How would I go about proving the following:

For any odd positive integer $s$, there exists a sequence of nonnegative integers $( a_0, a_1, \cdots, a_{n-1})$ and a nonnegative integer $m$ such that,

$$ 3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$$


I am really stuck. I was thinking of using the ring $\mathbb{Z}_{2^m}$ in a proof by contradiction, but I cannot even get started reducing the LHS to something simpler.


**Note 1:** I have reason to believe there exists such a sequence where 

 1. $a_0=0$
 3. $\{a_n\}$ is strictly monotonically increasing

**Note 2:** I think an example might help. 



$$ n=3 \quad \{a_n\} = [0, 2g-1, 2g+3] \quad s = \frac{5\times 4^g -2}{6} \quad m= 2g+5 \quad g >0$$


**Note 3:** I originally asked this on the [Mathematics Stack Exchange](https://math.stackexchange.com/questions/3414613/proof-of-3ns-sum-k-0n-1-3n-k-12a-k-2m), but it seems to be a question more suited for this exchange.

**Note 4:** The full example for $n=3$. 

If $s$ is an odd positive integer such that its orbit contains exactly $3$ odd integers including $s$ and $1$ then $s$ has exactly one of two forms:

$$S = \frac{4^{3j+g-1} - 4^{g} -6}{18} \quad \text{ with } \quad a_{g,j}= \{ 0, 2g-1, 6j+2g-3 \} \quad g,j > 0$$

or


$$S = \frac{4^{3j+g+1} - 4^{g} -3}{9} \quad \text{ with } \quad a_{g,j}= \{ 0, 2g, 6j+2g+2 \} \quad g,j > 0.$$

Its possible to get the exact forms for bigger orbits in the sense that the orbits contain exactly $k$ odd numbers including $s$ and $1$, it just gets harder and more tedious. Also, this thing needs a proof.