Let $S$ be an infinite set of positive integers, $N_S(z)$ be the number of elements of $S$ less than or equal to $z$, and let

$$D_S(z, n, p)= \sum_{k\in S,k\leq z}\chi(k\equiv p\bmod{n}).$$

Here $\chi$ is the indicator function, and $z, p, n$ are positive integers, with $p<n$ and $n>1$. If

$$\lim_{z\rightarrow\infty} \frac{D_S(z,n,p)}{N_s(z)} = \frac{1}{n}$$

for all $n>1$, regardless of $p$, then the set $S$ is said to be congruentially equidistributed, or in other words, free of congruential restrictions.The exact same concept, referred to as "uniformly distributed in $Z$", is discussed in chapter 5 in the book Uniform Distribution of Sequences by Kuipers and Niederreiter (1974), see here.

Example: The set $S^*$ of all $k+p_k$ where $p_k$ is the $k$-th prime, seems to be congruentially equidistributed. But the set of all primes is not.

If $S$ contains enough elements, say

$$N_S(z) \sim \frac{a z^b}{(\log z)^c} \mbox{ as } z\rightarrow\infty$$

where $a, b, c$ are non-negative real numbers with $\frac{1}{2}< b \leq 1$, is it true that $S+S=\{x+y,$ with $x, y \in S\}$ contains all the positive integers except a finite number of them?

This statement would be true if $S$ was a random set having the same distribution of elements. More precisely, in that case, as a result of the Borel-Cantelli lemma, $S+S$ almost surely contains all the positive integers but a finite number of them. See the last paragraph in my answer to my previous MO question here, for a justification.

Context

If $a=1, b=1, c=1$, we are dealing with numbers that are distributed just like prime numbers, so this is connected to the Goldbach conjecture (GC). The set $S^*$ (see example above) seems congruentially equidistributed, thus proving that every large enough integer is the sum of two elements of $S^*$, might be much less difficult than proving GC. But he set $S$ of primes is NOT congruentially equidistributed, presumably making GC harder to prove.

But GC is equivalent to

GC[1]: all but a finite number of positive integers can be written as $\frac{x-1}{2} + \frac{y-1}{2}$ where $x, y$ are odd primes. In this case $S = \{\frac{z-1}{2}, z$ is an odd prime $\}$ is denoted as $S_1$.

With this transformation, we have eliminated the problem of lack of congruential equidistribution modulo $2$ (the fact that the sum of two odd primes is never an odd number), but there are still plenty of congruential equidistribution failures, actually we may have introduced new ones.

However, we don't need all odd primes for GC, but an infinitesimal fraction of them, to cover all but a finite number of integers. See my second answer to this MO question, for a strong heuristic justification, also discovered by Andrew Granville 13 years ago, here. See also here, where it says (also with convincing arguments, not just from me alone, see here) that you only need super-primes, for which $a=1, b=1, c=2$. Even super-super-primes work.

So the next step is to see whether we can remove a bunch of integers from $S_1$ and perform a similar transformation as in [GC1] to gain more congruential equidistribution, say modulo 2, 3 and 5 this time. The resulting thinned $S_1$ is denoted as $S_2$, and [GC1] is equivalent to

[GC2]: All but a finite number of positive integers can be written as $x+y$ with $x,y\in S_2$.

We can go on and on and create an infinite chain of equivalent conjectures: [GC], [GC1], [GC2] and so on. However, not sure if

• we can eventually get rid of all congruential restrictions (achieving full congruential equidistribution)

• the final set $S_\infty$ makes sense, exists, contains too few numbers, or the set of exceptions becomes infinite

• my original question has a positive answer (congruential equidistribution implies that the set of exceptions is finite)

Note that we could use different sets for the sum of two sets, say $S+S'$ where both are subsets of primes. This would still be OK to prove GC. But I haven't explored that path yet. Not sure if it would help address the congruential equidistribution problem.

I also posted the question on MSE, here.