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a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/4 + 1, prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 are all prime}|, where phi(.) is Euler's totient function.
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%I #7 Jan 20 2014 02:44:35

%S 0,0,0,0,0,1,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

%N a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/4 + 1, prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 are all prime}|, where phi(.) is Euler's totient function.

%C Conjecture: (i) a(n) > 0 for all n > 370.

%C (ii) For any integer n >= 700, there is a positive integer k < n such that p = phi(k) + phi(n-k)/3 - 1, prime(p)^2 + (p-1)^2 and p^2 + (prime(p)-1)^2 are all prime.

%C Clearly, part (i) of this conjecture implies that there are infinitely many primes p with prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 both prime, and part (ii) implies that there are infinitely many primes p with prime(p)^2 + (p-1)^2 and p^2 + (prime(p)-1)^2 both prime.

%H Zhi-Wei Sun, <a href="/A236192/b236192.txt">Table of n, a(n) for n = 1..10000</a>

%e a(10) = 1 since phi(2) + phi(8)/4 + 1 = 3, prime(3)^2 + (2*3)^2 = 5^2 + 6^2 = 61 and 3^2 + (2*prime(3))^2 = 3^2 + 10^2 = 109 are all prime.

%e a(1241) = 1 since phi(83) + phi(1241-83)/4 + 1 = 82 + 96 + 1 = 179, prime(179)^2 + (2*179)^2 = 1063^2 + 358^2 = 1258133 and 179^2 + (2*prime(179))^2 = 179^2 + 2126^2 = 4551917 are all prime.

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

%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/4+1

%t a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000010, A000040, A236097, A236138, A236193.

%K nonn

%O 1

%A _Zhi-Wei Sun_, Jan 20 2014