$n>2\rightarrow{2^{n}\leq(2n)^{\pi(2n)}\land n^{\pi(2n)-\pi(n)}<2^{2n}}$

Question

let n be an integer greater than 2. and let pi be thePrime-counting function. prove that:

1. for any n: $2^{n}\leq(2n)^{\pi(2n)}$
2. $n^{\pi(2n)-\pi(n)}<2^{2n}$

Answer 1

Lemma: Let $n \in \mathbb N$ and $p$ be a prime. Let $\alpha_p$ be such that $p^{\alpha_p} \le 2n \le p^{\alpha_p+1}$. If $p^{v_p}$ divides $\binom{2n}{n}$ then $v_p \le \alpha_p$.

Proof: The greatest possible $v_p$ is $v_p = \sum_{k=1}^{\alpha_p} \left( \lfloor \frac{2n}{p^k} \rfloor - 2 \lfloor \frac{n}{p^k} \rfloor \right)$, and the terms into the parentheses are either $0$ or $1$ (check it). So $v_p \le \alpha_p$, and the lemma is done.

1. Let $\binom{2n}{n} = \prod_{p < 2n} p^{v_p}$. Then $p^{v_p} \le 2n$ by Lemma. We also have $\binom{2n}{n} = \frac{2n(2n-1)\dots(n+1)}{n(n-1)\dots1} \ge 2^n$. Thus $2^n \le \binom{2n}{n} \le (2n)^{\pi(2n)}$.

2. Notice that $\binom{2n}{n}$ is divisible by every prime $p \in (n,2n]$, and $\binom{2n}{n} < \sum_{j=0}^n \binom{2n}{j} = 2^{2n}$, so the product of the primes in $(n,2n]$ is lower than $2^{2n}$. There are $\pi(2n) - \pi(n)$ such primes, thus $n^{\pi(2n) - \pi(n)} < 2^{2n}$.