A085398 - OEIS

A085398

Let Cn(x) be the n-th cyclotomic polynomial; a(n) is the least k>1 such that Cn(k) is prime.

3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 6, 2, 4, 3, 2, 10, 2, 22, 2, 2, 4, 6, 2, 2, 2, 2, 2, 14, 3, 61, 2, 10, 2, 14, 2, 15, 25, 11, 2, 5, 5, 2, 6, 30, 11, 24, 7, 7, 2, 5, 7, 19, 3, 2, 2, 3, 30, 2, 9, 46, 85, 2, 3, 3, 3, 11, 16, 59, 7, 2, 2, 22, 2, 21, 61, 41, 7, 2, 2, 8, 5, 2, 2

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OFFSET

1,1

COMMENTS

Conjecture: a(n) is defined for all n. - Eric Chen, Nov 14 2014

Existence of a(n) is implied by Bunyakovsky's conjecture. - Robert Israel, Nov 13 2014

LINKS

Jinyuan Wang, Table of n, a(n) for n = 1..5000 (terms 1..1500 from Eric Chen)

Wikipedia, Bunyakowsky conjecture

FORMULA

a(A072226(n)) = 2. - Eric Chen, Nov 14 2014

a(n) = A117544(n) except when n is a prime power, since if n is a prime power, then A117544(n) = 1. - Eric Chen, Nov 14 2014

a(prime(n)) = A066180(n), a(2*prime(n)) = A103795(n), a(2^n) = A056993(n-1), a(3^n) = A153438(n-1), a(2*3^n) = A246120(n-1), a(3*2^n) = A246119(n-1), a(6^n) = A246121(n-1), a(5^n) = A206418(n-1), a(6*A003586(n)) = A205506(n), a(10*A003592(n)) = A181980(n).

EXAMPLE

a(11) = 5 because C11(k) is composite for k = 2, 3, 4 and prime for k = 5.

a(37) = 61 because C37(k) is composite for k = 2, 3, 4, ..., 60 and prime for k = 61.

MAPLE

f:= proc(n) local k;

for k from 2 do if isprime(numtheory:-cyclotomic(n, k)) then return k fi od

end proc:

seq(f(n), n = 1 .. 100); # Robert Israel, Nov 13 2014

MATHEMATICA

Table[k = 2; While[!PrimeQ[Cyclotomic[n, k]], k++]; k, {n, 300}] (* Eric Chen, Nov 14 2014 *)

PROG

(PARI) a(n) = k=2; while(!isprime(polcyclo(n, k)), k++); k; \\ Michel Marcus, Nov 13 2014