# A conjectured asymptotics giving rise to the desired upper bound for alpha_n

Even though I meant to put an end to this blog publishing a purposed proof of the asymptotic Goldbach conjecture, I finally decided, after the attack that took place in Paris yesterday, to keep on writing new articles here. This is my way to support Charlie Hebdo.

So, I observed that the upper bound $\alpha_{n}\ll \sqrt{n}\log^{2}n$ would follow from both the inequality $r_{0}\ll \log^{4} n$ established in the previous article and the relation $\alpha_{n}\ll \sqrt{nr_{0}(n)}$ that came to my mind rather unexpectedly.

Maybe using once again the inequality of arithmetic and geometric means could help establish the latter rigorously.

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# Sketch of proof of the asymptotic Goldbach conjecture

This article is meant to be the last one on this blog, and aims at summing up the ideas developped in the previous articles to give a rather accurate sketch of proof of the asymptotic Goldbach conjecture.

First, let’s give all the relevant notations:

$n$: a positive integer strictly greater than $13$.

$r$: a primality radius of $n$, namely a non-negative integer less than $n-2$ such that both $n-r$ and $n+r$ are primes.

$C_{k}(n)$: the $k$-tuple $(n \ \ mod \ \ p_{1},...,n \ \ mod \ \ p_{k})$, where $p_{i}$ denotes the $i$-th prime.

$ord_{C}(n)$ : $\pi(\sqrt{2n-3})$ where $\pi(x)$ is the number of primes below $x$.

$P_{ord_{C}(n)}$: the product of the first $ord_{C}(n)$ primes.

$r_{k}(n)$: the $k$-th potential typical primality radius of $n$, i.e the $(k+1)$-th non-negative integer less than $P_{ord_{C}}(n)$ such that both $C_{ord_{C}(n)}(n-r_{k}(n))$ and $C_{ord_{C}(n)}(n+r_{k}(n))$ contain no $0$:

for all prime $p$ less than $\sqrt{2n-3}$, one has simultaneously:

$n-r_{k}(n)\ \ mod \ \ p\neq 0$

and

$n+r_{k}(n) \ \ mod \ \ p\neq 0$

The adjective « typical » means that:

$n-r_{k}(n)\ge p_{ord_{C}(n)}+1$

and the adjective « potential » is used because of the upper limit

$P_{ord_{c}}(n)$

A potential typical primality radius of $n$ less than $n$ will be simply called a typical primality radius of $n$.

$N_{1}(n)$: the total number of potential typical primality radii of $n$, an expression of which is:

$\displaystyle{\prod_{3\le p\le p_{ord_{C}(n)}}(p-2)\prod_{p\mid n}\dfrac{(p-1)}{(p-2)}}$.

$N_{2}(n)$: the total number of typical primality radii of $n$.

$e_{1}$: the quantity $2r_{0}(n)$.

$e_{k+2}$: the quantity

$r_{k+1}(n)-r_{k}(n)$ if $0\le k\le N_{1}(n)-1$

The quantities $e_{i}$ for $i$ ranging from $1$ to $N_{1}(n)$ will be called « Goldbach gaps of the first kind ».

$\varepsilon_{i}$: a quantity such that:

if $i$ < $j$ then $\varepsilon_{i}$ < $\varepsilon_{j}$

and such that there exists $k$ such that $\varepsilon_{i}=e_{k}$

The quantities $\varepsilon_{i}$ will be called « Goldbach gaps of the second kind » and the number of $k$ such that:

$\varepsilon_{i}=e_{k}$ that is, the multiplicity of $\varepsilon_{i}$, will be denoted by $w_{i}$.

$\eta(n)$: the number of Goldbach gaps of the second kind.

$pr_{i}$: the ratio $\dfrac{w_{i}}{N_{1}(n)}$.

$s_{i}$: the quantity equal to:

$\dfrac{1}{2}$ if $i\in\{1,\eta(n)\}$ and equal to $0$ otherwise.

$\varepsilon_{m}$: the quantity $\dfrac{\varepsilon_{1}+\varepsilon_{\eta(n)}}{2}$.

$\varepsilon_{moy}$: the ratio $\dfrac{P_{ord_{C}(n)}}{N_{1}(n)}$.

The famous Goldbach conjecture asserts that every even integer greater than $3$ is the sum of two primes. We 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 $n$.

Then we define the « $k$-order configuration » of an integer $m$, denoted by $C_{k}(m)$, as the following sequence:

$(m \ \ mod \ \ 2, \ \ m \ \ mod \ \ 3,...,m \ \ mod \ \ p_{k})$.

For example $C_{4}(10)=(10\ \ mod \ \ 2,\ \ 10 \ \ mod \ \ 3, \ \ 10 \ \ mod \ \ 5, \ \ 10 \ \ mod \ \ 7)=(0,1,0,3)$.

We call $C_{ord_{C}(n)}(n)$ the « natural configuration » of $n$.

An almost sufficient condition to make $r'$ be a primality radius of $n$ is:

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 double statement is true, $r'$ will be called a « potential typical primality radius » of $n$.

Moreover, if $r'\le n-3$, then $r'$ will be called a « typical primality radius » of $n$ and denoted simply by $r$.

In what follows we show that every large enough positive integer admits a typical primality radius, from which it follows that every large enough even integer is the sum of two primes.

The proof is based on two lemmas:

Lemma 1: $\dfrac{P_{ord_{C}(n)}}{N_{1}(n)}=O(\log^{2} n)$

Proof:

One has:

$\dfrac{N_{1}(n)}{P_{ord_{C}(n)}}=\displaystyle{\dfrac{1}{2}\prod_{3\le p\le p_{ord_{c}(n)}}\left(\dfrac{p-2}{p}\right)\prod_{p\mid n}\dfrac{p-1}{p-2}}$

But:

$\displaystyle{\prod_{3\le p\le p_{ord_{c}(n)}}\left(\dfrac{p-2}{p}\right)}=\displaystyle{\prod_{3\le p\le p_{ord_{c}(n)}}\dfrac{p(p-2)}{p^{2}}}$

$=\displaystyle{\prod_{3\le p\le p_{ord_{c}(n)}}\dfrac{p(p-2)}{(p-1)^{2}}\prod_{3\le p\le p_{ord_{c}(n)}}\left(\dfrac{p-1}{p}\right)^{2}}$.

Using Mertens’ formula, namely:

$\displaystyle{\prod_{p\le x}\dfrac{p-1}{p}}\sim\dfrac{e^{-\gamma}}{\log x}$

one gets:

$\displaystyle{\prod_{3\le p\le p_{ord_{C}(n)}}\left(\dfrac{p-1}{p}\right)^{2}}\sim\dfrac{4e^{-2\gamma}}{\log^{2} n}$

and thus:

$\displaystyle{\prod_{3\le p\le p_{ord_{c}(n)}}\left(\dfrac{p-2}{p}\right)\ge(1+o(1)) \dfrac{8e^{-2\gamma}C_{J}}{\log^{2}n}}$

where

$C_{J}:=\displaystyle{\prod_{3\le p}\dfrac{p(p-2)}{(p-1)^{2}}}=0.66016...$

is the so called twin prime constant.

Finally:

$\dfrac{N_{1}(n)}{P_{ord_{C}(n)}}\ge(1+o(1))\dfrac{8e^{-2\gamma}C_{J}}{\log^{2}n}$

hence:

$\dfrac{P_{ord_{C}(n)}}{N_{1}(n)}=O(\log^{2} n)$.

Lemma 2: Assume that $\eta(n)\ge 3$.

Then $\varepsilon_{m}\le 2\varepsilon_{moy}$.

Proof:

One has:

$\displaystyle{\sum_{i=1}^{\eta(n)}w_{i}\varepsilon_{i}}=P_{ord_{C}(n)}$

and:

$\displaystyle{\sum_{i=1}^{\eta(n)}w_{i}=N_{1}(n)}$

thus:

$\varepsilon_{moy}=\displaystyle{\sum_{i=1}^{\eta(n)}pr_{i}\varepsilon_{i}}$

Moreover, one has:

$\varepsilon_{m}=\dfrac{\varepsilon_{1}+\varepsilon_{\eta(n)}}{2}$

so that:

$\varepsilon_{m}=\displaystyle{\sum_{i=1}^{\eta_{n}}s_{i}\varepsilon_{i}}$

hence:

$\vert\varepsilon_{moy}-\varepsilon_{m}\vert=\vert\displaystyle{\sum_{i}^{\eta(n)}\varepsilon_{i}(pr_{i}-s_{i})\vert}$

$=\vert(pr_{1}-\dfrac{1}{2})\varepsilon_{1}+(pr_{\eta(n)}-\dfrac{1}{2})\varepsilon_{\eta(n)}+\displaystyle{\sum_{i=2}^{\eta(n)-1}pr_{i}\varepsilon_{i}}\vert$

Hence:

$\vert\varepsilon_{moy}-\varepsilon_{m}\vert\le\displaystyle{\sum_{i=2}^{\eta(n)-1}pr_{i}\varepsilon_{i}}$

and:

$\vert\varepsilon_{moy}-\varepsilon_{m}\vert\le \varepsilon_{moy}$

Therefore $\varepsilon_{m}\le 2\varepsilon_{moy}$

Theorem: $r_{0}(n)=O(\log^4 n)$

Proof:

One can use the inequality of the arithmetic and geometric mean to write:

$\varepsilon_{1}.\varepsilon_{\eta(n)}\le \varepsilon_{m}^{2}$

As $\varepsilon_{m}\le 2\varepsilon_{moy}$, one gets:

$\varepsilon_{1}.\varepsilon_{\eta(n)}\le 4\varepsilon_{moy}^{2}$

As obviously $\varepsilon_{1}\ge 2$ and $2r_{0}(n)\le\varepsilon_{\eta(n)}$, one finally gets:

$r_{0}(n)\le \left(\dfrac{P_{ord_{C}(n)}}{N_{1}(n)}\right)^{2}$

Hence $r_{0}(n)\ll \log^{4} n$.

All that remains to be done is to prove the inequality $\eta(n)\ge 3$, which should be true whenever $n\ge 26$. This will be the subject of a future article.

# A proof of the inequality $latex \varepsilon_{m}\le 2\varepsilon_{moy}$

Given $n\ge 14$ a positive integer, let’s call the difference between two consecutive potential typical primality radii of $n$ a « Goldbach gap ». There are exactly $N_{1}(n)$ such Goldbach gaps. Let’s denote by $\eta(n)$ the number of distinct Goldbach gaps, and let’s write $\varepsilon_{i}$, $1\le i\le \eta(n)$ the $i$-th such Goldbach gap in the increasing order. Thus $\varepsilon_{1}=\varepsilon_{min}$ and $\varepsilon_{\eta(n)}=\varepsilon_{max}$. Moreover, let’s denote by $w_{i}$ the multiplicity of $\varepsilon_{i}$. One has $\displaystyle{\sum_{i}^{\eta(n)}w_{i}\varepsilon_{i}=P_{ord_{C}(n)}}$ and $\displaystyle{\sum_{i}^{\eta(n)}w_{i}=N_{1}(n)}$. Writing $p_{i}:=\dfrac{w_{i}}{N_{1}(n)}$, one gets $\varepsilon_{moy}=\displaystyle{\sum_{i=1}^{\eta(n)}p_{i}\varepsilon_{i}}$. Let’s define $s_{i}$ for $i$ ranging from $1$ to $\eta(n)$ so that $s_{1}=\dfrac{1}{2}$, $s_{\eta(n)}=\dfrac{1}{2}$ and $s_{i}=0$ if $i\not\in\{1,\eta(n)\}$. One has $\vert\varepsilon_{moy}-\varepsilon_{m}\vert=\vert\displaystyle{\sum_{i}^{\eta(n)}\varepsilon_{i}(p_{i}-s_{i})\vert}=\vert(p_{1}-\dfrac{1}{2})\varepsilon_{1}+(p_{\eta(n)}-\dfrac{1}{2})\varepsilon_{\eta(n)}+\displaystyle{\sum_{i=2}^{\eta(n)-1}p_{i}\varepsilon_{i}}\vert$. Hence $\vert\varepsilon_{moy}-\varepsilon_{m}\vert\le\displaystyle{\sum_{i=2}^{\eta(n)-1}p_{i}\varepsilon_{i}}$, and $\vert\varepsilon_{moy}-\varepsilon_{m}\vert\le \varepsilon_{moy}$. Therefore $\varepsilon_{m}\le 2\varepsilon_{moy}$.

# A proof of the upper bound r0(n)=O(log^4 n)

Given a positive integer $n\ge 14$, let $\varepsilon_{min}$ (respectively $\varepsilon_{max}$) be the minimal (respectively maximal) distance between two consecutive potential typical primality radii of $n$. Moreover, let $\varepsilon_{m}:=\dfrac{\varepsilon_{min}+\varepsilon_{max}}{2}$ and $\varepsilon_{moy}:=\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}$. One has $\varepsilon_{m}\le 2\varepsilon_{moy}$ (this inequality will be proved in the next article).

Using the inequality of arithmetic and geometric means, one can write the following inequality: $\varepsilon_{min}.\varepsilon_{max}\le \varepsilon_{m}^{2}$. As  $\varepsilon_{m}\le 2\varepsilon_{moy}$, one gets $\varepsilon_{min}.\varepsilon_{max}\le 4\varepsilon_{moy}^{2}$. So that $\varepsilon_{max}\le\dfrac{4}{\varepsilon_{min}}\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}$. Obviously $2r_{0}(n)\le \varepsilon_{max}$, hence $r_{0}(n)\le\dfrac{2}{\varepsilon_{min}}\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}$. As $\varepsilon_{min}\ge 2$, one finally gets $r_{0}(n)\le\left(\dfrac{P_{ord_{C}}(n)}{N_{1}(n)}\right)^{2}\lesssim\dfrac{\log^{4} n}{4.C_{J}^{2}}$, where by $x\lesssim y$ we mean $x\le (1+o(1)) y$ and where $C_{J}=0.66016...$ is the twin prime constant. Hence there exists $n_{0}$ such that $n\ge n_{0}\Rightarrow r_{0}(n)\le n-1$. Therefore every large enough even integer is the sum of two primes.

# k-central numbers and bounded gaps between primes

As in the previous articles, let’s assume Goldbach’s conjecture so as to define properly the notion of $k$-central number. The number of $k$-central numbers less than $x$ $\pi_{C,k}(x)$ should verify the following relation:

$\pi_{C,k}(x)=\vert\{n\le x, k_{0}(n)=k\}\vert$

and thus $\pi_{C,k}(x)\le\dfrac{\pi(x+\max_{n\le x}r_{0}(n))}{k}(1+o(1))$.

I now formulate the following conjecture:

Negligible fundamental primality radius conjecture (NFPR conjecture for short):

$\forall\varepsilon>0,\forall x>2, \max_{n\le x}r_{0}(n)=O_{\varepsilon}(x^{\varepsilon})$

One can deduce from this conjecture and the prime number theorem that $\dfrac{\pi(x+\max_{n\le x}r_{0}(n))}{k}\sim\dfrac{\pi(x)}{k}$.

Thus $\pi_{C,k}(x)\le \dfrac{\pi(x)}{k}(1+o(1))$.

Hence, defining $\mathcal{N}_{k}(x)$ as the quantity equal to:

$\displaystyle{\sum_{l=0}^{k}\pi_{C,l}(x)}$,  one gets:

$\mathcal{N}_{k}(x)\le\pi(x)(1+H_{k})(1+o(1))$

where $H_{k}$ is the $k$-th harmonic number.

So that one should have:

$\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})\le k(1+H_{k})(1+o(1))$

hence:

$\mathcal{N}_{k}(x)=O(k\log k)$.

Now, from the prime number theorem, one has:

$\mathcal{N}_{k}(x)\sim x$ for $k$ large enough and less than $x$

So, it should be possible to prove rigorously that the conjunction of Goldbach’s conjecture and NFPR conjecture would entail that:

$\displaystyle{\liminf_{n\to\infty} p_{n+k}-p_{n}=O(k\log k)}$

This last relation, as stated in http://arxiv.org/pdf/1306.0948.pdf, follows from Hardy-Littlewood’s prime k-tuples conjecture.

Let’s now define the quantity $\alpha(x,k)$ as follows:

$\alpha(x,k):=\frac{\pi(x)}{k}-\pi_{C,k}(x)$

It seems that there exists $C>0$ (and possibly not much bigger than $1$) such that:

$\forall(x,k)\vert \alpha(x,k)\vert0$

Then there exists a unique $m$ such that $n=\frac{p_{m}+p_{m+k}}{2}$.

One has obviously $p_{m}, hence:

$m\leqslant\pi(n)\leqslant m+k$

and thus $m\geqslant \pi(n)-k$

Moreover $m\leqslant\pi(x)$, so that the total number of $k$-central numbers below $x$ verifies:

$\pi_{C,k}(x)=\delta\vert\{m', \frac{p_{m}+p_{m+k}}{2}\leqslant x\}\vert+h_{k}(x)$

where $\delta$ is the probability for $n'=\frac{p_{m'}+p_{m'+k}}{2}$ to be $k$-central,

and $0\leqslant \vert h_{k}(x)\vert\leqslant 1$

There are $k$ possibilities for the value of $k_{0}(n')$, namely $k_{0}(n')=1, 2, \cdots, k$

Since $n'$ is $k$-central if and only if $k_{0}(n')=k$, one gets:

$\delta=\frac{1}{k}$

Thus $\pi_{C,k}(x)\geqslant\frac{\pi(x)-k}{k}$

Since $\pi_{C,k}(x)=\frac{\pi(x)}{k}-\alpha(x,k)$ one finally gets:

$\vert\alpha(x,k)\vert\leqslant max(\vert h_{k}(x)\vert,1)$

hence $\vert\alpha(x,k)\vert\leqslant 1$.

Obviously the next step consists in showing that:

$\displaystyle{\liminf_{n\to+\infty} p_{n+k}-p_{n}=O(\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n}))}$

and hopefully:

$\displaystyle{\liminf_{n\to+\infty} p_{n+k}-p_{n}\sim \mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}$

One has:

$p_{n+k}-p_{n}=\mathcal{N}_{n+k}(p_{n+k})-\mathcal{N}_{n+k}(p_{n})$

and thus:

$\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})=p_{n+k}-p_{n}-(n+k)(H_{n+k}-H_{n})+n(H_{n+k}-H_n)+O(n+k)$

hence:

$\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})=p_{n+k}-p_{n}-k(H_{n+k}-H_{k})+O(n+k)$

Thus $\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}=1+O(\dfrac{n+k}{k\log k})+O(\dfrac{\log n}{\log k})$

As Maynard proved that the quantity

$\displaystyle{\liminf_{n\to\infty}p_{n+k}-p_{n}}$

only depends on $k$, one should obtain:

$\displaystyle{\liminf_{n\to\infty}\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}}$

by substracting the divergent part of the error term above.

So that we finally get:

$\displaystyle{\liminf_{n\to\infty}\dfrac{p_{n+k}-p_{n}}{\mathcal{N}_{k}(p_{n+k})-\mathcal{N}_{k}(p_{n})}=1+O(\dfrac{1}{\log k})}$

and thus:

$\displaystyle{\liminf_{n\to\infty}p_{n+k}-p_{n}\sim k\log k}.$

# k-central numbers and Cramer’s conjecture

Assuming Goldbach’s conjecture (that is, $r_{0}(n) whenever $n>1$), let’s write $k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$.

I call $k_{0}(n)$ the order of centrality of $n$ and say that $n$ is a $k$-central number if and only if $k_{0}(n)=k$.

Now let’s consider $\rho(n):=\dfrac{2r_{0}(n)}{k_{0}(n)}$ and $\sigma(n):=\dfrac{\log(\rho(n))}{\log\log n}$. Let $\sigma_{+}:=\lim\sup \sigma(n)$ and $\sigma_{-}:=\lim \inf\sigma(n)$ and $\sigma_{m}:=\dfrac{\sigma_{+}+\sigma_{-}}{2}$. Zhang’s theorem implies $\sigma_{-}=0$, while Cramer’s conjecture implies $\sigma_{+}=2$. The Prime Number Theorem asserts that, on average, $\sigma(n)=1$, so that $\sigma_{m}$ is very likely to be equal to $1$.

Let’s now define the following density: $\delta_{\varepsilon,x}(\sigma):=\dfrac{\vert\{n\leq x, \sigma(n)\in(\sigma-\varepsilon,\sigma+\varepsilon)\}\vert}{x}$.

I formulate the following conjectures:

Symmetric Density Conjecture:

$\displaystyle{\forall \sigma\in(\sigma_{-},\sigma_{+}), \lim_{\varepsilon\to 0}\lim_{x\to \infty}\dfrac{\delta_{\varepsilon,x}(\sigma)}{\delta_{\varepsilon,x}(2\sigma_{m}-\sigma)}=1}$

Increasing Density conjecture:

$\displaystyle{\sigma_{-}\leq \sigma_{1}\leq \sigma_{2}\leq \sigma_{m}\Longrightarrow\lim_{\varepsilon\to 0}\lim_{x\to \infty}\dfrac{\delta_{\varepsilon,x}(\sigma_{1})}{\delta_{\varepsilon,x}(\sigma_{2})}\leq 1}$

It may be worth noticing that these two conjectures are a consequence of a rather natural hypothesis of rotational invariance.

Would these last two conjectures imply Cramer’s conjecture?
By the way, I think that the fact that whenever $r$ is a potential typical primality radius of $n$, then so is $P_{ord_{C}(n)}-r$, implies symmetric density conjecture.

Moreover, the same kind of reasoning can be applied mutatis mutandis replacing the quantities $\rho(n)=\dfrac{2r_{0}(n)}{k_{0}(n)}$ by $r_{0}(n)$, $\sigma_{-}$ by $\varsigma_{-}:=\lim\inf\dfrac{\log r_{0}(n)}{\log\log n}$ and $\sigma_{+}$ by $\varsigma_{+}:=\lim\sup\dfrac{\log r_{0}(n)}{\log\log n}$ to get $\varsigma_{-}=0$ (as a consequence of Maynard’s work on bounded gaps between primes), $\varsigma_{+}=4$ and $\varsigma_{m}:=\dfrac{\varsigma_{-}+\varsigma_{+}}{2}=2$ (as a consequence of both the fact that on average $r_{0}(n)=\dfrac{P_{ord_{C(n)}}}{2N_{1}(n)}$ and that this last quantity is $O(\log^{2}(n)$), hence giving further evidence for the equality $r_{0}(n)=O(\log^{4} n)$ derived under a statistical heuristics in the previous article.

# Upper bound for r0(n) (2)

Writing:

$\mathcal{N}_{n}(x)=k_{n}x(1+\frac{\alpha_{n}(x)}{x})$

and using the same kind of statistical heuristics as in « Primality radius »,

one gets:

$\alpha_{n}(x)=O(\sqrt{x}\log^{2} n)$

hence:

$r_{0}(n)=\dfrac{P_{ord_{C(n)}}}{2N_{1}(n)}+O(\sqrt{r_{0}(n)}\log^{2} n)$

that is:

$r_{0}(n)=O(\log^{2} n)+O(\sqrt{r_{0}(n)}\log^{2} n)$

Writing:

$r_{0}(n)=O(\log^{m} n)$

one gets:

$r_{0}(n)=O(\log^{2} n)+O(\log^{2+m/2} n)$

so that:

$m=4$

and $r_{0}(n)=O(\log^{4} n)$.

# Average value of r0(n)

Let’s denote by $r_{0}(n)$ the smallest potential typical primality radius of a given positive integer $n>1$.

Writing:

$\mathcal{N}_{n}(x)\approx k_{n}x$

where $\mathcal{N}_{n}(x)$ stands for the number of potential typical primality radii of $n$ below $x$ and $k_{n}$ is a quantity depending only on $n$,

one gets:

$\mathcal{N}_{n}(r_{0}(n)+\varepsilon)=1$

and

$\mathcal{N}_{n}(r_{0}(n)-\varepsilon)=0$

where $\varepsilon$ is any positive real number less than $1$.

Hence $\mathcal{N}_{n}(r_{0}(n))$ should be taken as equal to $1/2$,

and one would thus get:

$r_{0}(n)\approx 1/2k_{n}$

Considering the definition of the quantity $\alpha_{n}$ in the article « Primality radius », the only possible value of $k_{n}$ such that:

$\mathcal{N}_{n}(x)\approx k_{n}x$

is:

$k_{n}=\frac{N_{1}(n)}{P_{ord_{C}}(n)}$

Therefore one should expect the following relation to hold:

$r_{0}(n)\approx\frac{P_{ord_{C}(n)}}{2N_{1}(n)}$

As it can be proven that:

$\frac{P_{ord_{C}(n)}}{2N_{1}(n)}=O(\log^{2} n)$

one gets the following average value for $r_{0}(n)$:

$r_{0}(n)\approx C\log^{2}n$ for some $C>0$.

This might help to establish the asymptotic Goldbach’s conjecture (every large enough even integer is the sum of two primes) and Cramer’s conjecture that says:

$p_{n+1}-p_{n}=O(\log^{2}p_{n})$.

Let’s start with the notion of primality radius of a positive integer, which I came to think of while working on Goldbach’s conjecture.

By definition, given a positive integer $n>1$, the non negative integer $u$ is a primality radius of $n$ if and only if both $n-u$ and $n+u$ are prime numbers.

A prime has trivially a primality radius equal to $0$, but the concept becomes really interesting when one considers a composite integer. For example, it is easy to check that $3$ is the smallest primality radius of $14$.

The famous twin prime conjecture is equivalent to the following statement:

« $1$ is a primality radius of infinitely many positive integers »

whereas Goldbach’s conjecture simply becomes:

« every positive integer $n>1$ admits a primality radius ».

Obviously, $u$ can only be a primality radius of $n$ provided $u.

Here comes an almost copy-paste version of a question of mine on MathOverflow:

« 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 $x$.

$(n+u)$ is a prime only if for all prime $p$ less or equal to $\sqrt{2n-3}$:

$p$ doesn’t divide $(n+u)$.

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 by $C_{k}(n)$, as the following 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 $n$.

To make a non negative integer $r < 2\times 3\times\dots\ p_{ord_{C}(n)}$ be a primality radius of $n$, one can require the following condition:

1) 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 1) is true, then $r$ will be called a « potential typical primality radius » of $n$.
Moreover, if $r\leq n-3$, then $r$ will simply be called a « typical primality radius » of $n$.

Now let’s define:

$N_{1}(n)$ as the number of potential typical primality radii of $n$ (hence less than $P_{ord_{C}(n)}$ but potentially greater than $n$)

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 $n$

$\alpha_{n}$ 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 wonder whether this last statement is equivalent to RH (Riemann Hypothesis) or not. If so, it would mean that RH implies that every large enough even number is the sum of two primes. »

*the statistical heuristics I refer to is:

$\vert p-f\vert <\frac{1}{\sqrt{n}}$

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 $n$« .

so that $f=\dfrac{N_{2}(n)}{n}$.

This gives:

$\alpha_{n}=O(\sqrt{n}\log^{2}n)$

This is, up to the implied constant, the error term in the explicit formula of $\psi(n)$ under RH.

I was told several years ago by a brilliant youngster that such an upper bound for $\alpha_{n}$ implies GRH (Generalized Riemann Hypothesis, that deals with Dirichlet L-functions). The big deal is thus to prove some kind of converse implication.