A093148 a(n) = gcd(Fibonacci(n+5), Fibonacci(n+1)).

1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1, 3, 1

Offset

0,4

Comments

From Klaus Brockhaus, May 30 2010: (Start)

Periodic sequence: Repeat [1, 1, 1, 3].

Continued fraction expansion of (9+sqrt(165))/14.

Decimal expansion of 371/3333. (End)

Final nonzero digit of n^n in base 4. - José María Grau Ribas, Jan 19 2012

Links

Table of n, a(n) for n=0..104.

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

Formula

G.f.: (1+x+x^2+3*x^3)/(1-x^4); a(n) = 3/2-sin(Pi*n/2)-cos(Pi*n)/2.

From Klaus Brockhaus, May 30 2010: (Start)

a(n) = a(n-4) for n > 3; a(0) = a(1) = a(2) = 1, a(3) = 3.

a(n) = (3-(-1)^n+(1-(-1)^n)*i*i^n)/2 where i = sqrt(-1). (End)

a(n) = 1 + 2*0^mod(n+1, 4). - Wesley Ivan Hurt, Oct 23 2014

Maple

A093148:=n->1+2*0^(n+1 mod 4): seq(A093148(n), n=0..100); # Wesley Ivan Hurt, Oct 23 2014

Mathematica

f[n_] := Switch[Mod[n, 4], 0, 1, 1, 1, 2, 1, 3, 3]; Array[f, 105, 0] (* Robert G. Wilson v, Jan 23 2012 *)
PadRight[{}, 120, {1, 1, 1, 3}] (* Harvey P. Dale, Sep 03 2021 *)

Prog

(Magma) [1+2*0^((n+1) mod 4) : n in [0..100]]; // Wesley Ivan Hurt, Oct 23 2014
(PARI) a(n)=if(n%4==3, 3, 1) \\ Charles R Greathouse IV, Oct 07 2015

Crossrefs

Cf. A000045, A061347, A167816, A177704, A178591.

Sequence in context: A339877 A030328 A176563 * A069292 A368336 A365633

Adjacent sequences: A093145 A093146 A093147 * A093149 A093150 A093151

Keyword

easy,nonn

Author

Paul Barry, Apr 02 2004

Status

approved