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a(n) = |{0 < k < n - 2: k + phi(n-k)/2 is a square}|, where phi(.) is Euler's totient function.
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%I #16 Feb 02 2014 05:58:54

%S 0,0,0,0,0,1,2,1,1,2,1,3,1,2,4,1,1,2,4,3,2,3,1,3,2,3,3,4,3,1,8,3,3,2,

%T 4,4,2,2,5,6,4,6,3,2,5,4,4,5,4,1,8,6,3,3,5,6,3,4,5,9,5,2,3,6,6,5,4,4,

%U 6,8,6,8,4,3,5,8,4,1,6,6,6,3,9,8,5,4,6,7,7,6,7,5,6,8,3,10,5,5,5,4

%N a(n) = |{0 < k < n - 2: k + phi(n-k)/2 is a square}|, where phi(.) is Euler's totient function.

%C Conjecture: (i) a(n) > 0 for all n > 5.

%C (ii) For n > 31, there is a positive integer k < n - 2 with phi(k) + phi(n-k)/2 a square. If n > 70 is not equal to 150, then phi(k) + phi(n-k) is a square for some 0 < k < n.

%C (iii) If n > 5, then phi(k) + phi(n-k)/2 is a triangular number for some 0 < k < n - 2. For each n = 20, 21, ... there is a positive integer k < n with (phi(k) + phi(n-k))/2 a triangular number.

%H Zhi-Wei Sun, <a href="/A236567/b236567.txt">Table of n, a(n) for n = 1..10000</a>

%e a(8) = 1 since 1 + phi(7)/2 = 1 + 3 = 2^2.

%e a(11) = 1 since 8 + phi(3)/2 = 8 + 1 = 3^2.

%e a(78) = 1 since 40 + phi(38)/2 = 40 + 9 = 7^2.

%t SQ[n_]:=IntegerQ[Sqrt[n]]

%t p[n_,k_]:=SQ[k+EulerPhi[n-k]/2]

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

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

%Y Cf. A000010, A000290, A234246.

%K nonn

%O 1,7

%A _Zhi-Wei Sun_, Feb 02 2014