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

Author

Zhi-Wei Sun, Mar 26 2013

Keywords

• nonn

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.

Examples

			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.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A000040, A217785, A217788.

Programs

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}]]