%I #22 Apr 07 2013 13:26:27

%S 2,3,2,3,17,7,47,3,53,5,137,109,79,11,37,7,59,13,53,251,251,101,467,

%T 149,79,3,83,61,239,31,79,73,73,373,199,5,337,167,17,683,523,269,37,

%U 163,431,163,163,7,487,7,167,163,197,1549,137,503,139,263,151,283

%N Least prime p such that the polynomial sum_{k=0}^n C_k*x^{n-k} is irreducible modulo p, where C_k denotes the Catalan number binomial(2k,k)/(k+1).

%C Conjecture: (i) a(n) does not exceed n^2+n+5 for each n>0, and the Galois group of sum_{k=0}^n C_k*x^{n-k} over the rationals is isomorphic to the symmetric group S_n.

%C (ii) For any positive integer n, the polynomial sum_{k=0}^n binomial(2k,k)*x^{n-k} is irreducible modulo some prime if and only if n is not of the form 2k(k+1), where k is a positive integer.

%C (iii) For any positive integer n, the polynomial sum_{k=0}^n T_k*x^{n-k} is irreducible modulo some prime not exceeding n^2+n+5, where T_k referes to the central trinomial coefficient A002426(k) which is the coefficient of x^k in the expansion of (x^2+x+1)^k.

%H Zhi-Wei Sun, <a href="/A224416/b224416.txt">Table of n, a(n) for n = 1..350</a>

%e a(10) = 5 since sum_{k=0}^{10} C_k*x^{n-k} irreducible modulo 5 but reducible modulo any of 2 and 3.

%e Note also that a(11) = 137 coincides with 11^2+11+5.

%t A[n_,x_]:=A[n,x]=Sum[Binomial[2k,k]/(k+1)*x^(n-k),{k,0,n}]

%t Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+n+5]}];

%t Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

%Y Cf. A000040, A000108, A224480, A224417, A224418, A220072, A223934, A224210, A217785, A217788, A224197, A002426.

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Apr 06 2013