  It is shown that every Collatz sequence $C(s)$ consists only of same
structured finite subsequences $C^h(s)$ for $s\equiv9\ (mod\ 12)$ or $C^t(s)$
for $s\equiv3,7\ (mod\ 12)$. For starting numbers of specific residue classes
($mod\ 12\cdot2^h$) or ($mod\ 12\cdot2^{t+1}$) the finite subsequences have the
same length $h,t$. It is conjectured that for each $h,t\geq2$ the number of all
admissible residue classes is given exactly by the Fibonacci sequence. This has
been proved for $2\leq h,t\leq50$.
  Collatz's conjecture is equivalent to the conjecture that for each
$s\in\mathbb{N},s>1$, there exists $k\in\mathbb{N}$ such that $T^k(s)<s$. The
least $k\in\mathbb{N}$ such that $T^k(s)<s$ is called the stopping time of $s$,
which we will denote by $\sigma(s)$. It is shown that Collatz's conjecture is
true, if every starting number $s\equiv3,7\ (mod\ 12)$ have finite stopping
time.
  We denote $\tau(s)$ as the number of $C^t(s)$ until $\sigma(s)$ is reached
for a starting number $s\equiv3,7\ (mod\ 12)$. Starting numbers of specific
residue classes ($mod\ 3\cdot2^{\sigma(s)}$) have the same stopping times
$\sigma(s)$ and $\tau(s)$. By using $\tau(s)$ it is shown that almost all
$s\equiv3,7\ (mod\ 12)$ have finite stopping time and statistically two out of
three $s\equiv3,7\ (mod\ 12)$ have $\tau(s)=1$. Further it is shown what
consequences it entails, if a $C(s)$ grows to infinity.
