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a(n) = Catalan(n) + 2^n - 0^n.
2

%I #15 Mar 19 2018 05:41:36

%S 1,3,6,13,30,74,196,557,1686,5374,17820,60834,212108,751092,2690824,

%T 9727613,35423206,129775862,477900844,1767787478,6565168996,

%U 24468364172,91486757944,343068002258,1289920924540,4861979955884

%N a(n) = Catalan(n) + 2^n - 0^n.

%C Hankel transform is A141354.

%F G.f.: c(x)+2x/(1-2x), where c(x) is the g.f. of A000108. [corrected by _Paul Barry_, Oct 18 2010]

%F Conjecture: (n+1)*a(n) + 2*(-4*n+1)*a(n-1) + 4*(5*n-7)*a(n-2) + 8*(-2*n+5)*a(n-3) = 0. - _R. J. Mathar_, Nov 15 2012

%t f[n_] := Binomial[2n, n]/(n + 1) + 2^n - 0^n; f[0] = 1; Array[f, 29, 0] (* or *)

%t CoefficientList[ Series[1 + 1/2 (-4 + 2/(1 - 2x) + (1 - Sqrt[1 - 4x])/x), {x, 0, 28}], x] (* _Robert G. Wilson v_, Mar 18 2018 *)

%o (PARI) a(n) = binomial(2*n,n)/(n+1) + 2^n - 0^n; \\ _Michel Marcus_, Mar 18 2018

%Y Cf. A000108 (Catalan numbers), A141351.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Jun 27 2008