A053032 Odd primes p with one zero in Fibonacci numbers mod p.
11, 19, 29, 31, 59, 71, 79, 101, 131, 139, 151, 179, 181, 191, 199, 211, 229, 239, 251, 271, 311, 331, 349, 359, 379, 419, 431, 439, 461, 479, 491, 499, 509, 521, 541, 571, 599, 619, 631, 659, 691, 709, 719, 739, 751, 809, 811, 839, 859, 911, 919, 941, 971
Offset: 1
Keywords
Examples
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)
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
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
Programs
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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 *)
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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
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