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 May 7th, 2015, 01:43 PM #1 Newbie   Joined: Nov 2014 From: Colombia Posts: 6 Thanks: 0 A property of the prime counting function. Greetings to all members of this forum. I share something that to me seems new. I appreciate any suggestion or comment. $\displaystyle \pi(x)$ is the nummber of prime less than or equal to $\displaystyle x$. Consider the following iterative process $\displaystyle a_{0}(x)=\displaystyle\frac{1}{2}x^2$ $\displaystyle a_{n+1}(x)=a_{n}(x)+\displaystyle\frac{x^2-a_{n}(x)}{2+\displaystyle\frac{x}{\pi (x)}(n+1)}$ CONJECTURE. $\displaystyle \displaystyle\lim_{x \to\infty}{\frac{a_{\pi (x)}(x)}{x^2}}=\frac{4}{5}$ Last edited by Jonas Castillo T; May 7th, 2015 at 01:46 PM. May 8th, 2015, 07:12 AM #2 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Let $f(x,k)$ be defined as follows: $$f_0(x)=1/2$$ $$f_i(x)=f_{i-1}(x)+\frac{1-f_{i-1}(x)}{2+ix}\text{, }i>0$$ $$f(x,k)=f_k(x/k)$$ Then your conjecture can be rephrased: $$\lim_{x\to+\infty}f(x,\pi(x))=\frac45.$$ But I don't think it's correct. Aitken acceleration suggests that the limit is above 0.82. Indeed, it is already above 4/5 for 10^7, and only seems to increase from there. Thanks from Jonas Castillo T May 8th, 2015, 02:06 PM #3 Newbie   Joined: Nov 2014 From: Colombia Posts: 6 Thanks: 0 @CRGreathouse It seems to me that you changed the wording of the algorithm and conjecture. Here what I have simplified $\displaystyle a_{0}(x)=\displaystyle\frac{1}{2}$ $\displaystyle a_{n+1}(x)=a_{n}(x)+\displaystyle\frac{1-a_{n}(x)}{2+\displaystyle\frac{x}{\pi (x)}(n+1)}$ CONJECTURE. $\displaystyle \displaystyle\lim_{x \to\infty}{a_{\pi(x)}(x)=\frac{4}{5}}$ I think you can develop a simple algorithm for calculating pi (x) with a negligible margin of error. See here https://drive.google.com/file/d/0B0a...Q5LThOczQ/view Thanks you! May 8th, 2015, 02:15 PM #4 Global Moderator   Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Your conjecture, which is the same as the one I wrote in my post, is very likely to be wrong. The value (for either $a_{\pi(x)}/x^2$ or $f(x,\pi(x))$) is more than 4/5 at 10^7, 10^8, 10^9, and 10^10. Each of these is larger than the one preceding it. If you'd like to compute the value at 10^11, feel free -- I imagine it's larger than the value at 10^10. I have further technical reasons (Aitken acceleration) for expecting that the limit will exist and be larger than 4/5. May 9th, 2015, 01:37 PM #5 Newbie   Joined: Nov 2014 From: Colombia Posts: 6 Thanks: 0 @CRGreathouse Thanks for taking your time. I have a small question: You used a high-speed processor? A friend of mine did me the favor of scheduling algorithm.I used it on my PC, but collapses for large values of $\displaystyle x$, yields false results. I do not know anything about programming. I have not a high-speed procesdor. Thank you for paying attention to this madman! May 9th, 2015, 02:09 PM   #6
Newbie

Joined: Nov 2014
From: Colombia

Posts: 6
Thanks: 0

Here attached executable file SIMPLE-ALGORITHM. When to ask us the value of $\displaystyle Q(x)$, we entered the value of $\displaystyle \pi (x)$.
Finally, I want to ask a favor to CRGreathouse, If you can climb a table with data for certain values of $\displaystyle x$.

I would appreciate it greatly.
Attached Files SIMPLE-ALGORITHM.zip (27.0 KB, 0 views)

Last edited by Jonas Castillo T; May 9th, 2015 at 02:14 PM. May 9th, 2015, 07:26 PM #7 Newbie   Joined: Sep 2014 From: Portland, Oregon Posts: 15 Thanks: 8 Math Focus: Number Theory I spent a few cycles and confirmed that the value is increasing, including the values at $10^{11}$ and $10^{12}$, each of which resulted in a larger $a_{\pi(x)}(x)$ than the previous power of 10. [edit: note this was with the formula, not the exe which I will not run] I also tried the approximation. With $x=10,000,000$, $\pi(x)=664,579$ and we can quickly calculate the Riemann R function as $664,667$ which is quite close. Using the process, $q(x)=662,592$ and hence we say $\lfloor{q^2(x)/Li(x)}\rfloor$ is a good approximation to $\pi(x)$. I get $660,273$. This is quite a bit worse than well known approximations like average bounds, Li(x), Li(x)-Li(sqrt(x))/2, or R(x). Edit: I tried and got similar results with $10^8$ In terms of practical use, the process seems to take quite a long time unless I'm mistaken. If my oracle gives me the correct $q(x)$ to start testing, calculating $a_{\pi(x)}(x)$ takes a multiple of $\pi(x)$ floating point operations. My program, lacking an oracle or perhaps more thought, has to try multiple $q(x)$ values which makes it take longer. Last edited by danaj; May 9th, 2015 at 07:31 PM. Reason: clarify this was using the algorithm, not the exe Tags counting, counting function, function, pi(x), prime, prime numbers, property Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jim198810 Number Theory 6 March 26th, 2015 08:31 PM fafa Number Theory 24 June 22nd, 2013 01:55 AM Bogauss Number Theory 22 October 17th, 2012 02:27 PM guynamedluis Number Theory 2 April 21st, 2012 01:48 PM fucktor Number Theory 3 April 13th, 2009 12:34 PM

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