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A220947 Least prime p such that sum_{k=0}^n F(k+1)*x^{n-k} is irreducible modulo p, where F(j) denotes the Fibonacci number A000045(j).

%I A220947 #16 Apr 07 2013 23:50:19
%S A220947 2,3,2,11,3,2,5,3,2,11,5,41,181,31,73,89,5,7,71,11,29,5,193,41,89,61,
%T A220947 2,43,3,31,13,191,2,61,103,97,103,47,383,367,89,17,191,1627,193,163,5,
%U A220947 337,349,23,149,193,199,233,173,617,593,59,113,151
%N A220947 Least prime p such that sum_{k=0}^n F(k+1)*x^{n-k} is irreducible modulo p, where F(j) denotes the Fibonacci number A000045(j).
%C A220947 Conjecture: a(n) <= n^2+12 for all n>0.
%C A220947 Such a phenomenon happens quite often. In fact, for many interesting integer sequences a(k) (k=1,2,3,...), each of the polynomials x^n + sum_{k=0}^n a(k)*x^{n-k} (n>0) is irreducible modulo some prime not exceeding a*n^2+b*n+c, where a, b, c are suitable nonnegative constants.
%H A220947 Zhi-Wei Sun, <a href="/A220947/b220947.txt">Table of n, a(n) for n = 1..450</a>
%e A220947 a(2) = 3 since x^2+x+2 is irreducible modulo 3 but reducible modulo 2.
%e A220947 Note also that a(13) = 181 = 13^2+12.
%t A220947 A[n_,x_]:=A[n,x]=Sum[Fibonacci[k+1]*x^(n-k),{k,0,n}]
%t A220947 Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+12]}];
%t A220947 Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]
%Y A220947 Cf. A000040, A000045, A224416, A224417, A224418, A224480, A224210, A220072, A223934, A218465.
%K A220947 nonn
%O A220947 1,1
%A A220947 _Zhi-Wei Sun_, Apr 07 2013