The arithmetic mean of prime gaps around $x$ is $\ln(x)$. What is the geometric mean of prime gaps around $x$ ?

Does that strongly depend on the conjectures about the smallest and largest gap such as Cramer's conjecture or the twin prime conjecture ?

  • $\begingroup$ Why the geometric average? $\endgroup$
    – draks ...
    May 6, 2015 at 10:51
  • 4
    $\begingroup$ Why not ? A very often used average. It would give some insight. $\endgroup$
    – mick
    May 7, 2015 at 6:13
  • $\begingroup$ ...the arithmetic one is always the first that comes to my mind, so I just wondered why you expect more insight from "geometrical" point of view... $\endgroup$
    – draks ...
    May 7, 2015 at 9:01
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    $\begingroup$ @draks: Looking at things from two sides gives more insight than just one viewpoint. $\endgroup$
    – MvG
    May 7, 2015 at 9:47
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    $\begingroup$ Of course we have by the AM-GM inequality that the geometric mean is smaller. $\endgroup$
    – wythagoras
    May 10, 2015 at 18:48

2 Answers 2


In 1976 Gallagher proved, under the assumption of a uniform version of the Hardy-Littlewood $k$-tuples conjecture, that for any fixed $\lambda>0$ and integer $k$ $$\#\{\text{ integers } x\leq X\ :\ \pi(x+\lambda \log x)-\pi(x)=k\}\sim e^{-\lambda}\frac{\lambda^k}{k!}X,$$ that is it follows a Poisson distribution.

Since the waiting times for a Poisson distribution is an exponential distribution, Gallagher's work also yields (on the assumption of a uniform Hardy-Littlewood conjecture) that for fixed $\alpha,\beta$ $$\frac{1}{\pi(x)}\#\{n\leq \pi(x):\ g_n\in \left(\alpha \log x, \beta \log x\right)\}\sim \int_\alpha^\beta e^{-t}.$$ Thus the geometric mean of the $g_n$ asymptotically will equal $$\exp\left(\frac{1}{\pi(x)}\sum_{n\leq \pi(x)} \log (g_n)\right)\sim \exp\left(\log \log x+\int_0^\infty \log t e^{-t}dt\right).$$ Since $\int_0^\infty \log t e^{-t}dt=-\gamma$ where $\gamma$ is the Euler-Mascheroni constant, and we find that the geometric mean is

$$\sim e^{-\gamma}\log x.$$

  • $\begingroup$ Would like to see the actual proof (free). $\endgroup$
    – mick
    Jun 1, 2015 at 0:35

I thought Hardy-Littlewood might come into it. Here is some numerical data following Erics great answer: x-axis: N ; y-axis: Geometric mean of the first 10000 prime gaps following $10^N$ divided by $\ln 10^N$

x-axis: N

y-axis: Geometric mean of the first 10000 prime gaps following $10^N$ divided by $\ln 10^N$.

$e^{-\gamma} \approx 0.56146$.

  • 1
    $\begingroup$ This is fantastic. I was wondering why my numerical data seemed a little off - I did not take $N$ to be anywhere near large enough! $\endgroup$ May 17, 2015 at 11:03
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    $\begingroup$ It does converge rather slow(ly?). The number of gaps for a given size has to be large as well. Looking at the graph, 10000 gaps is not enough to get a very accurate result, but the trend is clearly visible. $\endgroup$ May 18, 2015 at 12:14

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