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 $\alpha_{n}=(\sqrt{2n})^{1+\varepsilon}+O(\log^{2}n)=O_{\varepsilon}(n^{1/2+\varepsilon})$.

**Édit June 5th 2015:** it turns out that the previous assumption is false. Nevertheless I would like to know whether a suitable generalization of the central limit theorem could be useful to show that, if $\alpha_{n}=o(n)$, then $\alpha_{n}=O(\sqrt{n}\log^{2} n)$.
Indeed writing $N_{2}(n)=\sum_{i=1}^{n}X_{i}(n)$ with $X_{i}(n)\in\{0,1\}$ for all $i$, one should be able to define a variance $\sigma^{2}$ as $\dfrac{1}{n}(N_{2}(n)-\dfrac{n.N_{1}(n)}{P_{ord_{c}(n)}})^{2}$ which should tend to $1$ for $n$ large enough, entailing the desired upper bound. Any ideas/insights/references are welcome.

**Edit March 5th, 2016:** Writing as above $N_{2}(n)=\displaystyle{\sum_{i=1}^{n}X_{i}(n)}$ with $X_{i}(n)\in\{0,1\}$, there is, among all possible realizations of the Binomial distribution of parameters $n$ and $p=\dfrac{N_{1}(n)}{P_{ord_{C}(n)}}$, exactly one that coincides with the sequence $(u_{i})_{i\le n}$ of general term term $1_{i\ \ is\ \ a \ \ typical\ \ primality \ \ radius \ \ of \ \ n}$. Defining the quantity $\varepsilon_{i}$ as $\vert X_{i}-\frac{N_{2}(n)}{n}\vert$, then the norm $\| x\|_{1}$ of the vector $x$ whose $i$-th component is $u_{i}$ is $\displaystyle{\\ x\|_{1}=\sum_{i=1}^{n}\varepsilon_{i}}$, while $\displaystyle{\| x\|_{2}=\left(\sum_{i=1}^{n}\varepsilon_{i}^{2}\right)^{1/2}}$. From $\alpha_{n}=\dfrac{P_{ord_{C}(n)}}{N_{1}(n)}\left(N_{2}(n)-\dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}\right)$, it follows that $|\alpha_{n}|\le \dfrac{P_{ord_{C}(n)}}{N_{1}(n)}\| x\|_{1}\le\sqrt{n}\log^{2}n\| x\|_{2}$.

All that remains to be done is proving that $\| x\|_{2}=O(1)$.

**Edit January 22nd 2019:** it seems that the stronger assumption $ \vert p-f\vert\lesssim\frac{p}{\sqrt{n}} $ holds numerically, at least for small values of $ n $. A proof thereof would entail that $ \alpha_{n}\lesssim\sqrt{n} $, which may be stronger than RH.

Actually, writing $ \dfrac{\alpha_{n}}{n}=\dfrac{1}{R_{n}} $ one gets $ R_{n}=\dfrac{n.N_{1}(n)}{N_{2}.P_{ord_{C}(n)}-n.N_{1}(n)} $.

Replacing in the latter $N_{2}(n) $ by the approximation thereof derived from Hardy-Littlewood k-tuple conjecture times $ \frac{n-p_{ord_{c}(n)}}{n} $ to eliminate non typical primality radii should provide (conditionally) the desired result.

Edit May 14th 2019 : can one use the Theorem 1 in https://arxiv.org/abs/1809.01409 to establish rigorously that $\dfrac{\alpha_{n}}{n}=o(1)$? As the sequence of primes is arbitrarily close to arbitrary long arithmetic progressions, one can expect the same to hold for the sequence of primality radii of a given integer $n$. The idea is that the considered sequence behaves 'almost' like an arithmetic progression of gap size $\Delta:=\dfrac{P_{ord_{C}(n)}}{N_{1}(n)}$, and that, were it actually such an arithmetic progression, the quantity $\alpha_{n}$ would vanish.

Edit June 10th 2019: from $p=n-r$ and $q=n+r$ it follows that a prime $l$ dividing $2r$ is such that $p\equiv q\pmod l$. But as $p$ and $q$ are prime both of them are coprime with any prime dividing $P_{ord_{C}(n)}$ and thus if $l\mid 2r$ then $l\mid P_{ord_{C}(n)}$. Of course the $l$-adic valuation of $r_{0}(n)$ can be greater than 1, but this property of being divisible only by primes in a prescribed finite set makes me think that, maybe, $r_{0}(n)$ can be interpreted as the conductor of some 'deep' arithmetic object associated to $n$, like an L-function or an elliptic curve (and thus, via the modularity theorem, to the level of some arithmetic subgroup of the modular group). This has of course an interest in itself but may also be used to provide an upper bound of $r_{0}(n)$ in terms of $n$.

Edit June 12, 2020: can this preprint by Maynard: https://arxiv.org/abs/2006.06572 shed some light on the conjectured relation $\alpha_{n}=o(n)$?

Edit March 20th, 2021: I started learning Python 3 and wrote a program related to this question which supports the main conjecture, namely that $\alpha_{n}\ll n^{1/2}\log^{2}n$.

Here come two screenshots as examples (sorry for using French in it):

The code is far from being optimal as I'm not well-versed in computer science but I can share it with whoever is interested.

The arithmeticity coefficient of $n$ is defined as $1-\frac{\vert\alpha_{n}\vert}{n}$ and measures how close the sequence of primality radii of $n$ is to an arithmetic progression, whose arithmeticity coefficient would be $1$.

Edit May 7th 2024: can we deduce from the previous paragraph that $r_{0}(n)<\left(\frac{n}{N_{2}(n)}\right)^{1+o(1)}$?

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