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 $n$. Moreover, if $r\leq n-3$, 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 $P_{ord_{C}(n)}=2\times 3\times...\times p_{ord_{C}(n)}$, $N_{2}(n)$ 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 $\forall \varepsilon>0, \ \ \alpha_{n}=O_{\varepsilon}\left(n^{\frac{1}{2}+\varepsilon}\right)$.

I would like to know whether this is equivalent to the Riemann Hypothesis or not. If so, it would mean that RH implies that every large enough even number is the sum of two primes.

Thank you in advance for your feedback.

EDIT October 13th 2013: to answer Gerry Myerson's question below, the statistical heuristics I refer to is $\vert p−f\vert\leqslant\dfrac{1}{\sqrt{n}}$ with $p$ the "probability" of an integer less than $P_{ord_{C}(n)}$ to be a potential typical primality radius of $n$, hence $p=\dfrac{N_{1}(n)}{P_{ord_{C}(n)}}$ and $f$ the "frequency" of the event "being a typical primality radius of $n$", hence $f=\dfrac{N_{2}(n)}{n}$. This gives $\alpha_{n}=O(\sqrt{n}\log^{2}n)$, which is, up to the implied constant, the error term in the explicit formula of $\psi(n)$ under RH.

Edit August 6th 2014: denoting by $r_{0}(n)$ the smallest typical potential primality radius of $n$, is there a rather rigorous way to figure out what the probability of the event $r_{0}(n)=1$ should be?

Edit January 7th 2015: it appears that the considered equivalence might be obtained from the conjunction of the statements $r_{0}(n)\leq\left(\dfrac{P_{ord_c(n)}}{N_1(n)}\right)^{2}\ll \log^4 n$ and $\alpha_{n}\ll\sqrt{nr_{0}(n)}$.

I didn't manage to prove the latter but any help would be greatly appreciated.

Edit April 8th 2015: it appears that the upper bound $\alpha_{n}=O_{\varepsilon}(n^{1/2+\varepsilon})$ would follow from the following reasonable assumption: $N_{2}(n)$ is the nearest integer to $N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}}(n)-\sqrt{2n-3}}$, which follows from the very definition of what a typical primality radius is. Indeed, writing $N_{2}(n)=\dfrac{n.N_{1}(n)}{P_{ord_{C}}(n)}=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(1)$, one gets $\dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}(1+\dfrac{\alpha_{n}}{n})=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(1)$, hence $1+\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\left(\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}\right)+O(\dfrac{P_{ord_{C}(n)}}{n.N_{1}(n)})$, i.e. $\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}-\dfrac{n(P_{ord_{C}(n)}-\sqrt{2n-3})}{n(P_{ord_{C}(n)}-\sqrt{2n-3})}+O(\dfrac{\log^{2} n}{n})$.

Thus $\alpha_{n}=\dfrac{(n-P_{ord_{C}(n)})\sqrt{2n-3}}{P_{ord_{C}(n)}-′\sqrt{2n-3}}+O(\log^{2} n)$ so $\alpha_{n}=(\sqrt{2n})^{1+\varepsilon}+O(\log^{2}n)=O_{\varepsilon}(n^{1/2+\varepsilon})$.