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Define $N_k \geq 6$ to be the $k-th$ primorial number and let $\sigma(n)$ be the divisor function.

It seems that $u_k = \dfrac{\sigma(N_k)}{N_k \log\log N_k}$ is a decreasing function ?

By computation, we find that the first few values of $u_k$ (to 2dp) are $u_1 = 3.42, u_2 = 1.96, u_3 = 1.63, u_{4} = 1.46, u_5 = 1.38 \cdots$

Is there a known result from which this follows ?

As far as i'm aware, the only existing result along these lines is Gronwall's theorem, which says that $\lim$ $\sup u_k = e^{\gamma}$, where $\gamma$ is the Euler-Mascheroni constant.

EDIT: By direct arithmetic that $u_k > u_{k+1}$ and simplifying, the problem reduces to showing that $\dfrac{\log\log N_{k+1}}{\log\log N_k} > \dfrac{1+p_{k+1}}{p_{k+1}}$, where $p_n$ denotes the $n-th$ prime.

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  • $\begingroup$ Check out Merten's theorems. If the values stay above the oscillations, then it is decreasing. I don't know enough to say if there is a counterexample, but if there is k will likely be greater than 10^8. Gerhard "Ask WolframAlpha. Maybe It Knows" Paseman, 2016.02.11. $\endgroup$
    – Gerhard Paseman
    Commented Feb 11, 2016 at 19:31
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    $\begingroup$ You misquote Gronwall's theorem which is about $\limsup_{N\to\infty}\frac{\sigma(N)}{N\log\log N}$ rather than about $\limsup_{k\to\infty}\frac{\sigma(N_k)}{N_k\log\log N_k}$. In fact it is easy to show that the latter equals $\frac{6}{\pi^2}e^\gamma=1.0827\dots$. $\endgroup$
    – GH from MO
    Commented Feb 11, 2016 at 20:05
  • $\begingroup$ @G.H, thank you very much for both your comment and answer, which left me curious: is it then true that $\frac{\sigma(N_{k}}{N_{k}\log\log N_k }< e^{\gamma}$, for sufficiently large $N_k$ ? $\endgroup$
    – favoured
    Commented Feb 11, 2016 at 21:17
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    $\begingroup$ @GHfromMO I don't believe that preprint needed RH to prove that Robin's inequality holds for square-free numbers. I thought it proved it unconditionally. $\endgroup$
    – Pace Nielsen
    Commented Feb 11, 2016 at 21:53
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    $\begingroup$ @PaceNielsen: I see, I did not look at that preprint for lack of time. I just verified on the back of an envelope earlier that the limit is $\frac{6}{\pi^2}e^\gamma$, from which the inequality follows for large $k$. This was the original question. I understand now that it holds for all $k$. Thanks for pointing this out. $\endgroup$
    – GH from MO
    Commented Feb 11, 2016 at 21:58

1 Answer 1

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The answer is probably no. Assume that $u_{k+1}<u_k$ for all $k$. Then $$ \frac{1}{p_{k+1}}<\frac{\log\log N_{k+1}}{\log\log N_k}-1=\frac{\log\frac{\log N_{k+1}}{\log N_k}}{\log\log N_k}<\frac{\frac{\log N_{k+1}}{\log N_k}-1}{\log\log N_k}=\frac{\log p_{k+1}}{\log N_k\log\log N_k}.$$ In particular, $$ \theta(p_k)\log\theta(p_k)<p_{k+1}\log p_{k+1},$$ whence $$ \theta(p_k)<p_{k+1}. $$ By Littlewood's theorem, the left hand side exceeds $p_k+p_k^{1/2}$ infinitely often, while it is widely believed that the right hand side is smaller than $p_k+p_k^{1/3}$ for every large $k$. So, if we believe the last upper bound, there are infinitely many counterexamples (but these might occur very rarely).

Added. By quoting Littlewood's theorem a bit more precisely, we get a contradiction already by Legendre's conjecture that there is a prime number between any two large consecutive squares.

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  • $\begingroup$ @G.H, thank you very much for both your comment and answer, which left me curious: is it then true that $\frac{\sigma(N_{k}}{N_{k}\log\log N_k }< e^{\gamma}$, for sufficiently large $N_k$ ? $\endgroup$
    – favoured
    Commented Feb 11, 2016 at 21:19
  • $\begingroup$ @favoured: Yes, because the fraction tends to $6/\pi^2$ times the right hand side. $\endgroup$
    – GH from MO
    Commented Feb 11, 2016 at 21:37
  • $\begingroup$ @ GH : what is $\theta(p_k)$ ? $\endgroup$
    – reuns
    Commented Feb 24, 2016 at 20:47
  • $\begingroup$ @user1952009: $\theta(x)=\sum_{p\leq x}\log p$ is the first Chebyshev function. See at en.wikipedia.org/wiki/Chebyshev_function $\endgroup$
    – GH from MO
    Commented Feb 24, 2016 at 20:49
  • $\begingroup$ GH, I had not noticed this question. It appears that person(s?) are switching various versions of RH. In particular, see the enjoyable paper by Planat et al about the Nicolas criterion, arxiv.org/abs/1012.3613 which shows that a related sequence, if increasing forever, proves RH but disproves Cramer's conjecture. $\endgroup$
    – Will Jagy
    Commented Feb 24, 2016 at 20:58

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