proof that all powers of 3 are perfect totient numbers
Given an integer , it is always the case that
where is the iterated totient function and is the integer such that . That is, all integer powers of three are perfect totient numbers.
The proof of this is easy and even considered trivial. Here it goes anyway:
Accepting as proven that , we can plug in and see that , which falls short of by . Given the proof that Euler is a multiplicative function (http://planetmath.org/ProofThatEulerPhiIsAMultiplicativeFunction) and the fact that , it is obvious that . Therefore, each iterate will be twice a power of three, with the exponent
gradually decreasing as the iterator nears . To put it algebraically, for . Adding up in ascending order starting at the th iterate, we obtain .
References
- 1 D. E. Ianucci, D. Moujie & G. L. Cohen, “On Perfect Totient Numbers” Journal of Integer Sequences, 6, 2003: 03.4.5