cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

Original entry on oeis.org

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

Views

Author

M. F. Hasler, Jul 25 2013

Keywords

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

Crossrefs

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

Programs

  • 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 *)
  • 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. */

A225770 Least k > 0 such that k^8 + n is prime, or 0 if there is no such k.

Original entry on oeis.org

0, 1, 1, 4, 1, 12, 1, 2, 3, 110, 1, 6, 1, 2, 195, 2, 1, 6, 1, 40, 3, 2, 1, 66, 25, 2, 9, 2, 1, 180, 1, 22, 15, 58, 25, 408, 1, 2, 3, 10, 1, 12, 1, 4, 465, 4, 1, 12, 5, 10, 147, 2, 1, 6, 35, 2, 45, 2, 1, 570, 1, 2, 21, 4, 0, 6, 1, 6, 9, 100, 1
Offset: 0

Views

Author

M. F. Hasler, Jul 25 2013

Keywords

Comments

See A225768 for motivation and references.

Examples

			a(0) = 0 since k^8 is not prime for any k > 0.
a(4) = 1 since k^8 + 4 is prime for k = 1, although k^8 + 4 = (k^4 - 2k^2 + 2)(k^4 + 2k^2 + 2), but the first factor equals 1 for k = 1.
a(64) = 0 since k^8 + 64 = (k^4 - 4*k^2 + 8)(k^4 + 4k^2 + 8) which is composite for all integers k > 1.
		

Crossrefs

See A085099, A225765, ..., A225769 for the k^2, k^3, ..., k^7 analogs.

Programs

  • PARI
    A225770(a,b=8)={#factor(x^b+a)~==1&for(n=1,9e9,ispseudoprime(n^b+a)&return(n));a==0 || a==64 || 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 n=4, cf. example.
Showing 1-2 of 2 results.