A224210 Least prime p such that sum_{k=0}^n (k+1)^2*x^{n-k} is irreducible modulo p.

2, 11, 7, 17, 11, 3, 7, 97, 3, 89, 31, 113, 43, 7, 23, 23, 17, 67, 23, 109, 17, 277, 103, 283, 59, 101, 157, 127, 29, 79, 23, 223, 73, 269, 433, 137, 5, 659, 109, 401, 419, 7, 373, 131, 89, 269, 149, 61, 829, 881

Offset: 1

Author

Zhi-Wei Sun, Apr 01 2013

Keywords

• nonn

Comments

Conjecture: a(n) does not exceed the (4n-3)-th prime for each n>0. Moreover, for any integers m>1 and n>0 the polynomial sum_{k=0}^n (k+1)^m*x^{n-k} is irreducible modulo some prime, and its Galois group over the rationals is isomorphic to the symmetric group S_n. Also, for m,n=2,3,... there are infinitely many integers b > n^m such that [n^m,...,2^m,1^m] in base b is prime.

We have a similar conjecture with the above (k+1)^m replaced by (2k+1)^m.

Examples

			a(3) = 7 since f(x) = x^3+4x^2+9x+16 is irreducible modulo 7 but reducible modulo any of 2, 3, 5. Note that
   f(x)==x*(x-1)^2 (mod 2),  f(x)==(x-1)*(x^2-x-1) (mod 3)
and
          f(x)==(x+1)*(x-1)^2 (mod 5).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A000040, A217785, A217788, A218465, A220072, A223934, A224197.

Programs

Mathematica

A[n_,x_]:=Sum[(k+1)^2*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,Prime[4n-3]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]