Quote:
Originally Posted by carpetpool
Is anyone able to come up with a counterexample to the following test? I'm sure there are some, I just can't construct any. Please let me know.
Define a(n) (case where b=1 see below) to be https://oeis.org/A116201
<snip>

This paper may be pertinent regarding the existence of counterexamples. In particular, it proves that there are Carmichael numbers composed entirely of primes p for which the given polynomial (mod p) splits into linear factors.
See also
this paper.
Looking at the OEIS entry, I asked myself, "Why is this a divisibility sequence?"
The characteristic polynomial f = x^4  x^3  x^2  x + 1 for the recurrence is a quartic with Galois group D4 (the dihedral group with 8 elements, AKA the "group of the square").
I knew that the sequence could be expressed in the form a(k) = trace(y*x^k) where y = Mod(p, f), with p a nonzero polynomial in x of degree less than 4.
Luckily, I know how to compute y. And what the computation showed was that for the given sequence, y^2  13 = 0. I'm sure that this is the key to the sequence being a divisibility sequence, but I'm too lazy to work out the details.
The field k
_{1} = Q(sqrt(13)) is a subfield of the splitting field K of f; K/Q has degree 8. Also, f splits into two quadratic factors in k
_{1}[x], so K is a fourgroup extension of k
_{1}.
The splitting field K has two other quadratic subfields, k
_{2} = Q(sqrt(39)) and k
_{3} = Q(sqrt(3)). The polynomial f remains irreducible in k
_{2}[x] and k
_{3}[x], so K is a cyclic quartic extension of k
_{2} and k
_{3}.
The field k
_{3} is also the field of cube roots of unity.
The extension K/k
_{2} is interesting because the class group of k
_{2} is cyclic of order 4, and K is the Hilbert Class Field of k
_{2}.