A113506 -id:A113506 - cp's OEIS frontend

A213358 a(n) = maximum(continuedfraction(F(n+1)^n/F(n)^n)) - L(n) + (1-(-1)^n)/2, where F(n) is Fibonacci(n) and L(n) is the n-th Lucas number.

1, 1, 0, 4, 7, 16, 0, 0, 397, 17, 0, 8047, 0, 0, 0, 0, 0, 0, 0, 6633, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Art DuPre, Jun 10 2012

nonn

This is a variation on A113506 and A113500.
If we consider the sequence of quotients [1,1,1,1] of a simple continued fraction, we can express this by the continued fraction of F(4+1)/F(4), where we have chosen not to write the last quotient as 2 but as 1,1.
It is a very interesting fact that if we take the fifth power of F(n+1)/F(n), for varying n, we very soon observe extremely large values in its continued fraction expansion. That's why Benoit Cloitre, author of A113560, takes vecmax, the max of all these quotients, in order to capture the large rate of growth of these quotients.  It would be better, in order to emphasize the largeness of these big quotients, to somehow quantify the ratio between the max and the average of all the quotients, or the average of all the others.
Since Cloitre found a Fibonacci-like behavior of the vecmax, I thought to replace the power 5 by the power n, thus getting the sequence vecmax(contfrac(F(n+1)^n/F(n)^n)) instead of vecmax(contfrac(F(n+1)^5/F(n)^5)). After I noticed its behavior was almost that of the Lucas sequence, I took the sequence which subtracted the almost-Lucas sequence from it, thus eventually getting 0 after the 20th term.

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A113506, A113500.

Table[Max[ContinuedFraction[Fibonacci[n+1]^n/Fibonacci[n]^n]] - LucasL[n] + (1 - (-1)^n)/2, {n, 200}] (* T. D. Noe, Jun 28 2012 *)

F(n) = fibonacci(n);
L(n) = F(n-1) + F(n+1);
a(n) = vecmax(contfrac(F(n+1)^n/F(n)^n)) - L(n) + (1-(-1)^n)/2;

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

Original entry on oeis.org

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

Art DuPre, Jun 20 2012

nonn

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.

			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.

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

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.

cp's OEIS Frontend

A213358 a(n) = maximum(continuedfraction(F(n+1)^n/F(n)^n)) - L(n) + (1-(-1)^n)/2, where F(n) is Fibonacci(n) and L(n) is the n-th Lucas number.

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

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