阶乘 证明整除

hbghlyj

6514. Proposed by Richard Askey, University of Wisconsin, Madison.
Show that
$$\frac{(3m + 3n)!(3n)!(2m)!(2n)!}{(2m + 3n) !(m + 2n) !(m + n)!m!n!n!}\inZ$$for $m, n=0,1, \ldots$.

hbghlyj

$f(m,n)$没有递推
一般证法：由thread-7666-1-1.html只需证$\forall x,y\in[0,1):[3x+3y]+[3y]+[2x]+[2y]\ge[2x+3y]+[x+2y]+[x+y]+[x]+2[y]$