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a(n) = (2*0^n + 3^n*binomial(2*n,n))/3.
3

%I #16 Sep 08 2022 08:45:15

%S 1,2,18,180,1890,20412,224532,2501928,28146690,318995820,3636552348,

%T 41655054168,479033122932,5527305264600,63958818061800,

%U 741922289516880,8624846615633730,100454095876204620,1171964451889053900,13693479385229998200,160213708807190978940

%N a(n) = (2*0^n + 3^n*binomial(2*n,n))/3.

%C (1 + (k-1)*sqrt(1-4*k*x))/(k*sqrt(1-4*k*x)) is the g.f. for ((k-1)*0^n + k^n*binomial(2*n,n))/k.

%H Vincenzo Librandi, <a href="/A099044/b099044.txt">Table of n, a(n) for n = 0..900</a>

%F G.f.: 1/3 + 4*x/(sqrt(1-12*x)(1-sqrt(1-12*x))) = (1 + 2*sqrt(1-12*x))/(3*sqrt(1-12*x)).

%F n*a(n) +6*(-2*n+1)*a(n-1)=0. - _R. J. Mathar_, Nov 24 2012

%F E.g.f.: (2 + exp(6*x) * BesselI(0,6*x)) / 3. - _Ilya Gutkovskiy_, Nov 17 2021

%t Join[{1}, Table[3^(n-1)*binomial(2*n,n), {n,1,30}]] (* _G. C. Greubel_, Dec 31 2017 *)

%o (Magma) [(2*0^n + 3^n*Binomial(2*n, n))/3: n in [ 0..20]]; // _Vincenzo Librandi_, Nov 24 2012

%o (PARI) for(n=0, 30, print1((2*0^n + 3^n*binomial(2*n,n))/3, ", ")) \\ _G. C. Greubel_, Dec 31 2017

%Y Cf. A069723, A088218, A099045, A099046.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Sep 24 2004