This question came to my mind this afternoon while trying to figure out a possible way to tackle de Polignac's conjecture, which states that every even positive integer can be written as the difference of two consecutive primes in infinitely many ways, and is of course related to [About a possible generalization of Green-Tao's theorem](https://mathoverflow.net/questions/116730/about-a-possible-generalization-of-green-taos-theorem). 


So here it comes: Can **the proof** of Green-Tao's theorem be quite naturally slightly modified to show **unconditionally** that the sequence of $1$-central numbers, i.e positive integers equal to half the sum of two consecutive odd primes, contains arbitrarily long arithmetic progressions? 


Many thanks in advance.



Green-Tao theorem for 1-central numbers
The statement in the original post is true. This follows from a [recent result of Pintz](http://arxiv.org/abs/1509.01564), which is a common generalization of the Green-Tao theorem and the Maynard-Tao theorem:


**Theorem.** Let $m\geq 2$ be an integer. Then for any sufficiently large admissible set $\mathcal{H}$, there exist $h\_1,\dots,h\_m\in\mathcal{H}$ such that the set
$$\{n\in\mathbb{N}:\ \text{$n+h\_1,\dots,n+h\_m$ are consecutive primes}\}$$
contains arbitrary long arithmetic progressions.


Indeed, the special case $m=2$ is already sufficient to derive the statement in the original post.


**P.S.** Note that the main theorems in [Pintz's paper](http://arxiv.org/abs/1509.01564) are stated with a typo: the requirement that $p\_i^\*$ is prime should be dropped (it is the $n$ above). I notified János Pintz, and he agreed with this correction.


