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A156906
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Transform of Fibonacci(n+1) with Hankel transform (-1)^binomial(n+1,2) * Fibonacci(n+1).
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1
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1, 0, 1, -1, 0, 0, 2, -2, -3, 3, 11, -11, -31, 31, 101, -101, -328, 328, 1102, -1102, -3760, 3760, 13036, -13036, -45750, 45750, 162262, -162262, -580638, 580638, 2093802, -2093802, -7601043, 7601043, 27756627, -27756627, -101888163
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OFFSET
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0,7
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COMMENTS
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Hankel transform is (-1)^binomial(n+1,2) * Fibonacci(n+1).
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}.
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LINKS
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FORMULA
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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;
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);
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.
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.
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
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
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MAPLE
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cx := (1-sqrt(1-4*x))/2/x ;
1+x^2*subs(x=-x^2, cx)/(1+x) ;
coeftayl(%, x=0, n) ;
end proc:
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MATHEMATICA
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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 *)
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PROG
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(Sage)
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)) )
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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STATUS
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approved
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