OFFSET
1,1
COMMENTS
For (p^2+2)/33 to be an integer, p must be congruent to 8, 14, 19, or 25 (mod 33). Examples: 107 = 8, 47 = 14, 19 = 19, and 487 = 25 (mod 33). [From Michael B. Porter, Oct 20 2009]
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..1000
EXAMPLE
p=19: (19^2+2)/33=11; p=47: (47^2+2)/33=67; p=107: (107^2+2)/33=347
MATHEMATICA
Select[Prime[Range[1000]], PrimeQ[(#^2+2)/33]&] (* Harvey P. Dale, Feb 18 2012 *)
CROSSREFS
KEYWORD
nonn,less
AUTHOR
Vincenzo Librandi, Sep 24 2009
EXTENSIONS
Corrected and extended by Charles R Greathouse IV, Oct 09 2009
STATUS
approved