login
A063469
Numbers n such that tau(n)*3 = phi(n).
9
7, 21, 26, 56, 70, 78, 108, 126, 168, 210
OFFSET
1,1
COMMENTS
"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
MATHEMATICA
Select[Range[300], 3DivisorSigma[0, #]==EulerPhi[#]&] (* Harvey P. Dale, Sep 15 2016 *)
PROG
(PARI) for(n=1, 10^7, if(3*numdiv(n)==eulerphi(n), print(n)))
CROSSREFS
KEYWORD
fini,full,nonn
AUTHOR
Jason Earls, Jul 26 2001
EXTENSIONS
"full" keyword from Max Alekseyev, Mar 01 2010
STATUS
approved