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A100240 G.f. A(x) satisfies: 4^n/2 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: ((4+z)^n + z^n)/2 = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.

%I A100240 #11 Feb 07 2021 04:57:01
%S A100240 1,1,2,2,0,-4,-6,2,22,30,-26,-154,-172,288,1190,990,-3040,-9620,-4970,
%T A100240 31350,79120,12580,-318210,-649610,174150,3185686,5233514,-4273078,
%U A100240 -31452228,-40495600,64593386,305819154,290278982,-835918098,-2921409370,-1771072346,9995237616,27317409988
%N A100240 G.f. A(x) satisfies: 4^n/2 = Sum_{k=0..n} [x^k]A(x)^n and also satisfies: ((4+z)^n + z^n)/2 = Sum_{k=0..n} [x^k](A(x)+z*x)^n for all z, where [x^k]A(x)^n denotes the coefficient of x^k in A(x)^n.
%F A100240 G.f.: A(x) = 2*x+sqrt(1-2*x+5*x^2).
%F A100240 Recurrence: n*a(n) = (2*n-3)*a(n-1) - 5*(n-3)*a(n-2). - _Vaclav Kotesovec_, Feb 07 2021
%e A100240 From the table of powers of A(x), we see that
%e A100240 4^n/2 = Sum of coefficients [x^0] through [x^n] in A(x)^n:
%e A100240 A^1=[1,1],2,2,0,-4,-6,2,22,30,-26,...
%e A100240 A^2=[1,2,5],8,8,0,-16,-24,8,88,120,...
%e A100240 A^3=[1,3,9,19],30,30,2,-54,-84,20,288,...
%e A100240 A^4=[1,4,14,36,73],112,112,16,-176,-288,32,...
%e A100240 A^5=[1,5,20,60,145,281],420,420,90,-570,-988,...
%e A100240 A^6=[1,6,27,92,255,582,1085],1584,1584,440,-1848,...
%e A100240 A^7=[1,7,35,133,413,1071,2331,4201],6006,6006,2002,...
%e A100240 A^8=[1,8,44,184,630,1816,4460,9320,16305],22880,22880,...
%e A100240 the main diagonal of which is:
%e A100240 [x^n]A(x)^(n+1) = (n+1)*A081696(n) for n>=0.
%t A100240 CoefficientList[Series[2*x + Sqrt[1 - 2*x + 5*x^2], {x, 0, 40}], x] (* _Vaclav Kotesovec_, Feb 07 2021 *)
%o A100240 (PARI) a(n)=if(n==0,1,(4^n/2-sum(k=0,n,polcoeff(sum(j=0,min(k,n-1),a(j)*x^j)^n+x*O(x^k),k)))/n)
%o A100240 (PARI) a(n)=polcoeff(2*x+sqrt(1-2*x+5*x^2+x^2*O(x^n)),n)
%Y A100240 Cf. A100223, A057083, A007440, A081696.
%K A100240 sign
%O A100240 0,3
%A A100240 _Paul D. Hanna_, Nov 30 2004