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

1,1

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, Entry points and periods of Lucas sequences

Eric Weisstein's World of Mathematics, Gaussian Prime.

Eric Weisstein's World of Mathematics, Lucas Number.

Mathematica

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

Prog

(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

Crossrefs

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

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

Sequence in context: A357426 A152091 A272550 * A106535 A178150 A214784

Adjacent sequences: A122866 A122867 A122868 * A122870 A122871 A122872

Keyword

nonn

Author

Alexander Adamchuk, Sep 16 2006

Status

approved