A225766 Least k>0 such that k^4+n is prime, or 0 if k^4+n is always composite.
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
Keywords
Examples
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.
Programs
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PARI
{(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.
Comments