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Number of partitions of n^3 into distinct cubes.
18

%I #31 Sep 19 2020 03:39:42

%S 1,1,1,1,1,1,2,1,1,3,1,1,3,4,6,6,7,6,20,18,21,42,55,52,80,126,140,201,

%T 323,361,600,626,938,1387,1648,2310,3620,4575,5495,9278,11239,14229,

%U 23406,28780,38218,53987,73114,87568,134007,181986,233004,348230,432184

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

%H Alois P. Heinz, <a href="/A030272/b030272.txt">Table of n, a(n) for n = 0..180</a>

%F a(n) = [x^(n^3)] Product_{k>=1} (1 + x^(k^3)). - _Ilya Gutkovskiy_, Apr 13 2017

%F a(n) = A279329(n^3). - _Vaclav Kotesovec_, May 06 2019

%F a(n) ~ exp(2^(7/4) * 3^(-3/2) * ((2^(1/3)-1) * Gamma(1/3) * Zeta(4/3))^(3/4) * n^(3/4)) * ((2^(1/3)-1) * Gamma(1/3) * Zeta(4/3))^(3/8) / (2^(17/8) * 3^(1/4) * sqrt(Pi) * n^(21/8)). - _Vaclav Kotesovec_, May 06 2019

%e a(6) = 2: [27,64,125], [216].

%e a(9) = 3: [1,27,64,125,512], [1,216,512], [729].

%t nmax = 50; poly = ConstantArray[0, nmax^3 + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[Do[poly[[j + 1]] += poly[[j - k^3 + 1]], {j, nmax^3, k^3, -1}];, {k, 2, nmax}]; Table[poly[[1 + n^3]], {n, 0, nmax}] (* _Vaclav Kotesovec_, Sep 19 2020 *)

%o (PARI) apply( A030272(n)=A279329(n^3), [0..30]) \\ _M. F. Hasler_, Jan 05 2020

%Y Cf. A000009, A003108, A033461, A259792, A279329.

%K nonn

%O 0,7

%A _Warren D. Smith_

%E a(0)=1 prepended by _Ilya Gutkovskiy_, Apr 13 2017