论文标题
两个连续的素数之间的差距$ g_n $满足$ g_n = o（{p_n}^{2/3}）$
The gap $g_n$ between two consecutive primes satisfies $g_n=O({p_n}^{2/3})$
论文作者
论文摘要
以下的论点证明了以下内容，这些论点不会围绕Riemann假设或筛子理论。如果$ p_n $是$ n^{\ rm th} $ prime和$ g_n = p_ {n+1} -p_n $，则$ g_n = o（{p_n}^{2/3}）$。
The following is proven using arguments that do not revolve around the Riemann Hypothesis or Sieve Theory. If $p_n$ is the $n^{\rm th}$ prime and $g_n=p_{n+1}-p_n$, then $g_n=O({p_n}^{2/3})$.
