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

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, 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, 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

Offset: 1

Author

Art DuPre, Jun 20 2012

Keywords

• nonn

Comments

The maximum elements themselves are in A113506.

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.

Examples

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

Programs

Maple

A213809 := proc(n)
        local c,a,i;
        (combinat[fibonacci](n+1)/combinat[fibonacci](n))^5 ;
        c := numtheory[cfrac](%,quotients) ;
        a := 1 ;
        for i from 2 to nops(c) do
                if op(i,c) > op(a,c) then
                        a := i ;
                end if;
        end do:
        a ;
end proc: # R. J. Mathar, Jul 06 2012

Formula

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

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