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%I #12 May 13 2013 11:18:32
%S 1,6,12,60,264,1392,7392,41424,236640,1384512,8224896,49554816,
%T 301884672,1856878080,11514915840,71915838720,451938731520,
%U 2855705994240,18132621772800,115637702461440,740356410961920,4756888756101120,30662391191715840,198229520200704000,1285001080928845824
%N Expansion of (1-sqrt(1-4x-20x^2))/(2x).
%C Image of c(x), the g.f. of the Catalan numbers A000108 under the mapping g(x)->(1+5x)g(x(1+5x)). 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 L. 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="/A103972/b103972.txt">Table of n, a(n) for n = 0..200</a>
%F G.f. : (1-sqrt(1-4x(1+5x))/(2x); a(n)=sum{k=0..n, 5^(n-k)C(k)C(k+1, n-k)}.
%F Another recurrence formula: (n+1)*a(n)=2*(2n-1)*a(n-1)+20*(n-2)*a(n-2). [_Richard Choulet_, Dec 17 2009]
%F a(n) ~ sqrt(12+2*sqrt(6))*(2+2*sqrt(6))^n/(2*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 17 2012
%p n:=30:a(0):=1:a(1):=6 :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[1-4*x-20*x^2])/(2*x), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 17 2012 *)
%o (PARI) x='x+O('x^66); Vec((1-sqrt(1-4*x-20*x^2))/(2*x)) \\ _Joerg Arndt_, May 13 2013
%Y Cf. A000108, A025227, A025228, A025229, A025230, A025231, A025232. [From _Richard Choulet_, Dec 17 2009]
%K easy,nonn
%O 0,2
%A _Paul Barry_, Feb 23 2005
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