A026569 a(n) = T(n,n), T given by A026568. Also a(n) = number of integer strings s(0),...,s(n) counted by T, such that s(n)=0.

1, 1, 3, 5, 13, 27, 67, 153, 375, 893, 2189, 5319, 13089, 32155, 79479, 196573, 487833, 1212135, 3018355, 7525585, 18792303, 46980373, 117589689, 294613155, 738844719, 1854484305, 4658460165, 11710592711, 29458662005, 74151824271

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Comments

Number of grand Motzkin n-paths avoiding UF. - David Scambler, Jun 20 2013

Examples

			For a(3) = 5 the five grand Motzkin paths are FDU, DFU, FUD, UDF and FFF. The paths containing UF, namely UFD and DUF, are avoided. - _David Scambler_, Jun 20 2013
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Rui Duarte and António Guedes de Oliveira, Generating functions of lattice paths, Univ. do Porto (Portugal 2023).
• J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.

Crossrefs

Cf. A026568.

Programs

GAP

List([0..30], n-> Sum([0..Int(n/2)], k-> Binomial(2*k,k)*Binomial( n-k, k) )); # G. C. Greubel, Aug 03 2019

Magma

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 1/Sqrt((1-x)*(1-x-4*x^2)) )); // G. C. Greubel, Aug 03 2019

Mathematica

CoefficientList[Series[Sqrt[1/((1-x)(1-x-4x^2))],{x,0,30}],x] (* Harvey P. Dale, Oct 06 2011 *)

PARI

my(x='x+O('x^30)); Vec( 1/sqrt((1-x)*(1-x-4*x^2)) ) \\ G. C. Greubel, Aug 03 2019

Sage

(1/sqrt((1-x)*(1-x-4*x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(2*k, k)*binomial(n-k, k). - Paul Barry, Sep 09 2004

G.f.: sqrt(1/((1-x)*(1-x-4*x^2))). - Ralf Stephan, Jan 08 2004

D-finite with recurrence: a(n) = 1/n*((2*n-1)*a(n-1) + (3*n-3)*a(n-2) - (4*n-6)*a(n-3)). - Vladeta Jovovic, Mar 12 2005

a(n) = Sum_{k=0..n} C(k, n-k)*C(2*(n-k), n-k). - Paul Barry, Jul 30 2005

G.f.: 1/(1-x-2*x^2/(1-0*x-x^2/(1-x-x^2/(1-0*x-2*x^2/(1-x-x^2/.... (continued fraction). Paul Barry, Dec 07 2008

a(n) ~ sqrt((5+13/sqrt(17))/8) * ((1+sqrt(17))/2)^n/sqrt(Pi*n). - Vaclav Kotesovec, Aug 10 2013