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What is the percentage of integers $n$ such that $\frac{\sigma(n)}{n} \geq x$ where $\sigma(n)$ is the sum of all divisors of $n$? Are there any methods of improving these bounds (percentages) for certain $x$?

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$\phi(n)"$ usually refers to Euler's totient function, so that gets confusing. The sum of all divisors is generally written $\sigma_1(n)$. – Qiaochu Yuan Jul 29 '10 at 4:59
100% for $x=1$. I doubt one can improve this percentage. :-) – Wadim Zudilin Jul 29 '10 at 9:17
The problem smells an attempt to approach the RH. The only reasonable estimates from below known for $\sigma(n)/n$, where $\sigma(n)=\sum_{d\mid n}d$, are "in average". Assuming the RH, one has $\sigma(n)/n<e^\gamma \log\log n$ (Robin's criterion), so that any better bound from below (even for a single value of $n$) would disprove the RH. – Wadim Zudilin Jul 29 '10 at 12:13
One trivial thing one can do for lower bounds is note that if $\sigma(n)/n>x$ then any multiple of $n$ has the same property. So computing the first few values of $n$ gives a silly lower bound. – Daniel Litt Jul 29 '10 at 13:08
Wadim: I don't have the references I want right now, but I believe there are some very precise and interesting answers to this question. I think it is a result of Erdos that there is a function $F:[1, \infty) \to [0,1]$ such that, for any $1 \leq a < b$, the density of $n$ with $\sigma(n)/n \in (a,b)$ approaches $F(a) - F(b)$. Hopefully, someone will be able to give the details here. For now, I point you to section 3 of for some similar results. – David Speyer Jul 29 '10 at 13:55

LATER EDIT: The very nice survey article by Steuding that David Speyer mentions in his comment actually refers for greater detail to the book by Mark Kac, an M.A.A. Carus Monograph, called "Statistical Independence in Probability, Analysis and Number Theory." Chapter 4 is called "Primes play a game of chance" and section 2 is called "The statistics of the Euler $\phi$-function." That begins on page 54. In the section Problems, pages 62-64, we learn that $$ \frac{\sigma(n)}{n} $$ does in fact have a limiting distribution (proved by Davenport, methods improved by Erdos), and this density $$ D \left\{ \frac{\sigma(n)}{n} < \omega \right\} = \tau(\omega)$$ is a continuous function of $\omega$. There is not much more to hope for in details, as Erdos showed that the analogous density for $$ \log \frac{\phi(n)}{n} $$ is continuous but "singular," that is has derivative 0 almost everywhere. $$ $$ However, Davenport's result does show that the abundant and deficient numbers both have densities, while the perfect numbers have density 0. While no "variance" is mentioned, a mean for the distribution is given, $$ M \left\{ \frac{\sigma(n)}{n} \right\} = \frac{\pi^2}{6}$$ ORIGINAL: There is a nice survey on fairly elementary methods here by J. L. Nicolas, in a 1988 book called "Ramanujan Revisited." $$ $$ Meanwhile, there is an unconditional result which has not been mentioned, for $N \geq 3$ we have $$ \sigma(N) < e^\gamma \; N \log \log N + \frac{ 0.6482 N}{\log \log N} $$ I hope I am reporting this correctly, it is from a secondary source, attribution is to G.Robin, Grandes valeurs de la fonction somme des diviseurs et hypothese de Riemann, J. Math. Pures Appl. (9) 63 (1984) 187-213. $$ $$ The overall methodology is to consider the colossally abundant numbers of Alaoglu and Erdos (1944),
which were eventually discovered to have also been present in the original version of Ramanujan's paper Highly Composite Numbers (1915). There is some history about why that section was initially omitted, evidently a paper shortage. $$ $$ Here is a link to the first page of a related recent article, also apparently a survey, by Nicolas:

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Will, it seems that Steven did not disappear... I guess he has more questions to ask. :-) – Wadim Zudilin Jul 30 '10 at 1:29

You are asking about abundant numbers, and sieve theory does work to give percentages of the sort you are asking for, for any $x$. For $x=2$, wikipedia reports that the density is strictly between $0.24$ and $0.25$.

Dickson's History of the Theory of Numbers has a proof of upper and lower bounds (for $x=2$) that is boiled down to its essence.

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Oh I see. So for $x=2$, there are already upper and lower bounds. I will Will Jagy's references and see what I can do for $x \neq 2$. Thanks for the links. – Steven Jul 30 '10 at 3:51
The number 2 is historically important, but plays no essential role in the proof of the upper and lower bounds. That is, for any particular $x$, the same method (and a nontrivial amount of computation) gives upper and lower bounds that agree out to a digit or two. – Kevin O'Bryant Jul 30 '10 at 13:17

Let $F(x)$ be the proportion of integers $n$ such that $\frac{\sigma(n)}{n} \ge x$. Deleglise [Experiment. Math. 7 (1998), no. 2, 137-143] showed that $F(2)$, the density of abundant numbers, satisfies $0.2474 < F(2) < 0.2480$. His method can be adapted to get bounds for other fixed values of $x$.

As $x\to \infty$, $F(x)\to 0$ very rapidly. In [Proc. Amer. Math. Soc. 135 (9) (2007) 2677–2681] I showed that $$F(x) = \exp(-e^{x e^{-\gamma}} (1+O(x^{-2}))),$$ where $\gamma$ is Euler's constant. This result also holds for the distribution function of $\frac{n}{\varphi(n)}$, where $\varphi(n)$ is Euler's totient function.

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