2 things about the OEIS Welcome Page
N. J. A. Sloane
njas at research.att.com
Wed Jun 4 19:39:36 CEST 2008
Max
> On Tue, May 20, 2008 at 7:01 PM, koh <zbi74583 at boat.zero.ad.jp> wrote:
>
>> Peter Pein computed the period of A137607 and observed that it is the same as A003558.
>> S : {1, 2, 3, 3, 5, 6, 4, 4, 9, 6, 11, 10, 9, 14, 5, 5, 12, 18, 12, 10, 7, 12, 23, 21, 8, 26, 20, 9, 29, 30, 6, 6, 33, 22, 35, 9, 20, 30, 39, 27, 41, 8, 28, 11, 12, 10, 36, 24, 15, 50, 51, 12, 53, 18, 36, 14, 44, 12, 24, 55, 20, 50, 7, 7, 65, 18, 36, 34, 69, 46, 60, 14, 42, 74, 15, 24, 20, 26,
5
>> 2, 33, 81, 20, 83, 78, 9, 86, 60, 29, 89, 90, 60, 18, 40, 18, 95, 48, 12, 98, 99}
>> S(n)=A003558(n-1)
>>
>> And A137605(n)+1=S(n)
>> So, A137605(n) is {multiplicative suborder of two mod 2*n-1}-1.
>> But it is not proved.
>
> If "multiplicative suborder of two mod 2*n-1" means A003558(n-1) then
> the above formula is incorrect.
> The first counterexample is given by n=8: A137605(8)=3 while A003558(8-1)-1=1.
>
> 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.
Maximilian
>
> An easily answered question :
>
> When can there be a fixed point p>0 ?
> It must be odd, p=2m+1, (m=0,1,...)
> and p=k-(p+1)/2 <=> 3p=2k-1
> <=> k=(3p+1)/2=3m+2 ; i.e. k=2,5,8,11...
>
> For these k=2 mod 3 there is indeed a unique nonzero fixed point p=1+2[k/3]
>
I got this theorem too.
And the following one.
If all members of a cycle are odd then they must be the same number.
It means that it is the fixed point.
>
> The number and length of cycles is also of some relevance
> when considering the structure of digraphs under quadratic maps in Z/pZ
> as used in the Lucas-Lehmer tests for primality of Mersenne, Fermat
> and Wagstaff primes.
>
I don't understand well.
Could you explain more about the relationship between the number and length of cycles and quadratic maps or Lucas-Lehmer test?
> There is a paper by Shallit and Vasiga, "...iteration of certain
> quadratic maps..."
> with some nice pictures of digraphs.
> (but there are not only cycles, but also trees and exterior nodes
> attached to the cycles)
>
> Maximilian
>
Yasutoshi
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