Let$\ \sigma(n)$ be the sum-of divisors function, with the divisors raised to$\ 1$. If the Riemann Hypothesis is false, Robin proved there are infinitely many counterexamples to the inequality$$\ \sigma(n)<e^\gamma n \log \log n.$$ There are 27 small counterexamples, but the conjecture is that it holds for every$\ n>5040$. Akbary and Friggstad showed the least counterexample to it must be a superabundant number, i.e. a number$\ a$ such that$\ \frac{\sigma(a)}{a}>\frac{\sigma(b)}{b}$ for all$\ b<a$. Now, it is a virtual certainty that if the inequality fails (for some$\ n>5040$), the maximum of the ratio$\ \frac{\sigma(n)}{n \log \log n}$ will be reached by a colossally abundant number, namely a number$\ c$ such that$\ \frac{\sigma(c)}{c^{1+\epsilon}}>\frac{\sigma(d)}{d^{1+\epsilon}}$, for all$\ d<c$ and for some$\ \epsilon>0$. Since it could lead me to something on the subject, what I'm asking is: if the inequality fails, will only a finite number of colossally abundant numbers satisfy Robin's inequality?

  • $\begingroup$ Could you explain why CA non-counter examples are a problem? Or might be a problem? $\endgroup$
    – almagest
    Sep 14 '14 at 18:40
  • $\begingroup$ @almagest If there are only a finite number of them, I might have a clue. $\endgroup$ Sep 14 '14 at 18:47
  • $\begingroup$ Have you gone through Robin's proof to see if the counterexamples constructed are in fact colossally abundant numbers? (PS: Your post would probably get more attention if you took the trouble to include Robin's inequality and the definition of colossaly abundant.) $\endgroup$ Sep 14 '14 at 18:53
  • $\begingroup$ @almagest Not a single exception, infinitely many exceptions. He proved there would be infinitely many counterexamples, as I stated in the body of the question. I'm asking whether there are infinitely many CA non-counterexamples or not, since it might lead me to something. And afterall, why shouldn't it be of interest? At least, I would wonder anyway. $\endgroup$ Sep 14 '14 at 19:01
  • $\begingroup$ @GregMartin I have, and there is a statement about CA numbers in the case RH is false. But I can't quite understand it: fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-xpa1/v/t1.0-9/… (I've never used Big Omega notation) I'll edit the question as you suggest, thanks $\endgroup$ Sep 14 '14 at 19:05

For the benefit of those who may not be familiar with all this:

In 1915 Ramanujan proved that if the Riemman Hypothesis is true, then for all sufficiently large $n$ we have an inequality on $\frac{\sigma(n)}{n}$, where $\sigma(n)$ is the sum of the divisors of the positive integer $n$. The inequality was $$\sigma(n)<e^\gamma n \ln\ln n$$ In 1984 Robin elaborated this to show that if there is a single exception to this for $n>5040$ (the largest currently known exception and a "colossally abundant number" - hereafter a "CA"), then the Riemann Hypothesis is false.

Because of the importance of the RH, this attracted a good deal of attention. But 30 years later nothing seems to have come of it.

Obviously the most plausible candidates to break the inequality are numbers with lots of divisors. I believe, though I am weak on the history, that the concept, if not the name, of CA came from Ramanujan during his 1915 work.

To give a little perspective: few people are interested in CA per se. But vast numbers of people are interested in RH, even if only a tiny number do serious work on it (because of the risk to one's reputation). So the immediate interest of the inequality was that it provided another way, superficially at least totally different, to disprove the RH by computation. People had got fed up with results that the first zillion zeros were on the line, particularly when analysts quoted Littlewood's "Miscellany" on the Skewes' number (which is now a somewhat less compelling point :) ). So this was something else to try.

However, after 30 years nothing has so far come of that. In the meantime people have been working on CA as objects of interest in their own right.

The question is whether if the RH is false (so that the inequality fails - Robin's result was an iff type result), then only a finite number of CA will satisfy the Robin inequality.

[Added later - the precise question having been clarified]

If I had realised that would be the question, I would never have started to answer it! I had earlier understood it to be a quite different question. But there are a few points to be made.

I have never read Robin's paper - my interest is in RH, and I do not regard Robin's inequality as a useful way of tackling the RH (a judgment which of course is of zero interest to anyone else). So I am at a serious disadvantage - in not having read the paper and to compound that, I cannot immediately lay my hands on it.

It is fairly easy to show that if $a,b$ are coprime counterexamples to the inequality, then so is $ab$ provided $a,b$ are sufficiently big (which they would be). It is also fairly clear that unless something weird happens at huge values, counterexamples are likely to be CA. So it seems a fairly safe guess that if RH is false then there will be infinitely many CA not satisfying the Robin inequality.

But unfortunately the question asks for something much stronger than that, namely will all but finitely many CA fail to satisfy it?

Short answer: good question; I have no idea and should delete this entire answer. But pending a little digging early in the coming week I will leave it here until a better answer comes.

In my defence, I would only say that I have only been using this site for less than 3 weeks. I have answered lots of daft questions, and had fun competing putting up answers fast. I failed to adjust adequately when this one came along. But it does illustrate the wisdom of the concept of clarifying the question with comments before writing Answers. I had started to do that, but got impatient when I could not immediately grasp the clarifications. That was entirely my fault. I apologise unreservedly.

  • $\begingroup$ Yes, but there can't be only a single exception, that's what I'm saying. $\endgroup$ Sep 14 '14 at 19:14
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    $\begingroup$ @WillJagy Many thanks for that. 11 pages of bedtime reading, I think I shall go to bed early tonight (now 8:30pm here in London, UK) :) $\endgroup$
    – almagest
    Sep 14 '14 at 19:34
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    $\begingroup$ @WillJagy. Thanks also for your favourite (104pp) manuscript! Actually quite short by modern standards. I am still laughing about the reluctance of the "experts" to get to grips with Mochizuki's opus proving (?) the abc conjecture. When will people grasp that the whole delight (and burden) of maths is that experts are irrelevant. You simply have to understand it yourself. There is no substitute -certainly when anything important is at stake. It is outrageous for "experts" to be refusing to do their homework, because their time is too important! $\endgroup$
    – almagest
    Sep 14 '14 at 19:41
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    $\begingroup$ No! It is a repeat of Perelman. But this time little of the fault is with the author. Unlike Perelman he has been publishing (long, difficult) papers leading up to the final result for years. But no one could be bothered to get to grips with them. Yes, of course it is easier to be spoonfed at a seminar. And particularly juicy if the author only publishes a bare key idea, so the audience can pile in to enhance their reputations with the easy consequences. But that is no excuse. The brilliant guys just cannot stop themselves trying to prove the hard things. That is why I moved out. I could! $\endgroup$
    – almagest
    Sep 14 '14 at 19:54
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    $\begingroup$ @WillJagy You much kinder than I have any right to hope for. I am off to bed :) $\endgroup$
    – almagest
    Sep 14 '14 at 20:46

I just stumbled across this question many years after the posting and did not see what I believe to be the answer. If RIE fails the only requirement on colossally abundant numbers being exceptions is that there must be at least one.

There could be only one c.a. exception. Or there could be a finite number of c.a. exceptions. Or there could be an infinite number of c.a exceptions.

If there are only a finite number of ca exceptions, then that would require there be an infinite number of colossally abundant numbers that satisfy RIE.

If there are an infinite number of ca exceptions then there may be a finite or an infinite number of colossally abundant numbers that satisfy RIE.


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