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%I #20 Jun 09 2017 20:55:20
%S 1,2,12,80,616,5040,43296,384384,3500640,32517056,306896512,
%T 2934597120,28369508608,276810483200,2722537128960,26963147796480,
%U 268659456837120,2691301381401600,27089160416102400,273833161582632960,2778754123765002240,28296326851107594240
%N a(n) = A000108(n)*A002605(n+1), where A000108 are the Catalan numbers.
%C More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, S(0)=1, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
%C Conjecture: +n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -8*(2*n-1)*(2*n-3)*a(n-2)=0. - _R. J. Mathar_, Oct 08 2016
%H G. C. Greubel, <a href="/A185020/b185020.txt">Table of n, a(n) for n = 0..960</a>
%F G.f.: sqrt( (1-4*x - sqrt(1-8*x-32*x^2))/24 )/x.
%e G.f.: A(x) = 1 + 1*2*x + 2*6*x^2 + 5*16*x^3 + 14*44*x^4 + 42*120*x^5 + 132*328*x^6 +...+ A000108(n)*A002605(n+1)*x^n +...
%t MapIndexed[CatalanNumber[#2 - 1] #1 &, Rest@ RecurrenceTable[{a[n] == 2 (a[n - 1] + a[n - 2]), a[0] == 0, a[1] == 1}, a, {n, 22}]] // Flatten (* or *)
%t CoefficientList[Series[Sqrt[(1 - 4 x - Sqrt[1 - 8 x - 32 x^2])/24]/x, {x, 0, 21}], x] (* _Michael De Vlieger_, Oct 08 2016 *)
%o (PARI) {A000108(n)=binomial(2*n,n)/(n+1)}
%o {A002605(n)=polcoeff(x/(1-2*x-2*x^2 +x*O(x^n)),n)}
%o {a(n)=A000108(n)*A002605(n+1)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A002605.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Dec 26 2012
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