# How can we show $\pi (2n) \ge \frac{\log \begin{pmatrix} 2n \\n \end{pmatrix} }{\log 2n}$

Proposition: $\pi (2n) \ge \frac{\log \begin{pmatrix} 2n \\n \end{pmatrix} }{\log 2n}$

Since this is a follow up proposition to this one: How can we show that $\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$

We can use : $$\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$$

attempt :

$$n \log 2 \le \sum_{p\le 2n}[\frac{\log 2n}{\log p}]\log p \le \sum_{p \le 2n} \log 2n = \pi(2n) \log 2n$$

(don't see how to "finish", if this is correct at all... )

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What does $n\log 2$ have to do with $\log\binom{2n}{n}$? –  anon Mar 15 '12 at 14:57

$$\rm p\Big|\binom{2n}{n}\implies p|(2n)! \implies p\le 2n;$$
$$\rm m=\prod_{p|n} p^{\operatorname{ord}_p(m)}: \quad m=\binom{2n}{n}\implies \log \binom{2n}{n}=\sum_{p\le 2n}(\log p)\cdot\operatorname{ord}_p\binom{2n}{n}$$
Thanks, anon. So $\log m = \sum_{p\le 2n} \log p ord_{p}m \le \sum_{p\le 2n} \frac{\log 2n}{\log p} \log p \le \sum_{p\le 2n} \log 2n = \pi(2n)\log 2n$ , right ? –  VVV Mar 15 '12 at 16:25