A135052 Expansion of g.f.: (1-2*x-sqrt(1-4*x+8*x^3-4*x^4))/(2*x^2*(1-x)).

1, 1, 3, 7, 19, 51, 143, 407, 1183, 3487, 10415, 31439, 95791, 294191, 909823, 2830943, 8856255, 27839167, 87888767, 278545663, 885903743, 2826612095, 9045147391, 29022168063, 93350430975, 300949170431, 972271227647

Offset: 0

Author

Paul Barry, Nov 15 2007

Keywords

• easy
• nonn

Comments

Sequence is the binomial transform of the aerated large Schroeder numbers A006318. Hankel transform is A060656(n+1).

Number of lattice paths, never going below the x-axis, from (0,0) to (n,0) consisting of up steps U = (1,1), down steps D = (1,-1) and horizontal steps H(k) = (k,0) for every positive integer k. For instance, for n=3, we have the 7 paths: H(1)H(1)H(1), H(1)H(2), H(2)H(1), H(3), H(1)UD, UDH(1), UH(1)D. - Emanuele Munarini, Mar 14 2011

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Paul Barry, Conjectures on Somos 4, 6 and 8 sequences using Riordan arrays and the Catalan numbers, arXiv:2211.12637 [math.CO], 2022.
• Rodrigo De Castro, Andrés L. Ramírez, and José L. Ramírez, Applications in Enumerative Combinatorics of Infinite Weighted Automata and Graphs, arXiv:1310.2449 [cs.DM], 2013.
• Rodrigo De Castro, Andrés L. Ramírez, and José L. Ramírez, Applications in Enumerative Combinatorics of Infinite Weighted Automata and Graphs, Scientific Annals of Computer Science, 24(1)(2014), 137-171

Programs

Mathematica

CoefficientList[Series[(1 - 2 x - Sqrt[1 - 4 x + 8 x^3 - 4 x^4]) / (2 x^2 (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Apr 19 2015 *)

Formula

a(n) = Sum_{k=0..n, Sum_{j=0..k/2, C(k/2+j, 2j)*C(j)*(1+(-1)^k)/2}}, where C(n) is A000108(n).

G.f.: 1/(1-x-2x^2/(1-x-x^2/(1-x-2x^2/(1-x-x^2/(1-x-2x^2.... (continued fraction). - Paul Barry, Jan 02 2009

a(n) = Sum_{s=0..n} Sum_{m=0..n-2s} (C(s)*binomial(m+2s,m) * binomial(n-2s-1,m-1)), where C(n) is A000108(n). - José Luis Ramírez Ramírez, Apr 19 2015

Conjecture: (n+2)*a(n) +(-5*n-4)*a(n-1) +2*(2*n+1)*a(n-2) +4*(2*n-5)*a(n-3) +12*(-n+3)*a(n-4) +4*(n-4)*a(n-5)=0. - R. J. Mathar, Apr 19 2015

a(n) ~ (2+sqrt(2))^(n+1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Apr 20 2015