Generalized GO

[koh]

    Dear seqfans Neil


    %I A000001
    %S A000001 1,1,2,3,6,5,6,12,8,9,18,11,20,18,14,15,20
    %N A000001 Generate a sequence by the following rule.
               If b(n-1) is divisible by two then b(n) = b(n-1)/2.
               If b(n-1) isn't divisible by two then b(n) = k-(b(n-1)+1)/2.
               k is a integer.
               a(k) = {Number of cycles} * {The longest period} for each k

    %C A000001 For all integers i,j If k=i, b(0)=j then b(n) becomes periodic.       
    %e A000001 k=16
               b(n)             Period
               16 8 4 2 1            5
               14 7 12 6 3 14        5
               13 9 11 10 5          5 
               Three cycles exist.
               So, a(16) = 3 * 5 = 15
    %Y A000001 A000002
    %K A000001 none
    %O A000001 1,3
    %A A000001 Yasutoshi Kohmoto   zbi74583 at boat.zero.ad.jp
    


    %I A000002
    %S A000002 2,3,4,6,7,9,10,12,15,16
    %N A000002 k such that A000001(k) = k-1
    %C A000002 A137606(m) is a subset of A000002(n).
    %e A000002 A000001(16) = 15 = 16-1. So, 16 appears in the sequence.
    %Y A000002 A000001
    %K A000002 none
    %O A000002 2,1
    %A A000002 Yasutoshi Kohmoto   zbi74583 at boat.zero.ad.jp



    Dear Peter 
    >the sequence of those k for which the mean of all values of a period of
goseq[k] (k>=2) is an integer value is:

    >1.) starting with
    >2,4,6,10,11,12,16,22,24,26,30,34,35,36,37,39,40,42,52,53,54,64,66,70,71,82,84,90,96,100,106,107,110,111,114,119,120,127,132,136,143,151

    >and 2.) _slightly_ too artificial for my taste ;-)

    Once I agreed you.
    But it means k for which (k-1)/{Number of cycles} = Integer
    It is rather interesting.
    I recommend you to submit it to OEIS.

    I think these numbers are not correct.
    My computing is the following.

    1,2,3,4,5,6,7,9,10,12,13,15,16,17

    For example, k=9 two 4 cycles exist.
    (4 + 4)/2 = 4 is an integer , so 9 is in the sequence.

    Or, do I misunderstand something?

    Yasutoshi