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Guy Robin proved $$ \frac{\sigma(n)}{n} < \exp(\gamma)ln(ln(n))+ \frac{0.6482}{ln(ln(n))}, $$ for all integers $n>2,$

where $\gamma$ is Euler's constant, and $ln$ is the logarithm to the base $exp(1).$

Question: There are analogue formulas for the sum of the $k$-th powers of the positive divisors of $n.$

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up vote 3 down vote accepted

Oh, yes, I gave a fairly complete answer at:

A hierarchy of k-highly composite numbers

There is a procedure iniated by Ramanujan that gives a sequence of particularly large values of $$\frac{\sigma_k(n)}{n^k},$$ essentially by taking any $\delta > 0$ and finding the $n = n_\delta$ giving the maximum of $$\frac{\sigma_k(n)}{n^{k + \delta}},$$ and choosing the largest if more than one $n$ gives the maximum. The recipe gives a recipe for the prime factorization of $n = n_\delta.$ Once one has a value, it follows that the original $\frac{\sigma_k(n)}{n^k}$ is larger than for any $m < n.$

There is a fair amount of work involved in interpolating these values into a bound of the type you quote. So, as far as I know, it has only been carried out in entirety for $k=0,1,$ where the former refers to the raw count of divisors.

EDIT: I found the complete answer for $k=0,$

What is the lower bound for highly composite numbers?

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@Will: May you describe with some details how e.g., Robin's formula follows from interpolating these values ? – Luis H Gallardo Jan 24 '11 at 19:06
Take a look at $$ $$ $$ $$ and the link at the end to the preprint of Choie, Lichiardopol, Moree, and Sole. – Will Jagy Jan 24 '11 at 20:44

The sigma function sigma_s(n) is a lot simpler for s > 1. The upper bound is sigma_s(n) < zeta(k)n^s, see Grosswald book on Sequences. For s < 1 see Tanenbaum book on probabilistic number theory.

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@Jerald: Thanks for answer. For example when $k=3$ we have more sharp upper bounds ? – Luis H Gallardo Jan 26 '11 at 19:09

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