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a(n) = |{0 < k < n-2: p = 2*phi(k) + phi(n-k)/2 + 1, prime(p) + 2 and prime(prime(p)) + 2 are all prime}|, where phi(.) is Euler's totient function.
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%I #10 Jan 27 2014 02:55:23

%S 0,0,0,0,0,0,0,1,1,0,2,1,1,2,1,3,2,2,0,2,3,1,2,1,3,3,2,2,1,1,1,3,0,2,

%T 3,2,1,3,0,2,0,1,1,1,1,2,0,0,0,0,2,2,0,1,0,0,0,0,0,0,0,0,1,1,0,1,1,1,

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

%C Conjecture: a(n) > 0 for every n = 640, 641, ....

%C We have verified this for n up to 75000.

%C The conjecture implies that there are infinitely many primes p with prime(p) + 2 and prime(prime(p)) + 2 both prime.

%H Zhi-Wei Sun, <a href="/A236480/b236480.txt">Table of n, a(n) for n = 1..10000</a>

%e a(8) = 1 since 2*phi(3) + phi(5)/2 + 1 = 7, prime(7) + 2 = 17 + 2 = 19 and prime(prime(7)) + 2 = prime(17) + 2 = 61 are all prime.

%e a(667) = 1 since 2*phi(193) + phi(667-193)/2 + 1 = 384 + 78 + 1 = 463, prime(463) + 2 = 3299 + 2 = 3301 and prime(prime(463)) + 2 = prime(3299) + 2 = 30559 are all prime.

%t p[n_]:=PrimeQ[n]&&PrimeQ[Prime[n]+2]&&PrimeQ[Prime[Prime[n]]+2]

%t f[n_,k_]:=2*EulerPhi[k]+EulerPhi[n-k]/2+1

%t a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-3}]

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

%Y Cf. A000010, A000040, A001359, A006512, A236456, A236457, A236458, A236468, A236470, A236481.

%K nonn

%O 1,11

%A _Zhi-Wei Sun_, Jan 26 2014