A192618 Prime powers p^k with even exponents k > 0 such that (1 + p^k)/2 is prime.

9, 25, 81, 121, 361, 625, 841, 2401, 3481, 3721, 5041, 6241, 10201, 14641, 17161, 19321, 28561, 32761, 39601, 73441, 83521, 121801, 143641, 167281, 201601, 212521, 271441, 279841, 323761, 326041, 398161, 410881, 436921, 546121, 564001, 674041

Offset: 1

Author

Klaus Brockhaus, Jul 05 2011

Keywords

• nonn

Comments

Subsequence of A056798.

From R. J. Mathar, Jul 11 2011: (Start)

For odd k we first have the case k=1, where (1+p)/2 is either classified as A005383 or A176897.

For odd k >= 3, (1+p^k)/2 is not prime. [Sketch of proof: for p=2 it is not integer. Otherwise for odd k, (1+p^k)/(1+p) = Sum_{j=0..k-1} (-p)^j, an integer, so 1+p^k is a multiple of 1+p. For odd p, (1+p^k)/2 is a multiple of (1+p)/2 and therefore composite.] (End)

Links

• Klaus Brockhaus, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A056798.

Programs

Magma

e:=20; u:=1000; z:=Min(2^e, u^2); S:=[ q: p in PrimesUpTo(u), k in [2..e by 2] | q le z and IsEven(1+q) and IsPrime((1+q) div 2) where q is p^k ]; Sort(~S); S;

Mathematica

Select[Union[Flatten[Table[Prime[n]^k, {n, 142}, {k, 0, 32, 2}]]], PrimeQ[(# + 1)/2] &] (* Alonso del Arte, Jul 05 2011 *)