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The prime number theorem gives an increasingly (proportionally) accurate approximation for the number of primes less than a given integer $n$.

Can we use this to find an equally accurate approximation which maps the $k$th prime number to its value $p$?

On the x axis should be the number of primes below $n$, $\frac{n}{log(n)}$ and the y axis the value of the number, $p$. So this is the inverse function for $\frac{n}{log(n)}$.

So would the value of the $k$th prime number be approximately $p \approx \frac{k}{\log k}$?

Formula's using any principle other than the PNT are also welcome.

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closed as unclear what you're asking by Did, Daniel W. Farlow, Lee Mosher, Shailesh, user299912 Jul 28 '16 at 8:27

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  • $\begingroup$ You might want to repost the question you are really asking, which is simply: What is the inverse function of $x\mapsto\frac{x}{\log x}$? To which an answer is: Check Lambert W-function. $\endgroup$ – Did Jul 27 '16 at 9:59
  • $\begingroup$ What do you mean by "maps the $x$th prime number to its size"??? $\endgroup$ – barak manos Jul 27 '16 at 10:23
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Well, the Prime Number Theorem says that, where $g(x)=\frac{x}{\ln x}$, the $g(N)$th prime is approximately $N$. So the function you are looking for is $f(x)=g^{-1}(x)$. I don't know of any easy way to find the inverse of $g(x)$.

Asymptotically, however, we have $f(x)=O(x \ln x)$. This can be seen by letting $f(x)=x (\ln x + h(x))$, for some function $h$. We know that $x=f(\frac{x}{\ln x})$; substituting in the expression for $f$ yields $x=\frac{x}{\ln x}\left(\ln\frac{x}{\ln x} + h\left(\frac{x}{\ln x}\right)\right) = \frac{x}{\ln x}\left(\ln{x} - \ln \ln x + h\left(\frac{x}{\ln x}\right)\right)$. When this is simplified, we get that $h\left(\frac{x}{\ln x}\right) = \ln \ln x$. It should be clear now that the growth of $h(x)$ is slower than that of $\ln x$ (you can see this by plugging in $h(x)=c\ln x$ and getting that $h$ grows too fast to satisfy the equality). Thus, $f(x)=x (\ln x + h(x))$ is $O(x \ln x)$.

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  • $\begingroup$ I will accept this answer if you put a proof as to why the growth is $O(xln(x))$ $\endgroup$ – Dis-integrating Jul 27 '16 at 16:50
  • $\begingroup$ I added the proof $\endgroup$ – msinghal Jul 27 '16 at 22:33
  • $\begingroup$ Simpler: $g(2x\log x)=2x/(1+r(x))$ with $r(x)=(\log\log x+\log2)/\log x\to0$ hence $g(2x\log x)>x$ for every $x$ large enough, that is, $f(x)<2x\log x$. $\endgroup$ – Did Jul 27 '16 at 23:15

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