GO Sequence

[koh]

    Dear Seqfans
    I have been studying about the longest period of "Endless rule" on GO.

    Once I submitted A096259, but I think these early results are not correct.

    I got a formula as follows.

    For 4<=n
    a(n) = n * 2^p * ( Sum_{0<=k<=m} ( Sum_{0<=i<=h_k} n_k/2^i ) - 1 )
    Where p = m Mod 2 
          n_0 = n
          n_k = n - [n_{k-1}/2^(h_{k-1}+1)] - 1
          2^h_k is the highest power of two dividing n_k
          n_m/2^h_m = 1

    1<=n<=10
    a(n) : 1,2,6,24,70,180,294,112,270,900,330,792

    [Example]
    n = 5            t         
    + + + + b        0
    w + + + b        1
    w + + b b        2
    w w + b b        3
    w w b b b        4
    + + b b b        4 
    w + b b b        5
    w b b b b        6
    + b b b b        6
    w b b b b        7
    w + + + +        7
    .....             .....
    + + + b +        14
    .....
    Do the same more four times 
     .....
    + + + + b        70

    Where "b" means black stone, "w" means white stone.

    [Example]
    n = 12
    a(12) = 12*2^0*(12 +6 +3 +10 +5 +9 +7 +8 +4 +2 +1 -1)
         =792

    I tried to compute the case of real GO n = 361 by hand.
    I feel that I should use a computer.

    a(361) = 361 *  2^p * ( 361 + 180 + 90 + 45 + 338 + 169 + 276 + 138 + 69 + 326 + 163 + 279 + 221 + 250 + 125 + 298 + 149 + 286 + 143 + 289 + 116 + 58 + 29 + 346 + 173 + 274 + 137 + 292 + 146 + 73 + .....               )



    I think the number in the phalenthes is also an interesting sequence and it must become 1 at last.
    But I wonder if any n exists whose n_k becomes periodic 
    Could anyone prove that the number must become 1 

    For example, the case n=11 and if 9 appears then the sequence becomes 9,6,3,9,6,....but indeed 9 doesn't appears.



    n                       a(n)/n*2^p
    1                          1
    2 1                        2
    3 1                        3
    4 2 1                      6
    5 2 1                      7   
    6 3 4 2 1                 15
    7 3 5 4 2 1               21
    8 4 2 1                   14
    9 4 2 1                   15
    10 5 7 6 3 8 4 2 1        45
    11 5 8 4 2 1              30
    12 6 3 10 5 9 7 8 4 2 1   66

    b(n) : 1,2,3,6,7,15,21,14,15,45,30,66
           a(n)/(n*2^p)

         

    c(n) : 0,1,1,2,2,4,5,3,3,8,5,10
           Number of steps n to 1



    d(n) : 1,2,3,4,6,7,10,12,19
           n such that all number 1<=k<=n-2 appear.



    e(n) : 361,180,90,45,338,..
           The rule is the following
           n(0)=361 
    If e(n-1) is divided by two then    e(n) = e(n-1)/2
    If e(n-1) isn't divided by two then e(n) = n(0)-(e(n-1)+1)/2
    [Example]
           n(0)=19
           19 9 14 7 15 11 13 12 6 3 17 10 5 16 8 4 2 1



    I will soon edit them and submit.



    Yasutoshi