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In this article (Theorem 1.2) there is a proof for Robin's inequality for odd numbers, $\sigma(n)/n< e^{\gamma}\log(\log(n))$ where $\gamma$ is the Euler-Mascheroni constant and $\sigma(n)$ is the divisor function. The proof is not hard but uses Euler's Totient function and other considerations.

In this other preprint (Theorem 3.4), the same theorem is stated but a much simpler proof is presented. I think there must be some kind of mistake in the proof because it is too easy compared to the proof already published but I'm not sure.

The author uses the bound

$\sigma(n)/n<e^{\gamma}\log(\log(n))+\frac{0.6483}{\log(\log(n))}$

and the fact that the divisor function is multiplicative. In the following way:

$\sigma(n)/n=s(n)=s(2n)/s(2)<2/3(e^{\gamma}\log(\log(2n))+\frac{0.6483}{\log(\log(2n))})< e^{\gamma}\log(\log(n))$

Is there any mistake in this proof?

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    $\begingroup$ Looks fine to me. Of course you have to work harder for the first display (on which the second display depends). $\endgroup$
    – GH from MO
    Commented Oct 15, 2022 at 18:31
  • $\begingroup$ The derivative seems ok wolframalpha.com/… Choie's proof is much harder, it seems to me that Vojak's proof has to have a mistake. There are other interesting proofs in that article using the same technique. (The article is not published either which makes me skeptical). $\endgroup$ Commented Oct 15, 2022 at 19:38

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Choie et. al. had more concerns than just odd $n;$ if we ask when their argument kicks in, it is simpler in appearance. Small prime $p,$

$$s(n)=s(pn)/s(p)< \frac{p}{1+p}\left(e^{\gamma}\log(\log(pn))+\frac{0.64821365}{\log(\log(pn))}\right) \; ?< ? \; e^{\gamma}\log(\log(n))$$

or $$ \frac{p}{1+p}\left( \frac{\log(\log(pn))}{ \log \log n}+\frac{0.363945701}{\log(\log(pn))\log \log n}\right) \; ?< 1 ? $$

This decreases as $n$ increases, using simple $$ \log \log n < \log \log pn < \log \log n + \frac{\log p}{\log n} $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

For odd $n \geq 17,$ we find $s(n) < e^\gamma \log \log n.$

For $n \neq 0 \pmod 3$ and $n \geq 56, \; \;$ $s(n) < e^\gamma \log \log n.$

For $n \neq 0 \pmod 5$ and $n \geq 898, \; \; \;$ $s(n) < e^\gamma \log \log n.$

For $n \neq 0 \pmod 7$ and $n \geq 19479, \; \; \;$ $s(n) < e^\gamma \log \log n.$

For $n \neq 0 \pmod {11}$ and $n \geq 19913559, \; \; \;$ $s(n) < e^\gamma \log \log n.$

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  • $\begingroup$ Thats interesting, That means for any p for n sufficiently large if p doesn't divide n, then n is under robins bound? For instance this recent article, theorem 3.2 would be your second claim. The rest of the claims I've never seen them. sciencedirect.com/science/article/pii/S0022247X14007069 $\endgroup$ Commented Oct 16, 2022 at 22:17
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    $\begingroup$ @AsanovicTomas yes, sufficiently large $n.$ The overall pattern is that a counterexample to RH here must resemble a Colossally Abundant number; in particular have non-increasing exponents in the prime factorization, including no prime gaps. In brief, the product of primorials. $\endgroup$
    – Will Jagy
    Commented Oct 16, 2022 at 23:36
  • $\begingroup$ I wonder tabout that condition that allows for p<n (or p<p' where p' is the largest prime in the canonical expression of n), if not for all collosally abundant number there exist p that doesn't divide n. $\endgroup$ Commented Oct 16, 2022 at 23:51

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