Finding a closed form for coefficients in $x^{3n}=x_0\left(a_nx+b_n+\frac {c_n}{x}\right)$

Question

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.

Answer 1

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.