# [seqfan] Re: A (new) sequence connected with Fibonacci numbers and Golden ratio

Fri Jun 25 10:16:19 CEST 2010

More deep analysis shows that all terms of the sequence equals to 6. The sequence revives if to put:

a(n) (n>=1) is the maximal number of the first positive
Fibonacci numbers given by the sequence which is defined by the
recursion: A_n(0)=1, A_n(1)=1 , for 2<= m<=5,
A_n(m)=floor(log_2(x_n^A_n(m-1)+x_n^A_n(m-2)))
and , for m>=6,
A_n(m)=ceil(log_2(x_n^A_n(m-1)+x_n^A_n(m-2))).

I am interested to understand what is the growth of this sequence.

Regards,

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Wednesday, June 23, 2010 21:35
Subject: [seqfan] A (new) sequence connected with Fibonacci numbers and Golden ratio
To: seqfan at list.seqfan.eu

>
> Dear SeqFans,
>
> Is it interesting the following sequence:
> Consider consecutive decimal approximations of 2^{\phi}, where
> \phi is the Golden ratio:
> x_1=3, x_2=3.0, x_3=3.06, x_4=3.069, x_5=3.0695 etc.
> Then a(n) (n>=1) is the maximal number of the first positive
> Fibonacci numbers which are given by the sequence defined by the
> recursion:A_n(0)=1, A_n(1)=1 and, for m>=2,
> A_n(m)=floor(log_2(x_n^A_n(m-1)+x_n^A_n(m-2))).
> The sequence begins with 6,6,...
>
> Regards,
>