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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

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  • $\begingroup$ ??? 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 $-2n,n \in \mathbb{N}^*$)... $\endgroup$ – reuns Aug 31 '17 at 23:48
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    $\begingroup$ 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 math.stackexchange.com/questions/630902/… .... @reuns, if you don't know the Nicolas criterion, see this link. $\endgroup$ – Will Jagy Aug 31 '17 at 23:57
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    $\begingroup$ Tks, yes I know it, it is me you emailed to your copy of Robin's paper. @WillJagy $\endgroup$ – reuns Aug 31 '17 at 23:59
  • $\begingroup$ @reuns That there are no violators $\displaystyle n \ge 5041$ if and only if there are no non-trivial zeros $\displaystyle re[s] ≠1/2$ $\endgroup$ – user476467 Sep 1 '17 at 0:00
  • $\begingroup$ @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. $\endgroup$ – reuns Sep 1 '17 at 0:14
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I put a copy of Nicolas 1983 here

I put a copy of Robin 1984 here.

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  • $\begingroup$ and there some related papers $\endgroup$ – reuns Sep 1 '17 at 2:52
  • $\begingroup$ 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. $\endgroup$ – user476467 Sep 1 '17 at 3:49

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