La version faible du k-tuple conjecture

Voici mon travaille:

Let $b,q \in \mathbb{P}$ and $d \in \mathbb{N}$, Consider functions:
$$\Upsilon_{b,q}(d) := {\small \prod_{\substack{b \leq a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{d}{a}}\right)}.$$
$$F_b(x) := {\small \prod_{\substack{b \leq a \\ \text{a prime}}} \left({\normalsize 1-\frac{x^2}{a^2}}\right)} \,\, , \quad C_b(d) := {\small \prod_{\substack{b \leq a \\ \text{a prime}}} \left({\normalsize 1-\left(\frac{d-1}{a-1}\right)^2}\right)}.$$

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

Consider Mertens 3 theorem: $\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)} \sim \frac{e^{-\gamma}}{\ln(q)}$

Then, and if $K_b := \displaystyle{\small \prod_{\substack{a < b \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)}$, 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 gives for $d \geq 2$ :

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

In other hand, consider the k-tuple $G(d,c) = (c_1,c_2,\cdots,c_d)$, and consider the set:

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

Consider the function:

$$I_{G(d,c)}(n) := \# \left\{ (b+c_1,b+c_2,\cdots,b+c_d) \in \mathcal{B}_q^{d} \, | \, b+c_d \leq n \right\}$$

For $n = \displaystyle {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$ and using **Chinese remainder theorem** we have:

$$I_{G(d,c)}(n) = f(d) {\small \left( \prod_{\substack{b \leq a \leq q \\ \text{a prime, }a \neq d}} {\normalsize (a-d)} \right)}$$

For a certain function $f(d)$ and prime number $b$, for example $G(2,c)=(0,m)$ and $n = \displaystyle {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$ we have :

$$ I_{G(2,c)}(n) = \prod_{\substack{3 \leq a \leq q \\ \text{a prime, } a | m}} (a-1) \prod_{\substack{3 \leq a \leq q \\ \text{a prime, } a \nmid m}} {\normalsize (a-2)} $$

Then we have :
$$I_{G(2,c)}(n) = {\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \left( \prod_{\substack{3 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-2)} \right)}$$

QUESTION 1: Find the generale formula of $f(d)$, and give the the value of the prime $b$.

Now the most importante idea is the weak form of k-tuple conjecture, if we consider the function $\Upsilon_{b,q}(d)$, we can proof easly that:

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

And for $G(2,c)=(0,m)$ we have : $I_{G(2,c)}(n) \sim \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} 2 \, C_2 \, \dfrac{n}{\log(\log(n))^2} \, e^{-2 \gamma}$ and $C_2 = C_3(2) =\displaystyle{\small \left( \prod_{\substack{3 \leq a \\ \text{a prime}}} {\normalsize \frac{a (a-2)}{(a-1)^2}} \right)}$.

The seconde importante idea :

Let $q(n)$ be the greatest prime number verify $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)} < n$

If $I_n$ denote the number of elements less than $n$ and coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$, then : $$I_n \sim \dfrac{n}{\log\log(n)} \, e^{-\gamma}$$

Using prime number theorem : $\dfrac{n}{\log(\log(n))} \, e^{-\gamma} = \dfrac{n}{\log(n)} \dfrac{\log(n)}{\log(\log(n))} \, e^{-\gamma} \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$

Then we have this fondamental equivalence:

$$I_n \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$$

Now, we can conjecture that:

$$I_{G(d,c)}(n) \sim \pi_{G(d,c)}(n) \, (\pi(q(n)) e^{-\gamma})^d$$

With : $\pi_{G(d,c)}(n) = \# \left\{ (b+c_1,b+c_2,\cdots,b+c_d) \in \mathbb{P}^{d} \, | \, b+c_d \leq n \right\}$

And that gives imediatly:

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

hint : $\pi(n) e^{-\gamma} \sim \dfrac{n}{\log(n)} e^{-\gamma} \sim n \displaystyle{\small \prod_{\substack{a \leq n \\ \text{a prime}}} {\normalsize \left(1 - \frac{1}{a}\right)}}$


Examples of conjectures using results above: If $G(2,c) = (0,m)$ we have : $\pi_{G(2,c)}(n) = \pi_m(n) \sim \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} 2 \, C_2 \; \dfrac{n}{\log(n)^2}$, with $C_2 = C_3(2) = \displaystyle{\small \left( \prod_{\substack{3 \leq a \\ \text{a prime}}} {\normalsize \frac{a (a-2)}{(a-1)^2}} \right)}$

If $G(3,c) = (0, 2, 6)$ we have $f(3)=1$, and using C.R.T : $\displaystyle\#\{(b, b+2, b+6) \in \mathcal{B}_q^3 \, | \, b+6 \leq {\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}} \} = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)}$, then : $\pi_{G(3,c)}(n) \sim \dfrac{1}{6} \dfrac{C_5(3) \, n}{F_5(1) K_5^3 \log^{3}(n)}$

Develope the constantes : $\pi_{G(3,c)}(n) \sim \dfrac{9}{2} \displaystyle{\small \left( \prod_{\substack{5 \leq a \\ \text{a prime}}} {\normalsize \frac{a^2 (a-3)}{(a-1)^3}} \right)} \, \dfrac{n}{\log(n)^3}$