Multiplicative order not

Max Alekseyev maxale at
Thu Jun 5 07:13:04 CEST 2008

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.


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