In the Collatz function, why $3^{2k}-1$ and $3^{2k-1}-1$ always share the same trailing trajectory? Why are the trajectories always the same for numbers of the form $3^{2k}-1$ and $3^{2k-1}-1$ for the Collatz function?
For example, let $k = 3$. So, $3^6-1 = 728$ and $3^5-1 = 242$. The trajectories are 
         242
  728    121
  364    364    same trajectory starts
  182    182
  91     91 
  274    ...
  137
  412
  206
  103
  310
  155
  ...

 A: Calculating modulo 4 we have $3^{2k-1}-1 \equiv 2 \pmod 4$ for all $k$.
So starting at $3^{2k-1}-1$ we will always do exactly one halving step before the next "$\times\,3+1"$ step. The result of those two steps are
$$ \frac32 (3^{2k-1}-1) + 1 = \frac12 3^{2k} - \frac32 + 1 = \frac{3^{2k}-1}{2} $$
which is also the first step after $3^{2k}-1$, at which point the two sequences have converged.
A: Let $f$ be the Collatz function, i.e. $f(n)=3n+1$ if $n$ is odd and $f(n)=\frac n2$ if $n$ is even.
Then, $3^{2k}-1$ is clearly even, so we have $$f(3^{2k}-1)=\frac{3^{2k}-1}2=1+3+3^2+\ldots+3^{2k-1},$$ where the second equality is the usual formula for the sum of a finite geometric progression.
Now, $3^{2k-1}-1$ is also even, so we have $$f(3^{2k-1}-1)=\frac{3^{2k-1}-1}2=1+3+3^2+\ldots+3^{2k-2}.$$ This is the sum of an odd number of odd terms, so it must be odd. Therefore, $$f(f(3^{2k-1}-1))=3(1+3+3^2+\ldots+3^{2k-2})+1=1+3+3^2+\ldots+3^{2k-1}.$$
Therefore, $f(f(3^{2k-1}-1))=f(3^{2k}-1)$ and we are done.
