%I #39 Jun 25 2024 08:30:16

%S 3,7,23,41,43,47,67,83,103,107,127,163,167,223,227,241,263,281,283,

%T 307,347,367,383,401,409,443,449,463,467,487,503,523,547,563,569,587,

%U 601,607,641,643,647,683,727,743,769,787,823,827,863,881,883,887,907,929

%N Odd primes p with 2 zeros in Fibonacci numbers mod p.

%C Also, odd primes that divide Lucas numbers of even index. - _T. D. Noe_, Jul 25 2003

%C Primes in A053030. - _Jianing Song_, Jun 19 2019

%C From _Jianing Song_, Jun 16 2024: (Start)

%C Primes p such that A001176(p) = 2.

%C For p > 2, p is in this sequence if and only if 8 divides of A001175(p), and if and only if 4 divides A001177(p). For a proof of the equivalence between A001176(p) = 2 and 4 dividing A001177(p), see Section 2 of my link below.

%C This sequence contains all primes congruent to 3, 7 (mod 20). This corresponds to case (2) for k = 3 in the Conclusion of Section 1 of my link below.

%C Conjecturely, this sequence has density 1/3 in the primes. (End) [Comment rewritten by _Jianing Song_, Jun 16 2024 and Jun 25 2024]

%H T. D. Noe, <a href="/A053027/b053027.txt">Table of n, a(n) for n=1..1000</a>

%H C. Ballot and M. Elia, <a href="http://www.fq.math.ca/Papers1/45-1/quartballot01_2007.pdf">Rank and period of primes in the Fibonacci sequence; a trichotomy</a>, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B3).

%H Nicholas Bragman and Eric Rowland, <a href="https://arxiv.org/abs/2202.00704">Limiting density of the Fibonacci sequence modulo powers of p</a>, arXiv:2202.00704 [math.NT], 2022.

%H M. Renault, <a href="http://webspace.ship.edu/msrenault/fibonacci/fib.htm">Fibonacci sequence modulo m</a>

%H Jianing Song, <a href="/A053027/a053027.pdf">Lucas sequences and entry point modulo p</a>

%F A prime p = prime(i) is in this sequence if p > 2 and A001602(i)/2 is even. - _T. D. Noe_, Jul 25 2003

%Y Cf. A001175, A001177.

%Y Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).

%Y 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.

%Y | m=1 | m=2 | m=3

%Y -----------------------------+-----------+---------+---------

%Y The sequence {x(n)} | A000045 | A000129 | A006190

%Y The sequence {w(k)} | A001176 | A214027 | A322906

%Y Primes p such that w(p) = 1 | A112860* | A309580 | A309586

%Y Primes p such that w(p) = 2 | this seq | A309581 | A309587

%Y Primes p such that w(p) = 4 | A053028** | A261580 | A309588

%Y Numbers k such that w(k) = 1 | A053031 | A309583 | A309591

%Y Numbers k such that w(k) = 2 | A053030 | A309584 | A309592

%Y Numbers k such that w(k) = 4 | A053029 | A309585 | A309593

%Y * and also A053032 (primes dividing Lucas numbers of odd index) U {2}

%Y ** also primes dividing no Lucas number

%K nonn

%O 1,1

%A _Henry Bottomley_, Feb 23 2000