Timeline for Erdos-Kac for sum of divisors
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 13, 2011 at 2:15 | answer | added | Andreas Weingartner | timeline score: 5 | |
| Jul 30, 2010 at 19:57 | comment | added | anon | Also it follows rather easily that for almost all integers n <= x we have \sigma(n)/n \leq e^gamma * logloglog n, this follows simply from the fact that for almost all integers n <= x we have w(n) \leq 2 loglog n (where w(n) is the number of prime factors of n). But of course this says nothing about RH. | |
| Jul 30, 2010 at 19:56 | comment | added | anon | The behaviour of the limiting distribution function at infinity and at 0 is well-known. For the later I recommend an article by Tenenbaum and Toulemonde (their method can be adapted to the case of \sigma), for the first it is known that the distribution function decays doubly exponentially i.e exp(-c*exp(t)) by a result of Erdos and with a recent improvement (I don't remember by who). Of course any attempt at RH this way is rather unlikely to work. | |
| Jul 30, 2010 at 1:50 | history | edited | Will Jagy | CC BY-SA 2.5 |
title
|
| Jul 30, 2010 at 1:11 | history | edited | Will Jagy | CC BY-SA 2.5 |
sigma
|
| Jul 29, 2010 at 18:15 | answer | added | Will Jagy | timeline score: 8 | |
| Jul 29, 2010 at 14:10 | comment | added | Wadim Zudilin | David, you are right: there are results about the limit distributions (I don't have references to them either as this knowledge comes from talks attended many years ago). It's just puzzling to see 5 questions by one person in a row, this one looks most reasonable. I doubt that this is a simple curiosity question (especially, after the appearance of the FLT in an ellipse problem). | |
| Jul 29, 2010 at 13:55 | comment | added | David E Speyer | 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 hdebruijn.soo.dto.tudelft.nl/jaar2004/prob.pdf for some similar results. | |
| Jul 29, 2010 at 13:08 | comment | added | Daniel Litt | 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. | |
| Jul 29, 2010 at 13:02 | answer | added | Kevin O'Bryant | timeline score: 5 | |
| Jul 29, 2010 at 12:13 | comment | added | Wadim Zudilin | 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. | |
| Jul 29, 2010 at 9:17 | comment | added | Wadim Zudilin | 100% for $x=1$. I doubt one can improve this percentage. :-) | |
| Jul 29, 2010 at 6:46 | history | edited | Charles Matthews | CC BY-SA 2.5 |
copy edit
|
| Jul 29, 2010 at 5:03 | history | edited | Steven | CC BY-SA 2.5 |
added 8 characters in body
|
| Jul 29, 2010 at 4:59 | comment | added | Qiaochu Yuan | $\phi(n)"$ usually refers to Euler's totient function, so that gets confusing. The sum of all divisors is generally written $\sigma_1(n)$. | |
| Jul 29, 2010 at 4:58 | history | asked | Steven | CC BY-SA 2.5 |