2
$\begingroup$

Let $$H_n=1+\frac{1}{2}+\cdots+\frac{1}{n},$$ the nth harmonic number and $$\sigma(n)=\sum_{d\mid n}d,$$ the sum of divisor function, for example $\sigma(6)=12$.

I believe that this could be a nice problem in analysis.

Question. For which integer $M\geq 3$ can you claim easily, without assuming Riemann Hypothesis, that previous arithmetical functions satisfies $$\sigma(n)\leq H_n+M^{H_n}\cdot\frac{\log H_n}{\log M},$$ for all $n\geq 1$?

In a phrase, what is the minimum integer $M\geq e$ that satisfies a modified Lagarias' statement, and it is easily to prove, unconditionally, with analysis?

Thanks in advance.

$\endgroup$
1
1
$\begingroup$

Looking at the formulation you want with $M,$ $M=3$ works, indeed any real $M > e$ works.

Lemma 3.1 in Lagarias is, for $n \geq 3,$

$$ e^\gamma n \log \log n \leq \exp (H_n) \log (H_n) $$

Formula (2.2), Robin showed, unconditionally, for $n \geq 3,$ $$ \sigma(n) < e^\gamma n \log \log n + 0.6482 \frac{n}{\log \log n}. $$

Without using the extra $H_n$ at all, we get $$ \sigma(n) < \exp (H_n) \log (H_n) + 0.6482 \frac{n}{\log \log n}. $$

If we include the $H_n,$ formula (3.7) is, this time for $n \geq 20,$ $$ H_n + \exp (H_n) \log (H_n) \leq e^\gamma n \log \log n + \frac{7n}{ \log n}.$$

$\endgroup$
5
  • $\begingroup$ Thanks @WillJagy I understand, then that it is easy to compute numerically the least $M$ such that $$H_n +M^{H_n}\cdot\frac{\log H_n}{\log M} \leq e^\gamma n \log \log n + \frac{7n}{ \log n}?$$ $\endgroup$ – user243301 Jan 7 '16 at 20:26
  • $\begingroup$ @JuanLG sure, the only reason to have $M > e$ is very small $n.$ I would keep $M=e$ and say some formulation you like works for $n > n_0,$ then explicitly find $n_0$ by computer, as all these bounds are effective (we know when they begin to be true) $\endgroup$ – Will Jagy Jan 7 '16 at 20:31
  • 1
    $\begingroup$ Then very thanks much @WillJagy $\endgroup$ – user243301 Jan 7 '16 at 20:38
  • $\begingroup$ @JuanLG The conditions of Lagarias and of Robin got more popular as elementary equivalents of RH, bu the most interesting to compute with is the original condition of Nicolas, math.univ-lyon1.fr/~nicolas and item number 84 at math.univ-lyon1.fr/~nicolas/publications.html $\endgroup$ – Will Jagy Jan 7 '16 at 20:44
  • $\begingroup$ Very thanks much, when I read another time your answer I see more clearly. Very thanks much for references too @WillJagy $\endgroup$ – user243301 Jan 7 '16 at 20:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy