%I
%S 0,0,1,2,1,3,1,4,1,1,1,5,3,7,3,1,1,7,5,9,4,2,1,9,5,2,4,3,1,10,5,14,2,
%T 2,2,1,6,14,5,4,1,15,5,16,5,5,3,17,8,4,5,6,3,17,7,5,2,6,6,17,11,25,3,
%U 5,3,1,11,25,4,4,4,22,10,26,6,7,8,3,9,26,7,9,6,25,8,3,7,9,10,25,15,6,2,9,9,2,13,29,3,7
%N a(n) = {0 < k < n: p = k + phi(nk) and 2*(np) + 1 are both prime}, where phi(.) is Euler's totient function.
%C Conjecture: a(n) > 0 for all n > 2.
%C Clearly, this implies Lemoine's conjecture which states that any odd number 2*n + 1 > 5 can be written as 2*p + q with p and q both prime.
%C See also A234808 for a similar conjecture.
%H ZhiWei Sun, <a href="/A234809/b234809.txt">Table of n, a(n) for n = 1..10000</a>
%e a(5) = 1 since 1 + phi(4) = 3 and 2*(53) + 1 = 5 are both prime.
%e a(16) = 1 since 7 + phi(9) = 13 and 2*(1613) + 1 = 7 are both prime.
%e a(41) = 1 since 7 +phi(34) = 23 and 2*(4123) + 1 = 37 are both prime.
%e a(156) = 1 since 131 + phi(25) = 151 and 2*(156151) + 1 = 11 are both prime.
%t f[n_,k_]:=k+EulerPhi[nk]
%t p[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[2*(nf[n,k])+1]
%t a[n_]:=a[n]=Sum[If[p[n,k],1,0],{k,1,n1}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000040, A046927, A234470, A234475, A234514, A234567, A234615, A234694, A234808
%K nonn
%O 1,4
%A _ZhiWei Sun_, Dec 30 2013
