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%I #13 Apr 02 2016 01:55:38
%S 2,5,17,23,53,59,113,149,167,179,197,233,269,347,359,449,557,563,617,
%T 647,683,743,773,797,827,863,977,1049,1103,1187,1319,1367,1373,1409,
%U 1499,1583,1607,1733,1787,1877,1907,1913,1997,2003,2039,2267,2309,2339
%N Primes p such that (p + 1)/3 + p is prime.
%C 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.
%C Therefore the ceiling function is only necessary when the primality testing function requires an integer argument.
%C And so, aside from 2, all terms are congruent to 5 mod 6.
%C 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-Sierpiński conjecture regarding rational numbers. - _Alonso del Arte_, Mar 12 2016
%e 2 is in the sequence because (2 + 1)/3 + 2 = 1 + 2 = 3, which is prime.
%e 5 is in the sequence because (5 + 1)/3 + 5 = 2 + 5 = 7, which is prime.
%e 11 is not in the sequence because (11 + 1)/3 + 11 = 15 = 3 * 5.
%t Select[Prime[Range[350]], PrimeQ[(# + 1)/3 + #] &] (* _Harvey P. Dale_, Feb 24 2013, simplified by _Alonso del Arte_, Mar 12 2016 *)
%Y Cf. A158708, A158709, A158710, A158711, A158712, A158713, A158714, A158719, A158720, A270384.
%K nonn
%O 1,1
%A _Vladimir Joseph Stephan Orlovsky_, Mar 24 2009
%E Title simplified by _Alonso del Arte_, Mar 12 2016
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