[seqfan] "near-Fibonacci sequences" and interval [0,1]
Vladimir Shevelev
shevelev at bgu.ac.il
Sun Jul 4 15:35:16 CEST 2010
Dear Seq Fans,
I would like to discuss the following construction.
It is easy to prove that, for n>=1, the Fibonacci sequence {F(n)}={1,1,2,3,5,8,...} (A000045) one can reproduce by the recursion: F(n)=1 and, for n>=2, F(n)=floor(\phi*F(n-1)), if n is even, and F(n)=ceil(\phi*F(n-1)), if n is odd, where \phi=Golden ratio.
Let x belong to [0,1] with the binary expansion x=0.x_1x_2...x_n..., where x_k, k=0,1,..., is in {0,1}. Define sequence {F_x(n)} by the recursion F_x(0)=1 and, for n>=2, F_x(n)=foor((\phi*F_x(n-1)), if x_(n-1)=0, and F_x(n)=ceil((\phi*F_x(n-1)), if x_(n-1)=1.
Then sequence {F_(n)} of Fobonacci numbers corresponds to x=0.010101...=1/3.
In case of x=2/3=0.10101...we obtain sequence {1,2,3,5,8,...}, i.e. a shift of {F(n)}, while in case of x=1=0.111111...we obtain sequence {1,2,4,7,12,20,...}, i.e. {F(n+2)-1}_(n>=1)-cf. A000071. Thus sequences {F_x(n)}, when 1/3<=x<=1, it is natural to call "near-Fibonacci sequences". The growth of every near-Fibonacci sequence is O(\phi^n) with the constant C in O(...) in interval [1/sqrt(5), \phi^2/sqrt(5)].
On the other hand, for x decreases from 1/3 to 0, we have a more complicated law with essentially decreasing growth. What is this?
In case of the ternary expansion y=0.y_1y_2,,,y_n..., where y belongs to the Cantor's set, i.e. y_k is in {0,2}, we obtain "near-Fibonacci sequences" for the part of Cantor's set in [1/4,1].
Regards,
Vladimir
Shevelev Vladimir
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