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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.

A094250 Array, A(n, k) = ((n+2)^(k+1) + (k+1)*n*(n+1) - 1)/(n+1)^2, read by antidiagonals.

Original entry on oeis.org

1, 1, 3, 1, 3, 7, 1, 3, 8, 15, 1, 3, 9, 22, 31, 1, 3, 10, 31, 63, 63, 1, 3, 11, 42, 117, 185, 127, 1, 3, 12, 55, 199, 459, 550, 255, 1, 3, 13, 70, 315, 981, 1825, 1644, 511, 1, 3, 14, 87, 471, 1871, 4888, 7287, 4925, 1023, 1, 3, 15, 106, 673, 3273, 11203, 24420, 29133, 14767, 2047
Offset: 0

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Author

N. J. A. Sloane, Jun 02 2004

Keywords

Examples

			Array, A(n, k), begins:
  1, 3,  7, 15,  31,   63,   127,    255,     511, ... A000225;
  1, 3,  8, 22,  63,  185,   550,   1644,    4925, ... A047926;
  1, 3,  9, 31, 117,  459,  1825,   7287,   29133, ... A073724;
  1, 3, 10, 42, 199,  981,  4888,  24420,  122077, ... A094195;
  1, 3, 11, 55, 315, 1871, 11203,  67191,  403115, ... A094259;
  1, 3, 12, 70, 471, 3273, 22882, 160140, 1120941, ...
Antidiagonals, T(n, k), begins as:
  1;
  1, 3;
  1, 3,  7;
  1, 3,  8, 15;
  1, 3,  9, 22,  31;
  1, 3, 10, 31,  63,   63;
  1, 3, 11, 42, 117,  185,  127;
  1, 3, 12, 55, 199,  459,  550,  255;
  1, 3, 13, 70, 315,  981, 1825, 1644,  511;
  1, 3, 14, 87, 471, 1871, 4888, 7287, 4925, 1023;
		

Crossrefs

Programs

  • Magma
    A094250:= func< n,k | ((n-k+2)^(k+1) + (k+1)*(n-k)*(n-k+1) - 1)/(n-k+1)^2 >;
    [A094250(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Aug 18 2023
    
  • Mathematica
    A094250[n_, k_]:= ((n-k+2)^(k+1) + (k+1)*(n-k)*(n-k+1) - 1)/(n-k+1)^2;
    Table[A094250[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Aug 18 2023 *)
  • SageMath
    def A094250(n, k): return ((n-k+2)^(k+1) + (k+1)*(n-k)*(n-k+1) - 1)/(n-k+1)^2
    flatten([[A094250(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Aug 18 2023

Formula

A(n, k) = ((n+2)^(k+1) + (k+1)*n*(n+1) - 1)/(n+1)^2 (array).
T(n, k) = ((n-k+2)^(k+1) + (k+1)*(n-k)*(n-k+1) - 1)/(n-k+1)^2 (antidiagonals).
G.f. for row n: (1-(n+1)*x)/((1-(n+2)*x)*(1-x)^2).