A122869 Primes p that divide Lucas((p-1)/2), where Lucas is A000032.

11, 19, 31, 59, 71, 79, 131, 139, 151, 179, 191, 199, 211, 239, 251, 271, 311, 331, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 619, 631, 659, 691, 719, 739, 751, 811, 839, 859, 911, 919, 971, 991, 1019, 1031, 1039, 1051, 1091, 1151, 1171, 1231, 1259

Offset: 1

Author

Alexander Adamchuk, Sep 16 2006

Keywords

• nonn

Comments

Final digit of a(n) is 1 or 9.

A002145 is the union of this sequence and A122870, Primes p that divide Lucas((p+1)/2).

Conjecture: This sequence is just the primes congruent to 11 or 19 mod 20. - Charles R Greathouse IV, May 25 2011 [The conjecture is correct. - Jianing Song, Jun 20 2025]

Note that F(p-1) = F((p-1)/2)*Lucas((p-1)/2), where F = A000045. Since gcd(F(n),Lucas(n)) = 1 or 2 (because Lucas(n)^2 - 5*F(n)^2 = 4*(-1)^n), this sequence lists primes p such that p divides F(p-1) but does not divides F((p-1)/2). By Propositions 1.1 and 1.2 (the k = 3 case) of my link below, this is primes p == 11, 19 (mod 20). - Jianing Song, Jun 20 2025

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)
• Jianing Song, Lucas sequences and entry point modulo p
• Eric Weisstein's World of Mathematics, Gaussian Prime.
• Eric Weisstein's World of Mathematics, Lucas Number.

Crossrefs

Cf. A000032, A000045, A053032, A076518, A122870.

Subsequence of A002145, A003626, A040105, A040147 and A064739.

Programs

Mathematica

Select[Prime[Range[1000]],IntegerQ[(Fibonacci[(#1-1)/2-1]+Fibonacci[(#1-1)/2+1])/#1]&]

PARI

lista(kmax) = {my(lucas1 = 1, lucas2 = 3, lucas3, p); for(k = 3, kmax, lucas3 = lucas1 + lucas2; p = 2*k + 1; if(isprime(p) && !(lucas3 % p), print1(p, ", ")); lucas1 = lucas2; lucas2 = lucas3);} \\ Amiram Eldar, Jun 06 2024