%I #9 Jan 19 2014 20:57:52

%S 0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,2,0,3,0,1,3,1,4,0,5,2,2,3,4,1,4,2,5,3,

%T 1,6,4,3,0,4,5,3,3,4,5,2,4,2,2,4,3,4,1,2,2,3,5,3,0,3,2,4,1,2,2,4,0,4,

%U 1,3,3,2,0,4,1,3,2,3,1,5,3,5,1,4,2,3,5,4,4,5,4,1,2,2,3,3,7,3,2,3

%N a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/3 - 1, prime(p-1) - (p-1) and prime(p-1) - 2*prime((p-1)/2) are all prime}|, where phi(.) is Euler's totient function.

%C Conjecture: (i) a(n) > 0 for all n > 73.

%C (ii) For any integer n > 472, there is a positive integer k < n such that p = phi(k) + phi(n-k)/4 - 1, prime(p) - 2*prime((p-1)/2) and prime(p) - 2*prime((p+1)/2) are all prime.

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

%H Zhi-Wei Sun, <a href="/A236138/b236138.txt">Table of n, a(n) for n = 1..10000</a>

%e a(20) = 1 since phi(11) + phi(9)/3 - 1 = 11, prime(10) - 10 = 29 - 10 = 19 and prime(10) - 2*prime(5) = 29 - 2*11 = 7 are all prime.

%e a(293) = 1 since phi(267) + phi(26)/3 - 1 = 176 + 12/3 - 1 = 179, prime(178) - 178 = 1061 - 178 = 883 and prime(178) - 2*prime(89) = 1061 - 2*461 = 139 are all prime.

%t PQ[n_]:=n>0&&PrimeQ[n]

%t p[n_]:=n>2&&PrimeQ[n]&&PrimeQ[Prime[n-1]-(n-1)]&&PQ[Prime[n-1]-2*Prime[(n-1)/2]]

%t f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/3-1

%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, A234694, A235924, A236074, A236097.

%K nonn

%O 1,16

%A _Zhi-Wei Sun_, Jan 19 2014