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%I #21 Dec 08 2021 10:54:01
%S 1,5,10,45,190,930,4660,24445,131190,719830,4013260,22684370,
%T 129661740,748252580,4353379560,25508284445,150392391590,891549228430,
%U 5310994644060,31775749689670,190860711108740,1150473009844380
%N Expansion of (1 - sqrt(1 - 4x - 16x^2))/(2x).
%C Image of c(x), the g.f. of the Catalan numbers A000108 under the mapping g(x) -> (1+4x)g(x(1+4x)). In general, the image of the Catalan numbers under the mapping g(x)->(1+i*x)g(x(1+i*x)) is given by a(n) = Sum_{k=0..n} i^(n-k)C(k)C(k+1,n-k).
%C More generally, the sequence C for which C(0)=a, C(1)=b and C(n+1) = sum(C(k)*C(n-k),k=0..n) has the following g.f. f: f(z) = (1-sqrt(1-4*z*(a-(a^2-b)*z)))/(2*z). We obtain: C(n)=(sum(-1)^(p-1)*2^{n-p}a^{n-2*p-1}*(a^2-b)^p*((2*n-2*p-1)*...*5*3*1/(p!*(n-2*p+1)!)),p=0..floor((n+1)/2)). By following Comtet [Analyse Combinatoire Tomes 1 et 2, PUF, Paris 1970], we obtain also: (n+1)*C(n) - 2*a*(2*n-1)*C(n-1) + 4*(n-2)*(a^2-b)*C(n-2) = 0. - _Richard Choulet_, Dec 17 2009
%H Vincenzo Librandi, <a href="/A103971/b103971.txt">Table of n, a(n) for n = 0..200</a>
%F G.f.: (1-sqrt(1-4x(1+4x))/(2x);
%F a(n) = Sum_{k=0..n} 4^(n-k)*C(k)*C(k+1, n-k).
%F Another recurrence formula: (n+1)*a(n) = 2*(2*n-1)*a(n-1) + 16*(n-2)*a(n-2). - _Richard Choulet_, Dec 17 2009
%F a(n) ~ sqrt(10 + 2*sqrt(5))*(2 + 2*sqrt(5))^n/(2*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 17 2012
%F Equivalently, a(n) ~ 5^(1/4) * 2^(2*n) * phi^(n + 1/2) / (sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Dec 08 2021
%p n:=30:a(0):=1:a(1):=5: for k from 1 to n do a(k+1):=sum('a(p)*a(k-p)','p'=0..k):od:seq(a(k),k=0..n); # _Richard Choulet_, Dec 17 2009
%t CoefficientList[Series[(1-Sqrt[1-4x-16x^2])/(2x),{x,0,30}],x] (* _Harvey P. Dale_, Apr 02 2012 *)
%Y Cf. A000108, A025227, A025228, A025229, A025230, A025231, A025232. - _Richard Choulet_, Dec 17 2009
%K easy,nonn
%O 0,2
%A _Paul Barry_, Feb 23 2005
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