%I #13 May 03 2020 16:05:44

%S 0,1,1,0,-2,-5,-8,-8,0,21,55,89,89,0,-233,-610,-987,-987,0,2584,6765,

%T 10946,10946,0,-28657,-75025,-121393,-121393,0,317811,832040,1346269,

%U 1346269,0,-3524578,-9227465,-14930352,-14930352,0,39088169,102334155,165580141,165580141,0,-433494437,-1134903170

%N A transform of the Fibonacci numbers.

%C Apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x), -x/(1-x)) (expressed as Riordan arrays) to -Fibonacci(n). Conjecture: all elements in absolute value are Fibonacci numbers.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,2,-1).

%F G.f.: x*(1-x)^2/(1 - 3*x + 4*x^2 - 2*x^3 + x^4);

%F a(n) = 3*a(n-1) - 4*a(n-2) + 2*a(n-3) - a(n-4);

%F a(n) = (sqrt(5)/2 - 1/2)^n*(sqrt(2*sqrt(5)/25 + 1/5)*sin(2*Pi*n/5) - sqrt(5)*cos(2*Pi*n/5)/5) + (sqrt(5)/2 + 1/2)^n*(sqrt(5)*cos(Pi*n/5)/5 + sqrt(1/5 - 2*sqrt(5)/25)*sin(Pi*n/5));

%F a(n) = -Sum_{j=0..n} (-1)^j*binomial(n, j)*Sum_{k=0..floor(j/2)} (-1)^k*binomial(n-k, k)*Fibonacci(j-2*k).

%t LinearRecurrence[{3,-4,2,-1},{0,1,1,0},50] (* _Harvey P. Dale_, May 03 2020 *)

%Y Cf. A000045.

%K easy,sign

%O 0,5

%A _Paul Barry_, Jan 30 2005