%I #18 Dec 24 2015 02:45:34
%S 0,1,0,1,0,0,2,1,0,0,3,1,0,0,0,6,3,1,0,0,0,11,5,1,0,0,0,0,23,12,3,1,0,
%T 0,0,0,47,23,6,1,0,0,0,0,0,106,52,14,3,1,0,0,0,0,0,235,110,29,6,1,0,0,
%U 0,0,0,0,551,253,68,15,3,1,0,0,0,0,0,0,1301,570,148,31,6,1,0,0,0,0,0,0,0
%N Triangle giving the numbers of different forests of m trees of smallest order 2, i.e., without isolated vertices.
%C Forests of order N with m components, m > floor(N/2) must contain an isolated vertex since it is impossible to partition N vertices in floor(N/2) + 1 or more trees without giving only one vertex to a tree.
%H Alois P. Heinz, <a href="/A105820/b105820.txt">Rows n = 1..141, flattened</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Forest.html">Forest</a>
%F a(n) = sum over the partitions of N: 1K1 + 2K2 + ... + NKN, with exactly m parts and no part equal to 1, of Product_{i=1..N} binomial(A000055(i)+Ki-1, Ki).
%F G.f.: 1/Product_{i>=2}(1 - x*y^i)^A000055(i). - _Vladeta Jovovic_, Apr 27 2005
%e a(12) = 1 because 5 nodes can be partitioned into two trees only in one way: one tree gets 3 nodes and the other tree gets 2. Since A000055(3) = A000055(2) = 1, there is only one forest. (The forests of order less than or equal to 5 are depicted in the Weisstein link.)
%Y Cf. A033185, A095133, A105786, A105821.
%K nonn,tabl
%O 1,7
%A _Washington Bomfim_, Apr 25 2005