Prime numbers and sieving up to $q(x)=\log(x)(1+o(1))$

Let $$x \in \mathbb{R}_{+}$$.

For $$q \in \mathbb{P}$$, let : $$\mathcal{B}_q = \{b \in \mathbb{N}^{*} \, | \, \gcd(b, \displaystyle{\small \prod_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}})=1 \}$$

Let $$q(x)$$ be the largest prime verify $$\displaystyle{\small \prod_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}} \leq x$$

Let $$I(x) = \#\{b \in \mathcal{B}_{q(x)} \, | \, b \leq x\}$$

I wrote an article here: https://lagrida.com/conjecture_k_uple.html where I proved:

$$I(x) \sim e^{-\gamma}\dfrac{x}{\log(\log(x))} \tag{1}$$

But when I look at $$I(x)$$ (the number of integers less than $$x$$ and coprime to $$\displaystyle{\small \prod_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}$$) I thought if i can display $$q(x)$$ in the relation $$(1)$$

Using prime number theorem $$q(x) \sim \log(x)$$

We have : $$\dfrac{x}{\log(\log(x))} = \dfrac{x}{\log(x)}\dfrac{\log(x)}{\log(\log(x))}$$

Then using prime number theorem :

$$I(x) \sim \pi(x) \big( \pi(q(x)) e^{-\gamma} \big)$$

Let $$k \in \mathbb{N},k \geq 2$$, and consider the $$k$$-tuple $$\mathcal{H}_k = (0,h_1,h_2,\cdots,h_{k-1})$$ with $$0 < h_1 < \cdots < h_{k-1}$$.

Consider $$I_{\mathcal{H}_k}(x) = \#\{(b,b+h_1,\cdots,b+h_{k-1})\in \mathcal{B}_{q(x)}^k \, | \, b+h_{k-1} \leq x\}$$.

In the same way I proved:

$$x \to +\infty$$ $$I_{\mathcal{H}_k}(x) \sim \mathcal{G}_k \, e^{-\gamma k} \, \dfrac{x}{\log(\log(x))^k}$$

With : $$\displaystyle\mathcal{G}_k = \prod_{\text{p prime}}\frac{1-\frac{w(\mathcal{H}_k, p)}{p}}{(1-\frac1p)^{k}}$$ and $$w(\mathcal{H}_k, p)$$ is the number of distinct residues $$\pmod p$$ in $$\mathcal{H}_k$$.

And I conjecture that:

$$x \to +\infty$$ $$I_{\mathcal{H}_k}(x) \sim \pi_{\mathcal{H}_k}(x) \big( \pi(q(x)) e^{-\gamma} \big)^k$$

With: $$\pi_{\mathcal{H}_k}(x) = \#\{(p,p+h_1,\cdots,p+h_{k-1})\in \mathbb{P}^k \, | \, p+h_{k-1} \leq x\}$$

And that gives immediately:

$$x \to +\infty$$ $$\pi_{\mathcal{H}_k}(x) \sim \mathcal{G}_k \dfrac{x}{\log(x)^k}$$

We can do the same thing for binary Goldbach's conjecture or consecutive primes.

Any proof of $$I(x) \sim \pi(x) \big( \pi(q(x)) e^{-\gamma} \big)$$ without using prime number theorem can be extended and prove the $$k$$-tuple conjecture.

My question : What can we prove about the set having this asymptotic cardinality formula $$\dfrac{I_{\mathcal{H}_k}(x)}{(\pi(q(x)) e^{-\gamma})^k}$$ ?

Goldbach's conjecture:

Let $$n$$ be a even integer.

Let $$G(n)=\#\{n=k_1+k_2,\,(k_1,k_2)\in\mathcal{B}_{q(n)}^2, k_1 \leq \displaystyle{\small \prod_{\substack{p \leq q(n) \\ \text{p prime}}} {\normalsize p}}\}$$, I proved that:

$$n \to +\infty$$: $$G(n) \sim \displaystyle 2 \, C_2 \, {\small \left(\prod_{\substack{p | n \\ \text{p prime}}} {\normalsize\frac{p-1}{p-2}} \right)} \dfrac{n}{\log(\log(n))^2} e^{-2 \gamma}$$

Where $$C_2 = \displaystyle{\small \prod_{\substack{3 \leq p \\ \text{p prime}}} \left({\normalsize 1-\dfrac{1}{(p-1)^2}}\right)}$$

And i conjecture that:

$$n \to +\infty$$: $$G(n) \sim G_p(n) \big( \pi(q(n)) e^{-\gamma} \big)^2$$

Where $$G_p(n) = \#\left\{n=k_1+k_2; (k_1,k_2)\in\mathbb{P}^2\right\}$$

And that gives immediatly:

$$n \to +\infty$$: $$G_p(n) \sim \displaystyle 2 \, C_2 \, {\small \left( \prod_{\substack{p | n \\ \text{p prime}}} {\normalsize \frac{p-1}{p-2}} \right)} \dfrac{n}{\log(n)^2}$$

• As I read your post, I am reminded of an exercise (but I do not remember from where) which suggested your approach to bound from above the number of primes less than x by x/loglog x. Thus I suspect there is an elementary derivation of (1), especially as Euler knew about sum 1/prime growing as loglog. I don't know if the exercise gave the multiplicative constant factor. (I challenge your notion that such an argument sheds light on prime ktuple conjecture.) Gerhard "Might Have Been This Century" Paseman, 2019.08.14. – Gerhard Paseman Aug 14 '19 at 14:58
• @GerhardPaseman, There's a version of that exercise in Paul Pollack's "Not Always Buried Deep: A Second Course in Analytic Number Theory." – JoshuaZ Aug 31 '19 at 21:22
• @GerhardPaseman, it's not about bounding $\pi(x)$, but i think that i found the relation between prime numbers and numbers sieved up to $q(x)$, and the key is understanding the relation $I(x) \sim \pi(x) \big( \pi(q(x)) e^{-\gamma} \big)$. – LAGRIDA Sep 22 '19 at 18:02
• @GerhardPaseman, i don't know how sum 1/prime growing as loglog can help to show that $I(x) \sim \pi(x) \big( \pi(q(x)) e^{-\gamma} \big)$ without using prime number theorem (i don't talk about proving $I(x) \sim e^{-\gamma}\dfrac{x}{\log(\log(x))}$). The goal is proving that the number of elements less than $x$ and coprime to primorial of $q(x)$ is asymptoticaly given by the number of prime numbers less than $x$ multiplied by the number of primes we sieved by multplied by a factor = $e^{-\gamma}$ – LAGRIDA Nov 27 '19 at 11:16