A225770 -id:A225770 - cp's OEIS frontend

A225768 Least k > 0 such that k^6 + n is prime, or 0 if k^6 + n is never prime.

0, 1, 1, 2, 1, 18, 1, 2, 0, 2, 1, 54, 1, 28, 3, 2, 1, 18, 1, 2, 399, 26, 1, 6, 5, 2, 21, 0, 1, 288, 1, 4, 3, 2, 105, 6, 1, 2, 33, 2, 1, 546, 1, 2, 3, 2, 1, 6, 35, 2, 51, 20, 1, 12, 5, 28, 9, 4, 1, 18, 1, 4, 63, 2, 0, 18, 1, 2, 3, 28, 1, 6, 1, 2, 15, 2, 35, 24, 1, 12, 3, 4, 1, 42, 115, 2, 111, 2, 1, 18, 91, 6, 3, 2, 3, 6, 1, 28, 3, 2

Offset: 0

M. F. Hasler, Jul 25 2013

nonn

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

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.

See A085099, A225765, A225767, A225769, A225770 for the k^2, k^3, ..., k^8 analogs.

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

{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 *)

{(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. */

A225766 Least k>0 such that k^4+n is prime, or 0 if k^4+n is always composite.

Original entry on oeis.org

0, 1, 1, 2, 1, 6, 1, 2, 3, 10, 1, 6, 1, 2, 165, 2, 1, 12, 1, 20, 3, 2, 1, 6, 35, 2, 3, 2, 1, 90, 1, 2, 3, 8, 5, 12, 1, 2, 9, 10, 1, 60, 1, 2, 75, 2, 1, 18, 5, 20, 3, 2, 1, 12, 85, 2, 3, 2, 1, 30, 1, 4, 21, 2, 0, 6, 1, 2, 3, 10, 1, 6, 1, 2, 255, 4, 3, 6, 1, 10, 27, 2, 1, 72, 5, 2, 3, 2, 1, 570, 11, 2, 3, 2, 5, 18, 1, 2, 3, 10

Offset: 0

M. F. Hasler, Jul 25 2013

nonn

See A225768 for motivation and references.

			a(4)=1 because 1^4+4=5 is prime. (Although x^4+4 = (x^2-2*x+2)(x^2+2x+2), this is prime for x=1 when the first factor equals 1.)
a(5)=6 because 1^4+5=6, 2^4+5=21, 3^4+5=86, 4^4+5=261 and 5^4+5 are all composite, but 6^4+5=1301 is prime.
a(64)=0 because x^4+64 = (x^2-4*x+8)(x^2+4x+8) is composite for all integer values of x>0. Indeed, x^2-4x+8=(x-2)^2+4 > 1 for all x.

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

{(a,b=4)->#factor(x^b+a)~==1&for(n=1,9e9,ispseudoprime(n^b+a)&return(n));a==1 || a==4 || 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 but one factors equal 1. This happens for n=4, cf. Example.

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

Original entry on oeis.org

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

M. F. Hasler, Jul 25 2013

nonn

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

			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.

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.

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

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

{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 *)

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.

A225769 Least k>0 such that k^7+n is prime, or 0 if there is no such k.

Original entry on oeis.org

0, 1, 1, 2, 1, 6, 1, 16, 9, 2, 1, 2, 1, 6, 5, 22, 1, 8, 1, 10, 3, 2, 1, 2, 17, 12, 3, 4, 1, 2, 1, 6, 5, 4, 3, 2, 1, 4, 5, 2, 1, 6, 1, 4, 23, 2, 1, 24, 5, 4, 3, 2, 1, 2, 5, 6, 3, 28, 1, 8, 1, 22, 39, 2, 3, 2, 1, 4, 5, 2, 1, 2, 1, 6, 9, 34, 7, 6, 1, 10, 3, 16, 1, 2, 23, 22, 3, 14, 1, 14, 23, 12, 9, 4, 3, 2, 1, 4, 9, 2, 1, 2, 1, 4, 5, 2, 1, 8, 1, 4, 3, 2, 1, 2, 23, 12, 5, 40, 31, 92, 7

Offset: 0

M. F. Hasler, Jul 25 2013

nonn

See A225768 for motivation and references.

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

(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.

cp's OEIS Frontend

A225768 Least k > 0 such that k^6 + n is prime, or 0 if k^6 + n is never prime.

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A225766 Least k>0 such that k^4+n is prime, or 0 if k^4+n is always composite.

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

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Maple

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PARI

A225769 Least k>0 such that k^7+n is prime, or 0 if there is no such k.

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PARI