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How can we show that

$$\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$$ where $p$ is a prime number and $n$ is a natural number

My attempt:

$$2^{n} \le \prod _{k=1}^{n} \frac{k+n}{k} = \begin{pmatrix} 2n \\ n \end{pmatrix} = \frac{(2n)!}{n!n!} = \sum_{n\le 2n}p^{\operatorname{ord}_{p}(p)}$$

$$\Rightarrow \operatorname{ord}_{p}(p) \le \sum_{m=1}^{\infty}\left(\left\lfloor\frac{2n}{p^{m}}\right\rfloor-2\left\lfloor\frac{n}{p^{m}}\right\rfloor \right) \le \left\lfloor\frac{\log 2n}{\log p}\right\rfloor \le \frac{\log 2n}{\log p}$$

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Related: this question. –  Américo Tavares Mar 15 '12 at 15:05

1 Answer 1

up vote 3 down vote accepted

You haven't explained any reasoning in your attempt but it appears to be in the correct direction. By Legendre's formula, $$ \operatorname{ord}_p \left( \binom{2n}{n} \right) = \sum_{m=1}^{\infty}\left(\left\lfloor\frac{2n}{p^{m}}\right\rfloor-2\left\lfloor\frac{n}{p^{m}}\right\rfloor \right) .$$

Each term in the sum is either $0$ or $1$ and there are at most $\displaystyle \left\lfloor\frac{\log 2n}{\log p}\right\rfloor$ non-zero terms since if $m > \log_p 2n $ then $ \displaystyle \frac{2n}{p^m}<1.$

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Thanks, Ragib Zaman. –  VVV Mar 15 '12 at 15:53

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