help needed in editing a bunch of submissions
N. J. A. Sloane
njas at research.att.com
Wed Jun 18 17:53:00 CEST 2008
submissions that did not use the web page
Best regards
Neil
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Date: Thu, 19 Jun 2008 10:00:14 +0900
Subject: Multiplicative order or not
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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 http://citeseer.ist.psu.edu/629149.html
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.
http://boat.zero.ad.jp/~zbi74583/ess0.htm
Max wrote :
On Wed, Jun 4, 2008 at 6:07 PM, koh <zbi74583 at boat.zero.ad.jp> 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.
Regards,
Max
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.
Yasutoshi
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