Jun 262012

Sylvester’s theorem  the binomial coefficient

\begin{equation}{n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\end{equation}

has a prime divisor \(p\) greater than \(k\). In other words, if \(n\geq 2k\), then the product of  \(k\) consecutive integers greater than \(k\) is divisible by a prime greater than \(k\). Note that for \(n=2k\) we obtain precisely Bertrand’s postulate.


  1. if \(n>6\) is an integer, then there always exists at least two prime numbers \(p\), one  is the form \(4k+1\) and the other \(4k+3\), such that \(n<p<2n\).
  2. if \(k\) is a positive  integer, then there always exists a positive  integer \(N\) such that  for every  integer \(n>N\),  there exist at least \(k\) prime numbers \(p\) such that \(n<p<2n\).
  • 对任意正整数 \(n>6\) ,  至少存在一个 \(4k+1\) 型和一个 \(4k+3\)  型素数 \(p\)  使得  \(n<p<2n\).
  • 对任意正整数 \(k\),  存在正整数 \(N\),  使得只要正整数 \(n>N\),  就存在 \(k\) 个素数 \(p\) 使得 \(n<p<2n\).

注意, 定理 \(2\) 已经隐含在 Ramanujan 给出的 Bertrand’s postulate 的证明中 Ramanujan’s proof of Bertrand’s postulate.

 Leave a Reply



This site uses Akismet to reduce spam. Learn how your comment data is processed.