[seqfan] Re: A371164 and A106309

[Neil Sloane]

Am I right in thinking that this means we have have two sequences that
agree for 104 terms, but then in one sequence the 105th term is 4259, and
in the other it is not?

Robert, can I ask you to be the editor for this job, and to prepare the two
sequences?

Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
Also Visiting Scientist, Math. Dept., Rutgers University,
Email: njasloane at gmail.com



On Sun, Mar 24, 2024 at 11:31 PM <israel at math.ubc.ca> wrote:

> Yes indeed, that works.  Let M be the companion matrix of the polynomial:
>
>     [ 0 0 0 0 1 ]
>     [ 1 0 0 0 1 ]
> M = [ 0 1 0 0 1 ]
>     [ 0 0 1 0 1 ]
>     [ 0 0 0 1 1 ]
>
> Thus the iteration corresponds to
>
> [ x(i+1), ... , x(i+5)] = [x(i), ..., x(i+4)] M mod p
>
> We can check directly that M^2129 == I mod p, so every initial condition
> [x(1), ..., x(5)] produces an orbit that is periodic with period 2129.
> Any smaller minimal period would have to divide 2129, but 2129 is prime,
> so the only other possibility is 1, and 1 is not an eigenvalue of M mod p.
>
> This indeed 4259 would satisfy the original definition of A106309 despite
> the polynomial not being irreducible mod 4259.
>
> Cheers,
> Robert
>
>
>
> On Mar 24 2024, D. S. McNeil wrote:
>
> >I can't find such a q,r pair either, but what about p=4259?  Then the
> >polynomial splits into five linear terms, and each term has the same least
> >k if I've done my arithmetic right (2129, or half p).
> >
> >The orbits seem to behave as expected (although since most cases seem to
> >have only a very tiny fraction of exceptions to the majority length
> anyway,
> >that appearance can't be trusted.)
> >
> >Is that enough symmetry to guarantee everything has the same orbit length?
> >
> >
> >Doug
> >
> >--
> >Seqfan Mailing list - http://list.seqfan.eu/
> >
> >
>
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