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%I #34 Mar 27 2024 23:55:52
%S 5,7,11,13,17,31,37,41,53,79,107,199,233,239,311,331,337,389,463,523,
%T 541,547,557,563,577,677,769,853,937,971,1009,1021,1033,1049,1061,
%U 1201,1237,1291,1307,1361,1427,1453,1543,1657,1699,1723,1747,1753,1759,1787,1801,1811,1861,1877,1997,1999
%N 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.
%C The first term not in A371566 is a(105) = 4259.
%H Robert Israel, <a href="/A106309/b106309.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>
%H Robert Israel, <a href="/A106309/a106309.pdf">Linear Recurrences With a Single Minimal Period</a>
%e a(3) = 11 is a term because the recurrence has period 16105 for all initial conditions except (0,0,0,0,0).
%p filter:= proc(p) local Q,q,F,i,z,d,k,kp,G,alpha;
%p Q:= z^5 - z^4 - z^3 - z^2 - z - 1;
%p if Irreduc(Q) mod p then return true fi;
%p F:= (Factors(Q) mod p)[2];
%p if ormap(t -> t[2]>1, F) then return false fi;
%p for i from 1 to nops(F) do
%p q:= F[i][1];
%p d:= degree(q);
%p if d = 1 then
%p kp:= numtheory:-order(solve(q,z),p);
%p else
%p G:= GF(p,d, q);
%p alpha:= G:-ConvertIn(z);
%p kp:= G:-order(alpha);
%p fi;
%p if i = 1 then k:= kp
%p elif kp <> k then return false
%p fi;
%p od;
%p true
%p end proc:
%p select(filter, [seq(ithprime(i),i=1..1000)]);
%Y Cf. A106287 (orbits of 5-step sequences). Contains A371566.
%K nonn,more
%O 1,1
%A _T. D. Noe_, May 02 2005, revised May 12 2005
%E 4259 found by D. S. McNeil.
%E Edited by _Robert Israel_, Mar 27 2024
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