A213809 - OEIS

%I #9 Jul 06 2012 09:40:10

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

%C The maximum elements themselves are in A113506.

%C 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 a(10k+m)=3+2k if m=0,1,3,4,5,6,7,8,9, k>0.

%F a(10k+2)=1+2k, k>0.

%e The continued fraction of the fraction corresponding to [1,1,1,1,1,1,1,1,1,1,1,1,1]^5 is

%e [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 := proc(n)

%p         local c,a,i;

%p         (combinat[fibonacci](n+1)/combinat[fibonacci](n))^5 ;

%p         c := numtheory[cfrac](%,quotients) ;

%p         a := 1 ;

%p         for i from 2 to nops(c) do

%p                 if op(i,c) > op(a,c) then

%p                         a := i ;

%p                 end if;

%p         end do:

%p         a ;

%p end proc: # _R. J. Mathar_, Jul 06 2012

%K nonn

%O 1,5

%A _Art DuPre_, Jun 20 2012