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%I #29 Apr 12 2023 11:23:35
%S 1,2,14,100,854,7644,72204,703560,7037030,71772844,743844452,
%T 7810307960,82909630972,888316731800,9593823377880,104332819539600,
%U 1141523825614470,12556761952114380,138785264158902900,1540516430396559000,17165754516697206420,191944345934966132040
%N a(n) = A000108(n)*A015518(n+1), where A000108 are the Catalan numbers and A015518(n) = 2*A015518(n-1) + 3*A015518(n-2).
%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
%C 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.
%H G. C. Greubel, <a href="/A185010/b185010.txt">Table of n, a(n) for n = 0..925</a>
%H S. B. Ekhad and M. Yang, <a href="http://sites.math.rutgers.edu/~zeilberg/tokhniot/oMathar1maple12.txt">Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences</a>, (2017).
%F G.f.: sqrt( (1-4*x - sqrt(1-8*x-48*x^2))/32 )/x.
%F G.f.: sqrt( M(4*x) ), where M(x) is g.f. of A001006. - _Werner Schulte_, Aug 10 2015
%F Conjecture: n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -12*(2*n-1)*(2*n-3)*a(n-2)=0. - _R. J. Mathar_, Oct 08 2016
%F G.f: B(m(4z)/4), where B(x) is the g.f. of A000984 and m(x) is the g.f. of A086246. - _Alexander Burstein_, May 20 2021
%e G.f.: A(x) = 1 + 1*2*x + 2*7*x^2 + 5*20*x^3 + 14*61*x^4 + 42*182*x^5 + 132*547*x^6 +...+ A000108(n)*A015518(n+1)*x^n +...
%t CoefficientList[Series[Sqrt[(1 - 4*x - Sqrt[1 - 8*x - 48*x^2])/32]/x, {x, 0, 50}], x] (* _G. C. Greubel_, Jun 09 2017 *)
%o (PARI) {A000108(n)=binomial(2*n,n)/(n+1)}
%o {A015518(n)=polcoeff(x/(1-2*x-3*x^2 +x*O(x^n)),n)}
%o {a(n)=A000108(n)*A015518(n+1)}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A000108, A001006, A015518.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Dec 26 2012
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