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The Riemann Hypothesis is true if and only if Robin's Inequality is true for all n>5040. It has also been shown by Akbary and Friggstad that the smallest counterexample greater than 5040, if it exists, must be a Superabundant Number.

These two facts suggest that Robin's Inequality might provide a more elementary way to show the Riemann Hypothesis is true or help provide a counterexample in the form of a Superabundant Number which violates the inequality. Robin's Inequality is elegantly stated:

$$f(n)=\frac{\sigma (n)}{e^{\gamma} n \log \log n}<1$$

Below is a plot of the LHS for the first 2000 Superabundant Numbers above 5040 from A004394.

enter image description here

Looking at this plot and knowing how close the values get to 1 make it seem plausible that RH could be false, but if true it appears to be an asymptotically an increasing function bounded above by 1. Examining the remaining terms provided, the function behavior looks similar and for the largest term listed $f(a[1000000])\approx0.9998655$. No terms before the millionth cross the threshold of one and $a[1000000]\approx10^{103082}$.

Is this the highest known verification of RH? It is my understanding that ZetaGrid and others working with the zeta function directly have only verified zeros up to $10^{13}$. Also, from my understanding of Briggs he appears to have only verified completely up to $10^{154}$ using SA numbers?

In fact, based on this curve and the possible relationship shown here, I am willing to venture a conjecture.

Goldbug
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    appears you want Briggs https://projecteuclid.org/download/pdf_1/euclid.em/1175789744 – Will Jagy Jul 15 '19 at 20:17
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    the bigger check was on CA numbers; note how he is not saying "the first" counterexample is CA. And he checked those very, very high. I show computation of CA numbers at https://math.stackexchange.com/questions/3288519/method-to-find-colossally-abundant-numbers/3290227#3290227 – Will Jagy Jul 15 '19 at 20:51
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    That the zeros of $\zeta(s)$ of imaginary part $< N=10^{18}$ are on the critical line implies $\psi(x)= x + O(x^{1/2} \log^4 x)$ effective bound is true for $x < f(N)$ and that the Robin inequality is true for $\log n < f(N)$ with $f(N)$ being something like $N^a$. @WillJagy – reuns Jul 15 '19 at 23:41
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    A "more elementary" way to show the Riemann Hypothesis is true?!? Seriously, there is no reason at all to think RH should be accessible to a proof by Robin's inequality. If anything, the relation between them means Robin's inequality is going to be very hard to prove. Accumulating numerical evidence is not likely to lead to any actual proof. – user1728 Jul 19 '19 at 18:48
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    Numerical evidence will never lead to a proof obviously, but observation is the first step in the scientific method and is often overlooked in mathematics. By "more elementary" I just mean it doesn't on the surface involve analytic number theory. You must be aware that some conjectures that were originally proven with complex machinery later turned out to have simpler elementary proofs. In short, I don't find your comment very constructive @user1728, perhaps this is why you choose to remain anonymous? As the immortal poet TS once said, "haters shall continueth to hate". – Goldbug Jul 19 '19 at 19:57
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    I think you are misunderstanding the philosophy of Robin's inequality. The "obvious" criterion for the RH are based or on the zeros crossing of Hardy Z function or on things like $\pi(x) = Li(x)+O(x^{1/2+\epsilon})$. The latter can be made effective : there is $C$ such that the RH is true iff $\pi(x) \in Li(x) + [-C x^{1/2} \log x,C x^{1/2} \log x]$. The point of Robin's inequality is to replace $\pi(x)$ by $\sigma(n)$ which at first seems much more "innocuous" being the coefficients of $\zeta(s)\zeta(s-1)$ not of $\log \zeta(s)$ or $1/\zeta(s)$, showing the RH is "hiding" in many places. – reuns Jul 20 '19 at 01:04
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    @goldbug it's curious to be guessing why I am anonymous when you are "goldbug". In any event, observation is used a lot in mathematics, both for numerical data and looking at special cases of general questions. The point of my previous comment was that it is not realistic to think the Robin's inequality viewpoint is going to lead to a proof of RH. – user1728 Jul 21 '19 at 21:05
  • @user1728 Perhaps my language is a bit optimistic, but I don't think we should completely dismiss the possibility that examining Robin's will help prove RH. I was surprised when I plotted the values to see how close they come to one and simply thought other people might find it interesting as well. In fact my efforts here are to see the feasibility of finding a counterexample to RH, which I also still don't dismiss is possible. And here is a link if you don't know what a Goldbug is. You can tell me your opinion of that idea as well! – Goldbug Jul 22 '19 at 13:52
  • To return to the original question, is this approach using Robin's Inequality and data from T.D. Noe the largest known verification of RH? Briggs looked at CA numbers but it is not clear that this provides complete verification up to some threshold, although it does show that these CA numbers he examined do not seem to provide a counterexample. – Goldbug Jul 22 '19 at 14:00
  • @reuns Is the inverse also true? i.e., does the truth of Robin's inequality up to a given value $x$ imply a zero-free region of zeta of a particular size? As it stands, I think that's one piece that should be fleshed out in the question itself, because the claim/question is 'highest known verification of RH' but the question itself doesn't have anything describing the mapping between what values of $N$ Robin's conjecture is true for and what regions can be shown zero-free, so currently it's an apples-to-oranges comparison. – Steven Stadnicki Aug 02 '19 at 19:27
  • @StevenStadnicki Sure. I think everything is encoded in the effective explicit formula $\psi(x) = x + \sum_{|\Im(\rho)| \le T}\frac{x^\rho}{\rho} + O(1)+O(\frac{ x\log^2(xT) }{T})$ thus if the zeros up to $N$ are on the critical line then $\psi(x) = x + \sum_{|t| \le N}\frac{x^{1/2+it}}{1/2+it} + O(x^{1/2}\log^2(x))=x+O(x^{1/2}\log^2(x))$ "the RH is true for $x < N^2$". The $O$ constant can be made effective and it is from there that they obtain the Robin criterion. – reuns Aug 03 '19 at 12:27
  • https://mathoverflow.net/questions/110944/what-does-the-numerically-verified-part-of-the-riemann-hypothesis-tell-about-pri/110959 – reuns Aug 03 '19 at 12:50
  • @reuns Not sure if I completely understand, are you saying there is a one-to-one correspondence between the zeros on the critical line up to $N$ of RH and the N of RI and that verifying RI up to $N$ validates RH up to $N^2$? – Goldbug Aug 05 '19 at 16:32

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