Lagrida




date: 30/12/2018
Author: LAGRIDA Yassine


The fondamentale conjectures of prime numbers





We conserve all notations in previous Article.

Consider Mertens Théorème:
${\small \prod_{\substack{a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)} \sim \frac{e^{-\gamma}}{\ln(q)}$


Let $b$ a prime number, and $d \in \mathbb{N}$

For $x \in \mathbb{R}$ consider the function : $F_b(x) = {\small \prod_{\substack{b \leq a \\ \text{a prime}}} \left({\normalsize 1-\frac{x^2}{a^2}}\right)}$

$F_b$ is convergente, and $F_b(0) = 1$, $F_b(1) = \frac{1}{\zeta(2)}{\small \prod_{\substack{a < b \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a^2}}\right)}^{-1}$
Consider the function :

$\Upsilon_{b,q}(d) = {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{d}{a}}\right)}$


Let the constante $K_b = {\small \prod_{\substack{a < b \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)}$

We have : $\Upsilon_{b,q}(0) = 1$ and $\Upsilon_{b,q}(1) \sim \frac{e^{-\gamma}}{K_b \ln(q)}$

We can easly show that :

$\Upsilon_{b,q}(d) = {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\left(\frac{d-1}{a-1}\right)^2}\right)} {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)}^2 {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1+\frac{d-2}{a}}\right)}^{-1}$


Returning To $F$ function we have :

$F_b(x)={\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1+\frac{x}{a}}\right)} {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{x}{a}}\right)} {\small \prod_{\substack{q < a \\ \text{a prime}}} \left({\normalsize 1-\frac{x^2}{a^2}}\right)}$


Then For $q$ growth to infinity :

${\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1+\frac{x}{a}}\right)}^{-1} \sim \frac{1}{F_b(x)} {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{x}{a}}\right)}$


We have ${\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\left(\frac{d-1}{a-1}\right)^2}\right)} \sim C_b(d)$ a constante.

Show that $C_3(2) = C_2$ is called twin primes constante, and that is false, the correct name is the constant of consecutif primes, and $C_5(3)$ is the constante of $3$ consecutif primes,...

Then we have :

$\Upsilon_{b,q}(d) \sim \left\{ \begin{array}{cl}1 & \text{ if } \ d = 0 \\\dfrac{e^{-\gamma}}{K_b \ln(q)} & \text{ if } \ d = 1 \\ \\\dfrac{C_b(d) \, e^{-2\gamma}}{K_b^2 \, F_b(d-2) \, \ln^{2}(q)} \Upsilon_{b,q}(d-2) & \text{ if } \ d \geq 2\end{array} \right.$


This recursive Formula give for $d \geq 2$ :

$\Upsilon_{b,q}(d) \sim \left\{ \begin{array}{cl}\dfrac{C_b(2) C_b(4) \cdots C_b(d) \, e^{- d \gamma}}{F_b(0) F_b(2) F_b(4) \cdots F_b(d-2) K_b^d} \,\, \dfrac{1}{\ln^d(q)} & \text{ if } \ \text{d is even} \\ \\\dfrac{C_b(3) C_b(5) \cdots C_b(d) \, e^{- d \gamma}}{F_b(1) F_b(3) \cdots F_b(d-2) K_b^d} \,\, \dfrac{1}{\ln^d(q)} & \text{ if } \ \text{d is odd}\end{array} \right.$


Let $G_d$ define the set of vectors of gaps between $d$ consucutif primes :

Example : $G_2$ is the set of distance between $2$ consecutif primes, we conjoncture that $G_2 = 2 \mathbb{N}^{*}$ (I proof that $\{8, 14 \} \cup \{a-1 \, | \, a \in \mathbb{P} , a \geq 3\}$ is Incluse in distances in $\mathcal{B}_q$ with $q$ growth to infinity - in generale i proov that if a gap appear then it stay infinitly-)

Example : $G_3$ is the set of distances between 3 primes, $G_3 = \{(b,b+2,b+4);(b,b+4,b+6);\cdots\}$

Let $g$ a particular distance between $d$ consucutif primes in $G_d$, example : $g=(b,b+2,b+4)$ in $G_3$

Let $\pi_{g,d}(n)$ be the primes less than $n$ and verify the distance $g$ in $G_d$.

We consider elements in $\mathcal{B}_q$ verifing the gap $g$ first, suppose that we have : $\#\{(\beta_1,\cdots,\beta_d) \in \mathcal{B}_q^d \text{ and } (\beta_1,\cdots,\beta_d) \text{ in gap g }\} = {\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)} + C$, with $C$ is a constante.

Show that the terme ${\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)}$ is always present(he is the dominonte terme in facte) because of the recursive formula $d_q = q d_p - d d_p + \text{someterms}$ (Show gaps demonstrations in previous article).

Then I conjoncture that :

$\pi_{g}(n) \sim \left\{ \begin{array}{cl}\dfrac{1}{\displaystyle{\small \left( \prod_{\substack{a < b \\ \text{a prime}}} {\normalsize a} \right)}} \dfrac{C_b(2) C_b(4) \cdots C_b(d)}{F_b(0) F_b(2) F_b(4) \cdots F_b(d-2) K_b^d} \,\, \dfrac{n}{\ln^d(n)} & \text{ if } \ \text{d is even} \\ \\\dfrac{1}{\displaystyle{\small \left( \prod_{\substack{a < b \\ \text{a prime}}} {\normalsize a} \right)}}\dfrac{C_b(3) C_b(5) \cdots C_b(d)}{F_b(1) F_b(3) \cdots F_b(d-2) K_b^d} \,\, \dfrac{n}{\ln^d(n)} & \text{ if } \ \text{d is odd}\end{array} \right.$


Consider those constantes :

$C_2 = C_3(2) = {\small \prod_{\substack{3 \leq a \\ \text{a prime}}} \left({\normalsize 1-\left(\frac{1}{a-1}\right)^2}\right)}$

$C_3 = C_5(3) = {\small \prod_{\substack{5 \leq a \\ \text{a prime}}} \left({\normalsize 1-\left(\frac{2}{a-1}\right)^2}\right)}$

$C_4 = C_5(4) = {\small \prod_{\substack{5 \leq a \\ \text{a prime}}} \left({\normalsize 1-\left(\frac{3}{a-1}\right)^2}\right)}$


Example, we have $\#\{(b,b+2) \in \mathcal{B}_q^2\} = {\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} - 1$ , then :

$\pi_2(n) \sim \frac{1}{2} \frac{4 C_2 \, n}{\ln^{2}(n)}$


we have $\#\{(b,b+4) \in \mathcal{B}_q^2\} = {\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$, then :

$\pi_4(n) \sim \frac{1}{2} \frac{4 C_2 \, n}{\ln^{2}(n)}$


Let $g = (b, b+2, b+6)$ We have $\#\{(b, b+2, b+6) \in \mathcal{B}_q^3 \} = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)}$, then :

$\pi_{g,3}(n) \sim \frac{1}{6} \frac{C_3 \, n}{F_5(1) k_5^3 \ln^{3}(n)}$


Show that $F_5(1) = \frac{1}{\zeta(2)} \frac{3}{2}$, then :

$\pi_{g,3}(n) \sim \frac{\pi^2 \, C_3 \, n}{2 \ln^{3}(n)}$


Let $g = (b, b+2, b+6, b+8)$ We have $\#\{(b, b+2, b+6, b+8) \in \mathcal{B}_q^4 \} = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-4)} \right)}$, then :

$\pi_{g,4}(n) \sim \frac{27}{2} \frac{C_5(2) C_4\, n}{F_5(2) \, \ln^4(n)}$


Reason of this conjoncture :

Show that the densite of elements in $\mathcal{B}_q$ verifing a gap $g$ :

$\dfrac{\displaystyle{\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)}+C}{\displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}} \sim \dfrac{\Upsilon_{b,q}(d)}{\displaystyle{\small \prod_{\substack{a < b \\ \text{a prime}}} {\normalsize a}}}$


Let $n \in \mathbb{N}^{*} \, , \quad n \geq 3$

Let $q(n)$ be the small prime verify $n < {\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ ($q(n)$ is not the n-th prime number !)

Then :

${\small \left( \prod_{\substack{a \leq p(n) \\ \text{a prime}}} {\normalsize a} \right)} \leq n < {\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$


with $q(n)$ is the next prime to $p(n)$

Using Mertens Theoreme :

${\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)} \sim \frac{e^{-\gamma}}{\ln(q(n))}$


Using Prime number theoreme :

$\ln {\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)} \sim q(n)$


Using Prime number theoreme :

$\ln(p(n)) \sim \ln(q(n)) \sim \ln(\ln(n))$


Using the function $I_p(n)$ defined here : Prime Numbers Construction, We have :

$I_{q(n)}(n) \sim e^{-\gamma} \frac{n}{\ln\ln(n)}$

Show that $I_{q(n)}(n)$ is the number of elements less than $n$ and coprime to ${\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$.

And we have the Prime number Theoreme :

$\pi(n) \sim \frac{n}{\ln(n)}$


Generale gaps :

Now we will show the conjecture for gaps that there cardinality more complexe in $\mathcal{B}_q$

Consider the article : Calculate Gaps Cardinality and In everything that follows consider it notations.

Consider generally a gap of $d$ consecutif elements in $\mathcal{B}_q$

Let $H_{b_1,q}(d) = \#\{(\beta_1,\cdots,\beta_d) \in \mathcal{B}_q^d \text{ and } (\beta_1,\cdots,\beta_d) \in g\}$

The gap $g$ appear the first time at $q=b_1$ and it cardinality $H_{b_1,b_1}(d) = s$.

For $q \geq b$ :

$H_{b_1,q}(d) = s {\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)} + \left( \sum_{k=1}^{n} E_{p,k} \right) + \sum_{\substack{b_1 \leq \alpha_1 < p \\ \alpha_1\text{ prime}}} \left( \sum_{k=1}^{n} E_{\alpha_1,k} \right) {\small \left( \prod_{\substack{\alpha < a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)}$


with $\sum_{k=1}^{n} E_{p,k}$ is a series depend to $p$ and represent the cardinality of gaps $\geq d+1$ that elimnation of some elements of there écarts give the gap $g$.

Then We have :

$s {\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)} \leq H_{b_1,q}(d) \leq f(d){\small \left( \prod_{\substack{a \leq q \\ \text{a prime , }a \neq d}} {\normalsize (a-d)} \right)}$


This inequality implies that $\frac{H_{b_1,q}(d)}{\displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}} \sim \frac{K_{b_1,q}(d)}{\ln^d(q)}$

And $H_{b_1,q}(d)$ is related with what we show with gaps with simple cardinality, $K_{b_1,q}(d)$ is the infinte sum of the constantes see in the top, show that :

$H_{b,q}(d) = s {\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)} + \sum_{\substack{b_1 \leq \alpha_1 < p \\ \alpha_1\text{ prime}}} \left( \sum_{k=1}^{n} E_{\alpha_1,k} \right) {\small \left( \prod_{\substack{\alpha < a \leq q \\ \text{a prime}}} {\normalsize (a-d)} \right)} + \left( \sum_{k=1}^{n} E_{p,k} \right)$


If we consider $d=2$ and the gap $(b,b+m)$ we have :

$s {\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} \leq H_{b,q}(2) \leq {\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$


Then we can conjecture that:
$\#\{(p,p+m)\in\mathbb{P}^2\, , \, p+m \leq n\} \sim 2 \theta_m \; C_2 \; \dfrac{n}{\ln^2(n)}$

with $\theta_m$ is a constante verify:
$\frac{s}{\displaystyle{\small \left( \prod_{\substack{3 \leq a \leq b_1 \\ \text{a prime}}} {\normalsize (a-2)} \right)}} \leq \theta_m \leq {\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)}$