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

1,1

Comments

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

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

Example

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

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)]);

Crossrefs

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

Sequence in context: A038958 A109416 A132170 * A371566 A227576 A114262

Adjacent sequences: A106306 A106307 A106308 * A106310 A106311 A106312

Keyword

nonn,more

Author

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

Extensions

4259 found by D. S. McNeil.

Edited by Robert Israel, Mar 27 2024

Status

approved