A051792 a(n) = (-1)^(n-1)*(a(n-1) - a(n-2)), a(1)=1, a(2)=1.

1, 1, 0, 1, 1, 0, -1, 1, 2, -1, -3, 2, 5, -3, -8, 5, 13, -8, -21, 13, 34, -21, -55, 34, 89, -55, -144, 89, 233, -144, -377, 233, 610, -377, -987, 610, 1597, -987, -2584, 1597, 4181, -2584, -6765, 4181, 10946, -6765, -17711, 10946, 28657, -17711, -46368

Offset

1,9

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for two-way infinite sequences

Index entries for linear recurrences with constant coefficients, signature (0,-1,0,1).

Formula

a(3-n) = A053602(n).

From Michael Somos: (Start)

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

a(n) = -a(n-2) + a(n-4). (End)

a(n) = b(n-1) + b(n-2) + b(n-3) + 2*b(n-4), where b(n) = i^n * A079977(n). - G. C. Greubel, Dec 06 2022

Mathematica

LinearRecurrence[{0, -1, 0, 1}, {1, 1, 0, 1}, 60] (* Harvey P. Dale, May 08 2017 *)

Prog

(PARI) a(n)=fibonacci((3-n)\2+(3-n)%2*2)
(Sage)
def A051792():
    x, y, b = 1, 1, true
    while True:
        yield x
        x, y = y, x - y
        y = -y if b else y
        b = not b
a = A051792()
print([next(a) for _ in range(51)]) # Peter Luschny, Mar 19 2020
(Magma) [Fibonacci(1 -Floor((n-4)/2) -2*((n-4) mod 2)): n in [1..60]]; // G. C. Greubel, Dec 06 2022

Crossrefs

Cf. A000045, A053602.

Sequence in context: A132091 A262090 A239881 * A053602 A272912 A123231

Adjacent sequences: A051789 A051790 A051791 * A051793 A051794 A051795

Keyword

easy,sign

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 10 1999

Status

approved