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Primes of the form (k^2+2)/6.
2

%I #11 Mar 13 2015 22:17:49

%S 3,11,17,43,67,113,131,193,241,353,523,641,683,1291,1601,1667,1873,

%T 2017,2243,2731,3083,3361,3851,4483,4817,4931,5281,5521,7211,8363,

%U 8513,8971,9283,9923,10753,11971,13633,16433,17713,18371,18593,19267,21841,22571

%N Primes of the form (k^2+2)/6.

%H Chai Wah Wu, <a href="/A245045/b245045.txt">Table of n, a(n) for n = 1..1500</a>

%e When k=4, (k^2+2)/6 = 3 is prime, so 4 is a member of the sequence. since putting k = 0, 1, 2, or 3 does not give a prime, so 4 is the first term.

%o (Python)

%o import sympy

%o [(k**2+2)/6 for k in range(10**6) if sympy.ntheory.isprime((k**2+2)/6) & ((k**2+2)/6).is_integer()]

%Y Cf. A154616, A002327, A066436. First 5 terms equal to A078116. First 4 terms equal to A127996.

%K nonn

%O 1,1

%A _Chai Wah Wu_, Jul 10 2014