[seqfan] Re: A371164 and A106309

[Neil Sloane]

I was the one who merged them. My reasons for doing so:
See the History of A371164. There you will see the question:
Is this the same as A106309 <https://oeis.org/A106309>? - R. J. Mathar
<https://oeis.org/wiki/User:R._J._Mathar>, Mar 22 2024

and the reply

ri Mar 22 06:35

*Joerg Arndt*: yes, by definition; A106309 should be marked dup

06:57

*Hugo Pfoertner*: No, the problem is that A106309 has never been
extended. The new sequence is the duplicate, but my usual routine
comparison didn't find it, because of an insufficient number of terms
in the "old" sequence.

Since Joerg said they were the same, I merged them.

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 12:28 AM <israel at math.ubc.ca> wrote:

> I see A371164 and A106309 have been merged.  A371164 was
>
> Primes p such that the polynomial x^5 - x^4 - x^3 - x^2 - x - 1 is
> irreducible mod p.
>
> while A106309 was originally
>
> Primes that yield a simple orbit structure in 5-step recursions.
>
> explained in the Comment:
>
> Consider the 5-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4)+x(k-5) mod
> n. For any of the n^5 initial conditions x(1), x(2), x(3), x(4), and x(5)
> in Zn, the recursion has a finite period. When n is a prime in this
> sequence, all of the orbits, except the one containing (0,0,0,0,0), have
> the same length.
>
> My problem is that I don't see how to prove that these are equivalent.
> Certainly if the polynomial is irreducible mod p, all the orbits will have
> the same length, namely the least positive integer k such that x^k-1 is
> divisible by p(x) = x^5 - x^4 - x^3 - x^2 - x - 1 over the integers mod p.
> But what if p is reducible, say p(x) = q(x) * r(x) mod p with q and r
> distinct, and the least positive integer k such that x^k - 1 is divisible
> by q(x) mod p is the same as the least k such that x^k - 1 is divisible by
> r(x) mod p. Then, unless I am mistaken, all orbits will have length k. Now
> I haven't found an example where this occurs, but I don't know of any
> reason why it shouldn't happen.
>
> Can anyone enlighten me?
>
> Cheers,
> Robert
>
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