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Number of partitions of n^3 into distinct cubes > 1.
2

%I #4 Jan 28 2018 11:02:05

%S 1,0,1,1,1,1,2,1,1,1,1,1,3,2,5,4,3,4,13,11,15,20,23,34,52,49,97,118,

%T 164,192,296,330,525,745,825,1354,1820,1994,3356,4605,5543,8335,12319,

%U 13124,21133,28634,33209,51272,71154,85329,126806,174704,210157,310269,433490,511199

%N Number of partitions of n^3 into distinct cubes > 1.

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>

%H <a href="/index/Par#part">Index entries for related partition-counting sequences</a>

%F a(n) = [x^(n^3)] Product_{k>=2} (1 + x^(k^3)).

%F a(n) = A280130(A000578(n)).

%e a(6) = 2 because we have [216] and [125, 64, 27].

%Y Cf. A000578, A003108, A030272, A078128, A259792, A279329, A280130, A298641, A298671, A298672.

%K nonn

%O 0,7

%A _Ilya Gutkovskiy_, Jan 27 2018