16
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ Why the geometric average? $\endgroup$ – draks ... May 6 '15 at 10:51
  • 2
    $\begingroup$ Why not ? A very often used average. It would give some insight. $\endgroup$ – mick May 7 '15 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 '15 at 9:01
  • 4
    $\begingroup$ @draks: Looking at things from two sides gives more insight than just one viewpoint. $\endgroup$ – MvG May 7 '15 at 9:47
  • 1
    $\begingroup$ Of course we have by the AM-GM inequality that the geometric mean is smaller. $\endgroup$ – wythagoras May 10 '15 at 18:48
7
+50
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Would like to see the actual proof (free). $\endgroup$ – mick Jun 1 '15 at 0:35
7
$\begingroup$

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 $\ln 10^N$.

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

$\endgroup$
  • 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! $\endgroup$ – Eric Naslund May 17 '15 at 11:03
  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Stocker May 18 '15 at 12:14

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.