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A063470
Numbers n such that tau(n)*4 = phi(n).
9
34, 45, 52, 102, 140, 156, 252, 360, 420
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) <= n-1 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
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
KEYWORD
fini,nonn,full
AUTHOR
Jason Earls, Jul 26 2001
EXTENSIONS
"full" keyword from Max Alekseyev, Mar 01 2010
STATUS
approved