[seqfan] Re: Sequences appoximating Fibonacci numbers

[Vladimir Shevelev]

I immediately have not understood, but it is a simple problem: x=2^{\phi}=3.06956450...

Regards,
Vladimir

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Tuesday, June 22, 2010 19:27
Subject: [seqfan]  Sequences appoximating Fibonacci numbers
To: seqfan at list.seqfan.eu

> Dear SeqFans,
>  
> Calculating sequence a(0)=1, a(1)=1 and, for n>=2,
>  a(n)=floor(log_2(3^a(n-1)+3^a(n-2))),
> I (by handy) get: 1,1,2,3,5,8,12,...
> From the arguments of the continuity, there exists a constant x, 
> such that, for Fibonacci sequence {F_n}, we have
> F(n)=floor(log_2(x^F(n-1)+x^F(n-2))).
> Sequence b(0)=1, b(1)=1 and, for n>=2, 
>  b(n)=floor(log_2(\pi^b(n-1)+\pi^b(n-2))), is
> 1,1,2,3,5,8,13,21,34,56,...
> Thus 3<x<\pi.
> Can anyone  calculate x with some true signs?
>  
> Regards,
> Vladimir
>  
> 
>  Shevelev Vladimir‎
> 
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 Shevelev Vladimir‎