OFFSET
0,4
COMMENTS
Motivated by the "particularly poor polynomial" n^6+1091 (composite for n=1,...,3905) mentioned on Weisstein's page about prime generating polynomials.
We have a(n) = 0 if n is a cube n = g^3 with g > 1 because then k^6 + g^3 = (k^2 + g)*(k^4 - k^2*g + g^2), which can be prime only when n = g = 1. - T. D. Noe, Nov 18 2013
By the theorem of Brillhart, Filaseta and Odlyzko (see link), if a(n) > n > 1 then x^6 + n must be irreducible. If x^6 + n is irreducible, the Bunyakovsky conjecture implies a(n) is finite. - Robert Israel, Apr 25 2016
LINKS
Robert Israel, Table of n, a(n) for n = 0..10000
J. Brillhart, M. Filaseta, and A. Odlyzko, On an irreducibility theorem of A. Cohn, Canadian Journal of Mathematics 33(1981), 1055-1059.
Eric Weisstein's World of Mathematics, Prime-Generating Polynomial.
MAPLE
f:= proc(n) local exact, x, k, F, nf, F1, C;
iroot(n, 3, exact);
if exact and n > 1 then return 0 fi;
if irreduc(x^6+n) then
for k from 1+(n mod 2) by 2 do if isprime(k^6+n) then return k fi od
else
F:= factors(x^6+n)[2]; #
F1:= map(t -> t[1], F);
nf:= nops(F);
C:= map(t -> op(map(rhs@op, {isolve(t^2-1)})), F1);
for k in sort(convert(select(type, C, positive), list)) do
if isprime(k^6+n) then return k fi
od:
0
fi
end proc:
map(f, [$0..100]); # Robert Israel, Apr 25 2016
MATHEMATICA
{0, 1}~Join~Table[If[IrreduciblePolynomialQ[x^6 + n], SelectFirst[Range[1 + Mod[n, 2], 10^3, 2], PrimeQ[#^6 + n] &], 0], {n, 2, 120}] (* Michael De Vlieger, Apr 25 2016, Version 10 *)
PROG
(PARI) {(a, b=6)->#factor(x^b+a)~==1 & for(n=1, 9e9, ispseudoprime(n^b+a)&return(n)); a==1&return(1); print1("/*"a":", factor(x^b+a)"*/")} /* For illustrative purpose only. The polynomial x^6+a is factored to avoid an infinite loop when it is composite. But there could be x such that this is prime, when all factors but one are 1 (not for exponent b=6, but, e.g., x=4 for exponent b=4), see A225766. */
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
M. F. Hasler, Jul 25 2013
STATUS
approved