 \\ & = 1 + 1 + 1 + 1 + 1 + 1 = 6, \end{align}

while σ1(12) is the sum of all the divisors:

\begin{align} \sigma_{1}(12) & = 1^1 + 2^1 + 3^1 + 4^1 + 6^1 + 12^1 \\ & = 1 + 2 + 3 + 4 + 6 + 12 = 28, \end{align}

and the aliquot sum s(12) of proper divisors is:

\begin{align} s(12) & = 1^1 + 2^1 + 3^1 + 4^1 + 6^1 \\ & = 1 + 2 + 3 + 4 + 6 = 16. \end{align}

## Table of values

n Divisors σ0(n) σ1(n) s(n) = σ1(n) − n Comment
1 1 1 1 0 square number: σ0(n) is odd; power of 2: s(n) = n − 1 (almost-perfect)
2 1,2 2 3 1 Prime: σ1(n) = 1+n so s(n) =1
3 1,3 2 4 1 Prime: σ1(n) = 1+n so s(n) =1
4 1,2,4 3 7 3 square number: σ0(n) is odd; power of 2: s(n) = n − 1 (almost-perfect)
5 1,5 2 6 1 Prime: σ1(n) = 1+n so s(n) =1
6 1,2,3,6 4 12 6 first perfect number: s(n) = n
7 1,7 2 8 1 Prime: σ1(n) = 1+n so s(n) =1
8 1,2,4,8 4 15 7 power of 2: s(n) = n − 1 (almost-perfect)
9 1,3,9 3 13 4 square number: σ0(n) is odd
10 1,2,5,10 4 18 8
11 1,11 2 12 1 Prime: σ1(n) = 1+n so s(n) =1
12 1,2,3,4,6,12 6 28 16 first abundant number: s(n) > n
13 1,13 2 14 1 Prime: σ1(n) = 1+n so s(n) =1
14 1,2,7,14 4 24 10
15 1,3,5,15 4 24 9
16 1,2,4,8,16 5 31 15 square number: σ0(n) is odd; power of 2: s(n) = n − 1 (almost-perfect)

The cases x=2, x=3 and so on are tabulated in , , , , , ...

## Properties

For a non-square integer every divisor d of n is paired with divisor n/d of n and $\sigma_{0}(n)$ is then even; for a square integer one divisor (namely $\sqrt n$) is not paired with a distinct divisor and $\sigma_{0}(n)$ is then odd.

For a prime number p,

\begin{align} \sigma_0(p) & = 2 \\ \sigma_0(p^n) & = n+1 \\ \sigma_1(p) & = p+1 \end{align}

because by definition, the factors of a prime number are 1 and itself. Also,where pn# denotes the primorial,

$\sigma_0(p_n\#) = 2^n \,$

since n prime factors allow a sequence of binary selection ($p_{i}$ or 1) from n terms for each proper divisor formed.

Clearly, $1 < \sigma_0(n) < n$ and σ(n) > n for all n > 2.

The divisor function is multiplicative, but not completely multiplicative. The consequence of this is that, if we write

$n = \prod_{i=1}^r p_i^{a_i}$

where r = ω(n) is the number of distinct prime factors of n, pi is the ith prime factor, and ai is the maximum power of pi by which n is divisible, then we have

$\sigma_x(n) = \prod_{i=1}^{r} \frac{p_{i}^{(a_{i}+1)x}-1}{p_{i}^x-1}$

which is equivalent to the useful formula:

$\sigma_x(n) = \prod_{i=1}^r \sum_{j=0}^{a_i} p_i^{j x} = \prod_{i=1}^r (1 + p_i^x + p_i^{2x} + \cdots + p_i^{a_i x}).$

It follows (by setting x = 0) that d(n) is:

$\sigma_0(n)=\prod_{i=1}^r (a_i+1).$

For example, if n is 24, there are two prime factors (p1 is 2; p2 is 3); noting that 24 is the product of 23×31, a1 is 3 and a2 is 1. Thus we can calculate $\sigma_0(24)$ as so:

\begin{align} \sigma_0(24) & = \prod_{i=1}^{2} (a_i+1) \\ & = (3 + 1)(1 + 1) = 4 \times 2 = 8. \end{align}

The eight divisors counted by this formula are 1, 2, 4, 8, 3, 6, 12, and 24.

We also note s(n) = σ(n) − n. Here s(n) denotes the sum of the proper divisors of n, i.e. the divisors of n excluding n itself. This function is the one used to recognize perfect numbers which are the n for which s(n) = n. If s(n) > n then n is an abundant number and if s(n) < n then n is a deficient number.

If n is a power of 2, e.g. $n = 2^k$, then $\sigma(n) = 2 \times 2^k - 1 = 2n - 1,$ and s(n) = n - 1, which makes n almost-perfect.

As an example, for two distinct primes p and q with p < q, let

$n = pq. \,$

Then

$\sigma(n) = (p+1)(q+1) = n + 1 + (p+q), \,$
$\phi(n) = (p-1)(q-1) = n + 1 - (p+q), \,$

and

$n + 1 = (\sigma(n) + \phi(n))/2, \,$
$p + q = (\sigma(n) - \phi(n))/2, \,$

where φ(n) is Euler's totient function.

Then, the roots of:

$(x-p)(x-q) = x^2 - (p+q)x + n = x^2 - [(\sigma(n) - \phi(n))/2]x + [(\sigma(n) + \phi(n))/2 - 1] = 0 \,$

allows us to express p and q in terms of σ(n) and φ(n) only, without even knowing n or p+q, as:

$p = (\sigma(n) - \phi(n))/4 - \sqrt{[(\sigma(n) - \phi(n))/4]^2 - [(\sigma(n) + \phi(n))/2 - 1]}, \,$
$q = (\sigma(n) - \phi(n))/4 + \sqrt{[(\sigma(n) - \phi(n))/4]^2 - [(\sigma(n) + \phi(n))/2 - 1]}. \,$

Also, knowing n and either σ(n) or φ(n) (or knowing p+q and either σ(n) or φ(n)) allows us to easily find p and q.

In 1984, Roger Heath-Brown proved that

$\sigma_0(n) = \sigma_0(n + 1)$

will occur infinitely often.

## Series relations

Two Dirichlet series involving the divisor function are:

$\sum_{n=1}^\infty \frac{\sigma_{a}(n)}{n^s} = \zeta(s) \zeta(s-a),$

which for d(n) = σ0(n) gives

$\sum_{n=1}^\infty \frac{d(n)}{n^s} = \zeta^2(s),$

and

$\sum_{n=1}^\infty \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s) \zeta(s-a) \zeta(s-b) \zeta(s-a-b)}{\zeta(2s-a-b)}.$

A Lambert series involving the divisor function is:

$\sum_{n=1}^\infty q^n \sigma_a(n) = \sum_{n=1}^\infty \frac{n^a q^n}{1-q^n}$

for arbitrary complex |q| ≤ 1 and a. This summation also appears as the Fourier series of the Eisenstein series and the invariants of the Weierstrass elliptic functions.

## Approximate growth rate

In little-o notation, the divisor function satisfies the inequality (see page 296 of Apostol’s book[4])

$\mbox{for all }\epsilon>0,\quad d(n)=o(n^\epsilon).$

More precisely, Severin Wigert showed that

$\limsup_{n\to\infty}\frac{\log d(n)}{\log n/\log\log n}=\log2.$

On the other hand, since there are infinitely many prime numbers,

$\liminf_{n\to\infty} d(n)=2.$

In Big-O notation, Peter Gustav Lejeune Dirichlet showed that the average order of the divisor function satisfies the following inequality (see Theorem 3.3 of Apostol’s book[4])

$\mbox{for all } x\geq1, \sum_{n\leq x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt{x}),$

where $\gamma$ is Euler's constant. Improving the bound $O(\sqrt{x})$ in this formula is known as Dirichlet's divisor problem

The behaviour of the sigma function is irregular. The asymptotic growth rate of the sigma function can be expressed by:

$\limsup_{n\rightarrow\infty}\frac{\sigma(n)}{n\,\log \log n}=e^\gamma,$

where lim sup is the limit superior. This result is Grönwall's theorem, published in 1913 (Grönwall 1913). His proof uses Mertens' 3rd theorem, which says that

$\lim_{n\to\infty}\frac{1}{\log n}\prod_{p\le n}\frac{p}{p-1}=e^{\gamma},$

where p denotes a prime.

In 1915, Ramanujan proved that under the assumption of the Riemann hypothesis, the inequality:

$\ \sigma(n) < e^\gamma n \log \log n$ (Robin's inequality)

holds for all sufficiently large n (Ramanujan 1997). In 1984 Guy Robin proved that the inequality is true for all n ≥ 5,041 if and only if the Riemann hypothesis is true (Robin 1984). This is Robin's theorem and the inequality became known after him. The largest known value that violates the inequality is n=5,040. If the Riemann hypothesis is true, there are no greater exceptions. If the hypothesis is false, then Robin showed there are an infinite number of values of n that violate the inequality, and it is known that the smallest such n ≥ 5,041 must be superabundant (Akbary & Friggstad 2009). It has been shown that the inequality holds for large odd and square-free integers, and that the Riemann hypothesis is equivalent to the inequality just for n divisible by the fifth power of a prime (Choie et al. 2007).

A related bound was given by Jeffrey Lagarias in 2002, who proved that the Riemann hypothesis is equivalent to the statement that

$\sigma(n) \le H_n + \ln(H_n)e^{H_n}$

for every natural number n, where $H_n$ is the nth harmonic number, (Lagarias 2002).

Robin also proved, unconditionally, that the inequality

$\ \sigma(n) < e^\gamma n \log \log n + \frac{0.6483\ n}{\log \log n}$

holds for all n ≥ 3.