[seqfan] 8/5 Sequence

zbi74583.boat at orange.zero.jp zbi74583.boat at orange.zero.jp
Wed Sep 30 09:29:58 CEST 2009

    Hi, Seqfan

    8/5 Sequence :

    IF  x=Even  THEN  x=x/2
    IF  x=Odd  AND
        8*x=0  Mod 5  THEN  x=(8*x+5)/5
        8*x=1  Mod 5  THEN  x=(8*x+4)/5
        8*x=2  Mod 5  THEN  x=(8*x+3)/5
        8*x=3  Mod 5  THEN  x=(8*x+2)/5
        8*x=4  Mod 5  THEN  x=(8*x+1)/5

    S1  1,2,1,2,….
    S2  3,5,9,15,25,41,66,33,53,85,….
    S3  7,12,6,3,…
    S4  11,18,9,….

    Once Don Reble computed 2,000,000 th term = 852756... 564079; it has 46892
digits, where the first term is 107.
It seems to be divergent.
    I think the most interesting thing is that 8/5 is rather small but it seems
to go infinity.

    It is easy to generalize the sequence.

    m/ n Sequence :
    IF  x=Even  THEN  x=x/2
    IF  x=Odd  AND
        m*x=0    Mod n  THEN  x=(m*x+n)/n
        m*x=1    Mod n  THEN  x=(m*x+n-1)/n
        m*x=2    Mod n  THEN  x=(m*x+n-2)/n
        m*x=n-1  Mod n  THEN  x=(m*x+1)/n

    Example :
    3/2 sequence is the same as 3x+1 sequence.

    IF  x=Even  THEN  x=x/2
    IF  x=Odd  AND
        3*x=0    Mod 2  THEN  x=(3*x+2)/2    ….the case doesn’t exist
        3*x=1    Mod 2  THEN  x=(3*x+1)/2

    Periodic :
    4/3. 5/3
    These m/n Sequence seem to be all periodic.
    Only 8/5 Sequence is divergent in the area where m/n racio is around 3/2.

    Obviously 10/5 Sequence is divergent.
    Proof :
    IF  x=Even  THEN  x=x/2
    IF  x=Odd  AND
        10*x=0  Mod 5  THEN  x=(10*x+5)/5=2*x+1    ….it is Odd.
    So, S(n)=2*S(n-1)+1

   And if m/n becomes much larger than 3 then it becomes divergent.
   The number “3” is able to be a little smaller, it may be 2.7.


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