[seqfan] Re: A371164 and A106309

[Neil Sloane]

Robert,  Thank you very much for your excellent work on those two sequences.

And also, for noticing that the old entry, which said that the two versions
were the same, was wrong.

And a new sequence - as you suggest - for the difference sequence is a
great idea.

Best regards
Neil

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



On Wed, Mar 27, 2024 at 11:04 PM <israel at math.ubc.ca> wrote:

> I have done the editing: A106309 is now "Primes p such that for all
> initial
> conditions (x(0),x(1),x(2),x(3),x(4)) in [0..p-1]^5 except [0,0,0,0,0],
> the
> 5-step recurrence x(k) = x(k-1) + x(k-2) + x(k-3) + x(k-4) + x(k-5) (mod
> p)
> has the same period.". Since A371164 was recycled, A371566 is now "Primes
> p
> such that x^5 - x^4 - x^3 - x^2 - x - 1 is irreducible (mod p)." It did
> turn out that the first 104 terms were the same. Other terms of A106309
> that are not in A371566 include 61643, 94307, 110063, ... After searching
> for some more of those, I think I'll submit them as a new sequence.
>
> Cheers,
> Robert
>
> On Mar 25 2024, israel at math.ubc.ca wrote:
>
> >We have two sequences.  I'm not sure if the first 104 terms all agree.
> >I'll have to check.  OK, I'll be the editor.
> >
> >Cheers,
> >Robert
> >
> >On Mar 25 2024, Neil Sloane wrote:
> >
> >>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 11:31PM <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
> >>> >
> >>> >--
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> >>> >
> >>> >
> >>>
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> >>>
> >>
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> >>
> >
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