A331899 Number of compositions (ordered partitions) of n^3 into distinct cubes.

1, 1, 1, 1, 1, 1, 7, 1, 1, 127, 1, 1, 127, 769, 10945, 15961, 86641, 86521, 430717, 4140367, 4146751, 93669001, 1538834041, 663998665, 6883029151, 1014140647, 20591858857, 121532206567, 1637261351983, 2981530899847, 5950338797191, 47072230385425

Offset: 0

Author

Ilya Gutkovskiy, Jan 31 2020

Keywords

• nonn

Examples

			a(6) = 7 because we have [216], [125, 64, 27], [125, 27, 64], [64, 125, 27], [64, 27, 125], [27, 125, 64] and [27, 64, 125].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..100
• Index entries for sequences related to compositions
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A000578, A030272, A259792, A290247, A298641, A298672, A298848, A331845, A331884.

Programs

Maple

b:= proc(n, i, p) option remember;
      `if`((i*(i+1)/2)^2n, 0, b(n-i^3, i-1, p+1))+b(n, i-1, p)))
    end:
a:= n-> b(n^3, n, 0):
seq(a(n), n=0..33);  # Alois P. Heinz, Jan 31 2020

Mathematica

b[n_, i_, p_] := b[n, i, p] = If[(i(i+1)/2)^2 < n, 0, If[n == 0, p!, If[i^3 > n, 0, b[n - i^3, i - 1, p + 1]] + b[n, i - 1, p]]];
a[n_] := b[n^3, n, 0];
a /@ Range[0, 33] (* Jean-François Alcover, Nov 26 2020, after Alois P. Heinz *)

Formula

a(n) = A331845(A000578(n)).