Geometric mean of prime gaps?

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 ?

• Why the geometric average? – draks ... May 6 '15 at 10:51
• Why not ? A very often used average. It would give some insight. – mick May 7 '15 at 6:13
• ...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... – draks ... May 7 '15 at 9:01
• @draks: Looking at things from two sides gives more insight than just one viewpoint. – MvG May 7 '15 at 9:47
• Of course we have by the AM-GM inequality that the geometric mean is smaller. – wythagoras May 10 '15 at 18:48

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

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

• Would like to see the actual proof (free). – mick Jun 1 '15 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 $\ln 10^N$.

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

• 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! – Eric Naslund May 17 '15 at 11:03
• 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. – Michael Stocker May 18 '15 at 12:14