1,2

The sequence forms a subset of A016777, as explained below:

For each prime p<>3 we have p^2 =1 (mod 3), see A024700.

(i) For the k where 2*k=2 (mod 3), that is where k=1 (mod 3), this leads to 2*k+p^2=0 (mod 3), so the 2*k+p^2 are divisible by 3 (not prime) unless p=3.

The subcase where 2*k+3^2 is prime generates this sequence here; the subcase where it is not generates A138685.

(ii) For the k where 2*k=0 (mod 3), that is where k=0 (mod 3), one can select any p^2 =1 (mod 3)

to generate a prime 2*k+p^2 = 1 (mod 3), so these k generate many primes (of the form A002476).

(iii) For the k where 2*k=1 (mod 3), that is where k=2 (mod 3), one can select any p^2 =1 (mod 3)

to generate a prime 2*k+p^2 = 2 (mod 3), so these k generate many primes (of the form A003627).

The unique primes associated with each n are in A007528: n=1 associated with A007528(2)=11=2*1+3^2,

n=4 associated with A007528(3)=17=2*4+3^2 etc.

{This sequence here} Union {A138685} = {A016777}.

3 is not in the sequence because {6+2^2, 6+3^3, 6+5^2, 6+7^2,..} = {10, 15, 31, 55,..,127,..,367,..}

contains the primes 31, 127, 367,..., generated with p=5,11,19...

4 is in the sequence because {8+2^2, 8+3^3, 8+5^2, 8+7^2,..} = {12, 17, 33, 57,...} contains

only one prime (that is, 17), generated with p=3.

b = {}; Do[a = {}; Do[If[PrimeQ[2*k + Prime[n]^2], AppendTo[a, k]], {n, 1, 100}]; If[Length[a] < 2, AppendTo[b, a]], {k, 1, 500}]; Union[Flatten[b]]

Cf. A138479, A138685, A138686, A138691, A138692, A138693, A138694, A007528, A138696, A138697, A138698, A138699, A138700.

Sequence in context: A137379 A096676 A092863 this_sequence A115288 A153003 A128429

Adjacent sequences: A138691 A138692 A138693 this_sequence A138695 A138696 A138697

nonn

Artur Jasinski (grafix(AT)csl.pl), Mar 27 2008

Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 15 2009