
COMMENTS

For a generalized Lucas sequence {U(n)} = {(a^n  b^n)/(a  b)}, where a + b and ab are nonzero coprime integers and (a/b) is not a root of unity, a prime factor of the nth term of some Lucas sequence U(n) is called primitive if it does not divide U(r) for any r < n.
Let P = a + b > 0, Q = ab and D = P^2  4Q = (a  b)^2. In the case a, b are real (equivalently, D > 0), Carmichael shows that, if n <> 1, 2, 6, then U(n) has at least one primitive prime factor not dividing D except U(3) ((P, Q, D) = (1, 2, 9), (1, 1, 5)), U(5) ((P, Q, D) = (1, 1, 5)) and U(12) ((P, Q, D) = (1, 1, 5)).
Voutier determines the cases U(n) has no primitive prime factor for n = 5, 7 <= n <= 30. For n = 5, 7 <= n <= 30, any prime factor of U(n) divides D or U_r for some r < n in the following cases:
U(5): (P, Q, D) = (1, 1, 5)*, (1, 2, 7), (1, 3, 11), (1, 4, 15)*, (2, 11, 40)*, (12, 55, 76), (12, 377, 1364),
U(7): (P, Q, D) = (1, 2, 7)*, (1, 5, 19),
U(8): (P, Q, D) = (1, 2, 7), (2, 7, 24),
U(10): (P, Q, D) = (2, 3, 8), (5, 7, 3), (5, 18, 47),
U(12): (P, Q, D) = (1, 1, 5), (1, 2, 7), (1, 3, 11), (1, 4, 15), (1, 5, 19), (2, 15, 56),
U(13), U(18), U(30): (P, Q, D) = (1, 2, 7).
(The symbol * indicates that any prime factor of the corresponding U(n) divides D)
Bilu, Hanrot and Voutier shows that these are all for n = 5, n >= 7. Hence this sequence is complete.
From Jianing Song, Feb 23 2019: (Start)
Let {U(n)} be a generalized Lucas sequence. If p is a prime U(p) has no primitive prime factor, then U(p) = +1. In contrast, if k is a composite number such that U(k) has no primitive prime factor, then U(k) cannot be +1. As a result, the possible solutions to U(k) = +1 for some Lucas sequence are k = 1, 2, 3, 5, 7, 13.
From U(1) = 1, U(2) = P, U(3) = P^2  Q, U(4) = P*(P^2  2*Q), U(6) = P*(P^2  Q)*(P^2  3*Q) we can see that for k = 1, 2, 3, 4, 6, there are infinitely many Lucas sequences such that U(k) has no primitive prime factor. Also, for p = 2, 3, there are infinitely many Lucas sequences such that any prime factor of U(p) divides D.
This sequence lists also numbers k such that E(p) = k has no solution over the primes for some Lucas sequence {U(n)}, where E(p) is the entry point of {U(n)} modulo p, that is, the smallest m > 0 such that p divides U(m). (End)
