Lagrida




date: 03/01/2019
Author: LAGRIDA Yassine


Calculate Gaps Cardinality





You should read the article : Prime Numbers Construction to understand this article.

In everything that follows :

Let $b_1, b, \alpha_1, \alpha, p, q \in \mathbb{P}$ with $b$ is the next prime to $b_1$, $\alpha$ is next prime to $\alpha_1$, and $q$ the next prime to $p$.

Suppose that a gap $g$ of $d$ consécutif elements is appear the first time in $\mathcal{B}_{b_1}$

Let $s = \#\{(\beta_1,\beta_2,\cdots,\beta_d) \in g \text{ and } (\beta_1,\beta_2,\cdots,\beta_d) \in \mathcal{B}_{b_1}^d\}$

Let $q > b_1$ :

Let $C_q = \#\{(\beta_1,\beta_2,\cdots,\beta_d) \in g \text{ and } (\beta_1,\beta_2,\cdots,\beta_d) \in \mathcal{B}_{q}^d\}$

And $C_p = \#\{(\beta_1,\beta_2,\cdots,\beta_d) \in g \text{ and } (\beta_1,\beta_2,\cdots,\beta_d) \in \mathcal{B}_{p}^d\}$

Let $E_{p,k}\,\,k=1,2,\cdots,n$ a séries depende to $p$.

From the article "Prime Numbers Construction", we have the existing of some series $E_{p,k}\,\,k=1,2,\cdots,n$ that :

$C_q = q C_p - d C_p + \sum_{k=1}^{n} E_{p,k}$

If we desende to $b$ we have :

$C_q = 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)} $


Show that if we use Chinese remainder theorem, then we have :

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


With $f(d)$ a function depend to the gap.

If $d=2$ and we consider the gap $(b,b+m)$ then we have : $C_q \leq {\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$


Example: Let calculate $\#\{(b,b+8) \in \mathcal{B}_q^2\}$

We have $8 = 0+2+6 = 0+6+2 = 0+4 + 4$

Show that the gap $(b,b+4;b+8)$ is impossible in $\mathcal{B}_q$ because necesserly one of $b,b+4,b+8$ is divisible by $3$.

Then we should calculate $\#\{(b,b+2,b+8) \in \mathcal{B}_q^3\}$ and $\#\{(b,b+6,b+8) \in \mathcal{B}_q^3\}$.

Calculate $\#\{(b,b+2,b+8) \in \mathcal{B}_q^3\}$ :

First apear of the gap $(b,b+2,b+8)$ is for $q = 7$ and $s=5$

We Have : $0+2+8 = 0+2+6+8$, Then we should calculate $\#\{(b,b+2,b+6,b+8) \in \mathcal{B}_q^4\}$

We already do that, and we have $\#\{(b,b+2,b+6,b+8) \in \mathcal{B}_q^4\} = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-4)} \right)} = E_{p,1}$

Then $C_q = (q-3) C_p + E_{p,1}$

Then We have $C_q = 5$ if $q=7$ and if $q \geq 11$ :

$C_q = 5 {\small \left( \prod_{\substack{11 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)} + {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-4)} \right)} + \sum_{\substack{7 \leq \alpha_1 < p \\ \alpha_1\text{ prime}}} {\small \left( \prod_{\substack{5 \leq a \leq \alpha_1 \\ \text{a prime}}} {\normalsize (a-4)} \right)} {\small \left( \prod_{\substack{\alpha < a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)} $



Calculate $\#\{(b,b+6,b+8) \in \mathcal{B}_q^3\}$ :

We have distances are symetrique to center, and this gap not passing by the center of primorial of $q$, then we have :

$\#\{(b,b+6,b+8) \in \mathcal{B}_q^3\} = \#\{(b,b+2,b+8) \in \mathcal{B}_q^3\}$


Returning to calculate the number of elements verfing the gap $(b,b+8)$ in $\mathcal{B}_q$.

Let $p$ be the next prime to $p_1$.

Let :
$E_{p,1} = 2 \times 5 = 10$ if $p=7$, and if $p \geq 11$ :

$E_{p,1} = 2 \left( 5 {\small \left( \prod_{\substack{11 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)} + {\small \left( \prod_{\substack{5 \leq a \leq p_1 \\ \text{a prime}}} {\normalsize (a-4)} \right)} + \sum_{\substack{7 \leq \alpha_1 < p_1 \\ \alpha_1\text{ prime}}} {\small \left( \prod_{\substack{5 \leq a \leq \alpha_1 \\ \text{a prime}}} {\normalsize (a-4)} \right)} {\small \left( \prod_{\substack{\alpha < a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)} \right)$ (we adding the gaps $(b,b+6,b+8)$ and $(b,b+2,b+8)$)

The gap $(b,b+8)$ appear the first time for $q=7$ and $s = C_7 = 2$

And if $q \geq 11$:

$C_q = 2 {\small \left( \prod_{\substack{11 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} + E_{p,1} + \sum_{\substack{7 \leq \alpha_1 < p \\ \alpha_1\text{ prime}}} \left( E_{\alpha_1,1} \right) {\small \left( \prod_{\substack{\alpha < a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$