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A133485
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Integers n such that the polynomial x^(2n+2)+x+1 is primitive and irreducible over GF(2).
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0
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OFFSET
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1,3
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COMMENTS
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An integer n>1 belongs to this sequence iff A100730(n)=2^(2n+3)-2.
Also, an integer n belongs to this sequence iff 2n+2 belongs to A073639.
The polynomial x^(2n+2)+x+1 in the definition can be replaced by its reciprocal x^(2n+2)+x^(2n+1)+1.
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LINKS
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MAPLE
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select(n -> (Irreduc(x^(2*n+2)+x+1) mod 2) and (Primitive(x^(2*n+2)+x+1) mod 2), [$0..500]); # Robert Israel, Aug 17 2015
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PROG
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(PARI) polisprimitive(poli)=np = 2^poldegree(poli)-1; if (type((x^np-1)/poli) != "t_POL", return (0)); forstep(k=np-1, 1, -1, if (type((x^k-1)/poli) == "t_POL", return (0)); ); return(1);
lista(nn) = {for (n=0, nn, poli = Mod(1, 2)*(x^(2*n+2)+x+1); if (polisirreducible(poli) && polisprimitive(poli), print1(n, ", ")); ); } \\ Michel Marcus, May 27 2013
(Sage)
def is_primitive(p):
d = 2^(p.degree())-1
if not p.divides(x^d-1): return False
for k in (d//q for q in d.prime_factors()):
if p.divides(x^k-1): return False
return True
P.<x> = GF(2)[]
for n in range(1, 1000):
p = x^(2*n+2)+x+1
if p.is_irreducible() and is_primitive(p):
print(n)
# Modification of the A002475 Script by Ruperto Corso
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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