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:


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 $\alpha_{n}=o(n)$?

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    $\begingroup$ Aside from the fact that it has a $(1/2)+\epsilon$ in it, is there any reason why you think there may be a relation to the Riemann Hypothesis? There are known connections between RH and the Goldbach conjecture, typing the two names into Google will turn up quite a bit (but be careful as some of it is on the cranky side). $\endgroup$ – Gerry Myerson Apr 15 '11 at 23:38
  • $\begingroup$ Have you thought about connecting your heuristic with the Ramanujan Conjecture instead? If the $\alpha_n$ were the Fourier coefficients of a holomorphic cusp form of weight 2 (and some level), then $\alpha_n=O(n^{1/2+\epsilon})$ by the (proven parts of the) Ramanujan Conjecture. Or possibly $\alpha_n=a_nn^{1/2}$, and you want $a_n=O(n^{\epsilon})$. Then $a_n$ could be the normalized Fourier coefficients of a Maass cusp form or a holomorphic cusp form of any weight. $\endgroup$ – B R Apr 17 '11 at 7:04
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    $\begingroup$ @BR: Since $\alpha_{n}$ is rational for all $n$, I don't think that your second statement can be true, particularly if the map $n\mapsto a_{n}$ is multiplicative, but I might be wrong. $\endgroup$ – Sylvain JULIEN Apr 17 '11 at 11:59
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    $\begingroup$ Sylvain, there was some very badly mangled LaTeX which I tried to iron out, but you should recheck very carefully everything after the last "hence" to make sure it came out right. $\endgroup$ – Todd Trimble Apr 9 '15 at 2:26
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    $\begingroup$ Thank you very much Todd, both for your work and for increasing my English vocabulary (I didn't know the verbs to mangle and to iron out). I fixed the few minor remaining mistakes, hope everything's clear now. $\endgroup$ – Sylvain JULIEN Apr 9 '15 at 10:07

I think it could be a safe assumption that this is not equivalent to RH in a simple way (assuming the other assertions of the question are true).

Here is why: to show that RH implies Goldbach (at least asymptotically) is not at all an unnatural idea, which however as far as I know is open.

For example, in 'Refinements of Goldbach's Conjecture, and the Generalized Riemann Hypothesis' Granville discusses questions close to this. However, it seems to me that the asymptotic counts of the number of solutions to 'the Goldbach equations' are related to the RH (and GRH).

Another example would be Deshouillers, Effinger, te Riele, Zinoviev 'A complete Vinogradov $3$-primes theorem under the Riemann hypothesis. Electron. Res. Announc. Amer. Math. Soc. 3 (1997), 99–104.' who showed that ternary Goldbach follows from GRH. This is less directly related as ternary Goldbach is long known asymptotically, and this is thus about eliminating 'small' counterexamples.

So, it just seems more than a bit unlikely that 'RH implies asymptotic Goldbach' can be solved with a half-page argument and an equivalence argument direct enough that somebody might simply supply it here.

In addition, despite an explicit request (made a while ago) there is still no information/evidence provided why this should be equivalent to RH, which I interpret as the absence of any such evidence.

Finally, since being equivalent is a bit of a vague notion (if both were true they were equivalent even if totally unrelated) and since this is too long for a comment anyway, I thought I give these generalities as an answer.

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  • $\begingroup$ One should get from mathoverflow.net/questions/103968/… and Serre's error term for the Chebotarev theorem under GRH that the relative error term $\alpha_{n}/n$, that is the error term divided by the main term, is $O(n^{1/2}\log n/\pi(n))=O(n^{-1/2}\log^{2}n)$, hence the conjectured upper bound for $\alpha_{n}$. $\endgroup$ – Sylvain JULIEN Mar 5 at 21:27

This is too long for a comment and even though this is an old question of mine, as I decided to investigate a little now that I have some piece of evidence that $ r_{0}(n)=O(\log^{2}n) $, I post this numerical observation as an answer.

The conjectural upper bound I just mentioned suggests that one might have $ \alpha_{n}=O(n^{1/2}r_{0}(n)) $ or, as I expect a link with RH, that there is some non trivial zero $ \rho $ of $ \zeta $ such that $\alpha_{n}\approx n^{\rho}r_{0}(n)/f(n) $ with $ f(n)>1 $.

It turns out that for $ n=14 $ one has $ \alpha_{n}\approx n^{1/2+i\gamma_{3}}r_{0}(n)/f(n) $ where $ f(n)=2\dfrac{nN_{1}(n)}{P_{ord_{C}(n)}} $ , $ 1/2+i\gamma_{k} $ being the $ k $-th critical zero of $ \zeta $ with positive imaginary part.

So $ N_{2}(n) $ might be the closest integer to $ \dfrac{nN_{1}}{P_{ord_{c}(n)}}+\frac{r_{0}(n)}{2n}n^{1/2+i\gamma_{N_{1}(n)}} $ .

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This is only a partial attempt of an answer but I think other people might be able to make it definitive.

Following the March 2016 edit, and replacing $X_{i}$ by $i^{-1}\sum_{k=1}^{i}X_{k}$ one may think that the asymptotic Goldbach conjecture, namely that every large enough even integer is the sum of two primes, boils down to saying that the corresponding version of what I denote by $\| x\|_{2}$ converges as $n$ tends to $\infty$. But the asymptotic Goldbach conjecture is equivalent to $\alpha_{n}=o(n)$. So one would get that $\alpha_{n}=o(n)\Longrightarrow\alpha_n=O(\sqrt{n}\log^{2}n)$.

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    $\begingroup$ What is $\alpha(n)$? I don't see it anywhere else on this page. Is it the same as $\alpha_n$? $\endgroup$ – Gerry Myerson Dec 9 '19 at 0:59
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    $\begingroup$ Yes, typo that I just fixed. $\endgroup$ – Sylvain JULIEN Dec 9 '19 at 6:01

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