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May 7th, 2015, 12:43 PM   #1
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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 12:46 PM.
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May 8th, 2015, 06:12 AM   #2
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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.
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May 8th, 2015, 01:06 PM   #3
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@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!
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May 8th, 2015, 01:15 PM   #4
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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.
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May 9th, 2015, 12:37 PM   #5
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@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!
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May 9th, 2015, 01:09 PM   #6
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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
File Type: zip SIMPLE-ALGORITHM.zip (27.0 KB, 0 views)

Last edited by Jonas Castillo T; May 9th, 2015 at 01:14 PM.
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May 9th, 2015, 06:26 PM   #7
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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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