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.

A185785 Second accumulation array of A107985, by antidiagonals.

Original entry on oeis.org

1, 5, 5, 15, 24, 15, 35, 70, 70, 35, 70, 160, 200, 160, 70, 126, 315, 450, 450, 315, 126, 210, 560, 875, 1000, 875, 560, 210, 330, 924, 1540, 1925, 1925, 1540, 924, 330, 495, 1440, 2520, 3360, 3675, 3360, 2520, 1440, 495, 715, 2145, 3900, 5460, 6370, 6370, 5460, 3900, 2145, 715, 1001, 3080, 5775, 8400, 10290, 10976, 10290, 8400, 5775, 3080, 1001
Offset: 1

Views

Author

Clark Kimberling, Feb 03 2011

Keywords

Comments

See A185784.

Examples

			Northwest corner:
1....5....15....35....70
5....24...70....160...315
15...70...200...450...875
35...160..450...1000..1925
		

Crossrefs

Cf. A185784.

Programs

  • Mathematica
    f[n_, k_] := Binomial[k + 2, 3]*Binomial[n + 2, 3]*(n + k + 2)/4;
    Table[f[n - k + 1, k], {n, 50}, {k, n, 1, -1}] // Flatten (* G. C. Greubel, Jul 12 2017 *)

Formula

T(n,k) = (n+k+2)*C(n+2,3)*C(k+2,3)/4, k>=1, n>=1.

A185786 Third accumulation array of A107985, by antidiagonals.

Original entry on oeis.org

1, 6, 6, 21, 35, 21, 56, 120, 120, 56, 126, 315, 405, 315, 126, 252, 700, 1050, 1050, 700, 252, 462, 1386, 2310, 2695, 2310, 1386, 462, 792, 2520, 4536, 5880, 5880, 4536, 2520, 792, 1287, 4290, 8190, 11466, 12740, 11466, 8190, 4290, 1287, 2002, 6930, 13860, 20580, 24696, 24696, 20580, 13860, 6930, 2002, 3003, 10725, 22275, 34650
Offset: 1

Views

Author

Clark Kimberling, Feb 03 2011

Keywords

Comments

See A185784. The pattern established by the formulas for A185785, A185786, A185787, suggests that the H-th accumulation array of A107985 may be given by
T(n,k)=(n+k+H)C(n+H,H+1)C(k+H,H+1)/(H+2).

Examples

			Northwest corner:
1....6.....21.....56.....126
6....35....120....315....700
21...120...405....1050...2310
56...315...1050...2695...5880
		

Crossrefs

Cf. A185784.

Programs

  • Mathematica
    (See A185784.)
    f[n_, k_] := Binomial[k + 3, 4]*Binomial[n + 3, 4]*(n + k + 3)/5; Table[f[n - k + 1, k], {n, 10}, {k, n, 1, -1}] // Flatten (* G. C. Greubel, Jul 12 2017 *)

Formula

T(n,k) = (n+k+3)*C(n+3,4)*C(k+3,4)/5.
Showing 1-2 of 2 results.