COMMENT on A137605

[koh]

    Dear Robert G Wilson v

    Thank you for computing b-file of A137605.

    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.


    Max Alekseyev wrote that A137606 is related with multiplicative order of 2 mod 2*m-1.



    Yasutoshi
    



Hi,

* koh <zbi74583 at boat.zero.ad.jp> [May 21. 2008 09:56]:
>     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.

_where_

>                k is a integer.

>                a(k) = {Number of cycles} * {The longest period} for each k

Why obfuscating by packing?
Are all cycles always of the same length?
What if not?


> 
>     %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

Do not forget the cycle 0->0!
Then, _surprise_, the sum of all lengths is 16!

Can you say anything about the cycle structure?
I bet this is equivalent to something well researched
(cyclotomic cosets, suborder, or so).



>     %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

There the omission of zero pops up as the artificial difference one.



>     %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.

The definition is not clear to.
Give code by all means!



>     I recommend you to submit it to OEIS.

While your sequence not not seem overly contrived to me you should do
basic research.  I bet analysis will show that your construction



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

CODE???


> 
>     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
>     

regards,   jj