[seqfan] 8/5 Sequence

kohmoto zbi74583.boat at orange.zero.jp
Mon Jun 11 11:01:40 CEST 2012


    Hi, Neil and Seqfan


    I think A082010 is not correct. it and the sequence of definition are different.
    If you started from 0 then it must be the following.
    0,1,2,1,2,....

    It is not so interesting.
    The first interesting case is the following.

    3,5,9,15,25,41,66,33,53,85,137,220,110,55,89,143,229,367,588,294,147,236,118,59,95,153,245,393

    http://list.seqfan.eu/pipermail/seqfan/2009-September/009818.html

    Ask Don Reble.
    He has 2000000 terms of the sequence.

    Once Franklin wrote
    >This can be simplified: if x is odd, then next x is floor(8/5*x)+1.
    >No need for separate m and n; let r be the ratio (m/n), and if x is 
    >odd, the next x is floor(r*x)+1. And now we don't even need r to be 
    >rational.


    You are right.
    8/5 sequence is one of the K-Sequence.

    a(n)=[A*a(n-1)+B]/p^m
    Where [N] is integer part of N, p^m is the highest p power dividing [A*a(n-1)+B]

    http://boat.zero.ad.jp/~zbi74583/ess0.htm

    3 examples :
    See A028948, A029580, A036982

    >When r is the golden ratio, (sqrt(5)+1)/2, starting with 3, we get 
    >A001595, a sequence of all odd numbers, so there is unlimited growth. 
    >This may shed some light on what is happening with 8/5.

    I think your opinion is interesting.
    My observation is the following.

    p=2, a(0)=3, A=1.6, 1<=B<2
    In this area all sequences are unlimited. 

    You say the case of A=(1+root(5))/2 is also unlimited.
    How did you find this fact?
    I think it is your discovery. 

    Once I conjectured as follows.

    "K-Sequence which represents Fibonacci Sequence exists"

    I knew that A001959 is one of the K-Sequence.
    So my conjecture is almost right.



    Yasutoshi
    


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