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Number of partitions of 3^n into powers of n.
2

%I #6 Mar 30 2012 17:37:35

%S 1,1,10,23,72,335,2220,19166,217862,3428059,71688050,1884401480,

%T 63363038340,2929516409504,178211319638172,13290584617658383,

%U 1240111930777216192,158642776309162956097,26642849845285577276244,5432337767302682299726906

%N Number of partitions of 3^n into powers of n.

%H Alois P. Heinz, <a href="/A196890/b196890.txt">Table of n, a(n) for n = 0..100</a>

%F a(n) = [x^(3^n)] 1/Product_{j>=0}(1-x^(n^j)) for n>1.

%e a(2) = 10 because there are 10 partitions of 3^2=9 into powers of 2: [1,8], [1,4,4], [1,2,2,4], [1,1,1,2,4], [1,1,1,1,1,4], [1,2,2,2,2], [1,1,1,2,2,2], [1,1,1,1,1,2,2], [1,1,1,1,1,1,1,2], [1,1,1,1,1,1,1,1,1].

%Y Row n=3 of A196879.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Oct 07 2011