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%I #20 Apr 22 2014 02:49:52
%S 1,1,2,2,4,2,6,4,8,4,10,6,12,6,14,8,16,8,18,10,20,10,22,12,24,12,26,
%T 14,28,14,30,16,32,16,34,18,36,18,38,20,40,20,42,22,44,22,46,24,48,24,
%U 50,26,52,26,54,28,56,28,58,30,60,30,62,32,64,32,66,34,68,34
%N Length of the continued fraction for floor(Fibonacci(n)*(1+sqrt(5))/2) / Fibonacci(n).
%C A240828 is an essentially identical sequence. - _Bruno Berselli_, Apr 18 2014
%H Bruno Berselli, <a href="/A185048/b185048.txt">Table of n, a(n) for n = 1..20000</a>
%F a(1)=a(2)=1, for k>=2 we have a(2k)=2*floor(k/2) and a(2k-1)=2*k-2.
%F G.f.: x*(x^7+x^6-x^5+x^4+x^3+x^2+x+1) / ((x-1)^2*(x+1)^2*(x^2+1)). - _Colin Barker_, Jun 20 2013
%t Table[Length[ContinuedFraction[Floor[Fibonacci[n]*GoldenRatio]/Fibonacci[n]]], {n, 70}]
%o (PARI) a(n)=if(n<3,1,if(n%2,n-1,2*floor(n/4)))
%Y Cf. A000045, A001622, A240828.
%K nonn,easy
%O 1,3
%A _Benoit Cloitre_, Feb 15 2011
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