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 Oct26 awarded Yearling Aug24 awarded Popular Question Jul2 awarded Curious Jul2 awarded Inquisitive May23 awarded Famous Question May21 awarded Notable Question Mar9 awarded Popular Question Oct26 awarded Yearling Apr15 awarded Notable Question Feb20 awarded Popular Question Dec13 awarded Popular Question Oct26 awarded Yearling Mar15 revised How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes? added 1 characters in body Mar15 comment How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes? Yea, exactly TonyK. Mar15 revised How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes? deleted 114 characters in body Mar15 revised How can we show that $\pi (x+y) - \pi(y) \le \frac{1}{3} x + C$ using the sieve of eratosthenes? edited title Mar15 accepted How to show $\pi (2n) \ge \log \binom{ 2n }{ n} / \log 2n$? Mar15 comment How to show $\pi (2n) \ge \log \binom{ 2n }{ n} / \log 2n$? 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 ? Mar15 accepted How can we show that $\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$ Mar15 comment How can we show that $\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$ Thanks, Ragib Zaman.