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

1,1

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.

Links

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

Example

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).

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}]

Crossrefs

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

Sequence in context: A323783 A306537 A170873 * A344776 A368640 A290572

Adjacent sequences: A224207 A224208 A224209 * A224211 A224212 A224213

Keyword

nonn

Author

Zhi-Wei Sun, Apr 01 2013

Status

approved