A106309 Primes p such that for all initial conditions (x(0),x(1),x(2),x(3),x(4)) in [0..p-1]^5 except [0,0,0,0,0], the 5-step recurrence x(k) = x(k-1) + x(k-2) + x(k-3) + x(k-4) + x(k-5) (mod p) has the same period.

5, 7, 11, 13, 17, 31, 37, 41, 53, 79, 107, 199, 233, 239, 311, 331, 337, 389, 463, 523, 541, 547, 557, 563, 577, 677, 769, 853, 937, 971, 1009, 1021, 1033, 1049, 1061, 1201, 1237, 1291, 1307, 1361, 1427, 1453, 1543, 1657, 1699, 1723, 1747, 1753, 1759, 1787, 1801, 1811, 1861, 1877, 1997, 1999

Offset: 1

Author

T. D. Noe, May 02 2005, revised May 12 2005

Keywords

• nonn

Comments

The first term not in A371566 is a(105) = 4259.

Examples

			a(3) = 11 is a term because the recurrence has period 16105 for all initial conditions except (0,0,0,0,0).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
• Robert Israel, Linear Recurrences With a Single Minimal Period

Crossrefs

Cf. A106287 (orbits of 5-step sequences). Contains A371566.

Programs

Maple

filter:= proc(p) local Q,q,F,i,z,d,k,kp,G,alpha;
  Q:= z^5  - z^4 - z^3 - z^2 - z - 1;
  if Irreduc(Q) mod p then return true fi;
  F:= (Factors(Q) mod p)[2];
  if ormap(t -> t[2]>1, F) then return false fi;
  for i from 1 to nops(F) do
    q:= F[i][1];
    d:= degree(q);
    if d = 1 then
       kp:= numtheory:-order(solve(q,z),p);
    else
       G:= GF(p,d, q);
       alpha:= G:-ConvertIn(z);
       kp:= G:-order(alpha);
    fi;
    if i = 1 then k:= kp
    elif kp <> k then return false
    fi;
  od;
  true
end proc:
select(filter, [seq(ithprime(i),i=1..1000)]);

Extensions

4259 found by D. S. McNeil.

Edited by Robert Israel, Mar 27 2024