A067336 a(0)=1, a(1)=2, a(n) = a(n-1)*9/2 - Catalan(n-1) where Catalan(n) = binomial(2n,n)/(n+1) = A000108(n).

1, 2, 8, 34, 148, 652, 2892, 12882, 57540, 257500, 1153888, 5175700, 23231864, 104335376, 468766292, 2106773874, 9470787588, 42583186476, 191494694352, 861248485884, 3873850923288, 17425765034376, 78391476387672, 352670161180884, 1586672665700328, 7138737091504152

Offset: 0

Author

Henry Bottomley, Jan 15 2002

Keywords

• nonn

Comments

Note that while a(n) is even (for n > 0), it is a multiple of 4 except when n = 2^m-1, i.e., when Catalan(n) is odd.

Result of applying the Riordan matrix ((1+sqrt(1-4*x))/2, (1-sqrt(1-4*x))/2) (inverse of (1/(1-x), x*(1-x))) to 3^n. - Paul Barry, Mar 12 2005

Hankel transform is A001787(n+1). - Paul Barry, Mar 15 2010

Examples

			a(2) =   2*9/2 -  1 =   8;
a(3) =   8*9/2 -  2 =  34;
a(4) =  34*9/2 -  5 = 148;
a(5) = 148*9/2 - 14 = 652.
		

Links

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

Crossrefs

Cf. A000108, A001045, A039599, A067337, A088218.

Programs

Mathematica

CoefficientList[Series[(1+Sqrt[1-4*x])/(3*Sqrt[1-4*x]-1), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)

Formula

a(n) = A067337(2n, n).

G.f.: (1+sqrt(1-4*x))/(3*sqrt(1-4*x)-1). - Paul Barry, Mar 12 2005

a(n) = Sum_{k=0..n} A039599(n,k)*A001045(k+1). - Philippe Deléham, Jun 10 2007

G.f.: (1-x*c(x))/(1-3*x*c(x)), where c(x) is the g.f. of A000108. - Paul Barry, Mar 15 2010

Conjecture: 2*n*a(n) + (-17*n+12)*a(n-1) + 18*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012

The above conjecture is true. - Nguyen Tuan Anh, Mar 15 2025

G.f.: 1 + 2*x/(Q(0)-3*x), where Q(k) = 2*x + (k+1)/(2*k+1) - 2*x*(k+1)/(2*k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 03 2013

a(n) ~ 3^(2*n-1) / 2^n. - Vaclav Kotesovec, Feb 13 2014