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Is $\ p_n^{\pi(n)} < 4^n$ where $p_n$ is the largest prime $\leq n$? Where $\pi(n)$ is the prime counting function. Using PMT it seems asymptotically $\ p_n^{\pi(n)} \leq x^n$ where $e \leq x$

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Yes, but this statement is often used as an ingredient in the proof of PMT, so it would be nice to see a simple proof. –  Grumpy Parsnip Jan 2 '11 at 4:32

2 Answers 2

up vote 7 down vote accepted

Using $$\pi(x) \le 1.25066 \frac{x}{\log x}$$ for all $x>1$ (from Rosser and Schoenfeld), you have $$(p_n)^{\pi(n)} \le e^{1.25066 n} < 3.5^n$$ for all $n\ge 2$.

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@arjang: This answers your question (in comments to my answer) better I suppose. Please feel free to tick this, instead of mine. –  Aryabhata Jan 2 '11 at 5:10
    
@Moron : both answers are correct, but i see what you mean –  Arjang Jan 2 '11 at 5:18
    
Just so it's clear for future reference, the paper by Rosser and Schoenfeld is Approximate formulas for some functions of prime numbers, equation 3.6. –  lhf Apr 29 '11 at 12:37

Yes,

Asymptotically you have

$$(p_n)^{\pi(n)} \leq n^{n/\log n} = e^n$$

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Thank you, so does this imply math.stackexchange.com/questions/15902/… asymptotically as well? now one can only wonder whether $\ p_n^{\pi(n)} < 4^n$ for $2 \leq n$ –  Arjang Jan 2 '11 at 4:36
    
@arjang: Yes, for any $c > 1$, have $\pi(n) \le c\ n/\log n$ for sufficiently large $n$. –  Aryabhata Jan 2 '11 at 4:40

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