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Let $\sigma$ be the divisor sum function, $\gamma$ the Euler-Mascheroni constant and $n>5040$. Robin showed that if the inequality$$\displaystyle \sigma(n)<e^{\gamma}n\log\log n$$ ever fails, it does infinitely often. It is quite intuitive that infinitely many of the counterexamples (if not all) would be superabundant, i.e. natural numbers $a$ such that $\displaystyle \frac{\sigma(a)}{a}>\frac{\sigma(b)}{b}$ for all $b<a$. My proof, which I'd like to have verified:

Let $SA_k$ be the $k$-th superabundant number. Then assume, without loss of generality,

$$ \left\{ \begin{array}{c} \sigma(SA_k)<e^{\gamma}SA_k\log\log SA_k \\ \sigma(m)\ge e^{\gamma}m \log \log m \\ SA_l < SA_k<m<SA_{k+1} \ , \end{array} \right. $$ where $SA_l$ is the largest superabundant counterexample. So we must have $$\displaystyle \frac{\sigma(m)}{\sigma(SA_k)}>\frac{m \log\log m}{SA_k\log\log SA_k} \\ \ \ \ \ \ \ \frac{\sigma(m)}{m \log\log m} >\frac{\sigma(SA_k)}{SA_k\log\log SA_k}. \ \ \ \ \ \ (1)$$ But since $m$ is between two consecutive superabundant numbers, it is not superabundant itself, hence it is by definition $\displaystyle \frac{\sigma(m)}{m}\le\frac{\sigma(SA_k)}{SA_k}$. Given that $ \log\log m >\log\log SA_k$, we easily find $$\displaystyle \frac{\sigma(m)}{m \log\log m} <\frac{\sigma(SA_k)}{SA_k\log\log SA_k},$$ contradicting $(1)$. As a result, we have no counterexamples $> SA_l$, which is an absurdity. $ \ \ \ \ \ \ \ \ \ \ \square$

Any comment, suggestion for improvement or alternative proof would be highly appreciated.

Vincenzo Oliva
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1 Answers1

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Looks right, although you have thrown in subscripts and extra letters that are not needed. Note that Akbary and Friggstad pointed out in 2009 that the first counterexample to Robin would be superabundant; given your argument, the reason that not all counterexamples need be superabundant is that a superabundant counterexamples can be followed by a trickle of not-quite-superabundant counterexamples that are smaller than the next superabundant number.

Shorter: $A$ is largest superabundant counterexample. There are infinitely many counterexamples (if any), so there is a counter example $n$ larger than some superabundant number $B$ that is not a counterexample, that is consecutive superabundant $B,C $ with $B < n < C.$ We do not really care about $C,$ can also forget $A.$

We have, since we are assuming $n$ is not superabundant, indeed is smaller than the very next SA number, $$ \frac{\sigma(n)}{n} \leq \frac{\sigma(B)}{B}. $$ As $B$ is not a counterexample, $$ \frac{\sigma(B)}{B \log \log B} < e^\gamma. $$ Since $n > B,$ we also have $$ \frac{1}{\log \log n} < \frac{1}{\log \log B}. $$ Combine with $ \frac{\sigma(n)}{n} \leq \frac{\sigma(B)}{B} $ to arrive at $$ \frac{\sigma(n)}{n \log \log n} < \frac{\sigma(B)}{B \log \log B} < e^\gamma. $$ This contradicts $n$ being a counterexample, so, if there are any counterexamples, infinitely many are superabundant.

Note that Robin showed in his section 3 that at least one colossally abundant number is a counterexample if there are any. I'm not sure it is known about infinitude of CA counterexamples, if any.

EEDDIITT: I once did something computational here with CA numbers, incredible amount of work for a disappointing outcome How would I find a number where $\sum_{d\mid n}d > 100n$? The conclusion is that RH is not going to be decided by something that can be done on a computer.

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Will Jagy
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  • So many thanks! As for using $m$ instead of $n$, it's due to a kind of tendency of mine to avoid using letters which I used in the presentation of the general problem. Well, now I agree with you, no need of $m$. And indeed, I like those capital letters more than $SA$ with some subscript, somewhat more essential (hence, nicer). Besides that, your proof is better because it doesn't depend on Robin's second theorem, thank you. I recall that result about a CA counterexample of Robin.

    Yeah, almost certainly RH needs a proof, regardless of its direction. Speaking of which, is the object of...

     – Vincenzo Oliva Nov 12 '14 at 22:05
  • my question obvious? If I'm not mistaken, Robin didn't prove it on his paper. This is actually a lemma I was planning to add to a broader paper. Is it instead not necessary? (also, I noticed I said "last" and not "largest". Totally better the latter, strange that I thought only of the former, not the first time I use "largest") – Vincenzo Oliva Nov 12 '14 at 22:06
  • *in the first comment I meant to say that although it uses Robin's second theorem, it doesn't depend on it as much as my proof. And afterall, it simply is more straightforward. – Vincenzo Oliva Nov 12 '14 at 22:12
  • @Vin, there are many papers that follow up on Robin, some elementary. I do not know the full contents of Akbary Friggstad, which is about three pages. It is certainly possible that some short article points out that there would be infinitely many SA counterexamples if any. For example, I found that reference in this: http://arxiv.org/abs/1211.2147v3 – Will Jagy Nov 12 '14 at 22:19
  • No, A.F. don't prove that, I had already checked. I took a look at N.-J. but as far as I can see XA numbers can't help me, as they are mostly dealt with if RH is true. Well, I guess the lemma will do no harm. – Vincenzo Oliva Nov 12 '14 at 22:37
  • So in the lemma it is enough to assume $n$ is a non-SA number being a counterexample greater than a SA number not being a counterexample, since you just need that (and Robin's second theorem) to derive the contradiction, namely $n$ not being a counterexample. Correct? – Vincenzo Oliva Nov 12 '14 at 23:04
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    @Vin that seems right. – Will Jagy Nov 12 '14 at 23:22