# Relation of Boubaker Polynomials to Fermat Polynomials

## Theorem

The Boubaker polynomials are related to Fermat polynomials by:

$\forall n \in \N: \map {B_n} x = \dfrac 1 {\paren {\sqrt 2}^n} \map {F_n} {\dfrac {2 \sqrt 2 x} 3} + \dfrac 3 {\paren {\sqrt 2}^{n - 2} } \map {F_{n - 2} } {\dfrac {2 \sqrt 2 x} 3}$

## Proof

(Using Riordan Matrix)

Since the ordinary generating function of the Boubaker Polynomials can be expressed in terms of Riordan matrices:

$\ds \sum_{n \mathop \geqslant 0}^{} \map {B_n} t x^n = \frac {1 + 3x^2 }{1 - x t + x^2} = \paren {1 + 3 x^2 \biggl\lvert 1 + x^2} \paren {\frac 1 {1 - x t} }$

then, by writing differently:

$\ds \map T {1 + 3x^2 \biggl\lvert 1 + x^2} = \map T {1 + 3x^2 \biggl \lvert 1} \map T {\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2} \map T {2 \biggl\lvert 2}$

and:

$\ds \map T {1 + 3x^2 \biggl\lvert 1 + x^2} = \map T {1 + 3x^2 \biggl\lvert 1} \map T {1 \biggl\lvert \sqrt 2} \map T {\frac 1 3 \biggl\lvert \frac 1 3 + \frac 2 3 x^2} \map T {3 \biggl\lvert \frac 3 {\sqrt 2} }$

and taking into account the Riordan matrix of Fermat polynomials:

$\ds \map T {\frac 1 3 \biggl\lvert \frac 1 3 + \frac 1 3 x^2}$

we obtain finally:

$\forall n \in \N: \map {B_n} x = \dfrac 1 {\paren {\sqrt 2}^n} \map {F_n} {\dfrac {2 \sqrt 2 x} 3} + \dfrac 3 {\paren {\sqrt 2}^{n - 2} } \map {F_{n - 2} } {\dfrac {2 \sqrt 2 x} 3}$

$\blacksquare$

## Sources

A. Luzon et al.: Recurrence Relations for Polynomial Sequences via Riordan Matrices