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.

4, 6, 8, 13, 54, 21, 56, 141, 282, 335, 132, 82, 3752, 93, 40, 5141, 774, 204, 60790, 27366, 270, 31591, 60, 247, 1976, 4848, 7112, 4954, 62808, 84, 17912, 78441, 3696, 8083, 5754, 19210, 21154, 17973, 59580

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

Cf. A000040, A217785, A217788.

Sequence in context: A023560 A340960 A261680 * A359676 A050902 A374190

Adjacent sequences: A218462 A218463 A218464 * A218466 A218467 A218468

Keyword

nonn

Author

Zhi-Wei Sun, Mar 26 2013

Status

approved