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May 7th, 2015, 12:43 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 $\displaystyle a_{0}(x)=\displaystyle\frac{1}{2}x^2$ $\displaystyle a_{n+1}(x)=a_{n}(x)+\displaystyle\frac{x^2a_{n}(x)}{2+\displaystyle\frac{x}{\pi (x)}(n+1)}$ 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 12:46 PM. 
May 8th, 2015, 06:12 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: $$f_0(x)=1/2$$ $$ f_i(x)=f_{i1}(x)+\frac{1f_{i1}(x)}{2+ix}\text{, }i>0 $$ $$ f(x,k)=f_k(x/k) $$ 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. 
May 8th, 2015, 01:06 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 $\displaystyle a_{0}(x)=\displaystyle\frac{1}{2}$ $\displaystyle a_{n+1}(x)=a_{n}(x)+\displaystyle\frac{1a_{n}(x)}{2+\displaystyle\frac{x}{\pi (x)}(n+1)}$ 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, 01:15 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, 12:37 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 highspeed 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 highspeed procesdor. Thank you for paying attention to this madman! 
May 9th, 2015, 01:09 PM  #6 
Newbie Joined: Nov 2014 From: Colombia Posts: 6 Thanks: 0 
Here attached executable file SIMPLEALGORITHM. 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. Last edited by Jonas Castillo T; May 9th, 2015 at 01:14 PM. 
May 9th, 2015, 06:26 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 06:31 PM. Reason: clarify this was using the algorithm, not the exe 

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counting, counting function, function, pi(x), prime, prime numbers, property 
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