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

Sun Jun 27 13:50:51 CEST 2010

Finally, I just have submitted a best defined sequence:

%I A179057
%S A179057 9,9,13,19,23,29,33,42
%N A179057 a(n) (n>=1) is the maximal number of the first Fibonacci numbers given by the sequence which is defined by the recursion: A_n(i)=A000045(i),i=0,1,2,3,4, and,for m>=5, A_n(m)=nint(log_2(x_n^A_n(m-1)+x_n^A_n(m-2))), where x_0=3 and, for n>=1, x_n is the nth decimal sign of 2^\phi (\phi=golden ratio)
%F A179057 For n>=5, F(n)=nint(log_2(2^{\phi*F(n-1)}+2^{\phi*F(n-2)})), where F(n) is the nth Fibonacci number and nint denote, as usually, the nearest integer.
%e A179057 For n=0 and m>=5, we have A_0(m)=nint(log_2(3^A_0(m-1)+3^A_0(m-2))). By this formula with the initial conditions, A_0(5)=5, A_0(6)=8, A_0(7)=13, A_0(8)=21 and A_0(9)=33. Since F(9)=34, then A_(m) gives the first 9 Fibonacci numbers: F(0),...,F(8). Thus a(0)=9.
%Y A179057 A000045
%K A179057 nonn
%O A179057 0,1

More terms?

Regards,

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Friday, June 25, 2010 11:27
Subject: [seqfan] Re: A (new) sequence connected with Fibonacci numbers and Golden ratio
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> 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,
> >
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>