

A063470


Numbers n such that tau(n)*4 = phi(n).


9




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
For n > 2, tau(n) > 2 and phi(n) <= n1 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


LINKS

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


FORMULA

a(1) = A175667(4)
a(A112954(4)) = A112955(4).  Enrique Pérez Herrero, May 12 2012.


PROG

(PARI) for(n=1, 10^6, if(numdiv(n)*4==eulerphi(n), print(n)))


CROSSREFS

Cf. A062516.
Cf. A112954, A020488, A063469.
Sequence in context: A248527 A206262 A302457 * A089715 A327132 A260284
Adjacent sequences: A063467 A063468 A063469 * A063471 A063472 A063473


KEYWORD

fini,nonn,full


AUTHOR

Jason Earls, Jul 26 2001


EXTENSIONS

"full" keyword from Max Alekseyev, Mar 01 2010


STATUS

approved



