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

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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). –  Gerry Myerson Apr 15 '11 at 23:38
    
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. –  B R Apr 17 '11 at 7:04
    
@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. –  Sylvain JULIEN Apr 17 '11 at 11:59
    
I may be wrong, too. (In fact, I think it would be a little weird if one cusp form solves Goldbach and there aren't similar problems for other cusp forms to solve, but it seems at least worth ruling out as a technique!) –  B R Apr 17 '11 at 16:38
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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. –  Todd Trimble Apr 9 at 2:26

1 Answer 1

up vote 16 down vote accepted

I think it could be a save 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 unatural 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 there 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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By the way, does the Law of Large Numbers entails that the quantity $\dfrac{N_{2}(n)/n-N_{1}(n)/P_{ord_{C}(n)}}{N_{1}(n)/P_{ord_{C}(n)}}$ tends to $0$ as $n$ tends to $\infty$? –  Sylvain JULIEN Sep 20 '11 at 18:19

protected by Todd Trimble Jan 7 at 14:27

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