Erdos-Kac for sum of divisors - MathOverflow

Erdos-Kac for sum of divisors

2010-07-29T04:58:01Z

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 $x$?

Answer by Kevin O'Bryant for Erdos-Kac for sum of divisors

2010-07-29T13:02:54Z

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.

Answer by Will Jagy for Erdos-Kac for sum of divisors

2010-07-29T18:15:12Z

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), http://en.wikipedia.org/wiki/Colossally_abundant_number
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: http://www.springerlink.com/content/p8311481mh32145v/

Answer by Andreas Weingartner for Erdos-Kac for sum of divisors

2011-02-13T02:15:33Z

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.

Answer by kukuriku for Erdos-Kac for sum of divisors

2011-02-13T03:26:55Z

@Weingartner: I'm happy you posted ! I mentioned your result in the comment to the OP's question, but I couldn't remember your name and/or the journal where your paper appeared.

P.S: I don't have the privileges to comment, since I lost my account.