通过递推式可以自己推出来L(2)=1L(3)=2L(4)=L(3)*C(3,1)+L(2)*C（3,1）=9......L(n+1)=L(n)*C(n,1)+L(n-1)*C(n,1).....
Dn=n!(1-1/1！+1/2！-1/3!+...+(-1)^n*1/n!)
=n!*[1-1/(1!)+1/(2!)-1/(3!)+1/(4!)+...+(-1)^n/(n!)]
