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.

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

Author

Zhi-Wei Sun, Apr 07 2013

Keywords

• nonn

Comments

Conjecture: a(n) <= (n+4)*(n+5)+1 for all n>0.

Examples

			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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A000040, A220947, A220072, A223934, A224210, A224416, A224417, A224418, A217788, A224197, A218465, A217785.

Programs

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}]