cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A298934 Number of partitions of n^2 into distinct cubes.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 3, 3, 1, 0, 3, 0, 2, 4, 0, 0, 1, 0, 0, 2, 3, 1, 1, 0, 6, 3, 6, 1, 6, 0, 3, 9, 0, 6, 6, 7, 0, 10, 3, 3, 6, 0, 8, 6, 13, 2, 10, 9, 10, 19, 2, 14, 21, 7, 2, 25
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(15) = 2 because we have [216, 8, 1] and [125, 64, 27, 8, 1].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1,
          `if`(n>i^2*(i+1)^2/4, 0, b(n, i-1)+
          `if`(i^3>n, 0, b(n-i^3, i-1))))
        end:
    a:= n-> b(n^2, n):
    seq(a(n), n=0..100);  # Alois P. Heinz, Jan 29 2018
  • Mathematica
    Table[SeriesCoefficient[Product[1 + x^k^3, {k, 1, Floor[n^(2/3) + 1]}], {x, 0, n^2}], {n, 0, 84}]

Formula

a(n) = [x^(n^2)] Product_{k>=1} (1 + x^(k^3)).
a(n) = A279329(A000290(n)).

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

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Jan 31 2020

Keywords

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].
		

Crossrefs

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)).
Showing 1-2 of 2 results.