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}).

Primality radius

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<n-2.

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.