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Primes p with prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 both prime.
5

%I #5 Jan 20 2014 02:44:53

%S 3,139,179,233,491,929,1217,1429,1597,1613,1987,2243,3061,3499,3529,

%T 4507,5737,5779,6329,7247,7823,8263,8839,9941,10259,11317,11383,12157,

%U 12421,13093,13219,13367,14449,14669,15101,15877,17449,18523,18593,19051

%N Primes p with prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 both prime.

%C By part (i) of the conjecture in A236192, this sequence should have infinitely many terms.

%H Zhi-Wei Sun, <a href="/A236193/b236193.txt">Table of n, a(n) for n = 1..10000</a>

%e a(1) = 3 since prime(2)^2 + (2*2)^2 = 25 is composite, but prime(3)^2 + (2*3)^2 = 5^2 + 6^2 = 61 and 3^2 + (2*prime(3))^2 = 3^2 + 10^2 = 109 are both prime.

%t p[n_]:=PrimeQ[Prime[n]^2+(2*n)^2]&&PrimeQ[n^2+(2*Prime[n])^2]

%t n=0;Do[If[p[Prime[k]],n=n+1;Print[n," ",Prime[k]]],{k,1,10^6}]

%Y Cf. A000040, A236119, A236143, A236192.

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Jan 20 2014