OFFSET
1,1
COMMENTS
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.
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.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..100
EXAMPLE
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.
MATHEMATICA
A[n_, x_]:=A[n, x]=Sum[(2k+1)*x^(n-k), {k, 0, n}]
B[n_, x_]:=B[n, x]=Sum[(2k+1)*x^k, {k, 0, n}]
Do[Do[Do[If[PrimeQ[A[n, b]]==True&&PrimeQ[B[n, b]]==True, Print[n, " ", b]; Goto[aa]], {b, 2n+2, 10^7}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 1, 20}]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 26 2013
STATUS
approved