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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 ?

wlad
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mick
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  • Why the geometric average? – draks ... May 06 '15 at 10:51
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    Why not ? A very often used average. It would give some insight. – mick May 07 '15 at 06:13
  • ...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 07 '15 at 09:01
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    @draks: Looking at things from two sides gives more insight than just one viewpoint. – MvG May 07 '15 at 09:47
  • @MvG ok; _at_mick +1... – draks ... May 07 '15 at 10:46
  • what do you mean by "around x"? – wlad May 09 '15 at 10:45
  • do you mean "up to x"? – wlad May 09 '15 at 11:19
  • I would imagine that this, as you ask, does depend strongly on certain unknown properties of the primes - the arithmetic mean is easy because we sum up a lot of differences and the sum "telescopes" - so if we average $n$ consecutive differences, the average is $\frac{p_{m+n}-p_m}n$ - which relies on a more global property of prime numbers, given by the prime number theorem. The geometric mean won't telescope and tells more about the distribution of prime gaps - e.g. if gaps tend to be either really small or really big, the geometric mean will be dragged down far from the arithmetic mean. – Milo Brandt May 10 '15 at 04:10
  • would this Q be better: if g- is the gap before a prime p, and g+ the gap after p, can we say anything about g-/g+? – JMP May 10 '15 at 10:23
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    Of course we have by the AM-GM inequality that the geometric mean is smaller. – wythagoras May 10 '15 at 18:48

2 Answers2

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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

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

Eric Naslund
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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 $\ln 10^N$.

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

Michael Stocker
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    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
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    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