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Define $\sigma(n):=\sum_{d\mid n} d$ and $G(n)=\frac{\sigma(n)}{n\log \log n}$. A positive integer $N$ is a GA1 number if $N$ is composite and the inequality $G(N)\geq G(N/p)$ holds for all prime factors $p$ of $N$. An integer $N>1$ is a GA2 number if $G(N) \geq G(aN)$ for all multiples $aN$ of $N$. Finally, a composite number is extraordinary if it is both GA1 and GA2.

A young friend of mine is asking whether there exists some extraordinary number with more than 2 prime factors (counting multiplicity).

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  • $\begingroup$ What is $G(N)$? $\endgroup$
    – Nick Gill
    Sep 11, 2020 at 11:11
  • $\begingroup$ @NickGill, sorry, had forgotten to define it. Sorted now. $\endgroup$
    – user123305
    Sep 11, 2020 at 11:14
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    $\begingroup$ I'm voting to close this question because from what I can tell it's claiming to give a proof of the Riemann hypothesis. $\endgroup$
    – Sam Hopkins
    Sep 14, 2020 at 3:19
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    $\begingroup$ @SamHopkins, that's one of the main reasons i had deleted the answer. However, some OP's voted for it to be re-opened, so i did just that. But i'm ready to re-delete it if necessary. $\endgroup$
    – user123305
    Sep 14, 2020 at 3:22

1 Answer 1

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The Riemann Hypothesis is true if and only if 4 is the only extraordinary number, see theorem 5 of Robin's theorem, primes, and a new elementary reformulation of the Riemann Hypothesis [arxiv.org/abs/1110.5078]. Since 4 has only 2 prime factors, counting multiplicity, in all likelihood the answer to the question in the OP is "no", although we do not yet have a proof.

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    $\begingroup$ My young friend has just sent me his proof that the answer is ''no'' (see his answer copied and pasted below). However, he doesn't seem to think that this result constitutes any significant progress towards RH. $\endgroup$
    – user123305
    Sep 11, 2020 at 12:24
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    $\begingroup$ for the record, this comment refers to a deleted answer by the OP and an extensive discussion at chat.stackexchange.com/rooms/112926/… $\endgroup$
    – Carlo Beenakker
    Sep 12, 2020 at 14:43
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    $\begingroup$ the first version of the answer had a typo on the formula for $\sigma(Nq^3)$, which affected all subsequent calculations, so had to delete over half of the post. Also, i thought it might not be very proper for the post to remain here (considering what it entails). But some OP's voted for it to be re-opened, so i did just that. $\endgroup$
    – user123305
    Sep 14, 2020 at 3:33