EDIT : I copy-paste the beginning of a previous question since Gerry Myerson suggested this question should be self-contained.

"let's consider a composite natural number $n$ greater or equal to $4$. Goldbach's conjecture is equivalent to the following statement: "there is at least one natural number $r$ such as $(n-r)$ and $(n+r)$ are both primes". For obvious reasons $r\leq n-3$. Such a number $r$ will be called a "primality radius" of $n$.

Now let's define the number $ord_{C}(n)$, which depends on $n$, in the following way: $ord_C(n):=\pi(\sqrt{2n-3})$, where $\pi(x)$ is the number of primes less or equal to $x$. $(n+r)$ is a prime only if for all prime $p$ less or equal to $\sqrt{2n-3}$, $p$ doesn't divide $(n+r)$. There are exactly $ord_{C}(n)$ such primes. The number $ord_{C}(n)$ will be called the "natural configuration order" of $n$. Now let's define the "$k$-order configuration" of an integer $m$, denoted $C_{k}(n)$, as the sequence $(m \ \ mod \ \ 2, \ \ m \ \ mod \ \ 3,...,m \ \ mod \ \ p_{k})$. For example $C_{4}(10)=(10\ \ mod \ \ 2,\ \ 10 \ \ mod \ \ 3, \ \ 10 \ \ mod \ \ 5, \ \ 10 \ \ mod \ \ 7)=(0,1,0,3)$. I call $C_{ord_{C}(n)}(n)$ the "natural configuration" of $n$.

A sufficient condition to make $r$ be a primality radius of $n$ is that for all integer $i$ such that $1\leq i\leq ord_{C}(n)$, $(n-r) \ \ mod \ \ p_{i}$ differs from $0$ and $(n+r) \ \ mod \ \ p_{i}$ differs from $0$. If this statement is true, $r$ will be called a "potential typical primality radius" of $n$."

Is it true that Dirichlet's theorem about primes in arithmetic progression implies that there exists $k_n$ depending only on $n$ such that the number $\mathcal{N}_{n}(x)$ of potential typical primality radii of $n$ less than $x$ is such that $\mathcal{N}_{n}(x)\sim k_{n}x$?
If so, the expression of $k_{n}$ is quite easy to find out.

Thanks in advance.