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%I #15 Dec 15 2017 17:35:21
%S 34,45,52,102,140,156,252,360,420
%N Numbers n such that tau(n)*4 = phi(n).
%C Phi(n) = k*tau(n) has at most finitely many solutions for any constant k or even for slowly increasing functions like k(n) = n^(epsilon). - _Labos Elemer_, Jul 20 2001
%C For n > 2, tau(n) > 2 and phi(n) <= n-1 so the least solution a(1) to tau(n)*k = phi(n), must be a(1) >= 2*k+1, for the case k=4, a(1) >= 2*4+1 = 9. - _Enrique Pérez Herrero_, May 12 2012
%F a(1) = A175667(4)
%F a(A112954(4)) = A112955(4). - _Enrique Pérez Herrero_, May 12 2012.
%o (PARI) for(n=1,10^6, if(numdiv(n)*4==eulerphi(n),print(n)))
%Y Cf. A062516.
%Y Cf. A112954, A020488, A063469.
%K fini,nonn,full
%O 1,1
%A _Jason Earls_, Jul 26 2001
%E "full" keyword from _Max Alekseyev_, Mar 01 2010
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