A331845 Number of compositions (ordered partitions) of n into distinct cubes.

1, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 6, 24

Offset: 0

Author

Ilya Gutkovskiy, Jan 29 2020

Keywords

• nonn
• look

Examples

			a(36) = 6 because we have [27,8,1], [27,1,8], [8,27,1], [8,1,27], [1,27,8] and [1,8,27].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..65536
• Index entries for sequences related to compositions
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A000578, A023358, A032020, A032021, A032022, A218396, A219107, A279329, A331843, A331844, A331846, A331847.

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, iroot(n, 3), 0):
seq(a(n), n=0..100);  # Alois P. Heinz, Jan 30 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, Floor[n^(1/3)], 0];
a /@ Range[0, 100] (* Jean-François Alcover, Oct 31 2020, after Alois P. Heinz *)