# Generating Function for Boubaker Polynomials

## Theorem

The Boubaker polynomials, defined as:

$\map {B_n} x = \begin{cases} 1 & : n = 0 \\ x & : n = 1 \\ x^2 + 2 & : n = 2 \\ x \map {B_{n - 1} } x - \map {B_{n - 2} } x & : n > 2 \end{cases}$

have as an ordinary generating function:

$\ds \map {f_{B_n, \operatorname {ORD} } } {x, t} = \sum_{n \mathop = 0}^{\infty} \map {B_n} x t^n = \frac {1 + 3 t^2} {1 + t \paren {t - x} }$

## Proof

 $\ds \map f {x, t}$ $=$ $\ds 1 + x t + \paren {x^2 + 2} t^2 + \sum_{n \mathop = 3}^\infty \paren {x \map {B_{n - 1} } x t^n - \map {B_{n - 2} } x t^n}$ Recursive definition of $\map {B_n} x$ for $n > 2$ $\ds$ $=$ $\ds 1 + x t + \paren {x^2 + 2} t^2 + x t \sum_{n \mathop = 3}^\infty \map {B_{n - 1} } x t^{n - 1} - t^2 \sum_{n \mathop = 3}^\infty \map {B_{n - 2} } x t^{n - 2}$ $\ds$ $=$ $\ds 1 + x t + \paren {x^2 + 2} t^2 + x t \paren {\map f {x, t} - 1 - x t} - t^2 \paren {\map f {x, t} - 1}$

 $\ds \leadsto \ \$ $\ds \paren {1 - x t + t^2} \map f {x, t}$ $=$ $\ds 1 + x t + x^2 t^2 + 2 t^2 - x t - x^2 t^2 + t^2$ $\ds$ $=$ $\ds 1 + 3 t^2$ $\ds \leadsto \ \$ $\ds \map f {x, t}$ $=$ $\ds \frac {1 + 3 t^2} {1 + t \paren {t - x} }$

$\blacksquare$