OFFSET
1,1
COMMENTS
Also, odd primes that divide Lucas numbers of odd index. - T. D. Noe, Jul 25 2003
From Charles R Greathouse IV, Dec 14 2016: (Start)
It seems that this sequence contains about 1/3 of the primes. In particular, members of this sequence constitute:
35 of the first 10^2 primes
330 of the first 10^3 primes
3328 of the first 10^4 primes
33371 of the first 10^5 primes
333329 of the first 10^6 primes
3333720 of the first 10^7 primes
33333463 of the first 10^8 primes
etc. (End)
Of the Fibonacci-like sequences modulo a prime p that are not A000004, one of them has a period length less than A001175(p) if and only if p = 5 or p is in this sequence. - Isaac Saffold, Dec 18 2018
Odd primes in A053031. - Jianing Song, Jun 19 2019
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
C. Ballot and M. Elia, Rank and period of primes in the Fibonacci sequence; a trichotomy, Fib. Quart., 45 (No. 1, 2007), 56-63 (The sequence B1).
Nicholas Bragman and Eric Rowland, Limiting density of the Fibonacci sequence modulo powers of p, arXiv:2202.00704 [math.NT], 2022.
M. Renault, Fibonacci sequence modulo m
FORMULA
EXAMPLE
From Michael B. Porter, Jan 25 2019: (Start)
The Fibonacci numbers (mod 7) repeat the pattern 0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1. Since there are two zeros, 7 is not in the sequence.
The Fibonacci numbers (mod 11) repeat the pattern 0, 1, 1, 2, 3, 5, 8, 2, 10, 1 which has only one zero, so 11 is in the sequence.
(End)
MATHEMATICA
Prime@ Rest@ Position[Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, Prime@ Range@ 168}], 1][[All, 1]] (* Michael De Vlieger, Aug 08 2018 *)
PROG
(PARI) fibmod(n, m)=(Mod([1, 1; 1, 0], m)^n)[1, 2]
is(n)=my(k=n+[0, -1, 1, 1, -1][n%5+1]); k>>=valuation(k, 2)-1; fibmod(k, n)==0 && fibmod(k/2, n) && isprime(n) \\ Charles R Greathouse IV, Dec 14 2016
CROSSREFS
Cf. A000204 (Lucas numbers), A001602 (index of the smallest Fibonacci number divisible by prime(n)).
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 this sequence U {2}
** also primes dividing Lucas numbers of even index
*** also primes dividing no Lucas number
KEYWORD
nonn
AUTHOR
Henry Bottomley, Feb 23 2000
STATUS
approved