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?
1 Answers
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.

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Yes, but there can't be only a single exception, that's what I'm saying. – Vincenzo Oliva Sep 14 '14 at 19:14
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Robin's theorem is an "iff" theorem; see here: http://en.wikipedia.org/wiki/Divisor_function – Christian Blatter Sep 14 '14 at 19:25
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@ChristianBlatter Yes. Sorry, I am biased because of my much greater interest in RH :) Of course, there is no reason not to be interested in CA numbers per se. – almagest Sep 14 '14 at 19:31
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When I see RH Robin things, I try to mention the Nicolas criterion, much easier to calculate primorials than colossally abundants. http://arxiv.org/abs/1202.0729 Nicolas was the adviser of Robin, Robin's criterion is based on that of Nicolas in any case. There ought to be a wikipedia page and a link to it at the Robin criterion page. See also http://arxiv.org/abs/1012.3613 – Will Jagy Sep 14 '14 at 19:31
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1@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) :) – almagest Sep 14 '14 at 19:34
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@WillJagy Is the moral of this story that we don't know the answer to my question? – Vincenzo Oliva Sep 14 '14 at 19:37
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Good. I have the Robin paper if you want to see, also the Nicolas original, Ramanujan original. I am not big on RH but enjoy the Ramanujan 'superior' recipe. – Will Jagy Sep 14 '14 at 19:38
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@numb, I have seen snippets about related ideas. I vaguely recall something about how the first counterexample to RH need only be superabundant, but that probably means also a CA counterexample. I don't recall anything about numbers finite or infinite, Nicolas is far more satisfactory in that regard. – Will Jagy Sep 14 '14 at 19:41
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1@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! – almagest Sep 14 '14 at 19:41
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Almagest, glad you like it. It started out as a permanent record of what Kap and I did together, with proofs, gradually got bigger. I did finally get about ten pages of it published, but I'm not sure anyone capable of proving the conjectures is interested; however, maybe give it a few years... My understanding is that friends are holding a major Mochizuki conference in November; the trouble seems to be his lack of interest in travel, giving talks, and in general explaining himself in person. – Will Jagy Sep 14 '14 at 19:46
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1No! 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! – almagest Sep 14 '14 at 19:54
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I'm pointing out this only because almagest might mislead someone who reads his comment. Did no one notice that he is erroneously thinking of$\ \sigma(n)$ as$\ \sigma_0(n)$? Robin's inequality concerns the sum-of-divisors function, with the divisors having exponent$\ 1$. – Vincenzo Oliva Sep 14 '14 at 19:55
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No, actually he is thinking of$\ \omega(n)$, since he said "factors" and not "divisors". – Vincenzo Oliva Sep 14 '14 at 19:57
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@Numberlover. You are correct! Many thanks for picking up the embarrassing typo. – almagest Sep 14 '14 at 19:59
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@Numberlover. Please note that you asked a question which was hard for me to understand, and still is. I spent some time trying to help you. Yes, I wrote this more carelessly (and much faster) than I perhaps should have, because I am extremely busy doing other things at the same time. I will read it through. – almagest Sep 14 '14 at 20:02
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@almagest And I thank you for that. I think I have made the question quite clearer. – Vincenzo Oliva Sep 14 '14 at 20:05
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@Numberlover. I will change factor to divisor because I agree that is probably the more common term amongst those in the field. I never use $\omega(n)$, but isn't it the number of prime factors. – almagest Sep 14 '14 at 20:08
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@almagest Yes, but if one says "factors", I certainly think of prime factors and not divisors. Also, I've never seen$\ \sigma(n)$ defined as the sum of the factors of$\ n$, and more importantly, you said "number", which indeed made me initially think of$\ \sigma_0(n)$. – Vincenzo Oliva Sep 14 '14 at 20:12
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@Numberlover +1 for interesting question. I apologise. I am tired out this evening (I am in London UK). – almagest Sep 14 '14 at 20:12
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@Numberlover. All good points. I have changed it. Please point out any other improvements. My main problem is that I still don't understand your interest/question. I now take it to be that you are interested in CA numbers more than RH, so could people please look at it and explain it from that point of view? My other problem is that I rarely interact with mathematicians, whereas I am more used to trying to explain things to a lay audience, so I instinctively try to avoid technical terms. Clearly a mistake in this case! – almagest Sep 14 '14 at 20:15
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@WillJagy Does this theorem of Robin's answer my question? https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-xpa1/v/t1.0-9/10599117_811146212281022_3030317901621603793_n.jpg?oh=e1c93842a42f678bcaa56dd7f8047593&oe=54CBDC0B&gda=1419257779_5ddb9e3f090aee2d3d345e181f46141a – Vincenzo Oliva Sep 14 '14 at 20:32
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@Will So what does he exactly states in that? (I actually don't know Big Omega notation, besides its name) – Vincenzo Oliva Sep 14 '14 at 20:38
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Almagest, you should get some rest, revisit this if you have time and energy tomorrow. As to Mochizuki, all my impressions are completely secondhand, wikipedia for example. There was some discussion on MO, which I thought was premature. Oh, I did not see your comment for some 40 minutes, if you do not type in @Will (at least three initial letters) I will probably not be notified. As far as the question above, I sent numberlover the Robin original, I do not think he has gotten much out of it, and hopes for miracles. – Will Jagy Sep 14 '14 at 20:41
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@numb, until finding an explicit definition in the paper itself, I would expect that it means the difference is sometimes larger in absolute value than the given size, but it may be either positive or negative. Look here: http://en.wikipedia.org/wiki/Big_O_notation#The_Hardy.E2.80.93Littlewood_definition – Will Jagy Sep 14 '14 at 20:43
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1@WillJagy You much kinder than I have any right to hope for. I am off to bed :) – almagest Sep 14 '14 at 20:46
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@almagest I guess you haven't found the answer to this question. Might you want to take a look at something related: http://math.stackexchange.com/questions/1015425/proof-regarding-robins-inequality-ri ? – Vincenzo Oliva Nov 12 '14 at 17:31
(I've never used Big Omega notation) I'll edit the question as you suggest, thanks
– Vincenzo Oliva Sep 14 '14 at 19:05