A106535 Numbers k such that the smallest x > 1 for which Fibonacci(x) == 0 mod k is x = k - 1.

11, 19, 31, 59, 71, 79, 131, 179, 191, 239, 251, 271, 311, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 631, 659, 719, 739, 751, 839, 971, 1019, 1039, 1051, 1091, 1171, 1259, 1319, 1399, 1439, 1451, 1459, 1499, 1531, 1559, 1571, 1619, 1759, 1811, 1831

Offset: 1

Author

Peter K. Pearson (ppearson+att(AT)spamcop.net), May 06 2005

Keywords

• nonn

Comments

This is a sister sequence to A000057, because this sequence, since {k : A001177(k) = k-1}, might be called a subdiagonal sequence of A001177, and {k : A001177(k) = k+1}, which might be called a superdiagonal sequence of A001177. Sequences A000057 and A106535 are disjoint. Is this sequence the set of all divisors of some family of sequences, like A000057 is? - Art DuPre, Jul 11 2012

Are all members of this sequence prime? Using A069106, any composite members must exceed 89151931. - Robert Israel, Oct 13 2015

From Jianing Song, Jul 02 2019: (Start)

Yes, all terms are primes. See a brief proof below.

Also, if p == 1 (mod 4) then b(p) divides (p-Legendre(p,5))/2. So terms in this sequence are congruent to 11 or 19 modulo 20.

Primes p such that ord(-(3+sqrt(5))/2,p) = p-1, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer. (End)

Comments from Amiram Eldar, Jan 30 2022 (Start)

Sequence A003147, "Primes p with a Fibonacci primitive root", is defined in the paper: Daniel Shanks, Fibonacci primitive roots, Fibonacci Quarterly, Vol. 10, No. 2 (1972), pp. 163-168, and 181.

A second paper on this subject Daniel Shanks and Larry Taylor, An Observation of Fibonacci Primitive Roots, Fibonacci Quarterly, Vol. 11, No. 2 (1973), pp. 159-160,

deals with terms p == 3 (mod 4) of A003147, i.e., the intersection of A003147 and A002145 (or A004767).

It states that if g is a Fibonacci primitive root of a prime p such that p == 3 (mod 4) then g-1 and g-2 are also primitive roots of p.

The first 2000 terms of (A003147 intersect A002145) agree with the present sequence, although the definitions are quite different. Are these two sequences the same? (End)

Links

• Robert Israel, Table of n, a(n) for n = 1..2000
• Alfred Brousseau, Primes which are factors of all Fibonacci sequences, Fib. Quart., 2 (1964), 33-38.
• Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972.
• Jianing Song, Proof that all terms are prime

Crossrefs

Cf. A000057, A001175, A069106.

Similar sequences that give primes p such that A001177(p) = (p-1)/s: this sequence (s=1), A308795 (s=2), A308796 (s=3), A308797 (s=4), A308798 (s=5), A308799 (s=6), A308800 (s=7),A308801 (s=8), A308802 (s=9).

Programs

GAP

Filtered([2..2000], n -> Fibonacci(n-1) mod n = 0 and Filtered( [2..n-2], x -> Fibonacci(x) mod n = 0 ) = [] );

Maple

A106535 := proc(n)
        option remember;
        if n = 1 then
                11;
        else
                for a from procname(n-1)+1 do
                        if A001177(a) = a-1 then
                                return a;
                        end if;
                end do:
        end if;
end proc: # R. J. Mathar, Jul 09 2012
# Alternative:
fmod:= proc(a,b) local A;
  uses LinearAlgebra[Modular];
  A:= Mod(b, <<1,1>|<1,0>>,integer[8]);
  MatrixPower(b,A,a)[1,2];
end proc:
filter:= proc(n)
  local cands;
  if fmod(n-1,n) <> 0 then return false fi;
  cands:= map(t -> (n-1)/t, numtheory:-factorset(n-1));
  andmap(c -> (fmod(c,n) > 0), cands);
end proc:
select(filter, [$2..10^4]); # Robert Israel, Oct 13 2015

Mathematica

f[n_] := Block[{x = 2}, While[Mod[Fibonacci@ x, n] != 0, x++]; x];Select[Range@ 1860, f@ # == # - 1 &] (* Michael De Vlieger, Oct 13 2015 *)

PARI

isok(n) = {x = 2; while(fibonacci(x) % n, x++); x == n-1;} \\ Michel Marcus, Oct 20 2015

Formula

{n: A001177(n) = n-1}. - R. J. Mathar, Jul 09 2012

Extensions

Corrected by T. D. Noe, Oct 25 2006