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 $q_k = p_{k} + p_{k+1} = p_{k} + g_{k}$ 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 $lfloor alpha p_krfloor$ 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 $lfloor(1+epsilon)p_krfloor = p_k + lfloor epsilon p_krfloor$, if conjecture A is true you have this result:

Any large enough integer $n$ can be written as $n=p + q + lfloor epsilon prfloor + lfloor epsilon qrfloor$, 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.