The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A285314 Numbers k such that the k-th term of some (generalized) Lucas sequence has no primitive prime factor. 1
 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 13, 18, 30 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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 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) LINKS Y. Bilu, G. Hanrot, P. M. Voutier, Existence of primitive divisors of Lucas and Lehmer numbers J. reine Angew. Math. 539 (2001), 75--122, a preprint version available from here. P. M. Voutier, Primitive divisors of Lucas and Lehmer sequences, Math. Comp. 64 (1995), 869--888. M. Yabuta, A simple proof of Carmichael's theorem on primitive divisors, Fibonacci Quart., 39 (2001), 439--443. EXAMPLE If (P, Q, D) = (1, -1, 5) (giving the Fibonacci sequence), U(12) = 144 = 2^4 * 3^2, while U(4) = 3 and U(6) = 8 = 2^3. Hence U(12) with (P, Q, D) = (1, -1, 5) has no primitive prime factor and 12 belongs to this sequence. CROSSREFS Cf. A001578, A058036, A246556 (smallest primitive prime factor of Fibonacci(n), Lucas(n) and Pell(n)). Cf. A086597, A086600 (number of primitive prime factors in Fibonacci(n) and Lucas(n)). Sequence in context: A270427 A270189 A257672 * A080544 A178878 A175326 Adjacent sequences:  A285311 A285312 A285313 * A285315 A285316 A285317 KEYWORD nonn,fini,full AUTHOR Tomohiro Yamada, Apr 17 2017 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified May 19 11:38 EDT 2022. Contains 353833 sequences. (Running on oeis4.)