A307644 -id:A307644 - cp's OEIS frontend

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

Ilya Gutkovskiy, Apr 19 2019

nonn

			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.

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

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

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

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 *)

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

More terms from Vaclav Kotesovec, Apr 20 2019

A307739 Number of partitions of n^4 into at most n fourth powers.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 1, 5, 3, 5, 2, 27, 4, 78, 14, 152, 551, 1331, 7377, 15504, 87583, 190028, 768864, 1510305, 5413291, 12221733

Offset: 0

Ilya Gutkovskiy, Apr 25 2019

			10^4 =
4^4 + 4^4 + 6^4 + 8^4 + 8^4 =
2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 + 8^4,
so a(10) = 3.

Eric Weisstein's World of Mathematics, Biquadratic Number

Cf. A000583, A105152, A259793, A299169, A299195, A307644, A307738.

from sympy.solvers.diophantine.diophantine import power_representation
def a(n):
    if n < 2: return 1
    return sum(len(list(power_representation(n**4, 4, j))) for j in range(1, n+1))
print([a(n) for n in range(1, 20)]) # Michael S. Branicky, Jul 09 2024

a(21)-a(27) from Michael S. Branicky, Jul 09 2024

cp's OEIS Frontend

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

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