%I #14 Feb 07 2017 17:35:08
%S 17,31,347,6863,493043,1092731,1295033,21253937,22665191,38272757,
%T 54439943,115501307,904231067,1121622323,2738124203,3067586681,
%U 3301293173,3673650011,4549540397,4599141251,6507781367,7222633241
%N Primes of form p^2+q^3 where p and q are primes.
%C p and q cannot both be odd, thus p=2 or q=2. If q=2 then we want primes of form p^2+8. But 8=-1 mod 3. Since p is prime, p=3 or == 1 or 2 mod 3. If p=1 or 2 mod 3 then 3|p^2+8, so p=3. Therefore with the exception of the first entry (3^2+8=17) this sequence is really just primes of the form q^3+4.
%H Ray Chandler, <a href="/A045700/b045700.txt">Table of n, a(n) for n = 1..10000</a>
%F Primes in A045699.
%e a(4) = 6863 = 19^3 + 2^2.
%p for n from 1 to 1000 do if (isprime((ithprime(n))^3+4)) then print((ithprime(n))^3+4,4); fi; if (isprime((ithprime(n))^2+8)) then print((ithprime(n))^2+8,8); fi; od;
%t Join[{17},Select[Prime[Range[300]]^3+4,PrimeQ]] (* _Harvey P. Dale_, Jul 20 2011 *)
%o (PARI) list(lim)=my(v=List([17]), t); lim\=1; forprime(p=3,sqrtnint(lim\1-4,3), if(isprime(t=p^3+4), listput(v, t))); Set(v) \\ _Charles R Greathouse IV_, Feb 07 2017
%Y Cf. A045699.
%K nice,nonn,easy
%O 1,1
%A _Felice Russo_
%E Extension and comment from Joe DeMaio (jdemaio(AT)kennesaw.edu)