My Math Forum Prime counting. Meissel, Lehmer: is there a general formula?

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 April 6th, 2018, 12:45 PM #1 Newbie   Joined: Sep 2017 From: Belgium Posts: 19 Thanks: 7 Prime counting. Meissel, Lehmer: is there a general formula? I am looking for a general formula to count prime numbers on which the Meissel and Lehmer formulas are based: $$\pi(x)=\phi(x,a)+a-1-\sum\limits_{k=2}^{\lfloor log_2(x) \rfloor}{P_k(x,a)}$$ Wiki - prime counting - Meissel Lehmer More precisely, I am looking for the detailed description of the $P_k$ for $k>3$. $P_k(x,a)$ counts the numbers $<=x$ with exactly $k$ prime factors all greater than $p_a$ ($a^{th}$ prime), but in the full general formula, this last condition is not necessary. The Meissel formula stops at $P_2$ (and still uses some $\phi$/Legendre parts) Wolfram - Meissel The Lehmer formula stops at $P_3$ (and still uses some $\phi$/Legendre parts) Wolfram - Lehmer I don't find anything about the general formula (using all the $P_k$ terms). Is there any paper on it? Why stop at $P_3$? is it a performance issue? Lehmer vaguely talk about it in his 1959 paper On the exact number of primes less than a given limit Deleglise talks about performances here Prime counting Meissel, Lehmer, ... Thanks Last edited by skipjack; April 6th, 2018 at 02:46 PM.
 June 22nd, 2019, 12:24 PM #2 Newbie   Joined: Sep 2016 From: Tempe, Arizona Posts: 6 Thanks: 1 The answer I posted to a question titled "one to one mapping between the floor function and the Riemann prime counting function" at https://math.stackexchange.com/q/3177249 may perhaps provide some insight. I've also derived an alternate expression for $b_{\pi}(n)$ if you're interested. Thanks from Collag3n
 June 23rd, 2019, 09:46 AM #3 Newbie   Joined: Sep 2017 From: Belgium Posts: 19 Thanks: 7 Hi Steven. I already looked at $a_{\pi}(n)=\sum\limits_{d|n}\mu(d)\,\omega(\frac{ n}{d})$ and $\omega(n)=\sum\limits_{d|n}a_{\pi}(d)$ But never at $b_{\pi}(n)$, so yes, I am interested. I noticed that unless $n$ contains one square in the prime factorization (no need to be full square-free), $b_{\pi}(n)=-\mu(n)\omega(n)$ But in this case, I have $K(x)=\sum\limits_{n=1}^x b_{\pi}(n)\lfloor \frac{x}{n}\rfloor$ instead of $\pi(x)=\sum\limits_{n=1}^x b_{\pi}(n)\lfloor \frac{x}{n}\rfloor$ For my initial question, I found a formula, but I am still interested in any references/paper on the subject: https://mathoverflow.net/questions/2.../300060#300060 Thx Last edited by Collag3n; June 23rd, 2019 at 09:53 AM.
 July 21st, 2019, 11:39 AM #4 Newbie   Joined: May 2018 From: NYC Posts: 11 Thanks: 0 I have an exact formula, given by a power series In my paper, An Exact Formula for the Prime Counting Function (https://arxiv.org/abs/1905.09818) I derive the very first power series for $\displaystyle \pi(x)$, which is a function of the values of the zeta function at the positive integers (hence, they have closed forms) and which doesn't depend on the zeros of the Riemann zeta function (hence, it's easier to compute.) It's still a power series, whose calculation becomes difficult for large values of $\displaystyle x$. The formula is: $\displaystyle \pi(x)=-8\sum _{h=1}^{\infty} H_{-2h}(x)\sum _{i=1}^h \log\zeta(2i)\sum _{v=i}^{h}\frac{(-1)^{h-v}(4\pi )^{2h-2v}}{\zeta(2v-2i)(2h+2-2v)!}$ where $\displaystyle H_{-2h}(x)$ is the sum of the $2h$-th powers of the first $x$ positive integers. I'm still thinking about the problem of how to turn this power series into an asymptotic formula. If you know how to do this, be my guest.

 Tags counting, formula, general, lehmer, meissel, prime

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