Finding a closed form for coefficients in $x^{3n}=x_0\left(a_nx+b_n+\frac {c_n}{x}\right)$ Consider,
$$
x^3=x+1
$$
Let $x_0$ be a solution to the above equation. Now consider $x^{3n}$. For $n=2$ we have:
$$
x^6=(x+1)^2
$$
$$
=x^2+2x+1
$$
$$
=x\left(x+2+\frac {1}{x}\right)
$$
$$
=x_0\left(x+2+\frac {1}{x}\right)
$$
For $n=3$ we have:
$$
x^9=(x+1)^3
$$
$$
=x^3+3x^2+3x+1
$$
$$
=3x^2+4x+2
$$
$$
=x\left(3x+4+\frac {2}{x}\right)
$$
$$
=x_0\left(3x+4+\frac {2}{x}\right)
$$
Similarly for $n=4$ we have:
$$
x^{12}=x_0\left(7x+9+\frac {5}{x}\right)
$$
In general we have:
$$
x^{3n}=x_0\left(a_nx+b_n+\frac {c_n}{x}\right),
$$
Where,
$$
a_{n+1}=a_n+b_n, a_2=1,
$$
$$
b_{n+1}=a_n+b_n+c_n, b_2=2,
$$
$$
c_{n+1}=a_n+c_n, c_2=1.
$$
My question is: is there a closed form for $a_n,b_n$ and $c_n$?
Any help would be appreciated.
 A: We derive generating functions for the recurrence relation:
\begin{align*}
a_{n+1}&=a_n+b_n\tag{1}\\
b_{n+1}&=a_n+b_n+c_n\qquad\qquad (n\geq 2)\tag{2}\\
c_{n+1}&=a_n+c_n\tag{3}\\
a_2&=1,b_2=2,c_2=1\\
\end{align*}

Let $A(x)=\sum_{n\geq 2} a_nx^n, B(x)=\sum_{n\geq 2} b_nx^n, C(x)=\sum_{n\geq 2} c_n x^n$.
We obtain from (1)
  \begin{align*}
\sum_{n\geq 2}a_{n+1}x^n&=\sum_{n\geq 2}a_nx^n+\sum_{n\geq 2}b_nx^n\\
\frac{1}{x}\sum_{n\geq 3}a_nx^n&=A(x)+B(x)\\
A(x)-x^2&=xA(x)+xB(x)\\
\color{blue}{(1-x)A(x)-xB(x)}&\color{blue}{=x^2}\tag{4}\\
\end{align*}
Since (3) and (1) have the same structure and initial condition, we get
\begin{align*}
\color{blue}{(1-x)C(x)-xA(x)}&\color{blue}{=x^2}\qquad\qquad\qquad\qquad\tag{5}\\
\end{align*}
The relationship (2):
\begin{align*}
\sum_{n\geq 2}b_{n+1}x^n&=\sum_{n\geq 2}\left(a_n+b_n+c_n\right)x^n\\
\frac{1}{x}\sum_{n\geq 3}b_nx^n&=A(x)+B(x)+C(x)\\
B(x)-2x^2&=xA(x)+xB(x)+xC(x)\\
\color{blue}{(1-x)B(x)-xA(x)-xC(x)}&\color{blue}{=2x^2}\tag{6}
\end{align*}

We take (4) - (6) and derive from them the generating functions.
\begin{align*}
(1-x)A(x)-xB(x)&=x^2\\
-xA(x)+(1-x)C(x)&=x^2\\
-xA(x)+(1-x)B(x)-xC(x)&=2x^2
\end{align*}

Solving the equations above we obtain
  \begin{align*}
\color{blue}{A(x)}&\color{blue}{=\frac{x^2}{1-3x+2x^2-x^3}}\\
&=x^2 + 3 x^3 + 7 x^4 + 16 x^5 + 37 x^6 + 86 x^7+\cdots\\
\color{blue}{B(x)}&\color{blue}{=\frac{1-x}{x}A(x)-x}\\
&=2 x^2 + 4 x^3 + 9 x^4 + 21 x^5 + 49 x^6 + 114 x^7 +\cdots\\
\color{blue}{C(x)}&\color{blue}{=\frac{x}{1-x}A(x)+\frac{x^2}{1-x}}\\
&=x^2 + 2 x^3 + 5 x^4 + 12 x^5 + 28 x^6 + 65 x^7+\cdots
\end{align*}
  where the expansion was done with some help of Wolfram Alpha.

