Constantes in asymptotique formulas of consecutif primes

Question

Let $q$ be a prime number, and $m$ an even number.

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

Let $I_{q,m}(n)$ denote the number of consecutif elements $(b,b+m)$ less than $n$ and coprime to $\displaystyle{\small \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a}}$

Let $q(n)$ be the small prime verify $n+m < \displaystyle{\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a}}$

Let $b_1$ be the prime that the gap $(b,b+m)$ is appear the first time in $\mathcal{B}_{b_1}$ and $s = \#\{(b,b+m)\in\mathcal{B}_{b_1}^2\}$

I proove that $I_{q(n),m}(n) \sim \; 2 \, C_2 \, \theta_m \; \dfrac{n}{\ln(\ln(n))^2} e^{-2 \gamma}$

With $\theta_m$ is a constante verify $\dfrac{s}{\displaystyle {\small \left( \prod_{\substack{3 \leq a \leq b_1 \\ \text{a prime}}} {\normalsize (a-2)} \right)}} \leq \theta_m \leq \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)}$ , and $C_2 = \displaystyle{\small \left( \prod_{\substack{3 \leq a \\ \text{a prime}}} {\normalsize \frac{a (a-2)}{(a-1)^2}} \right)}$

We have $\theta_2 = \theta_4 = 1$

If we denote $I_n$ the number of elements less than $n$ and coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ ($q(n)$ the small prime verify $n < \displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$)

I proove that $I_n \sim \dfrac{n}{\ln(\ln(n))} \, e^{-\gamma}$

We have : $\dfrac{n}{\ln(\ln(n))} \, e^{-\gamma} = \dfrac{n}{\ln(n)} \dfrac{\ln(n)}{\ln(\ln(n))} \, e^{-\gamma} \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$

Let $\lambda_m(n) = \#\{p_{k+1}-p_k=m \, , \, p_{k+1} \leq n\}$

Then if we use the same analogy with the case $I_{q(n),m}(n)$ :

$I_{q(n), m}(n) \sim \lambda_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$

Then : $\lambda_m(n) \sim \; 2 \, C_2 \, \theta_m \; \dfrac{n}{\ln(n)^2}$

Well that's just a conjecture (i don't proove $I_{q(n), m}(n) \sim \lambda_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$)

We can also concluse the same thing about Hardy-Littlewood conjecture $\pi_m(n) \sim \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} 2 \, C_2 \; \dfrac{n}{\ln(n)^2}$

EDITED

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i fixe the confusion between $\lambda_m(n)$ and $\pi_m(n)$

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http://lagrida.com/prime_numbers_construction.html

http://lagrida.com/Calculate_Gaps_Cardinality.html

http://lagrida.com/Fondamentale_Conjonctures_Prime_Numbers.html

Answer 1

• Let $N_k = \prod_{i=1}^k p_i$. The CRT says $$\mathbb{Z}/N_k \mathbb{Z} \cong \prod_{i=1}^k \mathbb{Z}/p_i \mathbb{Z}$$

  direct product of finite rings.

  Proof : for each $i$ there is $e_i\bmod N_k$, $e_i \equiv 1 \bmod p_i, (j \ne i), e_i \equiv 0 \bmod p_j$ so the isomorphism is $$n = \sum_{i=1}^k e_i n_i \in \mathbb{Z}/N_k \mathbb{Z} , \qquad n_i = n \in \mathbb{Z}/p_i \mathbb{Z}$$

• Whence $$n +m= \sum_{i=1}^k e_i (n_i +m)\in \mathbb{Z}/N_k \mathbb{Z} , \qquad n_i+m = n+m \in \mathbb{Z}/p_i \mathbb{Z}$$

• Let $1_{gcd(n,N_k)=1} = \begin{cases}1 \text{ if } gcd(n,N_k)=1 \\ 0 \text{ otherwise} \end{cases}$.

  Then

$$\sum_{n \bmod N_k} 1_{gcd(n,N_k)=1} 1_{gcd(n+m,N_k)=1} = \sum_{(n_i) \in \prod_{i=1}^k \mathbb{Z}/p_i \mathbb{Z}} 1_{gcd(\sum_{i=1}^k e_i n_i,N_k)=1}1_{gcd(\sum_{i=1}^k e_i (n_i+m),N_k)=1}$$ $$ =\sum_{(n_i) \in \prod_{i=1}^k \mathbb{Z}/p_i \mathbb{Z}} \prod_{i=1}^k 1_{gcd(n_i,p_i)=1}1_{gcd( n_i+m,p_i)=1}=\prod_{i=1}^k \sum_{n_i \in \mathbb{Z}/p_i \mathbb{Z}} 1_{gcd(n_i,p_i)=1}1_{gcd( n_i+m,p_i)=1}$$ $$ = \prod_{i=1, p_i \,\nmid\, m}^k(p_i-2)\prod_{l=1, p_l\, |\, m}^k(p_l-1) \tag{1}$$

• Since you are letting $k \to \infty$, why do you care of $\sum_{n \le N_k - m} 1_{gcd(n,N_k)=1} 1_{gcd(n+m,N_k)=1}$ when $\sum_{n \le N_k} 1_{gcd(n,N_k)=1} 1_{gcd(n+m,N_k)=1}$ is much easier ?

• The twin prime counting function is $$\pi_2(k^2) = \sum_{n \le \color{red}{k^2}} 1_{gcd(n,N_k)=1} 1_{gcd(n+2,N_k)=1}$$ and $(1)$ gives no clue about that.

• Let $N$ square-free. The same idea works for $$\sum_{n \bmod N} 1_{gcd(n,N)=gcd(n+m_1,N)=gcd(n+m_2,N)=1} = \prod_{p | N}\sum_{n \bmod p} 1_{gcd(n,p)=gcd(n+m_1,p)=gcd(n+m_2,p)=1}$$

  and $$\sum_{n \bmod N} \prod_{l=0}^L 1_{gcd(n+l,N)=1} = \prod_{p | N}\sum_{n \bmod p} \prod_{l=0}^L 1_{gcd(n+l,p)=1} = \prod_{p | N} \max(0,p-L-1)\tag{2}$$

• Not sure why you'd wnt to do so, but you can also look at $$\sum_{n \bmod N_k} 1_{gcd(n,N_k)=gcd(n+6,N_k)=1}\prod_{l=1}^5 1_{gcd(n+l,N_k)\ne 1} $$ $$ =\sum_{n \bmod N_k} 1_{gcd(n,N_k)=gcd(n+6,N_k)=1}\prod_{l=1}^5 (1-1_{gcd(n+l,N_k)= 1})$$ $$ = \sum_{m \in \{0,1\}^7,m_0=m_6=1} (-1)^{\sum_l m_l} f(N_k,m)$$ where for $m \in \{0,1\}^{L+1}$, $f(N_k,m) = \sum_{n \bmod N_k} \prod_{l=0}^L (1_{gcd(n+l,N_k)=1})^{m_l}$. Since $2 | N_k$ then $f(N_k,m) \ne 0$ for only $3$ values of $m$ : $f(N_k,(1,0,0,0,0,0,1)) = 2\prod_{i=3}^k (p_i-2)$,$ f(N_k,(1,0,1,0,0,0,1)) = f(N_k,(1,0,0,0,1,0,1))=\prod_{i=3}^k (p_i-3)$ so that

$$\sum_{n \bmod N_k} 1_{gcd(n,N_k)=gcd(n+6,N_k)=1}\prod_{l=1}^5 1_{gcd(n+l,N_k)\ne 1} =2\prod_{i=3}^k (p_i-2)-2 \prod_{i=3}^k (p_i-3) \tag{3}$$

Answer 2

You should use more common notations for readability, especially when your variables are changing all the time ($t$ becomes $\alpha$, $b$ becomes $\beta_p(i)$ then $\beta$, $\alpha$ becomes $t$ again $b$ is not $b$ anymore, $\alpha$ becomes something else again...)

$I_{p_a}(n)=\phi(n,a)$ (Legendre formula)

$u_q(n)=\phi(\frac{n}{p_{a+1}},a)$

$\mathcal{B}_q = \mathcal{A}_q \smallsetminus \mathcal{C}_q$ is equivalent to $\phi(n,a)=\phi(n,a-1)-\phi(\frac{n}{p_a},a-1)$

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

is just $$\phi(s\cdot m_k + t,k)=s\cdot \varphi(m_k)+\phi(t,k)$$

with ${\small \big( \prod_{\substack{a \leq p \\ a\text{ prime}}} {\normalsize a} \big)}=p_a\#=m_k$

${\small \big( \prod_{\substack{a \leq p \\ a\text{ prime}}} {\normalsize (a-1)} \big)}=\varphi(p_a\#)=\varphi(m_k)$

$s = \alpha$ and $t = \beta$

Also, your gaps formulas are only valid for sieves up to $p\#$.

Anyway, on your third link, your first formula you set $K_a$ with a formula only valid at $\infty$. Because of that, things go wrong when you start mixing $K_a$ and $K_b$