[seqfan] Re: A371164 and A106309

[israel at math.ubc.ca]

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
>
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