Divisor function

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 (including 1 and the number itself). 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; these are treated separately in the article Ramanujan's sum.

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 σz(n), for a real or complex number z, is defined as the sum of the zth powers of the positive divisors of n. It can be expressed in sigma notation as


 * $$\sigma_z(n)=\sum_{d\mid n} d^z\,\! ,$$

where $${d\mid n}$$ 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. When z is 1, the function is called the sigma function or sum-of-divisors function, and the subscript is often omitted, so σ(n) is the same as 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) &minus; 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} $$

σ-1(n) is sometimes called the abundancy index of n, and we have:



\begin{align} \sigma_{-1}(12) & = 1^{-1} + 2^{-1} + 3^{-1} + 4^{-1} + 6^{-1} + 12^{-1} \\ & = \tfrac11 + \tfrac12 + \tfrac13 + \tfrac14 + \tfrac16 + \tfrac1{12} \\ & = \tfrac{12}{12} + \tfrac6{12} + \tfrac4{12} + \tfrac3{12} + \tfrac2{12} + \tfrac1{12} \\ & = \tfrac{12 + 6 + 4 + 3 + 2 + 1}{12} = \tfrac{28}{12} = \tfrac73 = \tfrac{\sigma_1(12)}{12} \end{align} $$

Table of values
The cases x = 2 to 5 are listed in through, x = 6 to 24 are listed in  through.

Formulas at prime powers
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. However, these are not in general the smallest numbers whose number of divisors is a power of two; instead, the smallest such number may be obtained by multiplying together the first n Fermi–Dirac primes, prime powers whose exponent is a power of two.

Clearly, $$1 < \sigma_0(n) < n$$ for all $$n > 2$$, and $$\sigma_x(n) > n $$ for all $$n > 1$$, $$x > 0$$.

The divisor function is multiplicative (since each divisor c of the product mn with $$\gcd(m, n) = 1$$ distinctively correspond to a divisor a of m and a divisor b of n), but not completely multiplicative:


 * $$\gcd(a, b)=1 \Longrightarrow \sigma_x(ab)=\sigma_x(a)\sigma_x(b).$$

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:


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

which, when x &ne; 0, is equivalent to the useful formula:


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

When x = 0, $$\sigma_0(n)$$ is:


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

This result can be directly deduced from the fact that all divisors of $$n$$ are uniquely determined by the distinct tuples $$(x_1, x_2, ..., x_i, ..., x_r)$$ of integers with $$0 \le x_i \le a_i$$ (i.e. $$a_i+1$$ independent choices for each $$x_i$$).

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:


 * $$\sigma_0(24) = \prod_{i=1}^{2} (a_i+1) = (3 + 1)(1 + 1) = 4 \cdot 2 = 8.$$

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

Other properties and identities
Euler proved the remarkable recurrence:


 * $$\begin{align}

\sigma_1(n) &= \sigma_1(n-1)+\sigma_1(n-2)-\sigma_1(n-5)-\sigma_1(n-7)+\sigma_1(n-12)+\sigma_1(n-15)+ \cdots \\[12mu] &= \sum_{i\in\N} (-1)^{i+1}\left( \sigma_1 \left( n-\frac{1}{2} \left( 3i^2-i \right) \right) + \sigma_1 \left( n-\frac{1}{2} \left( 3i^2+i \right) \right) \right), \end{align}$$

where $$\sigma_1(0)=n$$ if it occurs and $$\sigma_1(x)=0$$ for $$x < 0$$, and $$\tfrac{1}{2} \left( 3i^2 \mp i \right)$$ are consecutive pairs of generalized pentagonal numbers (, starting at offset 1). Indeed, Euler proved this by logarithmic differentiation of the identity in his pentagonal number theorem.

For a non-square integer, n, every divisor, d, of n is paired with divisor n/d of n and $$\sigma_{0}(n)$$ is even; for a square integer, one divisor (namely $$\sqrt n$$) is not paired with a distinct divisor and $$\sigma_{0}(n)$$ is odd. Similarly, the number $$\sigma_{1}(n)$$ is odd if and only if n is a square or twice a square.

We also note s(n) = σ(n) − n. Here s(n) denotes the sum of the proper divisors of n, that is, the divisors of n excluding n itself. This function is used to recognize perfect numbers, which are the n such that 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, $$n = 2^k$$, then $$\sigma(n) = 2 \cdot 2^k - 1 = 2n - 1$$ and $$s(n) = n - 1$$, which makes n almost-perfect.

As an example, for two primes $$p,q:p<q$$, let


 * $$n = p\,q$$.

Then


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

and


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

where $$\varphi(n)$$ is Euler's totient function.

Then, the roots of


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

express p and q in terms of σ(n) and φ(n) only, requiring no knowledge of n or $$p+q$$, as


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

Also, knowing $n$ and either $$\sigma(n)$$ or $$\varphi(n)$$, or, alternatively, $$p+q$$ and either $$\sigma(n)$$ or $$\varphi(n)$$ allows an easy recovery of p and q.

In 1984, Roger Heath-Brown proved that the equality


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

is true for infinitely many values of $n$, see.

Series relations
Two Dirichlet series involving the divisor function are:


 * $$\sum_{n=1}^\infty \frac{\sigma_{a}(n)}{n^s} = \zeta(s) \zeta(s-a)\quad\text{for}\quad s>1,s>a+1,$$

where $$\zeta$$ is the Riemann zeta function. The series for d(n) = &sigma;0(n) gives:


 * $$\sum_{n=1}^\infty \frac{d(n)}{n^s} = \zeta^2(s)\quad\text{for}\quad s>1,$$

and a Ramanujan identity


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

which is a special case of the Rankin–Selberg convolution.

A Lambert series involving the divisor function is:


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

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.

For $$k>0$$, there is an explicit series representation with Ramanujan sums $$ c_m(n) $$ as :
 * $$\sigma_k(n) = \zeta(k+1)n^k\sum_{m=1}^\infty \frac {c_m(n)}{m^{k+1}}.$$

The computation of the first terms of $$c_m(n)$$ shows its oscillations around the "average value" $$\zeta(k+1)n^k$$:
 * $$\sigma_k(n) = \zeta(k+1)n^k \left[ 1 + \frac{(-1)^n}{2^{k+1}} + \frac{2\cos\frac {2\pi n}{3}}{3^{k+1}} + \frac{2\cos\frac {\pi n}{2}}{4^{k+1}} + \cdots\right]$$

Growth rate
In little-o notation, the divisor function satisfies the inequality:
 * $$\mbox{for all }\varepsilon>0,\quad d(n)=o(n^\varepsilon).$$

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

where $$\gamma$$ is Euler's gamma 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. His proof uses Mertens' third 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, Robin's inequality
 * $$\ \sigma(n) < e^\gamma n \log \log n $$ (where γ is the Euler–Mascheroni constant)

holds for all sufficiently large n. The largest known value that violates the inequality is n=5040. In 1984, Guy Robin proved that the inequality is true for all n > 5040 if and only if the Riemann hypothesis is true. This is Robin's theorem and the inequality became known after him. Robin furthermore showed that if the Riemann hypothesis is false then there are an infinite number of values of n that violate the inequality, and it is known that the smallest such n > 5040 must be superabundant. 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.

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.

A related bound was given by Jeffrey Lagarias in 2002, who proved that the Riemann hypothesis is equivalent to the statement that:
 * $$ \sigma(n) < H_n + e^{H_n}\log(H_n)$$

for every natural number n &gt; 1, where $$H_n$$ is the nth harmonic number,.