a(n) = round(w^n/2/sqrt(5)) where w = (1+r)/(1-r) = 2.89005363826396... and r = sqrt(sqrt(5)-2) = 0.485868271756...; for n >= 3, a(n) =
A071101(n+3).
G.f.: -x*(x-1)*(1+x)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4). -
R. J. Mathar, Jun 03 2009
Define a Lucas sequence {U(n)} in the ring of Gaussian integers by the recurrence U(n) = (1 + i)*U(n-1) + U(n-2) with U(0) = 0 and U(1) = 1. Then a(n) = |U(n)|^2.
Let a, b denote the zeros of x^2 - (1 + i)*x - 1 and c, d denote the zeros of x^2 - (1 - i)*x - 1.
Then a(n) = (a^n - b^n)*(c^n - d^n)/((a - b)*(c - d)).
a(n) = (alpha(1)^n + beta(1)^n - alpha(2)^n - beta(2)^n)/(2*sqrt(5)), where alpha(1), beta(1) are the roots of x^2 - ( 1 + sqrt(5))*x + 1 = 0, and alpha(2), beta(2) are the roots of x^2 - (1 - sqrt(5))*x + 1 = 0.
The o.g.f. is the Hadamard product of the rational functions x/(1 - (1 + i)x - x^2) and x/(1 - (1 - i)x - x^2). (End)
a(n) = (1/sqrt(5))*(T(n,phi) - T(n,-1/phi)), where phi = 1/2*(1 + sqrt(5)) is the golden ratio and T(n,x) denotes the Chebyshev polynomial of the first kind. Compare with the Fibonacci numbers,
A000045, whose terms are given by the Binet formula 1/sqrt(5)*( phi^n - (-1/phi)^n ).
a(n) = top right (or bottom left) entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 1; 1, 1]; the off-diagonal elements of M^n give the sequence of Fibonacci numbers. Bottom right entry of the matrix T(n, M) gives
A138574. See the remarks in
A100047 for the general connection between Chebyshev polynomials and linear divisibility sequences of the fourth order. (End)
a(n) = (((phi + sqrt(phi))^n + (phi - sqrt(phi))^n)/2 - (-1)^n * cos(n*arctan(sqrt(phi))))/sqrt(5), where phi=(1+sqrt(5))/2. -
Vladimir Reshetnikov, May 11 2016
