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A010736 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-2,n).

%I A010736 #31 Jul 31 2024 11:36:58
%S A010736 1,3,12,52,237,1119,5424,26832,134913,687443,3541932,18421524,
%T A010736 96585597,509960223,2709067968,14469453632,77655751329,418567792899,
%U A010736 2264867271852,12298297439892,66993811842477,366009125766463
%N A010736 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-2,n).
%C A010736 Threefold convolution of A001003 with itself. Number of dissections of a convex polygon with n+4 sides that have a quadrilateral over a fixed side (the base) of the polygon. Example: a(1)=3 because the only dissections of the convex pentagon ABCDE (AB being the base), that have a quadrilateral over AB are the dissections made by the diagonals EC, AD and BD, respectively. - _Emeric Deutsch_, Dec 27 2003
%C A010736 a(n-1) = number of royal paths (A006318) from (0,0) to (n,n) with exactly 2 diagonal steps on the line y=x. - _David Callan_, Jul 15 2004
%H A010736 Vincenzo Librandi, <a href="/A010736/b010736.txt">Table of n, a(n) for n = 0..1000</a>
%F A010736 G.f.: (1+z-sqrt(1-6*z+z^2))^3/(64*z^3). - _Emeric Deutsch_, Dec 27 2003
%F A010736 a(n) = (3/n)*Sum_{k = 1..n} binomial(n, k)*binomial(n+k+2, k-1) = 3*hypergeom([1-n, n+4], [2], -1), n>=1, a(0)=1.
%F A010736 Recurrence: (n+3)*(2*n-1)*a(n) = (12*n^2+11*n-11)*a(n-1) - (n-3)*(2*n-1)*a(n-2) + (3-n)*a(n-3). - _Vaclav Kotesovec_, Oct 07 2012
%F A010736 a(n) ~ 3 * (1 + sqrt(2))^(2*n+3) / (2^(11/4) * sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Oct 07 2012, simplified Dec 24 2017
%F A010736 From _Peter Bala_, Jul 31 2024: (Start)
%F A010736 a(n) = (3/4) * Sum_{k = 0..n+1} binomial(n+1, k)*binomial(n+k+1, k)/(k+2) for n >= 1.
%F A010736 Second-order recurrence: (n + 3)*n^2*a(n) = (2*n + 1)*(3*n^2 + 3*n - 2)*a(n-1) - (n - 2)*(n + 1)^2*a(n-2), with a(0) = 1 and a(1) = 3.
%F A010736 a(n) = (3/8) * hypergeom([2, n + 2, - n - 1], [1, 3], -1) for n >= 1.
%F A010736 a(n) = (3/4) * Integral_{x = 0..1} x*Legendre_P(n+1, 2*x+1) for n >= 1. Note that A006318(n) = Integral_{x = 0..1} Legendre_P(n, 2*x+1). (End)
%t A010736 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} ]
%t A010736 CoefficientList[Series[(1+x-Sqrt[1-6x+x^2])^3/(64x^3),{x,0,30}],x] (* _Harvey P. Dale_, Apr 18 2012 *)
%o A010736 (PARI) my(x='x+O('x^66)); Vec((1+x-sqrt(1-6*x+x^2))^3/(64*x^3)) \\ _Joerg Arndt_, May 04 2013
%Y A010736 Cf. A001003, A006318.
%Y A010736 Right-hand column 3 of triangle A011117.
%Y A010736 Third column of convolution triangle A011117.
%K A010736 nonn,easy
%O A010736 0,2
%A A010736 Robert Sulanke (sulanke(AT)diamond.idbsu.edu)
%E A010736 More terms from _Emeric Deutsch_, Dec 27 2003