Generalized GO

[Joerg Arndt]

should just be a comment at some seq already in the database.
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Date: Wed, 21 May 2008 03:30:58 -0700
From: "Max Alekseyev" <maxale at gmail.com>
To: koh <zbi74583 at boat.zero.ad.jp>
Subject: Re: RE : COMMENT on A137605
Cc: seqfan at ext.jussieu.fr
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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.

Of course, if A002326(n-1) is odd then there is no other choice but
A137605(n) = A002326(n-1) - 1.
But if A002326(n-1) is even then there are two choices possible...

Regards,
Max