A220949 Least prime p such that sum_{k=0}^n (2k+1)*x^(n-k) is irreducible modulo p.

2, 2, 3, 2, 5, 3, 71, 23, 11, 2, 5, 2, 13, 23, 47, 47, 269, 2, 7, 19, 53, 101, 7, 53, 113, 11, 23, 2, 43, 347, 53, 283, 191, 17, 41, 2, 239, 677, 3, 281, 37, 641, 613, 41, 17, 269, 181, 137, 383, 41, 127, 2, 71, 739, 71, 353, 59, 2, 83, 2

Offset: 1

Author

Zhi-Wei Sun, Apr 07 2013

Keywords

• nonn

Comments

Conjecture: a(n) <= n^2+22 for all n>0.

We have similar conjectures with 2k+1 in the definition replaced by (2k+1)^m (m=2,3,...).

Examples

			a(5) = 5 since f(x) = x^5+3*x^4+5*x^3+7*x^2+9*x+11 is irreducible modulo 5, but f(x)==(x+1)*(x^2+x+1)^2 (mod 2) and f(x)==(x+1)^4*(x-1) (mod 3).
Note also that a(7) = 71 = 7^2+22.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..500
• Zhi-Wei Sun, A family of polynomials and a related conjecture on primes, a message to Number Theory List, March 30, 2013.

Crossrefs

Cf. A000040, A218465, A224210, A224416, A224417, A224418, A224480, A220072, A223934.

Programs

Mathematica

A[n_,x_] := A[n,x] = Sum[(2k+1)*x^(n-k), {k,0,n}]; Do[Do[If[IrreduciblePolynomialQ[A[n,x], Modulus->Prime[k]] == True, Print[n," ",Prime[k]]; Goto[aa]], {k,1,PrimePi[n^2+22]}]; Print[n," ",counterexample]; Label[aa]; Continue,{n,1,100}]