<s>[INST] Ordered partitions of an integer (with a twist) I would like to know how to prove (preferably algebraically) that $P_1(2,n)=F_{2n+1}$, where $P_1(2,n)$ is what I define to be the number of ordered partitions of an integer, where the number $1$ has 2 possible colours. For example, 
\begin{align}
2
&=2\\
&={\color\red{1}}+{\color\red{1}}\\
&={\color\green{1}}+{\color\green{1}}\\
&={\color\green{1}}+{\color\red{1}}\\
&={\color\red{1}}+{\color\green{1}}\\
\end{align}
so in this case, $P_1(2,2)=5$.
This is my (failed) attempt to derive a formula for $P_1(k,n)$. Consider a case where the number $n$ is expressed as a sum of $m$ natural numbers, out of which there are $j$ number of $1$s. The number of possible ways to do so is given by
$$\binom{m}{j}[x^n]x^j(x^2+x^3+...)^{m-j}$$
Now we look at how many partitions we can get from this particular case. Since there are $j$ number of $1$s, we would have to multiply by $k^j$, and since $0\le j \le m$ and $1 \le m \le n$ we have
\begin{align}
P_1(k,n)
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}[x^n]x^j(x^2+x^3+...)^{m-j}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}[x^{n-2m+j}](1-x)^{-(m-j)}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{m-j+n-2m+j-1}{n-2m+j}\\
&=\sum^{n}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
&=2^{n-1}+\sum^{n-1}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
\end{align}
I have tried using the "snake-oil" method to compute this sum, that is,
$$P_1(2,n)=2^{n-1}+[x^{n-1}]\sum_{n\ge 1}\sum^{n-1}_{m=1}\sum^{m}_{j=0}k^j\binom{m}{j}\binom{n-m-1}{m-j-1}x^{n-1}$$
but I, embarrassingly, have trouble with changing the order of summation and determining the limits of summation.

This is where I am stuck, and I would like to seek your help in finding a closed form for this sum, and in particular, proving that
\begin{align}
P_1(2,n)
&=2^{n-1}+\sum^{n-1}_{m=1}\sum^{m}_{j=0}2^j\binom{m}{j}\binom{n-m-1}{m-j-1}\\
&=F_{2n+1}
\end{align}
Help will be greatly appreciated. Thank you.
 [/INST] Thanks to @paw88789's insightful explanation, I was able to derive a closed form for $P_1(2,n)$ without using any prior knowledge of the answer. The recurrence relation is
\begin{align}
a_n
&=2a_{n-1}+a_{n-2}+...+a_1+1\\
&=1+a_{n-1}+\sum^{n-1}_{m=1}a_m
\end{align}
Applying the forward difference operator $\Delta_n$ to both sides yields
\begin{align}
a_{n+1}-a_n&=a_n-a_{n-1}+a_n\\
a_{n+2}&=3a_{n+1}-a_n
\end{align}
Define the generating function
$$A(x)=\sum_{n\ge 1}a_nx^n$$
In terms of $A(x)$, the recurrence relation is
$$\frac{A(x)-2x-5x^2}{x^2}=\frac{3A(x)-6x}{x}-A(x)$$
Solving for $A(x)$ gives
\begin{align}
A(x)
&=\frac{2x-x^2}{1-3x+x^2}\\
&=\frac{1-x}{1-3x+x^2}-1\\
\end{align}
Extract the coefficient
\begin{align}
a_n
&=[x^n]\left(\frac{1-x}{1-3x+x^2}\right)\\
&=\frac{1}{\sqrt{5}}[x^n]\left(\frac{\tau}{x-(\tau+1)}-\frac{\phi}{x-(\phi+1)}\right)\\
&=\frac{1}{\sqrt{5}}\left(\phi\left(\frac{1}{\phi +1}\right)^{n+1}-\tau\left(\frac{1}{\tau +1}\right)^{n+1}\right)\\
&=\frac{1}{\sqrt{5}}\left(\phi(1+\tau)^{n+1}-\tau(1+\phi)^{n+1}\right)\\
&=\frac{1}{\sqrt{5}}\left(\phi^{2n+1}-\tau^{2n+1}\right)\\
\end{align}
where $\phi=\dfrac{1+\sqrt 5}{2}$ and $\tau=\dfrac{1-\sqrt 5}{2}$.
</s>