1
$\begingroup$

This question is an analogue of Geometric mean of prime gaps?

Where primes have been replaced by prime twins.

Eric's answer :

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 $k$ $$\#\{\text{ integers } x\leq X\ :t\ \pi(x+\lambda \log x)-\pi(x)=k\}\sim e^{-\lambda}\frac{\lambda^k}{k!}X,$$ 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

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


But there is more.

I asked the same question to my mentor tommy1729.

He reached the same conclusion but a dubious method ?? He referred to Mertens' product. ( Mertens' third theorem about $\Pi_p (1-1/p)$ , see https://en.m.wikipedia.org/wiki/Mertens'_theorems )

So inspired by that , i wonder about the geometric mean and if it relates - analogue to Mertens' product - with an infinite product over odd primes such as $ 0,5 $ $ \Pi_p ( 1 - 2/p )$ , the prime twins constant or similar.

Ofcourse we assume the AM to be $C ln(x)^2$. I guess we might also need an analogue of hardy-littlewood or something. What we actually need is an issue of " inverse mathematics " i guess.

But apart from theorems based on assumptions , there is also supportive data to be collected.

Imho this also calls for a generalization , but it is probably best to settle the above first.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.