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A223942
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Least prime q such that (x^{p_n}-1)/(x-1) is irreducible modulo q, where p_n is the n-th prime.
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2
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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, 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
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OFFSET
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1,1
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COMMENTS
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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.
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.
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.
By the above "Reciprocity Law" in the case n=1, we have a(k) = A002233(k) for all k > 1.
Conjecture: a(n) <= sqrt(7*p_n) for all n>0.
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LINKS
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EXAMPLE
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a(9)=5 since f(x)=(x^{23}-1)/(x-1) is irreducible modulo 5, but reducible modulo either of 2 and 3, for,
f(x)==(x^{11}+x^9+x^7+x^6+x^5+x+1)
*(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1) (mod 2)
and
f(x)==(x^{11}-x^8-x^6+x^4+x^3-x^2-x-1)
*(x^{11}+x^{10}+x^9-x^8-x^7+x^5+x^3-1) (mod 3).
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MATHEMATICA
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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]]]}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 1, 100}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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