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A213809 Position of the maximum element in the simple continued fraction of Fibonacci(n+1)^5/Fibonacci(n)^5.

%I A213809 #9 Jul 06 2012 09:40:10
%S A213809 1,1,1,1,3,3,3,3,3,5,5,3,5,5,5,5,5,5,5,7,7,5,7,7,7,7,7,7,7,9,9,7,9,9,
%T A213809 9,9,9,9,9,11,11,9,11,11,11,11,11,11,11,13,13,11,13,13,13,13,13,13,13,
%U A213809 15,15,13,15,15,15,15,15,15,15,17,17,15,17,17,17,17,17,17,17,19,19,17,19,19,19,19,19,19,19,21,21,19,21,21,21,21,21,21,21,23,23,21,23,23,23,23,23,23,23
%N A213809 Position of the maximum element in the simple continued fraction of Fibonacci(n+1)^5/Fibonacci(n)^5.
%C A213809 The maximum elements themselves are in A113506.
%C A213809 The fractions F(n+1)^5/F(n)^5 are 1, 32, 243/32, 3125/243,... (see A056572). The continued fractions are [1], [32], [7,1,1,2,6], [12,1,6,6,1,4], [10,2,17,17,1,4]..., and for the first 4 of these the maximum element is the first, for the 5th of these the maximum element is the third.
%F A213809 a(10k+m)=3+2k if m=0,1,3,4,5,6,7,8,9, k>0.
%F A213809 a(10k+2)=1+2k, k>0.
%e A213809 The continued fraction of the fraction corresponding to [1,1,1,1,1,1,1,1,1,1,1,1,1]^5 is
%e A213809 [11,11,7,1,39282,2,5,11,11,1,11,11] and the maximum occurs at place 5, which according to the formula, should be 3+2k, and since 13=10k+3, k=1 and 3+2k=3+2=5.
%p A213809 A213809 := proc(n)
%p A213809         local c,a,i;
%p A213809         (combinat[fibonacci](n+1)/combinat[fibonacci](n))^5 ;
%p A213809         c := numtheory[cfrac](%,quotients) ;
%p A213809         a := 1 ;
%p A213809         for i from 2 to nops(c) do
%p A213809                 if op(i,c) > op(a,c) then
%p A213809                         a := i ;
%p A213809                 end if;
%p A213809         end do:
%p A213809         a ;
%p A213809 end proc: # _R. J. Mathar_, Jul 06 2012
%K A213809 nonn
%O A213809 1,5
%A A213809 _Art DuPre_, Jun 20 2012