This question is a follow up to my question About Goldbach’s conjecture whose beginning of the content I now copy paste so as to get a self contained question:

“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 $rleq n-3$. 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 $1leq ileq ord_{C}(n)$, $(n-r) mod p_{i}$ 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 $n$.

Moreover, if $rleq n-3$, then $r$ will be called a “typical primality radius” of $n$.

Now let’s define $N_{1}(n)$ as the number of potential typical primality radii of $n$ less than $P_{ord_{C}(n)}$, where $P_{ord_{C}(n)}=2times 3times…times p_{ord_{C}(n)}$, $N_{2}(n)$ as the number of typical primality radii of $n$, and $alpha_{n}$ by the following equality:

$N_{2}(n)=dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}left(1+dfrac{alpha_{n}}{n}right)$

It is quite easy to give an exact expression of $N_{1}(n)$ and to show that:

$dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}>left(c.dfrac{n}{log(n)^{2}}right)left(1+o(1)right)$, where $c$ is a positive constant.

A statistical heuristics makes me think that $forall varepsilon>0, alpha_{n}=O_{varepsilon}left(n^{frac{1}{2}+varepsilon}right)$.”

So let’s consider a generalization of the counting function of the primality radii of $n$ as follows:

Let $N_{2}(n;k)$ denote the number of primality radii of $n$ not exceeding $k$ and define $beta_{n,k}$ writing the following equality:

$$N_{2}(n;k)=dfrac{kN_{1}(n)}{P_{ord_{c}}(n)}left(1+beta_{n,k}right)$$

Of course $N_{2}(n;n)=N_{2}(n)$ and $beta_{n,n}=frac{alpha_{n}}{n}$.

Denoting by $r_{0}(n)$ the smallest primality radius of $n$, one has trivially $N_{2}(n;r_{0}(n))=frac{r_{0}(n)N_{1}(n)}{P_{ord_{c}}(n)}left(1+beta_{n,r_{0}(n)}right)=1$.

So, can one get bounds on $beta_{n,k}$ so as to provide an upperbound for $r_{0}(n)$?