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

Author

Art DuPre, Jun 10 2012

Keywords

• nonn

Comments

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.

Links

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

Crossrefs

Cf. A113506, A113500.

Programs

Mathematica

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

PARI

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;