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.

A067955 Number of dissections of a convex polygon by nonintersecting diagonals into polygons with even number of sides and having a total number of n edges (sides and diagonals).

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 3, 1, 8, 13, 15, 56, 79, 157, 399, 624, 1448, 3061, 5571, 12826, 25559, 51608, 113828, 227954, 482591, 1031681, 2117323, 4542331, 9591243, 20119244, 43164172, 91165297, 193826856, 415024053, 881294603, 1886458874, 4038398755
Offset: 1

Views

Author

Emeric Deutsch, Mar 06 2002

Keywords

Comments

Number of ordered trees with n-1 edges, all of whose nodes have odd outdegree greater than two.
Conjecture: Number of lattice paths that do not cross below the x-axis from (1,0) to (n,0) using up-step (1,1) and down-steps {(1,-z): z is a positive even integer}. For example, a(8) = 1: [(1,1)(1,1)(1,1)(1,1)(1,1)(1,1)(1,-6)]. - Nicholas Ham, Aug 24 2015

Examples

			a(7)= 3 because the only dissections with 7 edges are given by a hexagon dissected by any of the three halving diagonals.
		

Programs

  • Maple
    Order := 40: solve(series((G-G^3)/(1-G^2+G^3),G)=z,G);
    # Alternative:
    f:= gfun:-rectoproc({115*n*(n+1)*a(n)+(617*n+1236)*(n+1)*a(n+1)+(2*(569*n^2+2657*n+3006))*a(n+2)+(2*(436*n^2+2737*n+4254))*a(n+3)+(6*(32*n^2+267*n+554))*a(n+4)-(4*(29*n^2+260*n+570))*a(n+5)-(8*(n+6))*(11*n+53)*a(n+6)-(16*(n+7))*(n+6)*a(n+7) = 0,a(0)=0,a(1)=1,a(2)=0,a(3)=0,a(4)=1,a(5)=0,a(6)=1},a(n),remember):
    map(f, [$1..100]); # Robert Israel, Sep 01 2015
  • Mathematica
    CoefficientList[InverseSeries[(x-x^3)/(1-x^2+x^3) + O[x]^40, x], x] // Rest (* Jean-François Alcover, Sep 16 2022 *)
  • PARI
    Vec(serreverse((x-x^3)/(1-x^2+x^3)+O(x^44))) \\ Joerg Arndt, Sep 28 2015

Formula

a(n) = (1/n)Sum_{j=1..floor((n-1)/3)} binomial(n, j)binomial((n-3-j)/2, j-1). [formula seems wrong]
G.f. G(z) satisfies (1+z)*G^3 - z*G^2 - G + z = 0.
115*n*(n+1)*a(n)+(617*n+1236)*(n+1)*a(n+1)+(2*(569*n^2+2657*n+3006))*a(n+2)+(2*(436*n^2+2737*n+4254))*a(n+3)+(6*(32*n^2+267*n+554))*a(n+4)-(4*(29*n^2+260*n+570))*a(n+5)-(8*(n+6))*(11*n+53)*a(n+6)-(16*(n+7))*(n+6)*a(n+7) = 0. - Robert Israel, Sep 01 2015
G.f. is the series reversion of (x-x^3)/(1-x^2+x^3). - Joerg Arndt, Sep 28 2015