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

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Apr 01 2013

STATUS

approved