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A158721
Primes p such that (p + 1)/3 + p is prime.
3
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
1,1
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-SierpiƄski 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 *)
KEYWORD
nonn
AUTHOR
EXTENSIONS
Title simplified by Alonso del Arte, Mar 12 2016
STATUS
approved