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.

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