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][1]. 

**Examples** 

Here $p_k$ denotes the $k$-th prime, with $p_1=2$. The set $S_1$ of all $k+p_k$  seems to be congruentially equidistributed. But the set of all primes is not. The set of squares and the set of cubes are not. If $\alpha$ is irrational, then the set consisting of all $\lfloor \alpha p_k \rfloor$ where the brackets represent the floor function, is congruentially equidistributed: this is a known result. The set $S_2$ consisting of all $(p_{k+1}+p_{k+2})/2$ is also congruentially equidistributed, it seems. 

**Question**

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][2], for a justification. 

**Connection to Goldbach conjecture**

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 $S_1$ (see example above) seems congruentially equidistributed, thus proving that every large enough integer is the sum of two elements of $S_1$, might be much less difficult than proving GC. 
The set of primes is NOT congruentially equidistributed, presumably making GC harder to prove. Note that $S_1$ is more sparse than the set of primes. Both $S_1$ and $S_2$ (see example) also have $a=1,b=1, c=1$. So an alternative to GC, easier to prove, could be:

> All large enough integer $z$ can be written as $z=x+y$ with $x,y\in S_2$.


*I also posted a shorter version of this question on MSE, [here][7].* 


  [1]: https://web.maths.unsw.edu.au/~josefdick/preprints/KuipersNied_book.pdf
  [2]: https://mathoverflow.net/questions/364909/curious-inversion-formula-in-additive-combinatorics
  [3]: https://mathoverflow.net/questions/363055/goldbach-conjecture-and-other-problems-in-additive-combinatorics
  [4]: https://projecteuclid.org/download/pdf_1/euclid.facm/1229618748
  [5]: https://mathoverflow.net/questions/364772/paradox-in-additive-combinatorics
  [6]: https://www.emis.de//journals/INTEGERS/papers/n43/n43.pdf
  [7]: https://math.stackexchange.com/questions/3752511/congruential-equidistribution-in-infinite-sets-or-sequences-of-positive-integers