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%I #23 Aug 04 2019 18:22:52
%S 1,1,4,18,86,427,2180,11373,60380,325259,1773842,9776637,54380144,
%T 304905223,1721650832,9782051362,55888463214,320898932595,
%U 1850762866662,10717217871255,62287285235230,363212668363520,2124430957852380
%N G.f.: A(x) = F(x*F(x)^2) = F(F(x)-1) where F(x) = 1 + x*F(x)^2 is the g.f. of A000108 (Catalan).
%C Ignoring a(0), the series reversal gives A030267 with alternating signs: 1, -4, 14, -46, 145, -444, ... - _Vladimir Reshetnikov_, Aug 03 2019
%F a(n) = Sum_{k=0..n} C(2k+1,k)/(2k+1) * C(2n,n-k)*k/n for n>0 with a(0)=1.
%F G.f.: A(x) = [1 - sqrt(5 - 4*F(x))]/(2*F(x)-2) where F(x) = (1-sqrt(1-4x))/(2x).
%F G.f. satisfies: A(x) = 1 + x*F(x)^2*A(x)^2 where F(x) is the g.f. of A000108.
%F G.f. satisfies: A(x*G(x)) = F(x*G(x)^3) = F(G(x)-1) where G(x) = F(x*G(x)) is the g.f. of A001764 and F(x) is the g.f. of A000108.
%F For n > 0, a(n) = 2^(n-1)*(2*n-1)!!*((hypergeom([-1/2,-n-1], [n], -4) - 1)/(n+1)! + 2*hypergeom([1/2,-n], [n+1], -4)/(n*n!)). - _Vladimir Reshetnikov_, Nov 07 2015
%F a(n) ~ 5^(2*n + 1/2) / (sqrt(3*Pi) * n^(3/2) * 4^n). - _Vaclav Kotesovec_, Nov 07 2015
%F a(n) = (A243585(n) - A007856(n+1)) / n for n >= 1. - _Peter Luschny_, Aug 04 2019
%e G.f.: A(x) = F(x*F(x)^2) = 1 + x + 4*x^2 + 18*x^3 + 86*x^4 +... where
%e F(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 +...
%e F(x)^2 = 1 + 2*x + 5*x^2 + 14*x^3 + 42*x^4 + 132*x^5 + 429*x^6 +...
%e A(x)^2 = 1 + 2*x + 9*x^2 + 44*x^3 + 224*x^4 + 1170*x^5 + 6226*x^6 +...
%e F(x)^2*A(x)^2 = 1 + 4*x + 18*x^2 + 86*x^3 + 427*x^4 + 2180*x^5 +...
%e From _Peter Bala_, Jul 21 2015: (Start)
%e Let B(x) = (A(x) - 1)/x = Sum_{n >= 0} a(n+1)*x^n. Then 1 + x*B'(x)/B (x) = 1 + 4*x + 20*x^2 + 106*x^3 + ... is the o.g.f. for A243585.
%e x*sqrt(B(x)) = x + 2*x^2 + 7*x^3 + 29*x^4 + ... is the o.g.f for A007582. (End)
%t a[0] = 1; a[n_] := 2^(n-1) (2n-1)!! ((Hypergeometric2F1[-1/2, -n-1, n, -4] - 1)/(n+1)! + 2 Hypergeometric2F1[1/2, -n, n+1, -4]/(n n!)); Table[a[n], {n, 0, 20}] (* _Vladimir Reshetnikov_, Nov 07 2015 *)
%t Flatten[{1, Table[Sum[Binomial[2*k + 1, k]/(2*k + 1)*Binomial[2*(n-k) + 2*k, n-k]*2*k/(2*(n-k) + 2*k), {k, 0, n}], {n, 1, 20}]}] (* _Vaclav Kotesovec_, Nov 07 2015 *)
%t A153294[0] := 1; A153294[n_] := (A243585[n] - A007856[n+1])/n;
%t Table[A153294[n], {n, 0, 22}] (* _Peter Luschny_, Aug 04 2019 *)
%o (PARI) {a(n)=if(n==0,1,sum(k=0,n,binomial(2*k+1,k)/(2*k+1)*binomial(2*(n-k)+2*k,n-k)*2*k/(2*(n-k)+2*k)))}
%Y Cf. A001764, A000108; A030267, A153293, A153295, A007852, A007856, A243585.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 15 2009
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