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%I #21 Jun 15 2021 02:19:47
%S 1,0,1,-1,0,0,2,-2,-3,3,11,-11,-31,31,101,-101,-328,328,1102,-1102,
%T -3760,3760,13036,-13036,-45750,45750,162262,-162262,-580638,580638,
%U 2093802,-2093802,-7601043,7601043,27756627,-27756627,-101888163
%N Transform of Fibonacci(n+1) with Hankel transform (-1)^binomial(n+1,2) * Fibonacci(n+1).
%C Hankel transform is (-1)^binomial(n+1,2) * Fibonacci(n+1).
%C 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 G. C. Greubel, <a href="/A156906/b156906.txt">Table of n, a(n) for n = 0..1000</a>
%F 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 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 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 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 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 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 cx := (1-sqrt(1-4*x))/2/x ;
%p A156906 := proc(n)
%p 1+x^2*subs(x=-x^2,cx)/(1+x) ;
%p coeftayl(%,x=0,n) ;
%p end proc:
%p seq(A156906(n), n=0..40); # _R. J. Mathar_, Jul 28 2016
%t 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 (Sage)
%o 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(n) for n in (0..40)] # _G. C. Greubel_, Jun 14 2021
%Y Cf. A000045, A000108, A055587.
%K easy,sign
%O 0,7
%A _Paul Barry_, Feb 17 2009
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