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A224480
Least prime q such that x^n + sum_{k=1}^n p_k*x^{n-k} is irreducible modulo q, where p_k denotes the k-th prime.
4
2, 11, 2, 2, 2, 2, 2, 53, 13, 3, 5, 2, 2, 2, 2, 421, 29, 19, 7, 2, 29, 37, 2, 743, 41, 23, 13, 47, 5, 2, 269, 139, 211, 31, 73, 307, 2, 2, 5, 89, 23, 839, 181, 379, 173, 89, 2, 353, 101, 307, 3, 29, 389, 2, 863, 71, 503, 619, 193, 2
OFFSET
1,1
COMMENTS
Conjecture: a(n) <= (n+4)*(n+5)+1 for all n>0.
LINKS
EXAMPLE
a(10) = 3 since P(x) = x^{10} + 2*x^9 + 3*x^8 + 5*x^7 + 7*x^6
+ 11*x^5 + 13*x^4 + 17*x^3 + 19*x^2 + 23*x + 29 is irreducible modulo 3, but reducible modulo 2, for,
P(x)==(x+1)^2*(x^3+x+1)*(x^5+x^3+1) (mod 2).
Note also that a(16) = 421 = (16+4)*(16+5)+1.
MATHEMATICA
A[n_, x_]:=A[n, x]=Sum[x^n+Prime[k]*x^(n-k), {k, 1, n}]
Do[Do[If[IrreduciblePolynomialQ[A[n, x], Modulus->Prime[k]]==True, Print[n, " ", Prime[k]]; Goto[aa]], {k, 1, PrimePi[n^2+9n+21]}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 07 2013
STATUS
approved