[math-fun] Odd behavior in a sequence

I'm looking at the linear recurrence 0, 1, 3, 11, 39, 139, . with a(n) = 3a(n-1) + 2a(n-2) for n >= 2. I was interested in the question, for which primes p do {a(n)} == Z (mod p)? That is, for which primes p do all residues r appear in {a(n)} (mod p) (Forgive my notational abuse)? We'll shorten that to "a covers p". I expected this to have a simple answer, and it almost does. It turns out that whether or not a covers p most of the time depends on p mod 17. In general, a covers p if p == 0, 3, 5, 6, 7, 10, 11, 12, or 14 (mod 7). However, there seem to be a smattering of primes p: 683, 1217, 2731, 11299, 48817, . which, according to their residue mod 17, should be covered by a, but are not. Very curious.