A307643 Number of partitions of n^3 into exactly n positive cubes.

1, 1, 0, 0, 0, 0, 1, 0, 2, 6, 14, 23, 51, 108, 228, 511, 1158, 2500, 5603, 12304, 26969, 59222, 130115, 285370, 624965, 1368603, 2987117, 6517822, 14187920, 30823278, 66834822, 144671698, 312551894, 673913968, 1450292087, 3114720013, 6676277754, 14281662079

Offset: 0

Author

Ilya Gutkovskiy, Apr 19 2019

Keywords

• nonn

Examples

			9^3 =
1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 =
1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 =
1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 =
1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 =
1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 =
2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 7^3,
so a(9) = 6.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..79
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A000578, A030272, A259792, A298671, A298672, A307644, A319435, A320841.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0, `if`(t=0, 1, 0),
      `if`(i<1 or t<1, 0, b(n, i-1, t)+
      `if`(i^3>n, 0, b(n-i^3, i, t-1))))
    end:
a:= n-> b(n^3, n$2):
seq(a(n), n=0..25);  # Alois P. Heinz, Oct 12 2019

Mathematica

b[n_, i_, t_] := b[n, i, t] = If[n == 0, If[t == 0, 1, 0], If[i < 1 || t < 1, 0, b[n, i - 1, t] + If[i^3 > n, 0, b[n - i^3, i, t - 1]]]];
a[n_] := b[n^3, n, n];
a /@ Range[0, 25] (* Jean-François Alcover, Nov 07 2020, after Alois P. Heinz *)

Formula

a(n) = A320841(n^3,n).

Extensions

More terms from Vaclav Kotesovec, Apr 20 2019