# is there a link with the probabilistic model for prime numbers?

Let $$x \in \mathbb{R}_+$$ and $$k \in \mathbb{N}^{*}$$.

Let : $$\mathcal{A}(x)=\#\{(a_1, a_2, \ldots, a_k) \in \mathbb{P}^k \mid (a_1, a_2, \ldots, a_k \text{ verifying some properties}) \, , a_k \leqslant x\},$$

Example1: for $$n$$ even number, let : $$\mathcal{A}(n)=\#\{ (p,n-p)\in\mathbb{P}^2 \, | \, p \leqslant n \}$$.

Example2: $$\mathcal{A}(x)=\#\{ p=n^2+1\in\mathbb{P} \mid p \leqslant x \}$$

Example3: $$\mathcal{A}(x)=\#\{ (p,p+2)\in\mathbb{P}^2 \, | \, p+2 \leqslant x \}$$

The well known probabilistic model for prime numbers can give an heutistic argument to estimate $$\mathcal{A}(x)$$ for $$x \to +\infty$$.

I suggest an other model, let $$q \in \mathbb{P}$$ and consider : $$\mathcal{B}_q = \{ b \in \mathbb{N} \mid \gcd(b, { \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}})=1\}.$$ Consider : $$\mathcal{S}(x, q)=\#\{(a_1, a_2, \ldots, a_k) \in \mathcal{B}_{q}^k \mid (a_1, a_2, \ldots, a_k \text{ verifying some properties}) \, , a_k \leqslant x\},$$ Using chinese remainder theorem, let: $$\mathcal{S}\Big({ \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}}, q\Big)=R_q.$$ Let $$q(x)$$ be the largest prime number verify $${ \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}} \leqslant x.$$

Using Prime number theorem we have $$q(x)=(1+o(1))\log(x)$$.

Let $$\mathcal{S}(x) = \mathcal{S}(x, q(x)).$$

Theorem: I proved the following theorem

$$\frac{\mathcal{S}(x)}{x} \underset{x \to +\infty}\sim \dfrac{R_{q(x)}}{{ \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}}.$$

Conjecture:

$$\begin{array}{rcl} \mathcal{S}(x) & \underset{x \to +\infty}\sim & \mathcal{A}(x) \big( \pi(q(x)) e^{-\gamma} \big)^k \\ &\underset{x\to+\infty} \sim & \mathcal{A}(x) \dfrac{\log(x)^k}{\log(\log(x))^k}e^{- k \gamma} . \end{array}$$

This conjecture gives:

$$\mathcal{A}(x) \underset{x \to +\infty}\sim x \dfrac{R_{q(x)}}{{ \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}} \ \dfrac{\log(\log(x))^k}{\log(x)^k} \ e^{k \gamma}.$$

Let $$q \in \mathbb{P}$$

Example 1: Prime number theorem

Let $$\mathcal{A}(x) = \#\{p \in \mathbb{P} \mid p \leqslant x\}$$

We have $$\mathcal{S}({ \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}}, q) = R_q = { \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize (p-1)}} .$$

according to the conjecture: $$\begin{array}{rcl} \mathcal{A}(x) & \sim & \dfrac{x}{\displaystyle{\small \prod_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}} \ { \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize (p-1)}} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & x \displaystyle{\small \prod_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize \left(1-\frac{1}{p}\right)}} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & x \dfrac{e^{-\gamma}}{\log(q(x))} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & x \dfrac{e^{-\gamma}}{\log(\log(x))} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & \dfrac{x}{\log(x)}. \end{array}$$ We know that this result is true (prime number theorem).

Example 2: Goldbach's conjecture

Let $$n$$ be an even number, and $$\mathcal{A}(n) = \#\{ (p, n-p) \in \mathbb{P}^2 \, | \, p \leqslant n \}$$

Using C.R.T We have :

$$\mathcal{S}({\displaystyle\prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}}, q)= R_q = \displaystyle\prod_{\substack{3 \leq p \leq q \\ \text{p prime, } p | n}} (p-1) \prod_{\substack{3 \leq p \leq q \\ \text{p prime, } p \nmid n}} {\normalsize (p-2)}=\displaystyle{\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q}} {\normalsize \frac{p-1}{p-2}} \Big)} {\small \Big( \prod_{\substack{3 \leq p \leq q \\ \text{p prime}}} {\normalsize (p-2)} \Big)}.$$

according to the conjecture: $$\begin{array}{rcl} \mathcal{A}(n) & \sim & \dfrac{n}{\displaystyle{\small \prod_{\substack{p \leq q(n) \\ \text{p prime}}} {\normalsize p}}} \ \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} {\small \prod_{\substack{3 \leq p \leq q(n) \\ \text{p prime}}} {\normalsize (p-2)}} \ \dfrac{\log(\log(n))^2}{\log(n)^2}e^{2 \gamma} \\ & \sim & \dfrac{n}{2} \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \displaystyle {\small \prod_{\substack{3 \leqslant p \leqslant q(n) \\ \text{p prime}}} \Big({\normalsize 1-\dfrac{2}{p}}\Big)} \dfrac{\log(\log(n))^2}{\log(n)^2}e^{2 \gamma} \\ & \sim & \dfrac{n}{2} \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{4 C_2 e^{-2 \gamma}}{\log(\log(n))^2} \dfrac{\log(\log(n))^2}{\log(n)^2}e^{2 \gamma} \\ & \sim & 2 C_2 \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{n}{\log(n)^2}. \end{array}$$ With $$C_2 = \displaystyle{\small \prod_{\substack{3 \leq p \\ \text{p prime}}} \left({\normalsize 1-\dfrac{1}{(p-1)^2}}\right)}$$, and we prove that $$\displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \sim \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime}}} {\normalsize \dfrac{p-1}{p-2}} \Big)}$$

Then: $$\mathcal{A}(n) \sim 2 C_2 \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime}}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{n}{\log(n)^2}.$$

Example3: the k-tuple conjecture: Prime numbers and sieving up to $q(x)=\log(x)(1+o(1))$

Conclusion: I propose above a model to estimate $$\mathcal{A}(x)$$ using chinese remainder theorem and prime number theorem.

Question: Is there a connection between the probabilistic model for prime numbers and the conjecture above ?

• Maybe partial summation can be a useful tool to prove your conjecture. – Sylvain JULIEN Jan 12 '20 at 16:56
• To prove the conjecture we should prove that $I(x) \sim \pi(x) \big( \pi(q(x)) e^{-\gamma} \big)$ ($I(x)$ is the number of elements less than $x$ and coprime to $p_1, p_2, \cdots, q(x)$) without using prime number theorem and this very proof will be extended and prove the other cases. – LAGRIDA Jan 13 '20 at 13:21
• Why don't you want to use the prime number theorem as it is undoubtedly true? – Sylvain JULIEN Jan 13 '20 at 14:37
• For example, for $n$ even number and $\mathcal{A}(n)=\#\{ (p,n-p)\in\mathbb{P}^2 \, | \, p \leqslant n \}$, we have $\mathcal{S}(n) = \mathcal{S}(n, q(n))=\#\{ (p,n-p)\in\mathcal{B}_{q(n)}^2 \, | \, p \leqslant n \}$. according to the theorem we have $\mathcal{S}(n) \sim \displaystyle 2 \, C_2 \, {\small \left(\prod_{\substack{p | n \\ \text{p prime}}} {\normalsize\frac{p-1}{p-2}} \right)} \dfrac{n}{\log(\log(n))^2} e^{-2 \gamma}$, and if we can prove that $\mathcal{S}(n) \sim \mathcal{A}(n) \big( \pi(q(n)) e^{-\gamma} \big)^2$ we are done with Goldbach's conjecture. – LAGRIDA Jan 13 '20 at 15:12
• The conjecture above states that there is a non-trivial relation between prime numbers less than $x$ and numbers coprime to $p_1,p_2,\cdots,q(x)$ and less $x$. But we know that $I(x) \sim \pi(x) \big( \pi(q(x)) e^{-\gamma} \big)$ holds because of $\pi(x) \sim \dfrac{x}{\log(x)}$ and not because this non-trivial relation. – LAGRIDA Jan 13 '20 at 15:16