OFFSET
0,5
COMMENTS
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.
LINKS
Index entries for linear recurrences with constant coefficients, signature (3,-4,2,-1).
FORMULA
G.f.: x*(1-x)^2/(1 - 3*x + 4*x^2 - 2*x^3 + x^4);
a(n) = 3*a(n-1) - 4*a(n-2) + 2*a(n-3) - a(n-4);
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));
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).
MATHEMATICA
LinearRecurrence[{3, -4, 2, -1}, {0, 1, 1, 0}, 50] (* Harvey P. Dale, May 03 2020 *)
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Paul Barry, Jan 30 2005
STATUS
approved