Lagrida




date: 25/12/2018
Author: LAGRIDA Yassine


Prime Numbers Construction


Since a year, and i work to try understanding prime numbers, and it's very difficult if we consider the set $\mathbb{P}$ as determinated in $\mathbb{N}$, then i think to build the set $\mathbb{P}$ just like the crible of Erathostinos do, and i think this idea is work, the set $\mathbb{P}$ in fact is built with natural numbers just like when we build $\mathbb{N}$ by adding $1$ each time.

In this article we will understand the nature of prime numbers, and the raison of there irregularity.

The idea is that not study primes, but a large set that primes inherit all his properties.

I am very enjoy those discovers, and i think that a lot of achievement here are prooved the first time, and there is questions that i am not prooven, you can show them in "questions to solve".

In everything that follows $p$ and $q$ are primes that $q$ is the next prime to $p$.


Notations :

Let $\mathbb{P}$ denote the prime number set : $\mathbb{P}=\{2,3,5,7,\cdots\}$

Let $p, q \in \mathbb{P}$, with $q$ is the next prime to $p$. Example: $p=13 \implies q=17$

For a finite set $A$, $\#A$ denote the number of element in this set.

For $a, b \in \mathbb{N}^{*}$ $a \wedge b$ denote the gcd between $a$ and $b$

$\lfloor x \rfloor$ is the largest integer not greater than $x$

$\varphi(n)$ denote the Euler indicatrice.

The Ip function

For $n \in \mathbb{N}$

$I_p(n)$ is defined as $I_p(0) = 0$, and for $n \geq 1$ :

$I_p(n) = \#\{k \in \mathbb{N}^{*} \, | \, k \leq n \text{ and } k \wedge {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} a = 1\}$

Properties :
  • $I_p(0) = 0$
  • $I_p(1) = 1$
  • $\forall n \in \mathbb{N} \, , \, 1 \leq n < q \implies I_p(n) = 1$
  • $I_p(q) = 2$
  • $\dots$


Example: determine $I_7(36)$ ?

The numbers less than $36$ and coprime to $2 \times 3 \times 5 \times 7$ are :

$1, 11, 13, 17, 19, 23, 29, 31$


Then $I_7(36) = 8$

Objectif: Find a formula to calculate the value of $I_p(n)$ with $n \in \mathbb{N}$

Let $n \in \mathbb{N}^{*}$, the Euclidean division of $n$ by ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ is written as :

$n = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + b \, , \quad 0 \leq b < {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} a$


If $n \wedge {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} a = 1$ then $b \wedge {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} a = 1$.

Let $\varphi$ be the Euler indicatrice : $\varphi {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} = {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} (a-1)$.

We have exactly ${\small \prod_{\substack{a \leq p \\ \text{a prime}}}} (a-1)$ numbres coprime with ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$.

in everything that follows, we will denote $\beta_p(i)$ the i-th number coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$.

For $i \in \left\{1,2,3,\cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} (a-1) \right\}$ :

$\beta_p(1)=1,\,\beta_p(2)=q,\cdots$


In everything that follows, let $\mathcal{B}_p$ be the set :

$\mathcal{B}_p = \left\{\beta_p(i) \,\, | \,\, i \in \{1,2,3,\cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} (a-1) \} \right\}$


Note1: We have the property : $b \wedge {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} = 1 \iff {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} \wedge {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - b = 1$
(this is a generale property $\alpha \wedge \beta = 1 \iff (\alpha-\beta) \wedge \beta = 1$)

Note2: If $0 \leq b < {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ then $0 \leq {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - b < {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} a$

Coroller: Let $i,j \in \mathbb{N}^{*}$ :
$i+j-1 = \prod_{\substack{a \leq p \\ \text{a prime}}} (a-1) \iff \beta_p(i) + \beta_p(j) = \prod_{\substack{a \leq p \\ \text{a prime}}} a$

Proof:

Let $i, j \in \mathbb{N}^{*}$ with $i+j-1 = \prod_{\substack{a \leq p \\ \text{a prime}}} (a-1)$

Considere the i-th number $\beta_p(i)$ (we have $1 \leq i \leq \prod_{\substack{a \leq p \\ \text{a prime}}} (a-1)$)

From Note1 and Note2, $\left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right) - \beta_p(i)$ is also coprime to $\left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right)$ and less than $\left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right)$

So, $\beta_p(k) = \left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right) - \beta_p(i)$ with $k \in \left\{1,2,3,\cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}}} (a-1) \right\}$

For $i=1$ $\beta_p(i) = 1$, and $\beta_p(k) = \left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right) - 1$

In this case $k = {\prod_{\substack{a \leq p \\ \text{a prime}}} (a-1)}$, then $k = j$

For $i=2$ $\beta_p(i) = q$, and $\beta_p(k) = \left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right) - q$

In this case $k = {\prod_{\substack{a \leq p \\ \text{a prime}}} (a-1)} - 1$, then $k = j$

$\cdots$

And we will stop at $i = \frac{1}{2} {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}$

And if $\beta_p(i) + \beta_p(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ it's clear that $i+j-1 = \prod_{\substack{a \leq p \\ \text{a prime}}} (a-1)$


$\beta_p(i)$ are symétrique, if $p \geq 5$ :

$\beta_p(i) = \left\{ \begin{array}{cl}1 & \text{if : } \ i = 1 \\q & \text{if : } \ i = 2 \\\vdots & ~ \ ~ \\\displaystyle \dfrac{1}{2} {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 2 & \text{if : } \ i = \displaystyle \dfrac{1}{2} {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \\ \\\displaystyle \dfrac{1}{2} {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + 2 & \text{if : } \ i = \displaystyle \dfrac{1}{2} {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}+1 \\ \\\vdots & ~ \ ~ \\\displaystyle{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}-q & \text{if : } \ i = \displaystyle{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} - 1 \\\displaystyle{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} a \right)}-1 & \text{if : } \ i = \displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \\\end{array} \right.$


Property 1: $\forall i \in \{1, 2, \cdots , {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\}\, , \quad I_p(\beta_p(i)) = i$

Property 2: Let $n \in \mathbb{N}$ and $0 \leq n \leq {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and $i \in \{1, 2, \cdots , {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} - 1\}$, then :

$I_p(n) = \left\{ \begin{array}{cl}0 & \text{ if : } \ n = 0 \\i & \text{ if : } \ \beta_p(i) \leq n < \beta_p(i+1) \\ \\\displaystyle {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} & \text{ if : } \ n = \displaystyle {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1 \text{ or } n = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}\end{array} \right.$

Example: Let $p=5$, ${\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}} = 30$, ${\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} = 8$ :

$\beta_p(1)=1, \,\, \beta_p(2)=7, \,\, \beta_p(3)=11, \,\, \beta_p(4)=13, \,\, \beta_p(5)=17, \,\, \beta_p(6)=19, \,\, \beta_p(7)=23, \,\, \beta_p(8)=29$


Property 3: Let $n \in \mathbb{N}^{*}$

$n \wedge {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}} = 1$ iff the Euclidean devision of $n$ by ${\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}$ is written as :

$n = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \,\, , \quad (t, i) \in \mathbb{N} \times \{1,\cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \}$


Now we can calculate the value of $I_p(n)$ for any $n\in \mathbb{N}$ :

Property 4: Let $n \in \mathbb{N}$, and the euclidean division of $n$ by ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ is written as : $n = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} \alpha + \beta$, then :

$I_p(n) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \alpha + I_p(\beta)$


Proof:
We have :
$n = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} \alpha + \beta \, , \quad 0 \leq \beta < {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$

Consider the property 3 :

This table contain all numbers coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than $n$:

$t=$$0$$1$$\cdots$$\alpha-1$$\alpha$
$\beta_p(1)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}+\beta_p(1)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}(\alpha - 1) + \beta_p(1)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} \alpha + \beta_p(1)$
$\beta_p(2)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p(2)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}(\alpha - 1) + \beta_p(2)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} \alpha + \beta_p(2)$
$\vdots$$\vdots$$\cdots$$\vdots$$\vdots$
$\vdots$$\vdots$$\cdots$$\vdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} \alpha + \beta_p(I_p(\beta))$
$\vdots$$\vdots$$\cdots$$\vdots$
$\beta_p\left({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\right)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p\left({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\right)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} ( \alpha - 1) + \beta_p\left({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\right)$


It's clear that : $I_p(n) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \alpha + I_p(\beta)$


Proporty 5 : Show that $\alpha = \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}} \right\rfloor$, then :

$I_p(n) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \, \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}} \right\rfloor + I_p(\beta)$


We have $ 0 \leq \beta < {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$, so $I_p(\beta)$ can take the values : $0, 1, 2, 3, \cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}$ as shown in property2.

Property 6: For $n \in \mathbb{N}^{*}$ :

${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \, \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}} \right\rfloor \leq \,\, I_p(n) \,\, \leq{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \, \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}} \right\rfloor + {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}$


Note: Let $n \in \mathbb{N}$, and the function :

$f(n) = I_p(n) - {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \, \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}} \right\rfloor$


Then, the function $f$ is periodique, and the periode $T = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$

The q-points :

Remember that $p$ and $q$ are consécutif primes numbers.

Définition: Let $\alpha \in \mathbb{N}^{*}$,
$\alpha$ is called q-point iff $q$ divide $\alpha$ and $\alpha \wedge {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} = 1$
And $\alpha$ is 2-point iff $2$ devide $\alpha$:

Examples:
  • $15$ is 3-point
  • $14$ is 2-point
  • $17$ is 17-point
  • $25$ is 5-point

Property 7: $\forall n \in \mathbb{N} \, , n \geq 2 \, , \,\,\, n $ is d-point, with $d$ is the smallest prime dividing $n$ (the smallest divisor $\neq 1$ of a number is always prime).

Property 8 : Let $\alpha$ be a q-point then :

$\alpha = q b$ with $b \wedge {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} = 1$


From Property 3 we have :
$b = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \,\, , \quad (t, i) \in \mathbb{N} \times \{1,\cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \}$


Then, we have :

$\alpha = q \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right) \,\, , \quad (t, i) \in \mathbb{N} \times \{1,\cdots, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \}$


The Uq(n) function :

Définition: For $n \in \mathbb{N}$, $u_q(n)$ is defined as:

$u_q(n) = \#\{k \in \mathbb{N} \, | \, k \leq n \text{ and k is q-point }\}$


Examples:
  • 2-points less than $30$ are the even numbers less than $30$ then : $u_2(30) = \left\lfloor \frac{30}{2}\right\rfloor = 15$
  • 3-points less than $30$ are : $3, 9, 15, 21, 27$ then $u_3(30)=5$
  • 5-points less than $30$ are : $5, 25$ then $u_5(30)=2$
  • 7-points less than $30$ are : $7$ then $u_7(30)=1$


From Property 8 the first q-point is $q$ and the seconde is $q^2$ ...

And $\forall n \in \mathbb{N} , \, \, 1 \leq n < q \implies u_q(n) = 0$
Property 10:

$\forall n \in \mathbb{N}^{*} \, : \, n-1 = \sum_{\substack{a \leq n \\ \text{a prime}}} u_a(n)$


Proof:

Let $n \in \mathbb{N}^{*}$, then $n-1 = \sum_{2 \leq k \leq n} 1$

From Property7 every $k \leq n$ is a-point, with $a$ is the smalest prime divisor of $k$.

The number $k$ is a-point, with $a$ is a unique prime number less than $n$.

$\begin{array}{rcl}\displaystyle\sum_{\substack{a \leq n \\ \text{a prime}}} u_a(n) & = & \displaystyle\sum_{\substack{a \leq n \\ \text{a prime}}} \sum_{\substack{k \leq n \\ \text{k is a-point}}} 1 \\& = & \displaystyle\sum_{2 \leq k \leq n} 1 \\& = & n-1 \end{array}$



Example: Let $n = 30$

$\begin{array}{rcl}\displaystyle\sum_{\substack{a \leq 30 \\ \text{a prime}}} u_a(30) & = & u_2(30) + u_3(30) + u_5(30) + u_7(30) + u_{11}(30) \\& ~ & + \, u_{13}(30) + u_{17}(30) + u_{19}(30) + u_{23}(30) + u_{29}(30) \\& = & 15 + 5 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 \\& = & 29 \end{array}$


Property 11: Let $n \in \mathbb{N}^{*}$,
$u_q(n) = I_p \left( \left\lfloor \frac{n}{q} \right\rfloor \right)$

Proof:

$I_p(n)$ is the number of numbers $b$ coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than $n$.

$u_q(n)$ is the number of numbers writing as $q b$ ( $b$ coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ ) and less than $n$

Let $b$ the largest number comprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and verify that $b \wedge {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} = 1 $ and $qb \leq n$.

It's clear that $u_q(qb) = I_p(b) \, , \quad (1)$

We have $u_q(n) = u_q(qb) \, , \quad (2)$ (From the définition of $b$)

And $qb \leq n \implies b \leq \left\lfloor \frac{n}{q}\right\rfloor$

Then $I_p(b) = I_p\left( \left\lfloor \frac{n}{q}\right\rfloor \right) \, , \quad (3)$ because there is no number coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ between $b+1$ and $\left\lfloor \frac{n}{q}\right\rfloor - 1$.

From $(1)$, $(2)$ and $(3)$ we have : $u_q(n) = I_p\left( \left\lfloor \frac{n}{q}\right\rfloor \right)$.


Note:
$\forall i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}\} \quad u_q(q \, \beta_p(i)) = i$


Show that $u_q(q \, \beta_p(i)) = I_p(\beta_p(i))$

Property 12:

Let $n \in \mathbb{N}^{*}$,the Euclidean division of $n$ by $q$ then by ${\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}$ is written as:

$n = q \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta \right) + k \,\, , \quad t\in\mathbb{N}, \, 0 \leq \beta < {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} , \, 0 \leq k < q$


Then we have :

$u_q(n) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + I_p(\beta)$

Proof:

we have :
$n = q \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta \right) + k \,\, , \quad t\in\mathbb{N}, \, 0 \leq \beta < {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} , \, 0 \leq k < q$


From Property 9 we have :
$u_q(n) = I_p\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta \right)$


From Property 4 we have :
$I_p\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta \right) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + I_p(\beta)$


Then :
$u_q(n) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + I_p(\beta)$



Note: we have the property : $\forall n,a,b \in \mathbb{N}^{*} \, : \quad \left\lfloor \frac{\left\lfloor \dfrac{n}{a} \right\rfloor}{b} \right\rfloor = \left\lfloor \frac{n}{a b} \right\rfloor$

Then : $t = \left\lfloor \frac{\left\lfloor \dfrac{n}{q} \right\rfloor}{\displaystyle {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a}}} \right\rfloor = \left\lfloor \frac{n}{\displaystyle {\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}}} \right\rfloor$

Then we have :

$u_q(n) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \, \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}}} \right\rfloor + I_p(\beta)$


Note:Let $n \in \mathbb{N}$, and consider the function :

$f(n) = u_q(n) - {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} \, \left\lfloor \frac{n}{\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}}} \right\rfloor$


Then, the function $f$ is periodique, and the periode $T = {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$.

The relation between Ip(n) and Ua(n) :

Property 13:

$\forall n \in \mathbb{N}^{*} \, : \, I_p(n) = n - \sum_{\substack{a \leq p \\ \text{a prime}}} u_a(n)$


Proof:

$I_p(n)$ is the number of numbers b coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than $n$.

Then $I_p(n)$ équal $n$ mines numbers divisible by $2, 3, \cdots, p$

But every number should be counted one time (6 for example should be counted one time as divisible by 2)

Then : $I_p(n) = n - u_2(n) - u_3(n) - \cdots - u_p(n)$



Note: For $n \in \mathbb{N}^{*} \, , \quad n-1 = \sum_{\substack{a \leq n \\ \text{a prime}}} u_a(n) = \sum_{\substack{a \leq p \\ \text{a prime}}} u_a(n) + \sum_{\substack{q \leq a \leq n \\ \text{a prime}}} u_a(n)$

Then,
$I_p(n) = 1 + \sum_{\substack{q \leq a \leq n \\ \text{a prime}}} u_a(n)$


The relation between $\beta_p(i)$ and $\beta_q(j)$ :

Remember the set $\mathcal{B}_q$ :

$\mathcal{B}_q = \left\{\beta_q(i) \,\, | \,\, i \in \{1,2,3,\cdots, {\small \prod_{\substack{a \leq q \\ \text{a prime}}}} (a-1) \} \right\}$


In everything that follows, let $\mathcal{A}_q$ be the set :

$\mathcal{A}_q = \left\{{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \,\, , (t,i)\in\{0,1,\cdots,q-1\}\times \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} \right\}$


In everything that follows, let $\mathcal{C}_q$ be the set :

$\mathcal{C}_q = \{q \, \beta_p(i) \,\, | \,\, i \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} \}$


property 14: Considere sets $\mathcal{A}_q$, $\mathcal{B}_q$ and $\mathcal{C}_q$, then :

$\mathcal{A}_q = \mathcal{B}_q \cup \mathcal{C}_q \, \text{ and } \, \mathcal{B}_q = \mathcal{A}_q \smallsetminus \mathcal{C}_q$


Proof:

We have $\max \mathcal{A}_q = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1$

Then $\max \mathcal{A}_q = {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - 1$

From Property 3, $\mathcal{A}_q$ contain all numbers coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - 1$.

We have $\max\mathcal{B}_q = {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - 1$

We have $\mathcal{B}_q$ contain all numbers coprime to ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$ and less than ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - 1$

Then $\mathcal{B}_q \subset \mathcal{A}_q \quad (1)$

We have $\forall i \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\}$ : $q \, \beta_q(i)$ is coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$

And $\max\mathcal{C}_q = q \left({\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1 \right) = {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - q \leq \max\mathcal{A}_q$

Then $\mathcal{C}_q \subset \mathcal{A}_q \quad (2)$

All elements in $\mathcal{C}_q$ are divisible by $q$ then they not coprime with ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$

Then $\mathcal{B}_q \cap \mathcal{C}_q = \emptyset \quad (3)$

We have $\#\mathcal{A}_q = q \, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} } = {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)} + {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}$

And $\#\mathcal{B}_q = {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}$ and $\#\mathcal{C}_q = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}$

We have $\#\mathcal{A}_q = \#\mathcal{B}_q + \#\mathcal{C}_q \quad (4)$

From $(1)$, $(2)$, $(3)$ and $(4)$ we have $\mathcal{A}_q = \mathcal{B}_q \cup \mathcal{C}_q$

Then, and from $(3)$ we have $\mathcal{B}_q = \mathcal{A}_q \smallsetminus \mathcal{C}_q$



Note: for $j \in \{1,2,\cdots,q {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}\}$ and in everything that follows $\alpha_q(j)$ will denote an element in $\mathcal{A}_q$

Examples: we have $\mathcal{B}_2 = \{1\}$

Lets determine $\mathcal{B}_3$ ?

we have $\mathcal{C}_3 = \{3\}$

And we have $\mathcal{A}_3 = \{1, 3, 5\}$

Then $\mathcal{B}_3 = \{1, 3, 5\} \smallsetminus \{3\} = \{1, 5\}$

Lets determine $\mathcal{B}_5$ ?

we have $\mathcal{C}_5 = \{5, 25\}$

And $\mathcal{A}_5 = \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29\}$

Then $\mathcal{B}_5 = \{1, 7, 11, 13, 17, 19, 23, 29\}$


Property 15: For $j \in \{1,2,\cdots, q \, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \}$ :

$\alpha_q(j)$ is coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$

Then $\alpha_q(j)$ is q-point or coprime to ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$ :

$\alpha_q(j) = \left\{ \begin{array}{cl}\beta_q(k)\, , k\in \{1,2,\cdots,\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)}}\} & \text{ if } \ \alpha_q(j) \wedge q = 1 \\ \\q \, \beta_p(k) \, , k\in \{1,2,\cdots,\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} & \text{ if } \ q | \alpha_q(j)\end{array} \right.$


Property 16: conserving same notations in property15:

$j = \left\{ \begin{array}{cl}k + u_q(\alpha_q(j)) & \text{ if } \ \alpha_q(j) = \beta_q(k) \\ \\k + I_q(\alpha_q(j)) & \text{ if } \ \alpha_q(j) = q \, \beta_p(k)\end{array} \right.$

Proof:

From Property 13, we have :
$\forall n \in \mathbb{N}^{*} \, : \, I_q(n) = n - \sum_{\substack{a \leq q \\ \text{a prime}}} u_a(n)$


Then $I_q(n) + u_q(n) = n - \sum_{\substack{a \leq p \\ \text{a prime}}} u_a(n)$

We have : $n - \sum_{\substack{a \leq p \\ \text{a prime}}} u_a(n) = I_p(n)$

Then :

$\forall n \in \mathbb{N}^{*} \, : \, I_q(n) + u_q(n) = I_p(n)$


Then :
$I_q(\alpha_q(j)) + u_q(\alpha_q(j)) = I_p(\alpha_q(j))$


Show that $\alpha_q(j)$ is the j-th number coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - 1$, then :

$j = I_p(\alpha_q(j))$


Then :
$j = I_q(\alpha_q(j)) + u_q(\alpha_q(j))$


From property 15 :

$\alpha_q(j) = \left\{ \begin{array}{cl}\beta_q(k)\, , k\in \{1,2,\cdots,\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)}}\} & \text{ if } \ \alpha_q(j) \wedge q = 1 \\ \\q \, \beta_p(k) \, , k\in \{1,2,\cdots,\displaystyle{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} & \text{ if } \ q | \alpha_q(j)\end{array} \right.$


If $\alpha_q(j) = \beta_q(k)$ then $I_q(\alpha_q(j)) = I_q(\beta_q(k)) = k$

If $\alpha_q(j) = q \, \beta_p(k)$ then $u_q(\alpha_q(j)) = u_q(q \, \beta_p(k)) = k$

Then :

$j = \left\{ \begin{array}{cl}k + u_q(\alpha_q(j)) & \text{ if } \ \alpha_q(j) = \beta_q(k) \\ \\k + I_q(\alpha_q(j)) & \text{ if } \ \alpha_q(j) = q \, \beta_p(k)\end{array} \right.$


Property 17: Let $j \in \{1,2,\cdots, q \, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \}$ and $\alpha_q(j) \in \mathcal{A}_q$, then:
$\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \iff j = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + i$


Proof:

Let $j \in \{1,2,\cdots, q \, {\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}} \}$ and $\alpha_q(j) \in \mathcal{A}_q$, then:

$\exists (t,i)\in\{0,1,\cdots,q-1\}\times \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} \,\, : \,\, \alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$


Show that $\alpha_q(j)$ is the j-th number coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$ and less than ${\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} - 1$, then :

$j = I_p(\alpha_q(j))$


From Property 4, we have :

$I_p(\alpha_q(j)) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + I_p(\beta_p(i))$


Then : $I_p(\alpha_q(j)) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + i$

Then : $j = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + i$

And if we have $j = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)} t + i$ then $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$



Property 18:
$\forall i \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} \, \exists ! t \in \{0,1,\cdots,q-1\} \, \exists ! k \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\}$ :
$q \, \beta_p(k) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$, and $k \neq i$


Proof:

Let $i \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\}$

If we suppose $\exists (t_1, t_2) \in \{0,1,\cdots,q-1\}^2$ and $(k_1, k_2) \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\}^2$ that :

$\left\{ \begin{array}{cl}q \, \beta_p(k_1) & = \ \displaystyle{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t_1 + \beta_p(i) \\q \, \beta_p(k_2) & = \ \displaystyle{\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t_2 + \beta_p(i)\end{array} \right.$


Then :

$q \, (\beta_p(k_2) - \beta_p(k_1)) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (t_2 - t_1)$


we have $q$ divide ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (t_2 - t_1)$.

We have $q$ is coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$, then $q$ divide $t_2 - t_1$.

we have $|t_2 - t_1| < q$, and $q$ divide $t_2 - t_1$, then $t_2 - t_1 = 0$

Then $k_1 = k_2$

Then we proof the unicity of $t$ and $k$.

We have $k \neq i$ because if we suppose that $k = i$, then

$\beta_p(i) (q - 1) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t$

If $t = 0$ then $q = 1$ contradiction!

If $t \neq 0$, then :

$\beta_p(i)$ divide ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t$

$\beta_p(i)$ is coprime to ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$, then $\beta_p(i)$ devide $t$.

we have $0 \leq t < q$ then $t$ is coprime to $\beta_p(i)$.

The unique possibility is $\beta_p(i) = 1$, then $q - 1 = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t$ and that is contradiction because $q \ll {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}$

Then $k \neq i$

Example:

we have $B_5 = \{1, 7, 11, 13, 17, 19, 23, 29\}$

Then $C_7 = \{7, 49, 77, 91, 119, 133, 161, 203\}$

$\begin{array}{rcl}7 \cdot 1 & = & 30 \cdot 0 + 7 \\7 \cdot 7 & = & 30 \cdot 1 + 19 \\7 \cdot 11 & = & 30 \cdot 2 + 17 \\7 \cdot 13 & = & 30 \cdot 3 + 1 \\7 \cdot 17 & = & 30 \cdot 3 + 29 \\7 \cdot 19 & = & 30 \cdot 4 + 13 \\7 \cdot 23 & = & 30 \cdot 5 + 11 \\7 \cdot 29 & = & 30 \cdot 6 + 23\end{array}$

The gaps in $\mathcal{B}_q$


Property 19: Let $i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}-1\}$, $q \geq 3$, then

There are exactly ${\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} - 1$ elements in $\mathcal{B}_q$ verifing that $\beta_q(i+1)-\beta_q(i)=2$


Proof:

Let $d_p$ denote the number of elements in $\mathcal{B}_p$ verifing that $\beta_p(k+1)-\beta_p(k)=2$.

In this proof we will find a recursive formula between $d_p$ and $d_q$.

Consider the property 14 :

$\mathcal{A}_q = \mathcal{B}_q \cup \mathcal{C}_q \, \text{ and } \, \mathcal{B}_q = \mathcal{A}_q \smallsetminus \mathcal{C}_q$


We have :

$\alpha_q(j) \in \mathcal{A}_q \iff \exists (t,i)\in\{0,1,\cdots,q-1\}\times \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} \, , \, \alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$


Let write all elements of $\mathcal{A}_q$ in a table :

$t=$$0$$1$$\cdots$$q-1$
$\beta_p(1)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p(1)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + \beta_p(1)$
$\beta_p(2)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p(2)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + \beta_p(2)$
$\vdots$$\vdots$$\cdots$$\vdots$
$\beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}})$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}})$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + \beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}})$


Then we have $\underbrace{q d_p}_{(1)} + \underbrace{q - 1}_{(2)}$ elements in $\mathcal{A}_q$ verfing $\alpha_q(j+1) - \alpha_q(j) = 2$

For $(1)$ show that $\beta_p(i+1) - \beta_p(i) = \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$

That mean for every column we have $d_p$ numbers vérifing $\alpha_q(j+1)-\alpha_q(j) = 2$

We have $q$ columns $\implies $ total is $q d_p$

For $(2)$ Consider de last value in a column and the first in the next column :

The Last value in a column : ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1$

And the first value in the next column : ${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (t+1) + \beta_p(1) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (t+1) + 1$

The difference between those consecutif elements = 2, and we have exacly $q-1$ of those elements in $\mathcal{A}_q$.

Then : $(1)$ and $(2)$ implies :

we have exactly $q d_p + q - 1$ elements in $\mathcal{A}_q$ verfing that $\alpha_q(j+1) - \alpha_q(j) = 2$


To calculate $d_q$ we should eliminate the q-points from those $(q d_p + q - 1)$ elements.

Consider Property18, for every $\beta_p(i)$, there is one $t \in \{0,1, 2,\cdots,q-1\}$ that $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$ is a q-point.

We have $d_p$ elements in $\mathcal{B}_p$ with $\beta_p(i+1) - \beta_p(i) = 2$, then we should eliminate $2 d_p$ numbers from our $(q d_p + q - 1)$ elements, because it demande to eliminate one q-point to not consider $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$

We should also elimenate 2 elements, one for $\beta_p(1) = 1$ and one for $\beta_p({\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1$, because those $\beta_p(i) \, , \, i=1, i={\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)} \right)}$ are not counted with $2 d_p$, this $2$ will eliminate from the $q-1$ numbers.

Then we have :

$d_q = q d_p - 2 d_p + (q-1) - 2$


We can write :

$d_q = (q-2) d_p + (q-2) - 1$


We will denote $a_i$ the i-th prime number, with $a_n = q$

We have $\mathcal{B}_3 = \{1, 5\}$, Then $d_3 = 0$

We have $\mathcal{B}_5 = \{1, 7, 11, 13, 17, 19, 23, 29\}$, Then $d_5 = 2$

We will descende to $3 = a_2$ :

$\begin{array}{rcl}d_{a_{n}} & = & (a_{n} - 2) d_{a_{n-1}} + (a_{n} - 2) - 1 \\ & = & (a_{n} - 2) \Big((a_{n-1} - 2) d_{a_{n-2}} + (a_{n-1} - 2) - 1 \Big) + (a_{n} - 2) - 1 \\ & = & (a_{n} - 2)(a_{n-1} - 2) d_{a_{n-2}} + (a_{n} - 2)(a_{n-1} - 2) - 1 \\ & = & (a_{n} - 2)(a_{n-1} - 2) \Big((a_{n-2} - 2) d_{a_{n-3}} + (a_{n-2} - 2) - 1 \Big) + (a_{n} - 2)(a_{n-1} - 2) - 1 \\ & = & (a_{n} - 2)(a_{n-1} - 2)(a_{n-2} - 2) d_{a_{n-3}} + (a_{n} - 2)(a_{n-1} - 2)(a_{n-2} - 2) - 1 \\ & = & \cdots \\ & = & (a_{n} - 2)(a_{n-1} - 2)(a_{n-2} - 2)\cdots (a_{3} - 2) d_{a_{2}} + (a_{n} - 2)(a_{n-1} - 2)(a_{n-2} - 2)\cdots (a_{3} - 2) - 1 \\ \\ & = & \displaystyle\left(\prod_{k=3}^{n} (a_k - 2) \right) - 1\end{array}$


Show that $3 - 2 = 1$, Then for $q \geq 3$ :

$d_q = {\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} - 1$



Property 20: Let $i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}-1\}$, $q \geq 3$, then

There are exactly ${\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$ elements in $\mathcal{B}_q$ verifing that $\beta_q(i+1)-\beta_q(i) = 4 $


Proof:

Let $d_p$ denote the number of elements in $\mathcal{B}_p$ verifing that $\beta_p(k+1)-\beta_p(k)=4$.

In this proof we will find a recursive formula between $d_p$ and $d_q$.

Consider the property 14 :

$\mathcal{A}_q = \mathcal{B}_q \cup \mathcal{C}_q \, \text{ and } \, \mathcal{B}_q = \mathcal{A}_q \smallsetminus \mathcal{C}_q$


We have :

$\alpha_q(j) \in \mathcal{A}_q \iff \exists (t,i)\in\{0,1,\cdots,q-1\}\times \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}\} \, , \, \alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$


Let write all elements of $\mathcal{A}_q$ in a table :

$t=$$0$$1$$\cdots$$q-1$
$\beta_p(1)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p(1)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + \beta_p(1)$
$\beta_p(2)$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p(2)$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + \beta_p(2)$
$\vdots$$\vdots$$\cdots$$\vdots$
$\beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}})$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} + \beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}})$$\cdots$${\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} (q-1) + \beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}})$


Then we have $q d_p$ elements in $\mathcal{A}_q$ verfing $\alpha_q(j+1) - \alpha_q(j) = 4$

show that $\beta_p(i+1) - \beta_p(i) = \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$

That mean for every column we have $dp$ numbers vérifing $\alpha_q(j+1)-\alpha_q(j) = 4$

We have $q$ columns $\implies$ total is $q d_p$.

Show that impossible having the case $\beta(i+1) - \beta(i) = 2$ and $\beta(i+2) - \beta(i+1) = 2$, then we can't have the gap $4$ in this case.

Show that there is $2$ q-points with $\alpha_q(j_1) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t_1 + 1$ and $\alpha_q(j_2) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t_2 + {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1$

When we eliminate them we will have two gaps for minimum : $q+1$

For $p \geq 3$ then $q \geq 5$ we have $q+1 \geq 6$, so the 2 cases not given a distance equal a $4$.

Consider Property18, for every $\beta_p(i)$, there is one $t \in \{0,1, 2,\cdots,q-1\}$ that $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$ is a q-point.

We have $d_p$ elements in $\mathcal{B}_p$ with $\beta_p(i+1) - \beta_p(i) = 4$, then we should eliminate $2 d_p$ numbers from our $q d_p$ elements, because it demande to eliminate one q-point to not consider $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$

Then :

$d_q = q d_p - 2 d_p$


Then, we can write :

$d_q = (q - 2) d_p$


It's clear that $d_q = {\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$



Property 21: Let $i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}-2\}$, $q \geq 5$, then

There are exactly ${\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)}$ elements in $\mathcal{B}_q$ verifing that $\beta_q(i+1) - \beta_q(i) = 2$ and $\beta_q(i+2) - \beta_q(i+1) = 4$


Proof:

Let $d_p$ denote the number of elements in $\mathcal{B}_p$ verifing that $\beta_p(k+1)-\beta_p(k)=2$ and $\beta_p(k+2)-\beta_p(k+1)=4$.

In this proof we will find a recursive formula between $d_p$ and $d_q$.

Consider the property 14, and the table of elements of $\mathcal{A}_q$ in proofs of property20 and property21 we have :

There is $q d_p$ elements in $\mathcal{A}_q$ verifing $\alpha_q(j+1)-\alpha_q(j)=2$ and $\alpha_q(j+2)-\alpha_q(j+1)=4$


Consider Property18, for every $\beta_p(i)$, there is one $t \in \{0,1, 2,\cdots,q-1\}$ that $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$ is a q-point.

We have $d_p$ elements in $\mathcal{B}_p$ verifing $\beta_p(k+1)-\beta_p(k)=2$ and $\beta_p(k+2)-\beta_p(k+1)=4$, then we should eliminate $3 d_p$ numbers from our $q d_p$ elements, because it demande to eliminate one q-point to not consider $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$ and $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+2) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right)$

Then :

$d_q = q d_p - 3 d_p$


We can write :

$d_q = (q - 3) d_p$


And Finily we can proof that :

$d_q = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)}$



Property 22: Let $i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}-2\}$, $q \geq 5$, then

There are exactly ${\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)}$ elements in $\mathcal{B}_q$ verifing that $\beta_q(i+1) - \beta_q(i) = 4$ and $\beta_q(i+2) - \beta_q(i+1) = 2$


Proof:

The same proof of Property21



Property 23: Let $i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}-1\}$, and $b$ denote the next prime to prime number $b'$, then :



If $q=5$, there are $2$ elements in $\mathcal{B}_5$ verifing that $\beta_5(i+1) - \beta_5(i) = 6$, and if $q \geq 7$ :
There are exactly $2 \left( {\small \left( \prod_{\substack{7 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} + {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)} + \sum_{\substack{5 \leq b' < p \\ \text{b' prime}}} {\small \left( \prod_{\substack{5 \leq a \leq b' \\ \text{a prime}}} {\normalsize (a-3)} \right)} {\small \left( \prod_{\substack{b < a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} \right)$ elements in $\mathcal{B}_q$ verifing that $\beta_q(i+1) - \beta_q(i) = 6$


Proof:

Let $d_p$ denote the number of elements in $\mathcal{B}_p$ verifing that $\beta_p(k+1)−\beta_p(k) = 6$.

Consider the property 14, we have at least $q d_p$ elements verifing $\alpha_q(j+1)−\alpha_q(j) = 6$

We have $6 = 2 + 4 = 4 + 2$, then the gap $6$ can appear if we eliminate q-points in $\mathcal{A}_q$ verifing $\alpha_q(j+1)−\alpha_q(j) = 2$ and $\alpha_q(j+2)−\alpha_q(j+1)=4$ OR $\alpha_q(j+1)−\alpha_q(j) = 4$ and $\alpha_q(j+2)−\alpha_q(j+1)=2$.

From Property 21 we have exactly ${\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)}$ elements in $\mathcal{B}_p$ verifing that $\beta_p(i+1) - \beta_p(i) = 2$ and $\beta_p(i+2) - \beta_p(i+1) = 4$. AND From Property 22 we have exactly ${\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)}$ elements in $\mathcal{B}_p$ verifing that $\beta_p(i+1) - \beta_p(i) = 4$ and $\beta_p(i+2) - \beta_p(i+1) = 2$, we should add those elements to $q d_p$ because when the numbers $\beta_p(i+1)$ (verifing $\beta_p(i+1) - \beta_p(i) = 2$ and $\beta_p(i+2) - \beta_p(i+1) = 4$ or $\beta_p(i+1) - \beta_p(i) = 4$ and $\beta_p(i+2) - \beta_p(i+1) = 2$) are eliminated (q-points) we will have new 6-gaps. and every $\beta_p(i+1)$ will eliminated one time.

Show that there is $2$ q-points with $\alpha_q(j_1) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t_1 + 1$ and $\alpha_q(j_2) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t_2 + {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1$

When we eliminate them we will have two gaps for minimum : $q+1$

For $p \geq 5$ then $q \geq 7$ we have $q+1 \geq 8$, so the 2 cases not given a distance equal a $6$.

Then :

There is exactly $q d_p + 2 {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)}$ elements in $\mathcal{A}_q$ verifing $\alpha_q(j+1) - \alpha_q(j)=6$


Consider Property18, for every $\beta_p(i)$, there is one $t \in \{0,1, 2,\cdots,q-1\}$ that $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$ is a q-point.

We have $d_p$ elements in $\mathcal{B}_p$ with $\beta_p(i+1) - \beta_p(i) = 6$, then we should eliminate $2 d_p$ numbers from our $q d_p + 2 {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)}$ elements, because it demande to eliminate one q-point to not consider $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$

Then :

$d_q = (q-2) d_p + 2 {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)}$


Descending to $q=7$ :

$d_q = 2 \left( {\small \left( \prod_{\substack{7 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} + {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)} + \sum_{\substack{5 \leq b' < p \\ \text{b' prime}}} {\small \left( \prod_{\substack{5 \leq a \leq b' \\ \text{a prime}}} {\normalsize (a-3)} \right)} {\small \left( \prod_{\substack{b < a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} \right)$



Property 24 : Let $i \in \{1,2,\cdots,{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)} \right)}-3\}$, $q \geq 5$, then :

There are exactly ${\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-4)} \right)}$ elements in $\mathcal{B}_q$ verifing that $\beta_q(i+1) - \beta_q(i) = 2$ and $\beta_q(i+2) - \beta_q(i+1) = 4$ and $\beta_q(i+3)-\beta_q(i+2) = 2$


Proof:

Let $d_p$ denote the number of elements in $\mathcal{B}_p$ verifing that $\beta_p(k+1) - \beta_p(k) = 2$ and $\beta_p(k+2) - \beta_p(k+1) = 4$ and $\beta_p(k+3)-\beta_p(k+2) = 2$.

In this proof we will find a recursive formula between $d_p$ and $d_q$.

Consider the property 14, and the table of elements of $\mathcal{A}_q$ in proofs of property20 and property21 we have :

There are $q d_p$ elements in $\mathcal{A}_q$ verifing that $\alpha_q(j+1)-\alpha_q(j)=2$ and $\alpha_q(j+2)-\alpha_q(j+1)=4$ and $\alpha_q(j+3)-\alpha_q(j+2) = 2$


Consider Property18, for every $\beta_p(i)$, there is one $t \in \{0,1, 2,\cdots,q-1\}$ that $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$ is a q-point.

We have $d_p$ elements in $\mathcal{B}_p$ verifing that $\beta_p(k+1)-\beta_p(k)=2$ and $\beta_p(k+2)-\beta_p(k+1)=4$ and $\beta_p(k+3)-\beta_p(k+2)=2$, then we should eliminate $4 d_p$ numbers from our $q d_p$ elements, because it demande to eliminate one q-point to not consider $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i) \right)$ and $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+2) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1) \right)$ and $\left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+3) \right) - \left( {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+2) \right)$

Then :

$d_q = q d_p - 4 d_p$


We can write :

$d_q = (q - 4) d_p$


And Finaly we can proof that :

$d_q = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-4)} \right)}$



Note : We can calculate also calculate gaps of consecutif elements in $\mathcal{B}_q$ equals $8, 10, 12 , \cdots$ with same way.

Property 25 :

News gaps appear in $\mathcal{B}_q$ whene we eliminite q-points from $\mathcal{A}_q$


Proof:

Consider property14, we have :

$\mathcal{A}_q = \mathcal{B}_q \cup \mathcal{C}_q \, \text{ and } \, \mathcal{B}_q = \mathcal{A}_q \smallsetminus \mathcal{C}_q$


Show that we have the same gaps in sets $\mathcal{B}_p$ and $\mathcal{A}_q$ for $p \geq 5$

Let $i \in \{1,2,\cdots,{\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}-2\}$ and $\beta_p(i)$, $\beta_p(i+1)$ and $\beta_p(i+2)$, and $\alpha_q(j) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i)$ , $\alpha_q(j+1) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+1)$ and $\alpha_q(j+2) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} t + \beta_p(i+2)$

Consider Property 8, we have the minimum gap between $2$ consecutif q-points is $2q$, then if $\beta_p(i+2) - \beta_p(i) < 2q$ we are sure that at most one of $\alpha_q(j)$ or $\alpha_q(j+1)$ or $\alpha_q(j+2)$ is q-point.

Consider Property 18 $\exists ! t \in \{0,1,\cdots,q-1\}$ that $\alpha_q(j+1)$ is a q-point, then when $\alpha_q(j+1)$ is q-point, usually $\alpha_q(j+2)-\alpha_q(j)$ is a new gap.

And if $\beta_p(i+1)-\beta_p(i)$ is the maximum gap in $\mathcal{B}_p$, certenly $\beta_p(i+2)-\beta_p(i)$ is a new gap in $\mathcal{B}_q$


Property 26 :

If a new gap appeare in $\mathcal{B}_q$, this gap will continue and growth in every $\mathcal{B}_b$ with $b \in \mathbb{P}, b \geq q$


Proof:

If we suppose that a gap between two consecutif elements is appear in $\mathcal{B}_p$ and Let $d_p$ be the total of those elements.

Then we have at least $q d_p$ elements in $\mathcal{A}_q$ verifing this new gap.

The Property 18 is very very importante, we should eliminate $2 d_p$ (q-points) from $q d_p$, and also eliminate $2$ (if $\beta_p(1)=1$ or $\beta_p({\small \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize (a-1)}}) = {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - 1$ can influence).

Then we have $(q-2) d_p - 2 \leq d_q$

Then, For a prime number $b \geq q$

${\small \left( \prod_{\substack{q \leq a \leq b \\ \text{a prime}}} {\normalsize (a-2)} \right)} - 2 \leq d_b$


We can also prolonge this proof with gaps between more than $2$ numbers.


Property 27 : Let $p \geq 5$,

$\{a-1 \, | \, a \in \mathbb{P} , 3 \leq a \leq q\}$ is subset of set of gaps between consecutif elements in $\mathcal{B}_b \, , b \in \mathcal{P} \, , \, b \geq 5$


Proof:

Let $p \geq 5$, we have $\mathcal{B}_p=\{1, 7, 11, 13, 17, 19, 23, 29\}$

We have the set of gaps between consecutif elements in $\mathcal{B}_q$ equal $\{2, 4, 6\}$

We have $\{2,4\} \subset \{2,4,6\}$

Then From Property 26 gaps $2$ and $4$ will always exists for $p \geq 7$, that's is trivial and we calculate the value of number of those elements in Property 20 and property 19.

If we consider $p$ in générale : $\mathcal{B}_p = \{1, q, \cdots , {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)} - q, {\small \left( \prod_{\substack{a \leq p \\ \text{a prime}}} {\normalsize a} \right)}-1\}$

Show that we have the gap $q-1$ in $\mathcal{B}_p$, then :

$\{a-1 \, | \, a \in \mathbb{P} , 3 \leq a \leq q\}$ is subset of set of gaps between consecutif elements in $\mathcal{B}_b \, , b \in \mathcal{P} \, , \, b \geq 5$


Questions to solve


We do a lot of proofs in this article, but there are questions that i can't solve, and some of them:

  • Determine the maximum gap between two consecutif elements in $\mathcal{B}_q$
  • Proof or disproof that between two consecutif q-points in $\mathcal{C}_q$ there is at least an element in $\mathcal{B}_q$
  • Proof that for every even $n$ exists $\mathcal{B}_q$ that the gap $n$ as a difference of two consecutif elements of $\mathcal{B}_q$ appear.


About Riemann hypothesis

Considere Chebyshev First function $\vartheta(n) = \sum_{\substack{a \leq n \\ \text{a prime}}} \ln(a)$

Riemann hypothesis is Equivalent to :

$\forall \epsilon > 0 \, , \quad | \vartheta(n) - n | = O(n^{\frac{1}{2} + \epsilon})$


Consider Property 10, we have $n - 1 = \sum_{\substack{a \leq n \\ \text{a prime}}} u_a(n)$

Then, Riemann hypothesis is Equivalent to :

$\forall \epsilon > 0 \, , \quad \left| \sum_{\substack{a \leq n \\ \text{a prime}}} \big(\ln(a) - u_a(n)\big) \right| = O(n^{\frac{1}{2} + \epsilon})$


We should stady the function : $F(n) = \left| \sum_{\substack{a \leq n \\ \text{a prime}}} \big(\ln(a) - u_a(n) \big) \right|$

Show that we know a lot of things about $u_a(n)$ from properties above.

Conclusion

In conclusion, is more practical to study $\mathcal{B}_q$ than the set $\mathbb{P}$, because in $\mathcal{B}_q$ it's easy to manupilate elements than elements in $\mathbb{P}$.

Remembering the set $\mathcal{B}_q$ :

$\mathcal{B}_q=\{b \in \mathbb{N}^{*} \, | \, b \wedge {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}=1 \text{ and } b \leq {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} \}$


We have :

$\# \mathcal{B}_q = {\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)}}$


From property 19 and for $q \geq 3$ we proof:

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


From property 20 and for $q \geq 3$ we proof:

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


From property 23 and for $q \geq 5$ we proof:

$\#\{(b,b+6) \in \mathcal{B}_q^2\} = 2 \text{ if } q = 5$

if $q \geq 7$ and $b$ denote the next prime to prime number $b′$, then :

$\#\{(b,b+6) \in \mathcal{B}_q^2\} = 2 \left( {\small \left( \prod_{\substack{7 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} + {\small \left( \prod_{\substack{5 \leq a \leq p \\ \text{a prime}}} {\normalsize (a-3)} \right)} + \sum_{\substack{5 \leq b' < p \\ \text{b' prime}}} {\small \left( \prod_{\substack{5 \leq a \leq b' \\ \text{a prime}}} {\normalsize (a-3)} \right)} {\small \left( \prod_{\substack{b < a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)} \right)$



From property 24 and for $q \geq 5$ we proof:

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


Show that following the same proofs we can calculate the number of elements any gap in $\mathcal{B}_q$