

A062516


Numbers n such that tau(n)*2 = phi(n).


12




OFFSET

1,1


COMMENTS

Sequence is finite, since for large n and suitable constants and epsilon: phi(n)2*Tau[n]>c1*n^(2/3)4c2*n^(1/2)>0 if n>c3, so phi(n)2*Tau[n]>0, QED. Moreover, 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


LINKS

Table of n, a(n) for n=1..9.


PROG

(PARI) for(n=1, 1000000, if(numdiv(n)*2==eulerphi(n), print(n), ))


CROSSREFS

Cf. A112954, A020488, A063469, A063470.
Sequence in context: A023498 A256343 A210289 * A249331 A075133 A066081
Adjacent sequences: A062513 A062514 A062515 * A062517 A062518 A062519


KEYWORD

nonn,fini,full


AUTHOR

Jason Earls, Jul 13 2001


EXTENSIONS

"full" keyword from Max Alekseyev, Mar 01 2010


STATUS

approved



