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%I #21 Feb 14 2014 03:22:40
%S 1,1,4,13,52,214,928,4141,18940,88258,417616,2001058,9690184,47348812,
%T 233158144,1155900541,5764510060,28898899594,145556001136,
%U 736206912982,3737768204344,19042072755124,97313398530496,498737257238482,2562773039735896,13200732624526804,68148459129343648
%N Expansion of c(x(1+2x)), c(x) the g.f. of A000108.
%C Hankel transform of a(n) is A047656(n+1)=3^C(n+1,2). In general, the Hankel transform of the expansion of c(x(1+r*x)) is (r+1)^C(n+1,2).
%C Number of paths weakly above X-axis from (0,0) to (0,2n) using steps (1,1), (1,-1) and two colors of (3,1). - _David Scambler_, Jun 21 2013
%D Barry, Paul; Hennessy, Aoife Four-term recurrences, orthogonal polynomials and Riordan arrays. J. Integer Seq. 15 (2012), no. 4, Article 12.4.2, 19 pp.
%H Vincenzo Librandi, <a href="/A129147/b129147.txt">Table of n, a(n) for n = 0..200</a>
%F a(n)=sum{k=0..n, C(k,n-k)*2^(n-k)*C(k)};
%F a(n)=(1/(2*pi))*int(x^n*sqrt(8+4x-x^2)/(x+2),x,2-2*sqrt(3),2+2*sqrt(3));
%F Conjecture: (n+1)*a(n) +2*(2-n)*a(n-1) +4*(5-4n)*a(n-2) +16*(2-n)*a(n-3)=0. - _R. J. Mathar_, Dec 14 2011
%F G.f.: Q(0), where Q(k)= 1 + (4*k+1)*x*(1+2*x)/(k + 1 - x*(1+2*x)*(2*k+2)*(4*k+3)/(2*x*(1+2*x)*(4*k+3) + (2*k+3)/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 15 2013
%F a(n) ~ sqrt(3-sqrt(3)) * (2*(1+sqrt(3)))^n / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Feb 01 2014
%t CoefficientList[Series[(1-Sqrt[1-4*x*(1+2*x)])/(2*x*(1+2*x)), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Feb 01 2014 *)
%o (PARI) x='x+O('x^66);
%o C(x)=(1-sqrt(1-4*x))/(2*x);
%o Vec(C(x*(1+2*x))) \\ _Joerg Arndt_, May 15 2013
%K easy,nonn
%O 0,3
%A _Paul Barry_, Apr 01 2007
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