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A185020 a(n) = A000108(n)*A002605(n+1), where A000108 are the Catalan numbers.

%I A185020 #24 Dec 14 2024 07:08:41
%S A185020 1,2,12,80,616,5040,43296,384384,3500640,32517056,306896512,
%T A185020 2934597120,28369508608,276810483200,2722537128960,26963147796480,
%U A185020 268659456837120,2691301381401600,27089160416102400,273833161582632960,2778754123765002240,28296326851107594240
%N A185020 a(n) = A000108(n)*A002605(n+1), where A000108 are the Catalan numbers.
%C A185020 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 A185020 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 A185020 G. C. Greubel, <a href="/A185020/b185020.txt">Table of n, a(n) for n = 0..960</a>
%F A185020 G.f.: sqrt( (1-4*x - sqrt(1-8*x-32*x^2))/24 )/x.
%F A185020 G.f. A(x) satisfies A(x) = sqrt( 1 + 4*x*A(x)^2 + 12*x^2*A(x)^4 ). - _Paul D. Hanna_, Dec 14 2024
%e A185020 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 A185020 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 A185020 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 A185020 (PARI) {A000108(n) = binomial(2*n,n)/(n+1)}
%o A185020 {A002605(n) = polcoef(x/(1-2*x-2*x^2 +x*O(x^n)), n)}
%o A185020 {a(n)=A000108(n)*A002605(n+1)}
%o A185020 for(n=0,30,print1(a(n),", "))
%Y A185020 Cf. A002605.
%K A185020 nonn
%O A185020 0,2
%A A185020 _Paul D. Hanna_, Dec 26 2012