Applying Collatz function iterations to large integers Given the Collatz function and its iterates:
$$
  T(n) =
\begin{cases}
n/2,  & \text{if $n$ is even} \\
(3n+1)/2, & \text{if $n$ is odd}
\end{cases}
$$
and $$ T^{k}(n) = T(T^{k-1}(n)).$$
Question. How does one apply iterates to large integers such as
$$ 
n = \big(3756801695685 \; \times \;2^{666669}\big) – 1\: ?\label{1}\tag{1} $$
We can begin by observing that repeated application of the Collatz function to numbers of the form 
$$ m = (2^{k} \times p) – 1 $$ 
where p is an odd integer, results in the iterate:
$$ T^{k}(m) = ((3^{k} \times p) – 1), $$
Thus, the trajectory of m travels k steps to arrive at a step that is $(3/2)^{k}$ times greater than m.
Applying this to the example in \eqref{1} yields a trajectory that spans 666669 consecutive odd numbers to arrive at 
$$
T^{666669}(n) = \big(3756801695685 \times 3^{666669}-1\big). \label{2}\tag{2}
$$
NOTE:
$$ 3756801695685 = 3 \times 5 \times 43 \times 347 \times 16785299 $$
Rephrasing the question:  Are there similar ways to shortcut the evaluation of remaining iterates? 
 A: $\forall k, \forall \alpha \in \mathbb{N}:$

*

*$T^k(\alpha\,2^k) = \alpha$

*$T^k(\alpha\,2^k + 2^k - 1) = \alpha\,3^k + 3^k - 1\quad\equiv \alpha\ (\text{mod } 2)$
So $T^{k+1}(\alpha\,2^{k+1} + 2^k - 1) = \alpha\,3^k + \frac{3^k - 1}{2}\quad\equiv \alpha + k\ (\text{mod } 2)$
After that, I have some relations, but less general.
These formulas can be proved by induction. But with a mixed representation in base 2 and 3, these formulas become obvious.
Because

*

*$(1)_2:(1)_3 = 4 = (2)_3:(0)_2$

*$(1)_2:(2)_3 = 5 = (2)_3:(1)_2$

*$(0)_2:(2)_3 = 2 = (1)_3:(0)_2$
Where $(\dots)_2$ is a binary representation and $(\dots)_3$ a representation in base $3$. Something like $\alpha:(0101)_2:(2)_3$ represents $\alpha \times 2^4 \times 3 + (0101)_2 \times 3 + (2)_3 = \alpha \times 2^4 \times 3 + 5 \times 3 + 2 = \alpha \times 2^4 \times 3 + 17$. And it is possible to mix figures in base 2 and figures in base 3.
(A little online tool to manipulate this kind of mixed radix.)
$\forall k, \forall \alpha \in \mathbb{N}:$

*

*$T\big(\alpha:(0)_2\big) = \alpha$

*$T\big(\alpha:(1)_2\big) = \alpha:(2)_3$

*$T^k\big(\alpha:(\underbrace{0\dots00}_{k\text{ times}})_2\big) = \alpha$

*$T^k\big(\alpha:(\underbrace{1\dots11}_{k\text{ times}})_2\big) = \alpha:(\underbrace{2\dots22}_{k\text{ times}})_3\quad\equiv \alpha\ (\text{mod } 2)$
So $T^{k+1}\big(\alpha:(0\underbrace{1\dots11}_{k\text{ times}})_2\big) = \alpha:(\underbrace{1\dots11}_{k\text{ times}})_3\quad\equiv \alpha + k\ (\text{mod } 2)$
A: For the general problem it seems to me the following is the most useful property when looking at big numbers of a certain structure.
Let $a_0 = r $ be an odd number, then $ a_1 = {3a_0+1 \over 2^A} $ where $A$ is the exponent of the primefactor $2$ in $3a_0+1$.    
Now consider $b_0 = 2^M\cdot x + r $ where $M$ is large, at least larger than $A$. 
Then
$$ b_1={3b_0+1 \over 2^A} ={ 3\cdot 2^M\cdot x + 3r+1 \over 2^A} = 3\cdot 2^{M-A}\cdot x +{  3r+1 \over 2^A}  = 3\cdot 2^{M-A}\cdot x + a_1 $$
That means, we can use any $b_0$ as large as we like and which is $a_0 + 2^Mx $ with $x$ odd and $M$ greater than $A$ and compute easily $b_1$ . Further iterations are identical to that of the iterations of $r$ alone - as long as $M$ is large anough.
