# Geometric mean of prime twin gaps?

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

Where primes have been replaced by prime twins.

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.