A010849 Let S(x,y) = number of lattice paths from (0,0) to (x,y) that use the step set { (0,1), (1,0), (2,0), (3,0), ....} and never pass below y = x. Sequence gives S(n-3,n).

1, 4, 18, 84, 403, 1976, 9860, 49912, 255701, 1323292, 6907830, 36331500, 192339687, 1024140336, 5481165832, 29469454640, 159094662121, 862087135988, 4687164401114, 25562520325828, 139803777476859, 766578879858024

Offset: 0

Author

Robert Sulanke (sulanke(AT)diamond.idbsu.edu)

Keywords

• nonn

Comments

Number of dissections of a convex polygon with n+5 sides that have a pentagon over a fixed side (the base) of the polygon. Example: a(1)=4 because the only dissections of the convex hexagon ABCDEF (AB being the base), that have a pentagon over AB are the dissections made by the diagonals FD, EC, AE and BD, respectively. - Emeric Deutsch, Dec 27 2003

a(n-1) = number of royal paths (A006318) from (0,0) to (n,n) with exactly 3 diagonal steps on the line y=x. - David Callan, Jul 15 2004

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A001003.

Right-hand column 4 of triangle A011117.

Fourth column of convolution triangle A011117.

Programs

Mathematica

f[ x_, y_ ] := f[ x, y ] = Module[ {return}, If[ x == 0, return = 1, If[ y == x-1, return = 0, return = f[ x, y-1 ] + Sum[ f[ k, y ], {k, 0, x-1} ] ] ]; return ]; Do[ Print[ Table[ f[ k, j ], {k, 0, j} ] ], {j, 10, 0, -1} ] (* End *)
CoefficientList[Series[(1 + x - Sqrt[1 - 6 x + x^2])^4 / (256 x^4), {x, 0, 30}], x] (* Vincenzo Librandi, May 03 2013 *)

PARI

x='x+O('x^66); Vec((1+x-sqrt(1-6*x+x^2))^4/(256*x^4)) \\ Joerg Arndt, May 04 2013

Formula

G.f.: (1+z-sqrt(1-6*z+z^2))^4/(256*z^4). 4-fold convolution of A001003 with itself. Convolution of A010683 with itself. - Emeric Deutsch, Dec 27 2003

a(n) = (4/n)*sum(binomial(n, k)*binomial(n+k+3, k-1), k=1..n) = 4*hypergeom([1-n, n+5], [2], -1), n>=1, a(0)=1.

Recurrence: n*(n+4)*a(n) = (7*n^2+16*n-3)*a(n-1) - (7*n^2-2*n-12)*a(n-2) + (n-3)*(n+1)*a(n-3). - Vaclav Kotesovec, Oct 07 2012

a(n) ~ sqrt(1632+1154*sqrt(2))*(3+2*sqrt(2))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012

Recurrence (an alternative): (n+4)*a(n) = (8-n)*a(n-8) + 4*(2*n-13)*a(n-7) + 12*(5-n)*a(n-6) + 4*(7-2*n)*a(n-5) + 26*(n-2)*a(n-4) + 4*(1-2*n)*a(n-3) - 12*(n+1)*a(n-2) + 4*(2*n+5)*a(n-1), n>=8. - Fung Lam, Feb 18 2014

Extensions

More terms from Emeric Deutsch, Dec 27 2003