**Update on 7/20/2020: It appears that conjecture A is not correct, you need more conditions for it to be true. See here (an answer to a previous MO question).**

The general problem that I try to solve is this: if $S$ is an infinite set of positive integers, equidistributed in a sense defined here, and large enough as defined in the same post, then all large enough integers can be written as the sum of two elements of $S$. I call this **conjecture A**, and the purpose of my previous question (same link) was to find whether this is a conjecture, a known fact, or not very hard to prove.

Here I try to solve what I call **conjecture B**. Let $p_k$ be the $k$-th prime ($p_1 = 2$) and $q_k = (p_{k} + p_{k+1})/2 = p_{k} + g_{k}$ where $g_{k} =(p_{k+1}-p_{k})/2$ is the half-gap between $p_{k}$ and $p_{k+1}$. Let $S_1$ be the set of all the $q_k$'s, for $k=2,3,\cdots$. Is $S_1$ equidistributed in the same sense, that is equidistributed in all residue classes? For this to be true, it suffices to prove that the half-gaps are equidistributed in residue classes. There is an attempt to answer that question here, but it is not clear to me if the answer is yes, no, or unsure. What is your take on this?

Assuming conjectures A and B are true, then any large enough integer is the sum of two elements of $S_1$. Another interesting result is this: let $S_2$ be the set of all $\lfloor \alpha p_k\rfloor$ where the brackets represent the floor function, $k=1,2,\cdots$, and $\alpha > 0$ is an irrational number. Then any large enough integer is the sum of two elements of $S_2$.

The interesting thing about $S_2$ is that it is known to be equidistributed and furthermore, you can choose $\alpha=1+\epsilon$ with $\epsilon$ an irrational number as close to zero as you want, but NOT exactly zero. Since $\lfloor(1+\epsilon)p_k\rfloor = p_k + \lfloor \epsilon p_k\rfloor$, if conjecture A is true you have this result:

Any large enough integer $n$ can be written as $n=p + q + \lfloor \epsilon p\rfloor + \lfloor \epsilon q\rfloor$, with $p, q$ primes and $\epsilon>0$ an irrational number as close to zero as you want (but not zero).

With $\epsilon=0$, this would be equivalent to Goldbach conjecture, but of course it does not work with $\epsilon=0$ since no odd integer $n$ is the sum of two primes, unless $n=p+2$ and $p$ is prime.

**Two useful references**

Provided by Andrew Granville, who also mentioned the following.

*As to your question the answer is a little surprising and has been the subject of some recent publicity - there are two papers by Robert Lemke Oliver and Soundararajan who look at how often one has $p_n= a \bmod{q}$ and $p_{n+1} = b \bmod{q}$.
Turns out these counts are far from uniformly distributed though an analysis via the circle method reveals that they should be asymptotically the same, but there is a large secondary term which plays a significant role as far as one can ever hope to compute.*

- The distribution of consecutive prime biases and sums of sawtooth random variables
- Unexpected biases in the distribution of consecutive primes

Finally, I will try to prove that if $S$ is equidistributed in residue classes, then $S+S$ is also equidistributed. I posted this as a question on MSE, here.