
COMMENTS

A generalized Lucas sequence {U(m)} is a sequence defined by U(0) = 0, U(1) = 1 and U(m) = P*U(m1)  Q*U(m2) for m >= 2, where P > 0, gcd(P, Q) = 1 and (P, Q) != (1, 0), (1, 1) or (2, 1). The discriminant is D = P^2  4*Q.
A primitive factor of U(m), m > 0 is any prime that divides U(m) but does not divide U(r) for any 0 < r < m. Let's call a prime factor of U(m) "strongly primitive" if it is primitive and does not divide D.
Bilu, Hanrot and Voutier shows that for (P, Q, D) != (1, 2, 7), if U(m) has no strongly primitive factor, than m = 1..8, 10 or 12 (except for a few cases, m can only be 1, 2, 3, 4 or 6. See A285314). (P, Q, D) = (1, 2, 7) is the only case where U(13), U(18) and U(30) have no primitive factor.
This is also the only case where there are nine terms without a primitive factor. The case (P, Q, D) = (1, 3, 11) has five (m = 1, 2, 5, 6, 12). (1, 1, 5) has four (m = 1, 2, 6, 12). (1, 5, 19) has four (m = 1, 2, 7, 12). For all other Lucas sequences there can be at most three terms without a primitive factor (e.g. m = 1, 2, 6 for (3, 2, 1)).
Note that b(7) = 7 has only a primitive factor (7) that is not strongly primitive.


LINKS

Table of n, a(n) for n=1..9.
Y. Bilu, G. Hanrot, P. M. Voutier, Existence of primitive divisors of Lucas and Lehmer numbers J. reine Angew. Math. 539 (2001), 75122, a preprint version available from here.
R. D. Carmichael, On the numerical factors of the arithmetic forms a^n + b^n, Ann. of Math., 15 (1913), 3070.
P. M. Voutier, Primitive divisors of Lucas and Lehmer sequences, Math. Comp. 64 (1995), 869888.
M. Yabuta, A simple proof of Carmichael's theorem on primitive divisors, Fibonacci Quart., 39 (2001), 439443.


EXAMPLE

We have b(1) = b(2) = 1 and b(3) = b(5) = b(13) = 1, so obviously b(m) has no primitive factor if m = 1, 2, 3, 5, 13.
b(8) = 3 has only one prime factor 3, but 3 divides b(4) = 3, so 8 is a term here.
b(12) = 45 has two prime factors 3 and 5, but 3 divides b(4) = 3 and 5 divides b(6) = 5, so 12 is here.
b(18) = 85 has two prime factors 5 and 17, but 5 divides b(6) = 5 and 17 divides b(9) = 17, so 18 is here.
b(30) = 24475 has three prime factors 5, 11 and 89, but 5 divides b(6) = 5, 11 divides b(10) = 11 and 89 divides b(15) = 89, so 30 is also here.
According to Bilu, Hanrot and Voutier, b(m) has at least one primitive factor for any other m (and at least one strongly primitive factor if m != 7).
