OFFSET
1,1
COMMENTS
Primes p such that A322906(p) = 1.
For p > 2, p is in this sequence if and only if A175182(p) == 2 (mod 4), and if and only if A322907(p) == 2 (mod 4). For a proof of the equivalence between A322906(p) = 1 and A322907(p) == 2 (mod 4), see Section 2 of my link below.
This sequence contains all primes congruent to 3, 23, 27, 35, 43, 51 modulo 52. This corresponds to case (3) for k = 11 in the Conclusion of Section 1 of my link below.
Conjecturely, this sequence has density 1/3 in the primes. [Comment rewritten by Jianing Song, Jun 16 2024 and Jun 25 2024]
LINKS
Jianing Song, Table of n, a(n) for n = 1..1200
Jianing Song, Lucas sequences and entry point modulo p
PROG
(PARI) forprime(p=2, 900, if(A322906(p)==1, print1(p, ", ")))
CROSSREFS
Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n+2) = m*x(n+1) + x(n). Let w(k) be the number of zeros in a fundamental period of {x(n)} modulo k.
| m=1 | m=2 | m=3
-----------------------------+----------+---------+----------
* and also A053032 U {2}
KEYWORD
nonn
AUTHOR
Jianing Song, Aug 10 2019
STATUS
approved