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%I #6 Mar 30 2012 17:37:35
%S 1,1,4,3,6,9,16,36,85,210,586,1914,6930,28178,125440,603350,3083410,
%T 17362239,112403052,820563290,6632950912,58209665965,544071039000,
%U 5353538904357,58523908575096,730174875609318,10274727352967428,159586345364505768
%N Number of partitions of 2^n into powers of n.
%H Alois P. Heinz, <a href="/A196889/b196889.txt">Table of n, a(n) for n = 0..150</a>
%F a(n) = [x^(2^n)] 1/Product_{j>=0}(1-x^(n^j)).
%e a(3) = 3 because there are 3 partitions of 2^3=8 into powers of 3: [1,1,3,3], [1,1,1,1,1,3], [1,1,1,1,1,1,1,1].
%Y Row n=2 of A196879.
%K nonn
%O 0,3
%A _Alois P. Heinz_, Oct 07 2011
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