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A218465 Least integer b>2n+1 such that the numbers written as [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime.

%I A218465 #40 Apr 01 2013 12:59:30
%S A218465 4,6,8,13,54,21,56,141,282,335,132,82,3752,93,40,5141,774,204,60790,
%T A218465 27366,270,31591,60,247,1976,4848,7112,4954,62808,84,17912,78441,3696,
%U A218465 8083,5754,19210,21154,17973,59580
%N A218465 Least integer b>2n+1 such that the numbers written as [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime.
%C A218465 Conjecture: Let n be any positive integer. Then a(n) exists, moreover there are infinitely many integers b > 2n+1 such that [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime. Also, the polynomial S_n(x) = sum_{k=0}^n (2k+1)*x^{n-k} is irreducible modulo some prime p < (n+1)(n+2), and the Galois group of S_n(x) over the field of rational numbers is isomorphic to the symmetric group S_n.
%C A218465 This conjecture can be extended by replacing 2k+1 by (2k+1)^m. For example, [1^2,3^2,5^2,...,61^2,63^2] and [63^2,61^2,...,3^2,1^2] in base b=241784 are both prime.
%H A218465 Zhi-Wei Sun, <a href="/A218465/b218465.txt">Table of n, a(n) for n = 1..100</a>
%e A218465 a(2)=6, since [1,3,5] in base 6 is 1*6^2+3*6+5=59, [5,3,1] in base 6 is 5*6^2+3*6+1=199, and both 59 and 199 are prime.
%t A218465 A[n_,x_]:=A[n,x]=Sum[(2k+1)*x^(n-k),{k,0,n}]
%t A218465 B[n_,x_]:=B[n,x]=Sum[(2k+1)*x^k,{k,0,n}]
%t A218465 Do[Do[Do[If[PrimeQ[A[n,b]]==True&&PrimeQ[B[n,b]]==True,Print[n," ",b];Goto[aa]],{b,2n+2,10^7}];
%t A218465 Print[n," ",counterexample];Label[aa];Continue,{n,1,20}]]
%Y A218465 Cf. A000040, A217785, A217788.
%K A218465 nonn
%O A218465 1,1
%A A218465 _Zhi-Wei Sun_, Mar 26 2013