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A156906 Transform of Fibonacci(n+1) with Hankel transform (-1)^binomial(n+1,2) * Fibonacci(n+1).

%I A156906 #25 Feb 04 2025 09:00:20
%S A156906 1,0,1,-1,0,0,2,-2,-3,3,11,-11,-31,31,101,-101,-328,328,1102,-1102,
%T A156906 -3760,3760,13036,-13036,-45750,45750,162262,-162262,-580638,580638,
%U A156906 2093802,-2093802,-7601043,7601043,27756627,-27756627,-101888163
%N A156906 Transform of Fibonacci(n+1) with Hankel transform (-1)^binomial(n+1,2) * Fibonacci(n+1).
%C A156906 Hankel transform is (-1)^binomial(n+1,2) * Fibonacci(n+1).
%C A156906 Image of Fibonacci(n+1) by the Riordan array (1/(1 + x*c(-x^2)), x*c(-x^2)/(1 + x*c(-x^2))) =  (1/(1-x), x*(1-x)/(1-2*x))^{-1} = A055587^{-1}.
%H A156906 G. C. Greubel, <a href="/A156906/b156906.txt">Table of n, a(n) for n = 0..1000</a>
%F A156906 G.f.: (1 + 2*x + sqrt(1+4*x^2))/(2*(1+x)) = 1 + x^2*c(-x^2)/(1+x) where c(x) is the g.f. of A000108;
%F A156906 G.f.: 1/(1 - x^2/(1 + x + 2*x^2/(1 - x/2 + 3*x^2/4/(1 + x/6 + 10*x^2/9/(1 - x/15 + 24/25*x^2/(1 + ...)))))) (continued fraction);
%F A156906 In the continued fraction expansion of the g.f. the general term is 1 - x*if(n=0, 0, (-1)^n/(Fibonacci(n)*Fibonacci(n+1))) + x^2*(-0^n + Fibonacci(n)*Fibonacci(n+2) )/Fibonacci(n+1)^2.
%F A156906 a(n) = Sum_{k=0..n} (-1)^(n-k)*b(k), where b(n) = binomial(1,n) -A000108((n-2)/2) * (-1)^(n/2) * (1+(-1)^n)/2 and b(0) = 1.
%F A156906 n*a(n) = -n*a(n-1) -4*(n-3)*a(n-2) -4*(n-3)*a(n-3). - _R. J. Mathar_, Nov 14 2011
%F A156906 a(n) = (1/2)*( 2*[n=0] - (-1)^n + Sum_{j=0..floor(n/2)} (-1)^(n+j+1)*binomial(2*j, j)/(2*j-1) ). - _G. C. Greubel_, Jun 14 2021
%p A156906 cx := (1-sqrt(1-4*x))/2/x ;
%p A156906 A156906 := proc(n)
%p A156906         1+x^2*subs(x=-x^2,cx)/(1+x) ;
%p A156906         coeftayl(%,x=0,n) ;
%p A156906 end proc:
%p A156906 seq(A156906(n), n=0..40);  # _R. J. Mathar_, Jul 28 2016
%t A156906 a[n_]:= (1/2)*(2*Boole[n==0] -(-1)^n + Sum[(-1)^(n+j+1)*Binomial[2*j, j]/(2*j-1), {j, 0, Floor[n/2]}]); Table[a[n], {n, 0, 60}] (* _G. C. Greubel_, Jun 14 2021 *)
%o A156906 (Sage)
%o A156906 def A156906(n): return (1/2)*( 2*bool(n==0) - (-1)^n + sum( (-1)^(n+j+1)*binomial( 2*j, j)/(2*j-1) for j in (0..n//2)) )
%o A156906 [A156906(n) for n in (0..40)] # _G. C. Greubel_, Jun 14 2021
%Y A156906 Cf. A000045, A000108, A055587.
%K A156906 easy,sign
%O A156906 0,7
%A A156906 _Paul Barry_, Feb 17 2009