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First differs from A246556 at n = 17. Since Pell(17) = 1136689 = 137 * 8297, we find that 137 does not divide any earlier Pell number, and hence A246556(17) = 137, but 8297 is also prime, and so a(17) = 8297.

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 n-th 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)

Conjecture: The n-th Pell number A000129(n) has a primitive prime factor for all n > 1. (The n-th Fibonacci number A000045(n) has a primitive prime factor for all n except n = 0, 1, 2, 6, and 12.)

For prime p, all prime factors of Pell(p) are primitive. Hence the only primes in this sequence are the prime numbers in A096650, which gives the indices of prime Pell numbers.