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.

A090985 Triangle read by rows: T(n,k) is the number of dissections of a convex n-gon by nonintersecting diagonals, having exactly k triangles (n >= 2, k >= 0).

Original entry on oeis.org

1, 0, 1, 1, 0, 2, 1, 5, 0, 5, 4, 6, 21, 0, 14, 8, 35, 28, 84, 0, 42, 25, 80, 216, 120, 330, 0, 132, 64, 309, 540, 1155, 495, 1287, 0, 429, 191, 890, 2475, 3080, 5720, 2002, 5005, 0, 1430, 540, 3058, 7788, 16302, 16016, 27027, 8008, 19448, 0, 4862, 1616, 9580, 30108, 54964, 96005, 78624, 123760, 31824, 75582, 0, 16796
Offset: 2

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Author

Emeric Deutsch, Feb 28 2004

Keywords

Comments

T(n,n-2) = [binomial(2n-4, n-2)]/(n-1) = Catalan(n-2) (A000108).
T(n,n-4) = binomial(2n-5, n-4) (A002054).
T(n,n-5) = binomial(2n-6, n-5) (A002694).
T(n,0) = A046736(n).
Row sums give the little Schroeder numbers (A001003).

Examples

			T(5,1)=5 because the dissections of a convex pentagon having exactly one triangle are obtained by the placement of a diagonal between any pair of non-adjacent vertices.
T(6,0)=4 because the dissections of a convex hexagon with no triangles are obtained by the null placement and by placing one diagonal between any of the 3 pairs of opposite vertices.
Triangle starts:
  1;
  0,  1;
  1,  0,  2;
  1,  5,  0,  5;
  4,  6, 21,  0, 14;
  8, 35, 28, 84,  0, 42;
  ...
		

Crossrefs

Programs

  • Maple
    T := (n,k)->binomial(n+k-2,k)*sum(binomial(n-2+k+i,i)*binomial(n-3-k-i,i-1), i=0..floor((n-2-k)/2))/(n-1): seq(seq(T(n,k),k=0..n-2),n=2..14);
  • Mathematica
    T [n_, k_] := Binomial[n+k-2, k] Sum[Binomial[n-2+k+i, i] Binomial[n-3-k-i, i-1], {i, 0, (n-2-k)/2}]/(n-1);
    Table[T[n, k], {n, 2, 12}, {k, 0, n-2}] // Flatten (* Jean-François Alcover, Jul 29 2018 *)

Formula

T(n, k) = binomial(n+k-2, k)*(Sum_{i=0..floor((n-2-k)/2)} binomial(n-2+k+i, i)*binomial(n-3-k-i, i-1))/(n-1).
G.f.: G=G(t, z) satisfies (1-t)G^3 + (1+t)zG^2 - z^2*(1+z)G + z^4 = 0.