A constant-recursive sequence is $$x_{i+m} = a_{m-1}x_{i+m-1} + a_{m-2}x_{i+m-2} + \cdots + a_0x_i,\tag{*} $$ where m is the order of sequence and $a_i$ is integer. The sequence $\{2, 3, 4, 9, 8, 27, 16, 81, ...\}$ is a constant-recursive sequence. But I can't represent it in the form of $(*)$.
It can be noted that this sequence consists of two trivial subsequences $2x_{i}$ for $\{2, 4, 8, 16, 32, ...\}$ and $3x_{i}$ for $\{3, 9, 27, 81, 243, ...\}$.
The question is how the sequence $\{2, 3, 4, 9, 8, 27, 16, 81, ...\}$ can be represented as $( * )$. I would like to see general approaches for solving such problems (when there is an initial sequence consisting of two trivial subsequences).
This answer assumes you know how the closed formula for $x_n$ corresponds to the roots of the polynomial:
$$p(x)=x^m-a_{m-1}x^{m-1}-\cdots-a_0$$
Assuming indices start at $i=0.$
The sequence $$y=(2,0,4,0,8,0,\dots)$$ can be written as:
$$y_i=\frac{(\sqrt 2)^{i+2}+(-\sqrt 2)^{i+2}}{2}$$
Similarly, $$z=(0,3,0,9,0,\dots)$$ can be written:
$$z_i=\frac{(\sqrt 3)^{i+1}+(-\sqrt 3)^{i+1}}{2}$$
So your sequence, $y+z,$ corresponds to the polynomial $(x^2-2)(x^2-3)=x^4-5x^2+6,$ which has roots $\pm \sqrt 2,\pm\sqrt 3.$
More generally, if $x_i$ is a recurrence corresponding to $p(x)$ and $y_i$ aecurrence correspondind to $q(x)$ then $$(x_1,y_1,x_2,y_2,\dots)$$ corresponds to: $$p(x^2)q(x^2)$$
(Sometimes you can find a smaller common recurrence, but this one always works.)
Your case is the particularly easy case, $p(x)=x-2, q(x)=x-3.$
The nature of $p(x^2)q(x^2)$ is to have no odd terms, so the corresponding recurrence is has even terms depend only on previous even terms, and odd terms only on previous odd terms.