  Define the (accelerated) Collatz map on the positive integers by
$\mathsf{Col}_2(n)=\frac{n}{2}$ if $n$ is even and
$\mathsf{Col}_2(n)=\frac{3n+1}{2}$ if $n$ is odd. We show that $\mathsf{Col}_2$
can be approximated by multiplication with $\frac{3^{\frac{1}{2}}}{2}$ in the
sense that the set of $n$ for which
$(\frac{3^{\frac{1}{2}}}{2})^kn^{1-\epsilon}\leq \mathsf{Col}_2^k(n)\leq
(\frac{3^{\frac{1}{2}}}{2})^kn^{1+\epsilon}$ for all $0\leq k\leq
2(\log\frac{4}{3})^{-1}\log n\approx 6.952\log n$ has natural density $1$ for
every $\epsilon>0$. Let $\tau(n)$ be the minimal $k\in\mathbb{N}$ for which
$\mathsf{Col}_2^k(n)=1$ if there exist such a $k$ and set $\tau(n)=\infty$
otherwise. As an application of the above we show that
$\liminf_{x\rightarrow\infty}\frac{1}{x\log x}\sum_{m=1}^{\lfloor
x\rfloor}\tau(m)\geq 2(\log\frac{4}{3})^{-1}$, partially answering a question
of Crandall and Shanks. We show also that assuming the Collatz Conjecture is
true in the strong sense that $\tau(n)\in O(\log n)$, then
$\lim_{x\rightarrow\infty}\frac{1}{x\log x}\sum_{m=1}^{\lfloor
x\rfloor}\tau(m)= 2(\log\frac{4}{3})^{-1}$.
