A270384 -id:A270384 - cp's OEIS frontend

2, 5, 17, 23, 53, 59, 113, 149, 167, 179, 197, 233, 269, 347, 359, 449, 557, 563, 617, 647, 683, 743, 773, 797, 827, 863, 977, 1049, 1103, 1187, 1319, 1367, 1373, 1409, 1499, 1583, 1607, 1733, 1787, 1877, 1907, 1913, 1997, 2003, 2039, 2267, 2309, 2339

Offset: 1

Vladimir Joseph Stephan Orlovsky, Mar 24 2009

nonn

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-Sierpiński conjecture regarding rational numbers. - Alonso del Arte, Mar 12 2016

			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.

Cf. A158708, A158709, A158710, A158711, A158712, A158713, A158714, A158719, A158720, A270384.

Select[Prime[Range[350]], PrimeQ[(# + 1)/3 + #] &] (* Harvey P. Dale, Feb 24 2013, simplified by Alonso del Arte, Mar 12 2016 *)

Title simplified by Alonso del Arte, Mar 12 2016

cp's OEIS Frontend

A158721 Primes p such that (p + 1)/3 + p is prime.

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