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A234809 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*(n-p) + 1 are both prime}|, where phi(.) is Euler's totient function.

%I A234809 #9 Dec 31 2013 04:01:14
%S A234809 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 A234809 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 A234809 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 A234809 a(n) = |{0 < k < n: p = k + phi(n-k) and 2*(n-p) + 1 are both prime}|, where phi(.) is Euler's totient function.
%C A234809 Conjecture: a(n) > 0 for all n > 2.
%C A234809 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 A234809 See also A234808 for a similar conjecture.
%H A234809 Zhi-Wei Sun, <a href="/A234809/b234809.txt">Table of n, a(n) for n = 1..10000</a>
%e A234809 a(5) = 1 since 1 + phi(4) = 3 and 2*(5-3) + 1 = 5 are both prime.
%e A234809 a(16) = 1 since 7 + phi(9) = 13 and 2*(16-13) + 1 = 7 are both prime.
%e A234809 a(41) = 1 since 7 +phi(34) = 23 and 2*(41-23) + 1 = 37 are both prime.
%e A234809 a(156) = 1 since 131 + phi(25) = 151 and 2*(156-151) + 1 = 11 are both prime.
%t A234809 f[n_,k_]:=k+EulerPhi[n-k]
%t A234809 p[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[2*(n-f[n,k])+1]
%t A234809 a[n_]:=a[n]=Sum[If[p[n,k],1,0],{k,1,n-1}]
%t A234809 Table[a[n],{n,1,100}]
%Y A234809 Cf. A000010, A000040, A046927, A234470, A234475, A234514, A234567, A234615, A234694, A234808
%K A234809 nonn
%O A234809 1,4
%A A234809 _Zhi-Wei Sun_, Dec 30 2013