The stronger form of Dirichlet's conjecture states that, for example, $$\lim_{N\to\infty} \frac{\text{ the number of primes } \leq N \text{ of the form } 1+8k }{\text{ the number of primes } \leq N \text{ of the form } 3+8k } = 1.$$
The $1$ and $3$ were arbitrary; they each could have been any number from $\lbrace{1,3,5,7\rbrace}.$
In particular, due to the prime number theorem, we have:
$$\text{ the number of primes } \leq N \text{ of the form } i+8k \approx \frac{N}{4\log(N)}\quad \text{ for each } i \in \lbrace{1,3,5,7\rbrace} $$
And that this doesn't work for just $8.$ It works for every number $n$ and every set of numbers coprime to $n\quad $(for $n=8,\ $ this set is $\lbrace{1,3,5,7\rbrace}).$
So, the primes satisfy Property $(1)$ which is defined as:
$$\text{ for each } i \in S:=\lbrace{j\in [n]: j\text{ coprime to } n \rbrace},\ \text{ the number of primes } \leq N$$ $$ \text{ of the form } i+nk \approx \frac{N}{t\log(N)}\quad \text{ where } t = \vert S\vert. $$
This places serious restrictions on the equal/even distribution of the set of prime numbers. Isn't this, by itself, enough to prove the asymptotic Goldbach conjecture, that is, there exists $C$ such that every even number $n\geq C$ can be written as the sum of two primes? I'm not saying I have a proof right now, but I guess what I'm asking is:
Is there an example of odd numbers which satisfy Property $(1)$ which is not an asymptotic additive basis of even numbers?
I would expect such a group of odd numbers to exist, and I'll sketch a construction below. The details are getting annoying, and I'm surprised that I can't find a reference that already does this.
The idea is going to be the following: We will inductively construct a sequence $N_i$ of positive integers, approaching infinity. For each $i$, we will choose a subset $P_i$ of primes in $(N_{i-1}, N_i]$ such that $2 N_i$ cannot be written as $p+q$ with $p \in \bigcup_{j \leq i} P_j$ and $q$ a prime in $[N_i, 2 N_i]$. Then $P:=\bigcup_{i = 1}^{\infty} P_i$ will be our set violating the Goldbach conjecture and, if we don't omit too many primes when making $P_i$, then $P$ will have the same asymtotic behavior in arithmetic progressions that the primes do.
To start our induction off, take $N_0 = 0$, $N_1 = 1$ and $P_1 = \emptyset$. Now, suppose that we have found $N_0$, $N_1$, ..., $N_{i-1}$ and $P_1$, $P_2$, ..., $P_{i-1}$. Choose $N_i$ such that $2N_i - p$ is not prime for any prime $p$ in $\bigcup_{j < i} P_j$; this can be easily done by the chinese remainder theorem. Now, to make $P_i$, start with the set of all primes in $(N_{i-1}, N_i]$ and delete the primes of the form $2 N_i - q$ for $q$ a prime $\geq N_i$. We have succeeded in making $2N_i$ not be of the form $p+q$ for $p \in \bigcup_{j \leq i} P_j$ and $q$ a prime in $[N_i, 2 N_i]$.
It remains to see how many primes are missing from $\bigcup_{j \leq i} P_j$. The number of primes which we deleted from $(N_{i-1}, N_i]$ should be around $\tfrac{N_i}{(\log N_i)^2}$, because that is how many ways we expect to be able to write $2 N_i$ as a sum of two primes. In any particular arithmetic progression $\{ am+b \}$ with $(a,b)=1$, the number of primes in $(N_{i-1}, N_i]$ will be $\tfrac{N_i}{\phi(a) \log N_i}$, by the PNT in arithmetic progressions. As $N_i \to \infty$, the extra factor of $\log N_i$ will overwhelm the constant factor $\phi(a)$.
The reason that I am not calling this post a complete proof is that I do not know to what extent we have proved that $2N$ can be written in $O \left( \tfrac{N}{(\log N)^2} \right)$ ways as a sum of two primes. I'm also surprised that I couldn't find a published paper which did this work for me.