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.
Best Answer
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.