Given a prime number $p$ , only $p = 3$ satisfies the equation $$p = \sum_{i = 1}^{c + 1} \phi^i(n),$$ where $\phi^i(x)$ is the iterated totient function and $c$ is the integer such that $\phi^c(n) = 2$ . That is, 3 is the only perfect totient number that is prime.

The first four primes are most easily examined empirically. Since $\phi(2) = 1$ , 2 is deficient totient number. $\phi(3) = 2$ , so, per the previous remark, it is a perfect totient number. For 5, the iterates are 4, 2 and 1, adding up to 7, hence 5 is an abundant totient number. The same goes for 7, with its iterates being 7, 6, 2, 1.

It is for $p > 7$ that we can avail ourselves of the inequality $\phi(n) > \sqrt{n}$ (true for all $n > 6$ ). It is obvious that $\phi(p) = p - 1$ , and by the foregoing, $\phi^2(p) > 3.162278$ (that is, it is sure to be more than the square root of 10), so it follows that $\phi(p) + \phi^2(p) > p + 2.162278$ and thus it is not necessary to examine any further iterates to see that all such primes are abundant totient numbers.