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%I #18 May 15 2022 07:32:28
%S 1,9,81,756,7290,72171,729729,7505784,78298974,826489170,8811646074,
%T 94753804536,1026499549140,11192793160815,122744496427425,
%U 1352917116177840,14979996753469110,166542316847391870,1858400773709785470,20806975169765062200,233671377667405024620
%N G.f.: c(3x)^3, c(x) the g.f. of A000108.
%F G.f.: 8/(1+sqrt(1-12*x))^3.
%F a(n) = (3*n+3)/(n+3) * 3^n * C(n+1).
%F Conjecture: (n+3)*a(n) -3*(5*n+7)*a(n-1) +18*(2*n-1)*a(n-2)=0. - _R. J. Mathar_, Nov 15 2011
%F From _Amiram Eldar_, May 15 2022: (Start)
%F Sum_{n>=0} 1/a(n) = 51/121 + 964*arcsin(1/(2*sqrt(3)))/(121*sqrt(11)).
%F Sum_{n>=0} (-1)^n/a(n) = 57/169 + 1204*arcsinh(1/(2*sqrt(3)))/(169*sqrt(13)). (End)
%t terms = 18;
%t c[x_] = (1 - Sqrt[1 - 4x])/(2x) + O[x]^terms // Normal;
%t CoefficientList[c[3x]^3, x][[1 ;; terms]] (* _Jean-François Alcover_, Dec 15 2017 *)
%Y Cf. A050159, A101600, A101602.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Dec 08 2004
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