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%I #2 Mar 31 2012 13:21:57
%S 1,4,12,45
%N Number of ways of placing n balls into n boxes and then into n containers.
%e a(2): We can have either 2 balls in a box and the other box empty, or a ball in each box. We may place both the boxes into one container, or one in each.
%e So we have [(2)()][], [(1)(1)][], [(2)][()] and [(1)][(1)], hence a(2)=4.
%e For a(3) we have: (0 is an empty box, / is a new container, - is an empty container)
%e 300/-/-, 30/0/-, 3/00/-, 3/0/0
%e 210/-/-, 21/0/-, 20/1/-, 2/10/-, 2/1/0
%e 111/-/-, 11/1/-, 1/1/1
%e total 12
%Y Cf. A000041.
%K nonn
%O 1,2
%A _Jon Perry_, Apr 30 2005