OFFSET
1,3
COMMENTS
The continued fraction for floor(Fibonacci(n)*(1+sqrt(5))/2)/Fibonacci(n) consists of 1's except for the last term.
Also, continued fraction expansion of (33+4*sqrt(33))/33. - Bruno Berselli, Feb 15 2011
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).
FORMULA
a(1)=a(2)=1 followed by the 4-periodic sequence with period [2,3,2,2].
a(n) = ((n mod 4)+3*(n+1 mod 4)+3*(n+2 mod 4)+5*(n+3 mod 4))/8 - A019590(n). - Bruno Berselli, Feb 15 2011
a(n) = 2+(1+(-1)^n)*(1+i^n)/4 - A019590(n), where i=sqrt(-1). - Bruno Berselli, Mar 16 2011
From Colin Barker, Jun 29 2013: (Start)
a(n) = a(n-4) for n>6.
G.f.: -x*(x^5+x^4+3*x^3+2*x^2+x+1) / ((x-1)*(x+1)*(x^2+1)). (End)
MATHEMATICA
Table[ContinuedFraction[Floor[Fibonacci[n]*GoldenRatio]/Fibonacci[n]][[-1]], {n, 100}]
CoefficientList[Series[-x*(x^5 + x^4 + 3*x^3 + 2*x^2 + x + 1)/((x - 1)*(x + 1)*(x^2 + 1)), {x, 0, 50}], x] (* G. C. Greubel, Jun 20 2017 *)
PadRight[{1, 1}, 120, {2, 2, 2, 3}] (* Harvey P. Dale, Mar 24 2018 *)
PROG
(PARI) x='x+O('x^50); Vec(-x*(x^5+x^4+3*x^3+2*x^2+x+1)/((x-1)*(x+1)*(x^2+1))) \\ G. C. Greubel, Jun 20 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, Feb 15 2011
STATUS
approved