OFFSET
1,1
COMMENTS
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
MATHEMATICA
Select[Range[150], 2*DivisorSigma[0, #]==EulerPhi[#]&] (* Harvey P. Dale, Jun 28 2022 *)
PROG
(PARI) for(n=1, 1000000, if(numdiv(n)*2==eulerphi(n), print(n), ))
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Jason Earls, Jul 13 2001
EXTENSIONS
"full" keyword from Max Alekseyev, Mar 01 2010
STATUS
approved