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 $e \leq x$
Asked
Active
Viewed 389 times
10
-
1Yes, but this statement is often used as an ingredient in the proof of PMT, so it would be nice to see a simple proof. – Cheerful Parsnip Jan 02 '11 at 04:32
2 Answers
7
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 $n\ge 2$.
Matthew Conroy
- 11,234
-
1@arjang: This answers your question (in comments to my answer) better I suppose. Please feel free to tick this, instead of mine. – Aryabhata Jan 02 '11 at 05:10
-
-
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
4
Yes,
Asymptotically you have
$$(p_n)^{\pi(n)} \leq n^{n/\log n} = e^n$$
Aryabhata
- 82,206
-
Thank you, so does this imply http://math.stackexchange.com/questions/15902/show-that-x-1x-2x-3-cdots-x-y-4n-for-any-given-n-in-mathbbn asymptotically as well? now one can only wonder whether $\ p_n^{\pi(n)} < 4^n$ for $2 \leq n$ – jimjim Jan 02 '11 at 04:36
-
@arjang: Yes, for any $c > 1$, have $\pi(n) \le c\ n/\log n$ for sufficiently large $n$. – Aryabhata Jan 02 '11 at 04:40