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 A185049 Last term in the continued fraction for floor(Fibonacci(n)*(1+sqrt(5))/2) / Fibonacci(n). 1
 1, 1, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3 (list; graph; refs; listen; history; text; internal format)
 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 Sequence in context: A323761 A078832 A086410 * A186181 A324983 A147561 Adjacent sequences: A185046 A185047 A185048 * A185050 A185051 A185052 KEYWORD nonn,easy,changed AUTHOR Benoit Cloitre, Feb 15 2011 STATUS approved

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Last modified March 25 00:24 EDT 2023. Contains 361511 sequences. (Running on oeis4.)