# [seqfan] Re: rational sequence $$f_n= (\frac{n}{f_{n-1}}+1)(n+1) \; , f_0 = 1$$

Max Alekseyev maxale at gmail.com
Thu Mar 26 14:58:10 CET 2009

Let g(n) = f(n)/(n+1). Then

g(0) = 1 and g(n) = 1 + 1/g(n-1) for n>=1.

This formula implies that g(n) is a convergent to a continued fraction
[1;1,1,1,...] = (1+sqrt(5))/2,
implying that
g(n) = F(n+2)/F(n+1), where F(n) is the n-th Fibonacci number.
Therefore,
f(n) = (n+1)*F(n+2)/F(n+1).

Regards,
Max

On Thu, Mar 26, 2009 at 8:57 AM, Georgi Guninski <guninski at guninski.com> wrote:
> rational sequence:
> $$f_n= (\frac{n}{f_{n-1}}+1)(n+1) \; , f_0 = 1$$
>
> f[n]= (n / f[n-1]+1)*(n+1) , f=1
>
>
> i found possible doubling formulas for this so it seems efficiently
> computable.
>
> 1. period mod p is uninteresting to me
> 2. i suspect it may be a combination of exponential and rational
> functions, though Fricas (a fork of Axiom) can't find such.
> 3. can't find relations for the numerators or denominators and OEIS
> returns nil.
>
> thanks.
>
> --
> georgi
>
>
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