A103311 A transform of the Fibonacci numbers.

0, 1, 1, 0, -2, -5, -8, -8, 0, 21, 55, 89, 89, 0, -233, -610, -987, -987, 0, 2584, 6765, 10946, 10946, 0, -28657, -75025, -121393, -121393, 0, 317811, 832040, 1346269, 1346269, 0, -3524578, -9227465, -14930352, -14930352, 0, 39088169, 102334155, 165580141, 165580141, 0, -433494437, -1134903170

Offset: 0

Author

Paul Barry, Jan 30 2005

Keywords

• easy
• sign

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).

Crossrefs

Cf. A000045.

Programs

Mathematica

LinearRecurrence[{3,-4,2,-1},{0,1,1,0},50] (* Harvey P. Dale, May 03 2020 *)

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).