@Gerry Myerson Yes, thank you very much.

@quid: When n is not too large, My conjecture gives an approximate value of the prime maximal gap ,which is closed to the actual value.

@Steven Landsburg: $\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$ means that $E=\frac{logN(logN-2loglogN)+2}{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}\approx 1$. "E" is close to 1, but it is not always equal to 1. For example, E=09.5, E=0.98, E=1.02 or E=1.04, and so on. So we can proof that $$\limsup_n \frac{p_{n+1} - p_n}{(\log p_n)^2} \geq 1$$

@quid,@Charles: $\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$ do not mean that $$\lim_{n\rightarrow \infty }\frac{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}{ logN(logN-loglogN)+2}=1$$. Please think it carefully.

@Per Alexandersson: Thank you for you advice. It means a lot for me.

@Steven Landsburg: This is my first time asking a question in mathoverflow, I am sorry.

@The User: "≈" means "almost equal to". But, $\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$ do not mean that $$\lim_{n\rightarrow \infty }\frac{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}{ logN(logN-2loglogN)+2}=1$$.