cp's OEIS Frontend

Carmichael proved that a(n) > 1 if n > 12.

See A001578 (smallest primitive prime factor of F(n)) and A061446 (primitive part of F(n)) for more links.

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)

Thanks to squarefree terms of A058635, numbers of the form 2^k appear in this sequence for k > 1. However it is not proven yet whether it is always true.

From Jon E. Schoenfield, Aug 05 2017: (Start)

The difficulty in extending this sequence is that it becomes hard to obtain the complete prime factorization of Fibonacci(m) as m increases. However, since every number having an odd number of divisors is a square, and the largest Fibonacci number that is also a square is Fibonacci(12) = 144, we can confine the search for terms > 12 to even numbers only.

Even for values of m for which we are unable to completely factorize Fibonacci(m), we can determine with a high degree of confidence whether m is in the sequence by considering only the multiplicities of the smaller primes in those factorizations, because multiplicities greater than 1 in the prime factorizations of Fibonacci numbers rarely occur among the larger prime factors. If, in place of the actual complete factorization of Fibonacci(m) for each examined value of m, we were to use only the multiplicities of the prime factors of Fibonacci(m) that are less than 10000 (which are quickly and easily counted using trial division), the terms we would obtain for this sequence would begin with 1, 4, 8, 16, 24, 32, 60, 64, 72, 96, 128, 192, 256, 300, 336, 512, 576, 648, 900, 1008, 1024, 1080, 1250, 1500, 1536, 1620, 1920, 2048, 2352, 2500, 2592, 2700, 4096, 4608, 5000, 5184, 5400, 5832, 7500, 8100, 8192, 8448, 8640, 9072, 9600, 10000, 13608, 15000, ...

Perhaps surprisingly, we would get the same terms (up through at least a(141) = 960000) if, instead of the multiplicities of prime factors <= 10000, we were to use the multiplicities of just the prime factors <= 13. (End)