cp's OEIS Frontend

A236998 a(n) = |{0 < k < n/2: phi(k)*phi(n-k) is a square}|, where phi(.) is Euler's totient function.

%I A236998 #13 Feb 02 2014 10:48:15
%S A236998 0,0,1,0,0,1,2,0,2,2,1,1,2,1,1,1,1,3,3,2,2,4,3,1,3,1,3,1,1,2,2,1,4,4,
%T A236998 3,3,1,1,5,2,3,7,2,5,3,4,3,2,7,3,2,3,4,6,2,1,7,5,3,2,2,4,4,2,6,4,3,5,
%U A236998 5,7,4,3,2,6,4,2,7,5,5,4,4,2,4,8,2,7,5,7,3,3,8,6,7,5,7,3,9,3,7,5
%N A236998 a(n) = |{0 < k < n/2: phi(k)*phi(n-k) is a square}|, where phi(.) is Euler's totient function.
%C A236998 Conjecture: (i) a(n) > 0 for all n > 8.
%C A236998 (ii) If n > 20, then phi(k)*phi(n-k) + 1 is a square for some 0 < k < n/2.
%C A236998 (iii) If n > 1 is not among 4, 7, 60, 199, 267, then k*phi(n-k) is a square for some 0 < k < n.
%C A236998 We have verified part (i) of the conjecture for n up to 2*10^6.
%H A236998 Zhi-Wei Sun, <a href="/A236998/b236998.txt">Table of n, a(n) for n = 1..10000</a>
%e A236998 a(17) = 1 since phi(5)*phi(12) = 4*4 = 4^2.
%e A236998 a(24) = 1 since phi(4)*phi(20) = 2*8 = 4^2.
%e A236998 a(56) = 1 since phi(8)*phi(48) = 4*16 = 8^2.
%t A236998 SQ[n_]:=IntegerQ[Sqrt[n]]
%t A236998 p[n_,k_]:=SQ[EulerPhi[k]*EulerPhi[n-k]]
%t A236998 a[n_]:=Sum[If[p[n,k],1,0],{k,1,(n-1)/2}]
%t A236998 Table[a[n],{n,1,100}]
%Y A236998 Cf. A000010, A000290, A234246, A236567, A237016.
%K A236998 nonn
%O A236998 1,7
%A A236998 _Zhi-Wei Sun_, Feb 02 2014