Iterated Euler's totient function

Question

Let $\phi(n)$ be the Euler totient function: $$ \phi(2)=1 \;,\; \phi(11)=10 \;,\; \phi(12)=4\;,$$ etc. Define $\Phi(n)$ to be the number of iterations $k$ so that $\phi^k(n)$ reaches $1$. For example, $\Phi(25)=5$ because $\phi(25)=20$ and continuing, it takes $5$ applications to reach $1$: $$25,20,8,4,2,1 \;.$$ Another example: $\Phi(113)=7$: $$113,112,48,16,8,4,2,1 \;.$$ Here is a plot of $\Phi(n)$:

---

         
          ^{Red curve: $0.43 + 1.22 \ln( n )$.}

---

$\Phi(n)$ is fit quite well (and well beyond what's shown above) by $c \ln(n)$.

Two questions:

Q1. What explains the logarithmic growth, at a high-level?

Q2. What explains the constant $c \approx 1.22$?

Likely both of these questions are answered in the literature.

Answer 1

Note that $\phi(n)$ is even (for $n\ge3$), and if $n$ is even then $\phi(n)\le n/2$. This immediately gives you Pillai's logarithmic upper bound.

Answer 2

Erdős et al. say this in On the Normal Behavior of the Iterates Of some Arithmetic Functions:

[...] it is easy to see that the set of numbers of the form $k(n)/ \log n$ is dense in $[1/ \log 3,1/ \log 2]$. What is still in doubt about $k(n)$ is its average and normal behavior. We conjecture that there is some constant $\alpha$ such that $k(n) \sim \alpha \log n$ on a set of asymptotic density $1$.

Here, $k(n)$ is what the OP calls $\Phi(n)$.

The original paper was published in 1990. Perhaps it is still the state of the art.