divisor function

In the parent article there has been proved the formula

\sigma_{1}(n)\;=\;\sum_{0<d\mid n}\!d\;=\;\prod_{i=1}^{k}\frac{p_{i}^{m_{i}+1}% -1}{p_{i}-1}

giving the sum of all positive divisors of an integer $n$ ; there the $p_{i}$ ’s are the distinct positive prime factors of $n$ and $m_{i}$ ’s their multiplicities.

It follows that the sum of the $z$ ’th powers of those divisors is given by

\displaystyle\sigma_{z}(n)\;=\;\sum_{0<d\mid n}\!d^{z}\;=\;\prod_{i=1}^{k}% \frac{p_{i}^{(m_{i}+1)z}-1}{p_{i}^{z}-1}.

(1)

This complex function of $z$ is called divisor function (http://planetmath.org/DivisorFunction). The equation (1) may be written in the form

\displaystyle\sigma_{z}(n)\;=\;\prod_{i=1}^{k}(1+p_{i}^{z}+p_{i}^{2z}+\ldots+p% _{i}^{m_{i}z})

(2)

usable also for $z=0$ . For the special case of one prime power the function consists of the single geometric sum (http://planetmath.org/GeometricSeries)

\sigma_{z}(p^{m})\;=\;1+p^{z}+p^{2z}+\ldots+p^{mz},

which particularly gives $m\!+\!1$ when $p^{z}=1$ , i.e. when $z$ is a multiple of $2i\pi/\ln{p}$ .

A special case of the function (1) is the $\tau$ function (http://planetmath.org/TauFunction) of $n$ :

\sigma_{0}(n)\;=\;\sum_{0<d\mid n}\!1\;=\;\prod_{i=1}^{k}(m_{i}+1)\;=\;\tau(n)

Some inequalities

\sigma_{m}(n)\,\geq\;n^{\frac{m}{2}}\sigma_{0}(n)\quad\mbox{for}\quad m=0,\,1,% \,2,\,\ldots

\sigma_{1}(mn)>\sigma_{1}(m)+\sigma_{1}(n)\quad\forall\,m,n\in\mathbb{Z}

\sigma_{1}(n)\,\leq\;\frac{n\!+\!1}{2}\sigma_{0}(n)

n\!+\!\sqrt{n}\,<\sigma_{1}(n)<\frac{6}{\pi^{2}}\!\cdot\!n\sqrt{n}

Title	divisor function
Canonical name	DivisorFunction
Date of creation	2013-11-27 18:13:38
Last modified on	2013-11-27 18:13:38
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11A05
Classification	msc 11A25