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Let $d \in \mathbb{N}, d \geq 2$ and consider the tuple $G(d,c) = (c_1,c_2,\cdots,c_d)$, with $c_1<c_2<\cdots<c_d$ are evens positive integers.

I am trying to found an asymptotique formula for the set $\pi_{G(d,c)}(n) = \# \left\{ (h+c_1,h+c_2,\cdots,h+c_d) \in \mathbb{P}^{d} \, , h \in \mathbb{N} \, | \, h+c_d \leq n \right\}$ or the k-tuple conjecture as known.

For that i proved a weak form of this conjecture, and i need help to see what we can get.

1) Studying The function $\Upsilon_{b,q}$ :

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

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

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

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

Then from $(1)$ we have:

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

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

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


2) Chinese Romainder Theorem:

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

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

($\mathcal{B}_q$ is the set of integers coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}\right)}$)

Consider the function:

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

($I_{q, G(d,c)}(n)$ is the number of tuples $(h+c_1,h+c_2,\cdots,h+c_d)$ coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}\right)}$ and less than $n$)

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

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

For some function $f(d,c)$ and prime number $b$.

Proof: Consider this system for all primes $p_i$ less than $q$ and the unknown $h$:

$$ \begin{array}{rcl} h+c_1 & \equiv & r_{1,i} \bmod{p_i} \\ h+c_2 & \equiv & r_{2,i} \bmod{p_i} \\ \, & \cdots & \, \\ h+c_d & \equiv & r_{d,i} \bmod{p_i} \\ \end{array} $$ With $r_{j,i},j=1,\cdots,d;i=1,\cdots,\pi(q)$ avoid $\bar{0}$ in $\mathbb{Z}/p_i \mathbb{Z}$.

Chinese Romainder Theorem say that $h$ a solution of the system above is unique modulo $\displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}\right)}$ for every pattern given by $r_{j,i}$ with $j=1,\cdots,d;i=1,\cdots,\pi(q)$.

Then for every $p_i$ less than $q$ we have exactely $p_i - d + g(c_1,c_2,\cdots,c_d)$ possibilities that $h+c_1$ and $h+c_2$ and ... $h+c_d$ avoid $\bar{0}$ in $\mathbb{Z}/p_i \mathbb{Z}$, with $g(c_1,c_2,\cdots,c_d)$ is the number of $c_i$ equals $\bar{0}$ in $\mathbb{Z}/p_i \mathbb{Z}$.

Then for $n = \displaystyle {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$ we have :

$$I_{q,G(d,c)}(n) = \displaystyle {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a - d + g(c_1,c_2,\cdots,c_d))} \right)}$$

For $p_i > c_d$ we have $g(c_1,c_2,\cdots,c_d)=0$ then :

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

For example if $G(2,c)=(0,m)$ with $m$ is even number large than $2$ and $n = \displaystyle {\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}$ we have $g(0,m)=0$ if $m$ coprime to $p$ prime less than $q$, and $g(0,m)=1$ if $p$ devide $m$, then:

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

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

3) The prime q(n) :

We have :

$$\#\left\{h \in \mathbb{N}^{*} , \, h \leq {\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}} \, \middle| \, \gcd(h, {\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}})=1 \right \}=\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize (a-1)}}$$

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

Then Using Prime number Theorem:

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

And also we have $\log\left(\displaystyle{\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a}}\right) \sim \log(n)$

Let $I_n$ 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 : $$I_n \sim \dfrac{n}{\log\log(n)} \, e^{-\gamma}$$ Proof: is by reuns here : Interesting missing proof

See that : $\log(n) \sim q(n)$ then : $\log(q(n)) \sim \log\log(n)$

Then we have :

$$\dfrac{n}{\log(\log(n))} \, e^{-\gamma} = \dfrac{n}{\log(n)} \dfrac{\log(n)}{\log(\log(n))} \, e^{-\gamma} \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$$

Then we have this fondamental equivalent:

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

Let $I_{G(d,c)}(n) = I_{q(n), G(d,c)}(n)$

For $N=\displaystyle {\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ we have :

$$\begin{array}{rcl} \dfrac{I_{G(d,c)}(N)}{\displaystyle {\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a}}} & = & \left( \displaystyle \dfrac{f(d,c)}{\displaystyle {\small \prod_{\substack{a < b \\ \text{a prime}}} {\normalsize a}}}\right) \displaystyle{\small \prod_{\substack{b \leq a \leq q(n) \\ \text{a prime}}} \left({\normalsize 1-\frac{d}{a}}\right)}. \\ & = & \left( \displaystyle \dfrac{f(d,c)}{\displaystyle {\small \prod_{\substack{a < b \\ \text{a prime}}} {\normalsize a}}} \right)\Upsilon_{b,q(n)}(d) \\ \end{array}$$

The weak form of k-tuple conjecture : Then we have this beautiful and close result to the conjecture:

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

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

Now, we can conjecture that:

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

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

And that gives imediatly:

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


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 $G(3,c) = (0, 2, 6)$ and using C.R.T : $\displaystyle\#\{(h, h+2, h+6) \in \mathcal{B}_q^3 \, | \, b+6 \leq {\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}} \} = {\small \left( \prod_{\substack{5 \leq a \leq q \\ \text{a prime}}} {\normalsize (a-3)} \right)}$, and $f(3,c)=1$ 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 : $\pi_{G(3,c)}(n) \sim \dfrac{9}{2} \displaystyle{\small \left( \prod_{\substack{5 \leq a \\ \text{a prime}}} {\normalsize \frac{a^2 (a-3)}{(a-1)^3}} \right)} \, \dfrac{n}{\log(n)^3}$

The same result given by prime triplet conjecture!


Then I found the same results that k-tuple conjecture predicts without using heuristic method. And the key to prove this conjecture is founding another proof of the fondamental equivalence $I_n \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$ without using prime number theorem.

My question is how dificult to prove $I_{G(d,c)}(n) \sim \pi_{G(d,c)}(n) \, (\pi(q(n)) e^{-\gamma})^d$?

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  • $\begingroup$ You can consider moving your question to math overflow. $\endgroup$ – Elaqqad Apr 10 at 21:08
  • $\begingroup$ @Elaqqad, why.? $\endgroup$ – LAGRIDA Apr 11 at 8:59
  • 2
    $\begingroup$ Because, there you will find more specialized mathematicians who could help with your approach - but this is just my opinion-. $\endgroup$ – Elaqqad Apr 14 at 11:24

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