Where [imath]p_k[/imath] is the [imath]k[/imath]th prime number, the following ratio [imath]x[/imath] seems to settle toward some nonzero number:
[imath]\lim_{k \to \infty} \frac{k}{p_k} = x[/imath].
For example:
[imath]k = 1e0[/imath], [imath]p_k = 2[/imath], [imath]x = 0.5[/imath],
[imath]k = 1e1[/imath], [imath]p_k = 29[/imath], [imath]x \approx 0.344828[/imath],
[imath]k = 1e2[/imath], [imath]p_k = 541[/imath], [imath]x \approx 0.184843[/imath],
...
[imath]k = 1e10[/imath], [imath]p_k = 252097800623[/imath], [imath]x \approx 0.039667[/imath],
[imath]k = 1e11[/imath], [imath]p_k = 2760727302517[/imath], [imath]x \approx 0.036222[/imath],
[imath]k = 1e12[/imath], [imath]p_k = 29996224275833[/imath], [imath]x \approx 0.033338[/imath].
That is... For the integers 1 through 2 inclusive, 0.5 of them are prime. Likewise, for the integers 1 through 29996224275833, approximately 0.03 of them are prime.
Is anyone familiar with what [imath]x[/imath] settles on as [imath]k \to \infty[/imath]? What is this constant called? Does anyone have access to [imath]k > 1e12[/imath]?
Thanks for any help!
Number of primes...
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Number of primes...
Last edited by taby on Wed Feb 25, 2009 8:58 pm UTC, edited 1 time in total.
Re: Number of primes...
I believe the constant is called "zero", and its decimal expansion begins 0.00000....
It is odd that the ratio would appear to remain "high" for such large values, but primes do not have positive density on the naturals, so it cannot converge to a positive number.
It is odd that the ratio would appear to remain "high" for such large values, but primes do not have positive density on the naturals, so it cannot converge to a positive number.

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Re: Number of primes...
The Prime Number Theorem says that the relative error of approximating [imath]p_k[/imath] with [imath]k \log k[/imath] is asymptotically zero, so:
[math]\lim_{k\rightarrow\infty} \frac k {p_k} = \lim_{k\rightarrow\infty}\frac 1 {\log k} = 0.[/math]
It just goes to zero rather slowly.
[math]\lim_{k\rightarrow\infty} \frac k {p_k} = \lim_{k\rightarrow\infty}\frac 1 {\log k} = 0.[/math]
It just goes to zero rather slowly.
GENERATION 16 + 31i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation.
Re: Number of primes...
I see. Thank you stephentyrone for linking the concepts together for me.

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Re: Number of primes...
I don't think k/pk is the number of primes in the first pkintegers that are prime. pk is the k'th prime number, not the number of primes below k...or have I missed something?
Re: Number of primes...
Of the first p_{k} integers, k of them are prime, namely p_{1} through p_{k}.
Re: Number of primes...
Here [imath]p_k[/imath] is the [imath]k[/imath]th prime number. Here [imath]n[/imath] would denote any positive integer. In terms of the prime counting function [imath]\pi(n)[/imath], where [imath]n = p_k[/imath], it is simplified to [imath]\pi(p_k) = k[/imath]. The "long" version would be:
[imath]\lim_{k \to \infty} \frac{\pi(p_k)}{p_k} = 0[/imath].
So the equation I wrote was basically a ratio involving the number of primes and the number of positive integers. As pointed out by the others, this ratio is 0 for an infinitely large [imath]n[/imath] (or [imath]p_k[/imath], in this case).
I'm not sure if that's what you're referring to.
[imath]\lim_{k \to \infty} \frac{\pi(p_k)}{p_k} = 0[/imath].
So the equation I wrote was basically a ratio involving the number of primes and the number of positive integers. As pointed out by the others, this ratio is 0 for an infinitely large [imath]n[/imath] (or [imath]p_k[/imath], in this case).
I'm not sure if that's what you're referring to.

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Re: Number of primes...
Buttons wrote:Of the first p_{k} integers, k of them are prime, namely p_{1} through p_{k}.
oh, got it. thanks.
Re: Number of primes...
I recently posted a C program here that approximates the prime counting function, if you're interested. Can I have your prime number?
Here is the function compared to some actual values of Pi(x):
Here is the function compared to some actual values of Pi(x):
Spoiler:
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