(Andy Balaam and Miguel Garvie, 07/2003)
F(n,0) = 1
F(n,1) = n + 1
F(n,2) = n(n+1)/2 + n + 1
and F(n,3) = X + F(n,2) where X is the number of order 3 terms.
(Marieke Rohde, 23/11/2004)
F(n, m) = SUM(i=0 to m) |terms of order i in a polynomial of order n|
F(n, m) = SUM(i=0 to m) ( (n+i-1) choose i )