OFFSET
1,1
COMMENTS
Original title was "Primes p such that Ceiling[p/3] + p is prime." If p = 1 mod 6, then p/3 falls between 2 and 3 mod 6, and the ceiling function bumps it up to 3 mod 6. Therefore ceiling(p/3) + p = 4 mod 6, which is an even number greater than 2 and therefore obviously composite.
Therefore the ceiling function is only necessary when the primality testing function requires an integer argument.
And so, aside from 2, all terms are congruent to 5 mod 6.
Set q = (p + 1)/3 + p, then (p + 1)/(q + 1) = 3/4. If this sequence is proven infinite, that would prove two specific cases of the Schinzel-SierpiĆski conjecture regarding rational numbers. - Alonso del Arte, Mar 12 2016
EXAMPLE
2 is in the sequence because (2 + 1)/3 + 2 = 1 + 2 = 3, which is prime.
5 is in the sequence because (5 + 1)/3 + 5 = 2 + 5 = 7, which is prime.
11 is not in the sequence because (11 + 1)/3 + 11 = 15 = 3 * 5.
MATHEMATICA
Select[Prime[Range[350]], PrimeQ[(# + 1)/3 + #] &] (* Harvey P. Dale, Feb 24 2013, simplified by Alonso del Arte, Mar 12 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Joseph Stephan Orlovsky, Mar 24 2009
EXTENSIONS
Title simplified by Alonso del Arte, Mar 12 2016
STATUS
approved