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.

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A331900 Number of compositions (ordered partitions) of the n-th triangular number into distinct triangular numbers.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 3, 13, 3, 55, 201, 159, 865, 1803, 7093, 43431, 14253, 22903, 130851, 120763, 1099693, 4527293, 4976767, 7516897, 14349685, 72866239, 81946383, 167841291, 897853735, 455799253, 946267825, 5054280915, 3941268001, 17066300985, 49111862599
Offset: 0

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Author

Ilya Gutkovskiy, Jan 31 2020

Keywords

Examples

			a(6) = 3 because we have [21], [15, 6] and [6, 15].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, p) option remember; (t->
          `if`(t*(i+2)/3n, 0, b(n-t, i-1, p+1)))))((i*(i+1)/2))
        end:
    a:= n-> b(n*(n+1)/2, n, 0):
    seq(a(n), n=0..37);  # Alois P. Heinz, Jan 31 2020
  • Mathematica
    b[n_, i_, p_] := b[n, i, p] = With[{t = i(i+1)/2}, If[t(i+2)/3 < n, 0, If[n == 0, p!, b[n, i-1, p] + If[t > n, 0, b[n-t, i-1, p+1]]]]];
    a[n_] := b[n(n+1)/2, n, 0];
    a /@ Range[0, 37] (* Jean-François Alcover, Nov 17 2020, after Alois P. Heinz *)

Formula

a(n) = A331843(A000217(n)).

Extensions

More terms from Alois P. Heinz, Jan 31 2020

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.