[seqfan] Re: A (new) sequence connected with Fibonacci numbers and Golden ratio
Vladimir Shevelev
shevelev at bgu.ac.il
Mon Jun 28 17:17:33 CEST 2010
Very thanks, Richard! I accept your version. It should be indicated about your editing.
I would like only to add to %C that, more exactly, it could be proved that, for m=a(n), A_n(m)=Fibonacci(m)-1.
Regards,
Vladimir
----- Original Message -----
From: Richard Mathar <mathar at strw.leidenuniv.nl>
Date: Monday, June 28, 2010 15:30
Subject: [seqfan] Re: A (new) sequence connected with Fibonacci numbers and Golden ratio
To: seqfan at seqfan.eu
>
> The problem with this variant is:
> i) The formula apparently concerns the fibonacci numbers, so it
> should not be here but in A000045
> ii) "decimal sign" is not defined
> iii) "nint" is usually "round" (see the OEIS FAQ)
> iv) LaTeX backslash notation is useless. (This is not a Latex,
> nor a Mma database.)
> v) The base keyword is clearly missing.
> The more basic problem is that no answer is given to the
> question why one would
> try to define an "almost" Fibonacci sequence based on some
> almost arbitrary
> manipulation of the base-10 representation of the golden ratio.
> Anyway, a proposal for a refined version is:
>
>
> %I A179057
> %S A179057 9,9,13,19,23,29,33,42
> %N A179057 a(n) is the smallest argument m for which an
> auxiliary sequence A_n(m) differs from Fibonacci(m).
> %C A179057 Given n, an auxiliary sequence A_n(m) is defined by
> A_n(m)=A000045(m), 0<=m<5 and
> %C A179057 A_n(m)=round( log_2(x_n^A_n(m-1)+x_n^A_n(m-2))),
> m>=5, where x_n is a truncated
> %C A179057 approximation of 2^A001622 = 3.0695645076529..., namely
> %C A179057 x_n = floor(2^A001622*10^n)/10^n = 3, 3.0, 3.06,
> 3.069, 3.0695,... for n = 0, 1, 2, 3,...
> %C A179057 If one would replace x_n by the exact value of
> 2^(golden ratio), the A_n(m) would reproduce the Fibonacci sequence.
> %C A179057 The sequence shows the index where A_n(m) diverges
> first from Fibonacci(m): A_n(m) = Fibonacci(m) for
> 0<=m<a(n) and A_n(m) <> Fibonacci(m) for m=a(n).
> %e A179057 For n=0 and m>=5, we have A_0(m) =
> round(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 Cf. A000045
> %K A179057 nonn,base,less,new
> %O A179057 0,1
> %A A179057 Vladimir Shevelev (shevelev(AT)bgu.ac.il), Jun 27 2010
>
>
>
>
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>
Shevelev Vladimir
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