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Let $p_n$ be the $n$th prime.

Could someone please help me with the steps between $\pi(n)\sim\dfrac{n}{\log(n)}$ and $n=\pi(p_n)$, to the statement $p_n\sim n\log(n\log(n))$?

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    $\begingroup$ In Hardy's number theory book, it's $p_n \sim n \log n,$ without the extra occurrence of log inside the outer log. I don't know if that makes a difference... $\endgroup$ – coffeemath May 21 '14 at 10:42
  • $\begingroup$ Not much of a difference for small $n$, I don't know for large $n$. $\endgroup$ – martin May 21 '14 at 11:01
  • $\begingroup$ I am trying to find that in Hardy's book - do you have a chapter ref? $\endgroup$ – martin May 21 '14 at 11:09
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    $\begingroup$ In Hardy's "The Theory of Numbers" (fourth edition) on page 9 Theorem 6 (of section 1.8) is a statement of the prime number theorem, that $\pi(x) \sim x/\log x.$ [It isn't proved there, but in Ch XXII is a proof). More to the point, a fairly simple passage from the prime number theorem to $p_n \sim n \log n$ is presented right after that on the rest of page 9 into page ten. Basically he shows the function inverse to $x/\log x$ is asymptotic to $x \log x.$ $\endgroup$ – coffeemath May 21 '14 at 11:29
  • $\begingroup$ Great - that you :) $\endgroup$ – martin May 21 '14 at 11:31
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Here is a sketchy derivation.

Note that

$$ n \log(n \log n) = n \log n + n \log \log n \sim n \log n $$

so we just need to prove the simpler statement that $p_n \sim n \log n$.

The formulas

$$ \pi(n) \sim \frac{n}{\log n} $$ $$ \pi(p_n) \sim n $$

imply

$$ \begin{align} n &\sim \frac{p_n}{\log p_n} \\ p_n &\sim n \log p_n \\ &\sim n \log(n \log p_n) \\ &\sim n (\log n + \log \log p_n) \\ &\sim n \log n \end{align}$$

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  • $\begingroup$ Very clear - thank you! :) $\endgroup$ – martin May 21 '14 at 12:15
  • $\begingroup$ last steps resemble the back substitution method using to solve recurrences. i think as you discard the loglog part, you implicitly made the assumption that p_n does not grow crazy fast like e^(x^x) or smthing right? Then that assumption turns out to be true at the end. ok i see another thing, if we go substitute more, we get logn + loglogn + .. and hopefully after some point loglog..log(p_n) grows slower than logn and all of the stuff is just logn. $\endgroup$ – Ahmet Bilal Nov 3 '18 at 13:33

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