0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.