%I #20 Apr 06 2013 18:38:44
%S 2,2,2,3,2,2,3,2,5,2,3,2,7,3,5,2,2,2,2,7,5,3,2,3,5,2,5,2,11,3,3,2,3,2,
%T 2,7,5,2,5,2,2,2,19,5,2,3,2,3,2,7,3,7,7,11,3,5,2,43,5,3
%N Least prime q such that (x^{p_n}-1)/(x-1) is irreducible modulo q, where p_n is the n-th prime.
%C It is well known that (x^{p^n}-1)/(x^{p^{n-1}}-1) is irreducible over the rationals for any prime p and positive integer n.
%C We have the following "Reciprocity Law": For any positive integer n and primes p > 2 and q, the cyclotomic polynomial (x^{p^n}-1)/(x^{p^{n-1}}-1) is irreducible modulo q if and only if q is a primitive root modulo p^n.
%C This can be proved as follows: As any monic irreducible polynomial over F_q=Z/qZ of degree k divides x^{q^k}-x in the ring F_q[x], the polynomial f(x)= (x^{p^n}-1)/(x^{p^{n-1}}-1) in F_q[x] has an irreducible factor of degree k < deg f if and only if f(x) is not coprime to x^{q^k}-x for some k < p^n-p^{n-1}. Note that gcd(x^{p^n}-1,x^{q^k-1}-1) = x^{gcd(p^n,q^k-1)}-1. If p^n | q^k-1, then x^{p^n}-1 | x^{q^k}-x and hence f(x) divides x^{q^k}-x; if p^n does not divide q^k-1, then gcd(x^{p^n}-1,x^{q^k-1}-1) divides x^{p^{n-1}}-1 and hence f(x) is coprime to x^{q^k}-x. Thus, f(x) is irreducible modulo q, if and only if p^n | q^k-1 for no 0 < k < p^n-p^{n-1}, i.e., q is a primitive root modulo p^n.
%C By the above "Reciprocity Law" in the case n=1, we have a(k) = A002233(k) for all k > 1.
%C Conjecture: a(n) <= sqrt(7*p_n) for all n>0.
%H Zhi-Wei Sun, <a href="/A223942/b223942.txt">Table of n, a(n) for n = 1..450</a>
%e a(9)=5 since f(x)=(x^{23}-1)/(x-1) is irreducible modulo 5, but reducible modulo either of 2 and 3, for,
%e f(x)==(x^{11}+x^9+x^7+x^6+x^5+x+1)
%e *(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1) (mod 2)
%e and
%e f(x)==(x^{11}-x^8-x^6+x^4+x^3-x^2-x-1)
%e *(x^{11}+x^{10}+x^9-x^8-x^7+x^5+x^3-1) (mod 3).
%t Do[Do[If[IrreduciblePolynomialQ[Sum[x^k,{k,0,Prime[n]-1}],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[Sqrt[7*Prime[n]]]}];
%t Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]
%Y Cf. A000040, A002233, A122028, A217785, A217788, A218465, A220072, A223934.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Mar 29 2013
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