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Is the following statement true:

Let $\Bbb{A}$ be the set of all Natural numbers n, greater than or equal to 5041, for which the inequality $\displaystyle \sigma(n)<e^{\gamma}n\log\log n$ is not true (does not hold).

$\Bbb{A}$ = $\{\}$ $\Longleftrightarrow $ For all $\displaystyle s ≠ -2n:\zeta(s)=0∧0<Re[s]<1=1/2$

Consequently, is the following wikipedia article making the same explicit statement?

Why or why not?

Referencing https://en.wikipedia.org/wiki/Colossally_abundant_number#Relation_to_the_Riemann_hypothesis

user476467
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  • ??? The Robin's inequality is true for every $n > 5041$ if an only if all the non-trivial zeros of $\zeta(s)$ are on $\Re(s) = 1/2$ (the other are at $-2n,n \in \mathbb{N}^*$)... – reuns Aug 31 '17 at 23:48
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    His adviser, Jean-Louis Nicolas, made the first concrete RH equivalent. It is much easier to experiment with, as finding the colossally abundant numbers is a somewhat elaborate program. Anyway, see http://math.stackexchange.com/questions/630902/is-there-a-good-preferably-comprehensive-list-of-which-conjectures-imply-the-r/630970#630970 .... @reuns, if you don't know the Nicolas criterion, see this link. – Will Jagy Aug 31 '17 at 23:57
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    Tks, yes I know it, it is me you emailed to your copy of Robin's paper. @WillJagy – reuns Aug 31 '17 at 23:59
  • @reuns That there are no violators $\displaystyle n \ge 5041$ if and only if there are no non-trivial zeros $\displaystyle re[s] ≠1/2$ – user476467 Sep 01 '17 at 00:00
  • @user476467 Yes. Because $\log \zeta(s) = -\sum_{p \text{ prime}} \log(1-p^{-s})$ the RH is directly related to the asymptotic distribution of primes, and so is the Robin inequality (because it suffices to look at $\sigma(n)$ for primorials $n = \prod_{p \le k} p$). Explaining the Robin criterion more in detail needs some background, so you need to tell us your level. – reuns Sep 01 '17 at 00:14
  • I would like to clarify that $\Bbb{A}$ is an empty set. Apologies @reuns – user476467 Sep 01 '17 at 00:14
  • I'm a computer scientist and amateur number theorist, not classically trained in Number Theory or advanced mathematics. I've spent hundreds of hours studying the approaches to the Riemann Hypothesis because it fascinates me. I've read many papers and articles searching for the strongest statement of the relationship between the RH and Robin's inequality. I may have missed a Robin criterion emphasis on the feasibility of disproving a 28th violator of RI. Thank you for your answer. Yes, further enlightenment will be put to good use. Because of The RH, I aspire to a mathematics degree. @reuns – user476467 Sep 01 '17 at 00:56
  • @user476467 Robin's criterion is more a miscellaneous topic in the theory of $\zeta$. At first see the books mentioned there (you'll need a background in real, complex and Fourier analysis) – reuns Sep 01 '17 at 02:54

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I put a copy of Nicolas 1983 here

I put a copy of Robin 1984 here.

Will Jagy
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  • and there some related papers – reuns Sep 01 '17 at 02:52
  • Pertaining to the question at hand. I understand that the authors are indeed making it abundantly clear that in proving no further exceptions exist for either the RH or RI, directly proves the remaining statement. – user476467 Sep 01 '17 at 03:49