Generating function of ordered partitions

We can do this by generating functions which you want to.

For partition $n$,

If we separate $n$ into $n$ numbers, we can generate the function $x+x^2+\cdots+x^n$ and the coefficient of $x^n$ is the number of combination.

If we separate $n$ into $\left(n-1\right)$ numbers, we can generate the function $\left(x+x^2+\cdots+x^n\right)^2$ and the coefficient of $x^n$ is the number of combination.

...

If we separate $n$ into $2$ numbers, we can generate the function $\left(x+x^2+\cdots+x^n\right)^{n-1}$ and the coefficient of $x^n$ is the number of combination.

If we separate $n$ into $1$ number, we can generate the function $\left(x+x^2+\cdots+x^n\right)^n$ and the coefficient of $x^n$ is the number of combination.

Therefore the total number of combination of partition of $n$ is the coefficient of $x^n$ of this function $$\small f:\left(x+x^2+\cdots+x^n\right)+\left(x+x^2+\cdots+x^n\right)^2+\cdots+\left(x+x^2+\cdots+x^n\right)^{n-1}+\left(x+x^2+\cdots+x^n\right)^n$$

However, this function $$\small g:\left(x+x^2+\cdots+x^n+\cdots\right)+\left(x+x^2+\cdots+x^n+\cdots\right)^2+\cdots+\left(x+x^2+\cdots+x^n+\cdots\right)^n+\cdots$$ has the same coefficient of $x^n$ as $f$. Then, we can simplify $g$ like that: \begin{align}\small\dfrac{x}{1-x}+\left(\dfrac{x}{1-x}\right)^2+\cdots+\left(\dfrac{x}{1-x}\right)^n+\cdots&\small=\dfrac{\dfrac{x}{1-x}}{1-\dfrac{x}{1-x}}\\&\small=\dfrac{x}{1-2x}\\&\small=x\left(1+\left(2x\right)+\left(2x\right)^2+\cdots+\left(2x\right)^{n-1}+\cdots\right)\\&\small=x+2x^2+4x^3+8x^4+\cdots+2^{n-1}x^n+\cdots\end{align}

Therefore, the number of partition of $n=2^{n-1}$

The cases you have shown is $n=4$, which the number of partition is $8$, same as counting.