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A225766
Least k>0 such that k^4+n is prime, or 0 if k^4+n is always composite.
4
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
0,4
COMMENTS
See A225768 for motivation and references.
EXAMPLE
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.
PROG
(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.
CROSSREFS
See A085099, A225765--A225770 for the k^2, k^3, ..., k^8 analogs.
Sequence in context: A178254 A085099 A193807 * A334085 A249831 A304527
KEYWORD
nonn
AUTHOR
M. F. Hasler, Jul 25 2013
STATUS
approved