A062516 Numbers k such that 2*tau(k) = phi(k).

5, 9, 15, 28, 40, 72, 84, 90, 120

Offset: 1

Author

Jason Earls, Jul 13 2001

Keywords

• nonn
• fini
• full

Comments

Sequence is finite, since for large k and suitable constants and epsilon: phi(k) - 2*tau(k) > c1*k^(2/3) - 4*c2*k^(1/2) > 0 if k > c3, so phi(k) - 2*tau(k) > 0, QED. Moreover, phi(k) = m*tau(k) has at most finitely many solutions for any constant m or even for slowly increasing functions like m(k) = k^(epsilon). - Labos Elemer, Jul 20 2001

Crossrefs

Cf. A112954, A020488, A063469, A063470.

Programs

Mathematica

Select[Range[150],2*DivisorSigma[0,#]==EulerPhi[#]&] (* Harvey P. Dale, Jun 28 2022 *)

PARI

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

Extensions

"full" keyword from Max Alekseyev, Mar 01 2010