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

Richard Mathar mathar at strw.leidenuniv.nl
Thu Mar 26 15:36:25 CET 2009

gg> From seqfan-bounces at list.seqfan.eu Thu Mar 26 14:35:30 2009
gg> Date: Thu, 26 Mar 2009 14:57:18 +0200
gg> From: Georgi Guninski <guninski at guninski.com>
gg> To: seqfan at seqfan.eu
gg> Subject: [seqfan]  rational sequence $$f_n= (\frac{n}{f_{n-1}}+1)(n+1) \; , f_0 = 1$$
gg>
gg> rational sequence:
gg> $$f_n= (\frac{n}{f_{n-1}}+1)(n+1) \; , f_0 = 1$$
gg>
gg> f[n]= (n / f[n-1]+1)*(n+1) , f=1
gg>
gg> ...

ma> Max Alekseyev maxale at gmail.com
ma> ...
ma> Let g(n) = f(n)/(n+1). Then
ma>
ma> g(0) = 1 and g(n) = 1 + 1/g(n-1) for n>=1.
ma>
ma> This formula implies that g(n) is a convergent to a continued fraction
ma> [1;1,1,1,...] = (1+sqrt(5))/2,
ma> implying that
ma> g(n) = F(n+2)/F(n+1), where F(n) is the n-th Fibonacci number.
ma> Therefore,
ma> f(n) = (n+1)*F(n+2)/F(n+1).

The reason for g not to be found in the OEIS is that
gcd( n, F(n) ) = A104714(n) is not all-1, so in
f(n) = (n+1)*F(n+2)/F(n+1) = A023607(n+1)/A000045(n+1)
there are cancellations while computing (n+1)/F(n+1).

Richard J. Mathar