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The Riemann hypothesis is equivalent to

$\forall n\geqslant2,\: \sigma(n)<H_n+e^{H_n}\log{H_n}$,

where $\sigma(n)$ is the divisor sum of $n$ and $H_n$ is the nth harmonic number.

For large $n$, $H_n$ is small. The only $n\geqslant 2$ I have found for which $\sigma(n)>e^{H_n}\log{H_n}$ are 2, 3, 4, 6, 12, 24, and 60. Are any other such $n$ known?

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    $\begingroup$ Since $e^{H_n} \log H_n \geq e^\gamma n \log \log n$ for any $n \geq 3$, any counterexample to your inequality would violate Robin's inequality too, so if the Riemann hypothesis is true you shouldn't find any greater than $5040$. $\endgroup$
    – user41593
    Commented Aug 7, 2017 at 13:09
  • $\begingroup$ Many thanks. I've checked way past 5040, so the answer is no, and furthermore, there are no such $n$. Which is good, because it means the RH is true iff $\forall n\gt5040,\: \sigma(n)<e^{H_n}\log{H_n}$. If you write this into an answer, I will accept it. $\endgroup$
    – user113088
    Commented Aug 7, 2017 at 21:26

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According to Lemma 3.1 of Lagarias's paper in which he introduces his criterion for RH, one has $$ e^{H_n} \log H_n \geq e^{\gamma} n \log \log n$$ for any $n \geq 3$. This means that any $n$ which violates your inequality would also yield a counterexample to Robin's inequality $$ \sigma (n) < e^\gamma n \log \log n.$$ Since it is known that the truth of RH implies that $5040$ is the largest counterexample, no $n$ for which $\sigma(n) >e^{H_n}\log H_n$ greater than $5040$ should be found. A computer check then confirms that $n=60$ is the largest counterexample. The interest in Lagarias's criterion lies in its requirement being more restrictive than Robin's (and, of course, than the inequality in this question).

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    $\begingroup$ I guess you meant "no counterexamples to $\sigma(n)<e^{H_{n}}+\log H_{n} $"? $\endgroup$
    – Sylvain JULIEN
    Commented Sep 15, 2017 at 10:48
  • $\begingroup$ Of course, thanks for the correction. $\endgroup$
    – user41593
    Commented Sep 15, 2017 at 15:08

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