# Explicit upper bound for r0(n)

Numerical computations seem to show that $r_{0}(n)$ is always less than $0.07\log^{4} n$ whenever $n>28$. Solving the equation $x=0.07\log^{4} x$, one gets as a threshold the value $x0\approx 0.25216...$, hence every even integer greater than $3$ is the sum of two primes, as the small cases (less than $57$) can be easily checked by hand.

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# sharp upper bound for alpha_{n}

As mentioned in the previous article, 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 « rather close » to $N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}}(n)-\sqrt{2n-3}}$

This follows from the very definition of what a typical primality radius is. Indeed, writing

$N_{2}(n)=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n))$

one gets:

$\dfrac{n.N_{1}(n)}{P_{ord_{C}(n)}}(1+\dfrac{\alpha_{n}}{n})=N_{1}(n)\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n))$

Hence

$1+\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\left(\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}\right)+O(f(n)\dfrac{P_{ord_{C}(n)}}{n.N_{1}(n)})$

So $\dfrac{\alpha_{n}}{n}=\dfrac{P_{ord_{C}(n)}}{n}\dfrac{n-\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}-\dfrac{n(P_{ord_{C}(n)}-\sqrt{2n-3})}{n(P_{ord_{C}(n)}-\sqrt{2n-3})}+O(f(n)\dfrac{\log^{2} n}{n})$

Thus $\alpha_{n}=\dfrac{(n-P_{ord_{C}(n)})\sqrt{2n-3}}{P_{ord_{C}(n)}-\sqrt{2n-3}}+O(f(n)\log^{2} n)$

Finally $\vert\alpha_{n}\vert=(\sqrt{2n})^{1+\varepsilon}+O(f(n)\log^{2}n)$

Assuming $f(n)=\sqrt{n}$ (which remains to be proven), one gets $\vert\alpha_{n}\vert=O_{\varepsilon}(n^{1/2+\varepsilon})$. Maybe some kind of central limit theorem could shed a new light on the true nature of $f(n)$.

# A conjectural expression for N2(n)

The definition of a « typical » primality radius $r$ of $n$ leads to the following assumption:

the following ratio:

$\dfrac{N_{2}(n)}{N_{1}(n)}$

is asymptotically equal to the ratio:

$\dfrac{n-\sqrt{2n-3}}{P_{ord_{c}(n)}-\sqrt{2n-3}}$

Hence, for $n$ large enough, the quantity $N_{2}(n)$ should be « close » to the expression above times $N_{1}(n)$.

This implies $\alpha_{n}=o(n)$, as one can easily figure out replacing $N_{2}(n)$ by its exact expression given by:

$\frac{n.N_{1}(n)}{P_{ord_{C}(n)}}(1+\dfrac{\alpha_{n}}{n})$

in the following limit:

$\frac{N_{1}(n)}{N_{2}(n)}\left(\frac{n-\sqrt{2n-3}}{P_{ord_{C}}(n)-\sqrt{2n-3}}-\frac{N_{2}(n)}{N_{1}(n)}\right)\to 0$

We will show in the next article that one might even get:

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

# 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.

# 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.