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

%I A224210 #11 Apr 01 2013 13:00:04
%S A224210 2,11,7,17,11,3,7,97,3,89,31,113,43,7,23,23,17,67,23,109,17,277,103,
%T A224210 283,59,101,157,127,29,79,23,223,73,269,433,137,5,659,109,401,419,7,
%U A224210 373,131,89,269,149,61,829,881
%N A224210 Least prime p such that sum_{k=0}^n (k+1)^2*x^{n-k} is irreducible modulo p.
%C A224210 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.
%C A224210 We have a similar conjecture with the above (k+1)^m replaced by (2k+1)^m.
%H A224210 Zhi-Wei Sun, <a href="/A224210/b224210.txt">Table of n, a(n) for n = 1..300</a>
%e A224210 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
%e A224210    f(x)==x*(x-1)^2 (mod 2),  f(x)==(x-1)*(x^2-x-1) (mod 3)
%e A224210 and
%e A224210           f(x)==(x+1)*(x-1)^2 (mod 5).
%t A224210 A[n_,x_]:=Sum[(k+1)^2*x^(n-k),{k,0,n}]
%t A224210 Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,Prime[4n-3]}];
%t A224210 Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]
%Y A224210 Cf. A000040, A217785, A217788, A218465, A220072, A223934, A224197.
%K A224210 nonn
%O A224210 1,1
%A A224210 _Zhi-Wei Sun_, Apr 01 2013