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%I #17 May 21 2019 06:23:21
%S 0,1,1,1,2,1,2,2,2,2,3,3,2,2,2,2,3,3,1,2,3,4,4,3,4,5,4,6,5,5,5,5,5,5,
%T 4,5,4,4,2,5,5,3,6,6,3,7,6,6,6,5,6,6,4,5,5,5,5,6,4,5,8,7,5,9,6,7,8,8,
%U 7,6,6,8,5,7,7,6,5,6,8,8,8,10,6,10,13,10,10,9,6,11,9,7,3,9,6,6,9,7,5,12
%N Number of ways to write 2*n as phi(x^2) + phi(y^2) + phi(z^2), where x,y,z are positive integers with x <= y <= z, and phi(.) is Euler's totient function (A000010).
%C Conjecture 1: a(n) > 0 for all n > 1. In other words, the set {phi(x^2) + phi(y^2) + phi(z^2): x,y,z = 1,2,3,...} contains all even numbers greater than two.
%C Conjecture 2: For any integer n > 3, we can write 2*n+1 as phi(x^2) + phi(y^2) + sigma(z^2) with x,y,z positive integers, where the function sigma(.) is given by A000203.
%H Zhi-Wei Sun, <a href="/A308342/b308342.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/332067">Does phi(x^2) + phi(y^2) + phi(z^2) represent all even numbers greater than two?</a>, Question 332067 on MathOverflow, May 20, 2019.
%e a(2) = 1 with 2*2 = phi(1^2) + phi(1^2) + phi(2^2).
%e a(3) = 1 with 2*3 = phi(2^2) + phi(2^2) + phi(2^2).
%e a(4) = 1 with 2*4 = phi(1^2) + phi(1^2) + phi(3^2).
%e a(6) = 1 with 2*6 = phi(2^2) + phi(2^2) + phi(4^2).
%e a(19) = 1 with 2*19 = phi(3^2) + phi(5^2) + phi(6^2).
%t f[n_]:=f[n]=n*EulerPhi[n]
%t T={};Do[If[f[n]<=200,T=Append[T,f[n]]],{n,1,200}];
%t tab={};Do[r=0;Do[If[f[k]>2n/3,Goto[cc]];Do[If[f[m]<f[k]||f[m]>(2n-f[k])/2,Goto[bb]];If[MemberQ[T,2n-f[k]-f[m]],r=r+1];Label[bb],{m,1,(2n-f[k])/2}];Label[cc],{k,1,2n/3}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000010, A000203, A000290, A002618, A262311, A262982, A262985.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, May 20 2019