A225767 Least k>0 such that k^5+n is prime, or 0 if k^5+n is never prime.

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

Offset

0,4

Comments

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

Links

Robert Israel, Table of n, a(n) for n = 0..10000

J. Brillhart, M. Filaseta, A. Odlyzko, On an irreducibility theorem of A. Cohn. Canad. J. Math. 33(1981), 1055-1059.

Example

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.

Maple

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:
map(f, [$0..100]); # Robert Israel, Apr 25 2016

Mathematica

{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 *)

Prog

(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.

Crossrefs

See A085099, A225765--A225770 for the k^2, k^3, ..., k^8 analogs.

Sequence in context: A271547 A010154 A109011 * A019763 A307506 A179050

Adjacent sequences: A225764 A225765 A225766 * A225768 A225769 A225770

Keyword

nonn

Author

M. F. Hasler, Jul 25 2013

Status

approved