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%I #24 Apr 09 2013 12:16:12
%S 2,2,3,2,5,3,71,23,11,2,5,2,13,23,47,47,269,2,7,19,53,101,7,53,113,11,
%T 23,2,43,347,53,283,191,17,41,2,239,677,3,281,37,641,613,41,17,269,
%U 181,137,383,41,127,2,71,739,71,353,59,2,83,2
%N Least prime p such that sum_{k=0}^n (2k+1)*x^(n-k) is irreducible modulo p.
%C Conjecture: a(n) <= n^2+22 for all n>0.
%C We have similar conjectures with 2k+1 in the definition replaced by (2k+1)^m (m=2,3,...).
%H Zhi-Wei Sun, <a href="/A220949/b220949.txt">Table of n, a(n) for n = 1..500</a>
%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;83b8a70.1303">A family of polynomials and a related conjecture on primes</a>, a message to Number Theory List, March 30, 2013.
%e a(5) = 5 since f(x) = x^5+3*x^4+5*x^3+7*x^2+9*x+11 is irreducible modulo 5, but f(x)==(x+1)*(x^2+x+1)^2 (mod 2) and f(x)==(x+1)^4*(x-1) (mod 3).
%e Note also that a(7) = 71 = 7^2+22.
%t A[n_,x_] := A[n,x] = Sum[(2k+1)*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,PrimePi[n^2+22]}]; Print[n," ",counterexample]; Label[aa]; Continue,{n,1,100}]
%Y Cf. A000040, A218465, A224210, A224416, A224417, A224418, A224480, A220072, A223934.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Apr 07 2013
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