A063470 Numbers n such that tau(n)*4 = phi(n).

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

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