I have requested a pdf of Robin 1984 from campus scanning service. One highlight of the article that really should be mentioned is this:

For $n \geq 13,$ we have
$$ \sigma(n) \; < \; \; e^\gamma \; n \log \log n \; + \; \frac{ \; 0.64821364942... \; \; n \; }{\log \log n},$$ with the constant in the numerator giving equality for $n=12.$

see:

Which $n$ maximize $G(n)=\frac{\sigma(n)}{n \log \log n}$?

That, at least, rests on effective bounds of Rosser and Schoenfeld (1962), which can be downloaded from ROSSER

Well, maybe not so directly. R+S do the unconditional bound for $n/\phi(n)$ in Theorem 15, pages 71-72, formulas (3.41) and (3.42). The treatment for $\sigma(n)$ is quite similar in spirit, maybe Robin was the first to write it down. The analogue of the primorials PRIMO and $n^{1-\delta}/\phi(n)$ is the colossally abundant CA numbers and $\sigma(n)/ n^{1 + \delta}.$

Well, I am not sure where it is written down, but it is easy enough to show that the maximum value, for some $0 < \delta \leq 1, $ of
$$ \frac{ n^{1-\delta}}{\phi(n)} $$
occurs when the prime factor $p$ of $n$ has exponent
$$ v_p(n) = \left \lfloor \frac{p^{1-\delta}}{p-1} \right \rfloor.$$
Since, for a fixed $\delta,$ this expression is either 0 or 1 and nonincreasing in $p,$ it turns out that the optima occur at the primorials, the products of the consecutive primes from 2 to something...

From Alaoglu and Erdos,
the maximum value, for some $0 < \delta \leq 1, $ of
$$ \frac{\sigma(n)}{ n^{1+\delta}} $$
occurs when the prime factor $p$ of $n$ has exponent
$$ v_p(n) = \left\lfloor \frac{\log (p^{1 + \delta} - 1) - \log(p^\delta - 1)}{\log p} \right\rfloor \; - \; 1. $$

This is Theorem 10 on page 455. The results of this construction are the colossally abundant numbers. The construction is originally due to Ramanujan, but the part of his manuscript that dealt with ca numbers was not printed owing to paper shortages at the time.

Hardy and Wright use $d(n)$ for the number of divisors of $n.$ This is in the original paper by Ramanujan. For some $0 < \delta \leq 1, $ the maximum of $$ \frac{d(n)}{ n^{\delta}} $$ occurs when the prime factor $p$ of $n$ has exponent
$$ v_p(n) = \left\lfloor \frac{1}{p^\delta - 1} \right\rfloor. $$ The results are called the superior highly composite numbers SHC.

So, taking all three with $\delta = 1/2,$ we get lemmas
$$ \phi(n) \geq \sqrt{\frac{n}{2}}, \; \; d(n) \leq \sqrt{3n}, \; \; \sigma(n) \leq 3 \left( \frac{n}{2} \right)^{3/2}. $$

In all three cases, if $\delta$ is such that more than one number $n$ achieves the maximum value of the ratio specified, we are choosing the largest of these $n$'s.

savethe link (e.g. by right clicking on it). It's a PDF file, which you can then view in the normal way. $\endgroup$ – Tom Leinster Dec 25 '11 at 18:341more comment