[seqfan] Re: A (new) sequence connected with Fibonacci numbers and Golden ratio
Richard Mathar
mathar at strw.leidenuniv.nl
Mon Jun 28 13:56:23 CEST 2010
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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