Sylvester’s theorem  the binomial coefficient

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

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.

Theorem

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 $$n<p<2n$$.

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