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%I #17 Jun 28 2022 18:26:21
%S 5,9,15,28,40,72,84,90,120
%N Numbers k such that 2*tau(k) = phi(k).
%C Sequence is finite, since for large k and suitable constants and epsilon: phi(k) - 2*tau(k) > c1*k^(2/3) - 4*c2*k^(1/2) > 0 if k > c3, so phi(k) - 2*tau(k) > 0, QED. Moreover, phi(k) = m*tau(k) has at most finitely many solutions for any constant m or even for slowly increasing functions like m(k) = k^(epsilon). - _Labos Elemer_, Jul 20 2001
%t Select[Range[150],2*DivisorSigma[0,#]==EulerPhi[#]&] (* _Harvey P. Dale_, Jun 28 2022 *)
%o (PARI) for(n=1,1000000, if(numdiv(n)*2==eulerphi(n),print(n),))
%Y Cf. A112954, A020488, A063469, A063470.
%K nonn,fini,full
%O 1,1
%A _Jason Earls_, Jul 13 2001
%E "full" keyword from _Max Alekseyev_, Mar 01 2010
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