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A371569
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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, but x^5 - x^4 - x^3 - x^2 - x - 1 is reducible (mod p).
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1
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4259, 61643, 94307, 110063, 118171, 348149, 1037903, 1872587, 2149403, 2331859, 2450807, 2490263, 2500847, 2521823, 2534659, 2772179, 2788367, 2789939, 3271883, 3399707, 3550751, 3577487, 3640859, 3861899, 3904309, 4016219, 4063211, 4236719, 4245239, 4368739, 4441007, 4542779, 5033477, 5446283
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OFFSET
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1,1
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COMMENTS
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In each of the first 2000 terms, x^5 - x^4 - x^3 - x^2 - x - 1 splits into linear factors (mod p). Are there any where it does not?
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LINKS
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EXAMPLE
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a(3) = 94307 is a term because 94307 is prime, z^5 - z^4 - z^3 - z^2 - z - 1 = (z + 11827)*(z + 78583)*(z + 54610)*(z + 14536)*(z + 29057) (mod 94307), and the recurrence has period 47153 for all initial conditions except (0,0,0,0,0), as -11827, -78583, -54610, -14536, and -29057 all have multiplicative order 47153 (mod 94307).
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MAPLE
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filter:= proc(p) local Q, q, F, i, z, d, k, kp, G, alpha;
if not isprime(p) then return false fi;
Q:= z^5 - z^4 - z^3 - z^2 - z - 1;
if Irreduc(Q) mod p then return false 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:= NumberTheory:-MultiplicativeOrder(p+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(i, i=3 .. 10^7, 2)]);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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