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%I #9 Dec 22 2013 01:55:18
%S 0,0,0,1,1,0,2,1,1,3,2,1,1,2,3,4,5,4,2,2,2,5,4,1,5,4,4,3,2,8,5,2,1,3,
%T 9,5,9,4,4,6,2,4,9,5,5,7,9,3,1,10,6,8,3,6,4,5,7,8,3,5,5,4,6,6,10,14,8,
%U 3,3,6,9,5,7,7,9,2,8,8,9,5,6,6,6,8,9,7,9,4,5,9,10,8,8,7,14,9,5,7,6,10
%N a(n) = |{0 < k < n: k*phi(n-k) + 1 is a square}|, where phi(.) is Euler's totient function.
%C Conjecture: (i) a(n) > 0 if n is not a divisor of 6. The only values of n with a(n) = 1 are 4, 5, 8, 9, 12, 13, 24, 33, 49.
%C (ii) If n >= 60, then k + phi(n-k) is a square for some 0 < k < n. If n > 60, then sigma(k) + phi(n-k) is a square for some 0 < k < n, where sigma(k) is the sum of all positive divisors of k.
%C (iii) If n > 7 is not equal to 10 or 20, then phi(k)*phi(n-k) + 1 is a square for some 0 < k < n.
%C (iv) If n > 7 is not equal to 10 or 19, then (phi(k) + phi(n-k))/2 is a triangular number for some 0 < k < n.
%C Note that (n - 1)*phi(1) + 1 = n. So a(n) > 0 if n is a square.
%H Zhi-Wei Sun, <a href="/A234246/b234246.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4) = 1 since 3*phi(1) + 1 = 2^2.
%e a(5) = 1 since 3*phi(2) + 1 = 2^2.
%e a(8) = 1 since 4*phi(4) + 1 = 3^2.
%e a(9) = 1 since 8*phi(1) + 1 = 3^2.
%e a(12) = 1 since 2*phi(10) + 1 = 3^2.
%e a(13) = 1 since 4*phi(9) + 1 = 5^2.
%e a(14) = 2 since 2*phi(12) + 1 = 3^2 and 6*phi(8) + 1 = 5^2.
%e a(24) = 1 since 12*phi(12) + 1 = 7^2.
%e a(33) = 1 since 3*phi(30) + 1 = 5^2.
%e a(49) = 1 since 48*phi(1) + 1 = 7^2.
%t SQ[n_]:=IntegerQ[Sqrt[n]]
%t a[n_]:=Sum[If[SQ[k*EulerPhi[n-k]+1],1,0],{k,1,n-1}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000010, A000290, A233542, A233544, A233547, A233566, A233567, A233867, A233918, A234200
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, Dec 21 2013
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