OFFSET
1,2
COMMENTS
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).
n=4 associated with A007528(3)=17=2*4+3^2 etc.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
EXAMPLE
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.
MATHEMATICA
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]]
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Mar 27 2008
EXTENSIONS
Edited by R. J. Mathar, May 15 2009
STATUS
approved