This question might seem very novice, but i'm not sure about the solution.
We have domino puzzle of size $2 \times\ n$ and we get 4 points for every vertical block and 1 point for horizontal block, every block is size of $2 \times 1$. How many possibilities are there to get exactly $m$ points? For $m=6$ there are 3 possibilities, $\text{VHH}$,$\text{HHV}$,$\text{HHHHHH}$
My attempt: I know it's easy to make this task without generating functions, but it's just an exercise for G.F.
Let's make sequences$$a_n = <1, x^2, x^4, x^6, x^8,\dots >$$ $$b_n = <1,x^4,x^8,x^{12},x^{16},\dots>$$
Exponential generating function(egf) for $a_n$ will be $$\sum a_n\frac{z^n}{n!}$$ and for $b_n$ will be $$\sum b_n\frac{z^n}{n!}$$
And multiplying those two series should give us some new series and answer for $m$(excluding odd $m$ of course, answer here is always 0) is coefficient standing next to $z^m$, but i'm not getting true answer for any $m$. I'd appreciate some help on this task, thanks in advance.
To tile a $2 \times n$ board, horizontal ($1 \times 2$) blocks must always be used in pairs, one below the other.
As we care about both size (width) and "points", we need a bivariate generating function. Let's use the variable $z$ to mark size, and $u$ to mark points. Then a vertical block corresponds to $zu^4$, and a pair of horizontal block corresponds to $z^2u^2$. Any tiling is a sequence of these, so the class of all configurations is
$$\mathcal{C} = \operatorname{S\scriptsize EQ}(\mathcal{V} + \mathcal{HH}) \implies C(z, u) = \frac{1}{1 - (zu^4 + z^2u^2)}$$
The coefficient of $z^nu^m$ in $C(z)$ gives the answer (number of ways to tile a $2 \times n$ board, and get exactly $m$ points).
If you want to allow all possible values of $n$, just set $z = 1$: then we have
$$C(u) = \frac{1}{1-(u^4 + u^2)} = 1 + u^2 + 2u^4 +3u^6 + 5u^8 + O(u^{10})$$
and you look at the coefficient of $u^m$ in it: indeed the coefficient of $u^6$ is $3$.