help needed in editing a bunch of submissions

N. J. A. Sloane njas at
Wed Jun 18 17:53:00 CEST 2008

submissions that did not use the web page
 Best regards
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Date: Thu, 19 Jun 2008 10:00:14 +0900
Subject: Multiplicative order or not
From: "koh" <zbi74583 at>
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    Neil wrote : 
 >Max, Yasutoshi:

 >Yesterday I updated A137605 with Max's comment.
 >Is it ok now?

    Thank you Max, Neil, Robert.
    But what does the following mean?
      First occurrence of k=0..: 1, 2, 4, 8, 6, 7, 22, 26, 10, 13, 12, 18, 1366, 15, 50, 386, ..., . - Robert G. Wilson v.   

    Maximilian wrote : 
 >>    Could you explain more about the relationship between the number and length of cycles and quadratic maps or Lucas-Lehmer test?

 >It is not so easy. You should first have a look at the pictures in:

 >>> paper by Shallit and Vasiga, "...iteration of certain quadratic maps..."

 >available at

    I saw the paper.
    It is too  difficult for me to read.    .

 >Then, there is a link between taking the square, and multiplying by 2
 >modulo some m (somehow the inverse of your operation)

 >but it is a bit complicated to explain ; actually Max and I try to
 >write a paper about this since quite some time..

    I wonder if you are interested in 3x+1 sequence and further generalization of 3x+1.
    If so then see "Collatz's 3x+1 sequence" in my home page.

    Max wrote : 

On Wed, Jun 4, 2008 at 6:07 PM, koh <zbi74583 at> wrote:

>> In general, I can prove that either A137605(n) = A002326(n-1) - 1, or
>> A137605(n) = A002326(n-1)/2 - 1.
>> But at the moment I don't see an efficient way to determine which case
>> is taking place.
>    My understanding is the following.
>    If b(n-1) is odd then b(n)=k-(b(n-1)+1)/2
>         2*b(n)=2*k-b(n-1)-1
>         2*b(n)=-b(n-1)       Mod 2*k-1
>           b(n)=-b(n-1)/2     Mod 2*k-1
>    If the signature "-1" doesn't exist then A137605(n) = A002326(n) - 1
>    Indeed it exists.
>    The smallest case of A137605(n) = A002326(n)/2 - 1.
>    The existence of coefficient "1/2" depends on number of the signature.

Actually, it depends on the oddness of the number of -1's. If the
number of -1's is odd then there is the coefficient 1/2 in the
formula. But it seems to be no apparent way to compute this number (or
just its oddness) without computing the whole sequence of b()'s.


    I want to know the oddness.
    If I get the law of the oddness then the formula for A096259 becomes better.
    The "p" of "2^p" in the formula is the same thing as the oddness.

    I think the most interesting fact is that the GO sequence appears in game of GO.



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