login
Least k>0 such that k^5+n is prime, or 0 if k^5+n is never prime.
3

%I #19 Apr 26 2016 00:15:47

%S 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,

%T 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,

%U 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

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

%C See A225768 for motivation and references.

%C 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

%H Robert Israel, <a href="/A225767/b225767.txt">Table of n, a(n) for n = 0..10000</a>

%H J. Brillhart, M. Filaseta, A. Odlyzko, <a href="https://dx.doi.org/10.4153%2FCJM-1981-080-0">On an irreducibility theorem of A. Cohn</a>. Canad. J. Math. 33(1981), 1055-1059.

%e a(3)=8 because 1^5+3, 2^5+3, ..., 7^5+3 are all composite, but 8^5+3=32771 is prime.

%e 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.

%p f:= proc(n) local x,k,F,nf,F1,C;

%p if irreduc(x^5+n) then

%p for k from 1+(n mod 2) by 2 do if isprime(k^5+n) then return k fi od

%p else

%p F:= factors(x^5+n)[2]; #

%p F1:= map(t -> t[1],F);

%p nf:= nops(F);

%p C:= map(t -> op(map(rhs@op,{isolve(t^2-1)})),F1);

%p for k in sort(convert(select(type,C,positive),list)) do

%p if isprime(k^5+n) then return k fi

%p od:

%p 0

%p fi

%p end proc:

%p map(f, [$0..100]); # _Robert Israel_, Apr 25 2016

%t {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 *)

%o (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.

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

%K nonn

%O 0,4

%A _M. F. Hasler_, Jul 25 2013