Let $n \in \mathbb{N}$, a number large enough.

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

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

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

My Question is : there is an other proof that $I_n \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$ ? (not using the proof above! and just this formula without details about $\pi(n)$ or $\pi(q(n))$)

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

  • $\begingroup$ I don't see a proof in what you wrote. $\endgroup$ – reuns Jan 23 at 15:34
  • $\begingroup$ From the definition of $q(n)$ we have $\displaystyle{\small \left( \prod_{\substack{a \leq p(n) \\ \text{a prime}}} {\normalsize a} \right)} \leq n < {\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ with $q(n)$ the next prime to $p(n)$. And Mertens 3rd theorem give $\displaystyle{\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)} \sim \frac{e^{-\gamma}}{\ln(q(n))}$. $\endgroup$ – LAGRIDA Jan 23 at 15:42
  • $\begingroup$ And the prime number theorem give $\displaystyle \ln {\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)} \sim q(n)$ and $\ln(p(n)) \sim \ln(q(n)) \sim \ln(\ln(n))$ and that give $\displaystyle I_n \sim \dfrac{n}{\ln\ln(n)} \, e^{-\gamma}$ $\endgroup$ – LAGRIDA Jan 23 at 15:44
  • $\begingroup$ The question is how you evaluate $I_n$. The answer is in my post $\endgroup$ – reuns Jan 23 at 15:56

For $c> 0$ fixed, $P_k= \prod_{p \le k}p ,k \to \infty$ then $$\sum_{m \le c P_k} 1_{gcd(m,P_k) = 1} \sim c\, \varphi(P_k)\sim c P_k \frac{e^{-\gamma}}{\log k}$$

The proof shows it is also valid for $c$ depending on $k$ not decreasing too fast.

$$\sum_{m \le c P_k} 1_{gcd(m,P_k) = 1} = \sum_{d | P_k} \mu(d)\sum_{m \le c P_k} 1_{d | m} = \sum_{d | P_k} \mu(d) \lfloor \frac{c P_k}{d}\rfloor = \sum_{d | P_k} \mu(d) (\frac{c P_k}{d}+O(1)) \\ = c \prod_{p \le k} p (1-p^{-1}) + O(\tau(P_k))= c \varphi(P_k)+O(P_k^\epsilon)=c P_k \frac{e^{-\gamma}}{\log k}+O(P_k^\epsilon)$$ Where the last step is Mertens third theorem and $\tau(m) = \sum_{d | m} 1 = O(m^\epsilon)$

The $O$ constant is uniform in $ c$.

In your question $q(n) = k$ and $P_k$ is the least primorial above $n$ so $ n = c_n P_k, c_n \in (\frac{1}{k},1]$ then $ I_n = \sum_{m \le c_n P_k} 1_{gcd(m,P_k) = 1}$, $\frac1{c_n} = O(k) = O(\log P_k)$ so you get the result

$$I_n = \sum_{m \le c_n P_k} 1_{gcd(m,P_k) = 1}= c_n P_k \frac{e^{-\gamma}}{\log k}+O(P_k^\epsilon) = n \frac{e^{-\gamma}}{\log k}+O((kn)^\epsilon)= e^{-\gamma}\frac{n}{\log \log n}+O(n^{\epsilon'})$$

The PNT gives $q(n) \sim \log(P_k) \sim \log n$ and $\pi(n) \sim \frac{n}{\log n}$ so $I_n \sim e^{-\gamma}\frac{\pi(n) q(n)}{\log \log n} \sim e^{-\gamma}\frac{\pi(n) q(n)}{\log q(n)}\sim e^{-\gamma}\pi(n) \pi(q(n))$ but this is just an obfuscation of the result, to be avoided at all cost.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Aloizio Macedo Jan 26 at 3:13
  • $\begingroup$ We need the quantitative missing proof $\displaystyle I_n \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$, and that will give us the proof $\displaystyle I_{q(n), m}(n) \sim \pi_m(n) \big( \pi(q(n)) \, e^{- \gamma} \big)^2$ and that proove imediately Hardy-LittleWood conjecture : $\displaystyle \pi_m(n) \sim \displaystyle{\small \left( \prod_{\substack{a | m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} 2 \, C_2 \; \dfrac{n}{\ln(n)^2}$ $\endgroup$ – LAGRIDA Feb 13 at 15:38

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