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A225767
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Least k>0 such that k^5+n is prime, or 0 if k^5+n is never prime.
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2
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0, 1, 1, 8, 1, 2, 1, 4, 3, 2, 1, 2, 1, 6, 3, 2, 1, 6, 1, 10, 3, 2, 1, 14, 7, 4, 3, 2, 1, 2, 1, 22, 0, 8, 3, 2, 1, 4, 3, 2, 1, 2, 1, 10, 5, 4, 1, 2, 13, 10, 3, 2, 1, 6, 17, 12, 5, 2, 1, 12, 1, 12, 5, 4, 3, 2, 1, 4, 3, 2, 1, 2, 1, 4, 3, 2, 7, 2, 1, 4, 63, 2, 1, 18, 5, 4, 11, 32, 1, 14, 11, 6, 5, 4, 3, 2, 1, 6, 11, 2
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OFFSET
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0,4
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COMMENTS
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See A225768 for motivation and references.
By the theorem of Brillhart, Filaseta and Odlyzko (see link), if a(n) > n > 1 then x^5 + n must be irreducible. If x^5 + n is irreducible, the Bunyakovsky conjecture implies a(n) is finite. - Robert Israel, Apr 25 2016
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LINKS
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EXAMPLE
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a(3)=8 because 1^5+3, 2^5+3, ..., 7^5+3 are all composite, but 8^5+3=32771 is prime.
a(32)=0 because x^5+32 = (x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16) is composite for all integer values of x>0.
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MAPLE
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f:= proc(n) local x, k, F, nf, F1, C;
if irreduc(x^5+n) then
for k from 1+(n mod 2) by 2 do if isprime(k^5+n) then return k fi od
else
F:= factors(x^5+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^5+n) then return k fi
od:
0
fi
end proc:
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MATHEMATICA
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{0, 1}~Join~Table[If[IrreduciblePolynomialQ[x^5 + n], SelectFirst[Range[1 + Mod[n, 2], 10^2, 2], PrimeQ[#^5 + n] &], 0], {n, 2, 120}] (* Michael De Vlieger, Apr 25 2016, Version 10 *)
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PROG
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(PARI) A225767(a, b=5)={#factor(x^b+a)~==1&for(n=1, 9e9, ispseudoprime(n^b+a)&return(n)); a==1 || print1("/*"factor(x^b+a)"*/")} \\ For illustrative purpose only. The polynomial is factored to avoid an infinite search loop when it is composite. But a factored polynomial can yield a prime when all factors but one equal 1. This happens for b=4, n=4, cf. A225766.
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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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