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 1-10 of 17 results. Next

A331843 Number of compositions (ordered partitions) of n into distinct triangular numbers.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 2, 0, 2, 7, 2, 0, 2, 6, 1, 4, 6, 2, 12, 24, 3, 8, 0, 8, 32, 6, 2, 13, 26, 6, 34, 36, 0, 32, 150, 3, 20, 50, 14, 54, 126, 32, 32, 12, 55, 160, 78, 122, 44, 174, 4, 72, 294, 36, 201, 896, 128, 62, 180, 176, 164, 198, 852, 110, 320, 159, 212, 414
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2020

Keywords

Examples

			a(10) = 7 because we have [10], [6, 3, 1], [6, 1, 3], [3, 6, 1], [3, 1, 6], [1, 6, 3] and [1, 3, 6].
		

Crossrefs

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<1, 0,
          `if`(issqr(8*n+1), 1+h(n-1), h(n-1)))
        end:
    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, h(n), 0):
    seq(a(n), n=0..73);  # Alois P. Heinz, Jan 31 2020
  • Mathematica
    h[n_] := h[n] = If[n<1, 0, If[IntegerQ @ Sqrt[8n+1], 1 + h[n-1], h[n-1]]];
    b[n_, i_, p_] := b[n, i, p] = Function[t, 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]]]]][(i(i + 1)/2)];
    a[n_] := b[n, h[n], 0];
    a /@ Range[0, 73] (* Jean-François Alcover, Apr 27 2020, after Alois P. Heinz *)

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

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Jan 29 2020

Keywords

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].
		

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, 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 *)

A331846 Number of compositions (ordered partitions) of n into distinct squarefree parts.

Original entry on oeis.org

1, 1, 1, 3, 2, 3, 9, 5, 12, 16, 21, 41, 42, 49, 59, 79, 130, 231, 230, 295, 226, 495, 609, 699, 1472, 1042, 1377, 2308, 2982, 3425, 3879, 4877, 7156, 7189, 13531, 14797, 13570, 19551, 27667, 30327, 36382, 47519, 60783, 70561, 78330, 136988, 121659, 174851
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2020

Keywords

Examples

			a(7) = 5 because we have [7], [6, 1], [5, 2], [2, 5] and [1, 6].
		

Crossrefs

A331847 Number of compositions (ordered partitions) of n into distinct prime powers (1 excluded).

Original entry on oeis.org

1, 0, 1, 1, 1, 3, 2, 5, 3, 11, 10, 13, 18, 19, 52, 30, 61, 77, 114, 109, 146, 260, 318, 341, 356, 631, 666, 927, 848, 1849, 1978, 2305, 2213, 3560, 4302, 4748, 5588, 6779, 13952, 9044, 15534, 16897, 25084, 20731, 29524, 34882, 49360, 50765, 55112, 106903, 83652, 128552, 106638
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2020

Keywords

Examples

			a(10) = 10 because we have [8, 2], [7, 3], [5, 3, 2], [5, 2, 3], [3, 7], [3, 5, 2], [3, 2, 5], [2, 8], [2, 5, 3] and [2, 3, 5].
		

Crossrefs

A331884 Number of compositions (ordered partitions) of n^2 into distinct squares.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 7, 1, 31, 123, 151, 121, 897, 7351, 5415, 14881, 48705, 150583, 468973, 1013163, 1432471, 1730023, 50432107, 14925241, 125269841, 74592537, 241763479, 213156871, 895153173, 7716880623, 2681163865, 3190865761, 22501985413, 116279718801
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 30 2020

Keywords

Examples

			a(5) = 3 because we have [25], [16, 9] and [9, 16].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, p) option remember;
          `if`(i*(i+1)*(2*i+1)/6n, 0, b(n-i^2, i-1, p+1))+b(n, i-1, p)))
        end:
    a:= n-> b(n^2, n, 0):
    seq(a(n), n=0..35);  # Alois P. Heinz, Jan 30 2020
  • Mathematica
    b[n_, i_, p_] := b[n, i, p] = If[i(i+1)(2i+1)/6 < n, 0, If[n == 0, p!, If[i^2 > n, 0, b[n - i^2, i - 1, p + 1]] + b[n, i - 1, p]]];
    a[n_] := b[n^2, n, 0];
    a /@ Range[0, 35] (* Jean-François Alcover, Nov 08 2020, after Alois P. Heinz *)

Formula

a(n) = A331844(A000290(n)).

Extensions

a(24)-a(34) from Alois P. Heinz, Jan 30 2020

A332007 Number of compositions (ordered partitions) of n into distinct pentagonal numbers.

Original entry on oeis.org

1, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 2, 6, 0, 0, 0, 1, 2, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 2, 7, 2, 0, 0, 6, 26, 6, 0, 0, 0, 0, 0, 2, 6, 0, 0, 1, 8, 24, 0, 0, 2, 8, 6, 0, 0, 0, 6, 26, 6, 0, 0, 0, 6, 30, 25, 2, 0, 2, 30, 122, 6, 0, 6, 24
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 04 2020

Keywords

Examples

			a(18) = 6 because we have [12, 5, 1], [12, 1, 5], [5, 12, 1], [5, 1, 12], [1, 12, 5] and [1, 5, 12].
		

Crossrefs

A331918 Number of compositions (ordered partitions) of n into distinct odd squares.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 1, 2, 6, 24, 0, 0, 0, 0, 0, 2
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 01 2020

Keywords

Examples

			a(35) = 6 because we have [25, 9, 1], [25, 1, 9], [9, 25, 1], [9, 1, 25], [1, 25, 9] and [1, 9, 25].
		

Crossrefs

Programs

  • Maple
    N:= 200: # for a(0)..a(N)
    G:= mul(1+t*x^(i^2),i=1..floor(sqrt(N)),2):
    F:= proc(n) local R, k, v;
      R:= coeff(G,x,n);
      add(k!*coeff(R,t,k),k=1..degree(R,t))
    end proc:
    F(0):= 1:
    map(F, [$0..N]); # Robert Israel, Feb 03 2020

A331984 Number of compositions (ordered partitions) of n into distinct square pyramidal numbers.

Original entry on oeis.org

1, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 6, 24, 0, 0, 0, 0, 1, 2, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 6, 24, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 6, 25
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 03 2020

Keywords

Examples

			a(20) = 6 because we have [14, 5, 1], [14, 1, 5], [5, 14, 1], [5, 1, 14], [1, 14, 5] and [1, 5, 14].
		

Crossrefs

A332014 Number of compositions (ordered partitions) of n into distinct hexagonal numbers.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 2, 6, 1, 2, 0, 0, 6, 24, 2, 6, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 7, 26, 0, 0, 0, 0, 2, 8, 6, 0, 0, 0, 0, 6, 24
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 04 2020

Keywords

Examples

			a(22) = 6 because we have [15, 6, 1], [15, 1, 6], [6, 15, 1], [6, 1, 15], [1, 15, 6] and [1, 6, 15].
		

Crossrefs

A332015 Number of compositions (ordered partitions) of n into distinct heptagonal numbers.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 1, 2, 0, 0, 6, 24, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 6, 25, 2
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 04 2020

Keywords

Examples

			a(26) = 6 because we have [18, 7, 1], [18, 1, 7], [7, 18, 1], [7, 1, 18], [1, 18, 7] and [1, 7, 18].
		

Crossrefs

Showing 1-10 of 17 results. Next