Relation of Boubaker Polynomials to Fermat Polynomials

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Theorem

The Boubaker polynomials are related to Fermat polynomials by:

$B_n \left({x}\right) = \dfrac 1 {\left({\sqrt 2}\right)^n} F_n \left({\dfrac {2 \sqrt 2 x} 3}\right) + \dfrac 3 {\left({\sqrt 2}\right)^{n-2}} F_{n-2} \left({\dfrac {2 \sqrt 2 x} 3}\right): \quad n = 0, 1, 2, \ldots$


Proof

(Using Riordan Matrix)

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

$\displaystyle \sum_{n \geqslant 0}^{} B_n \left({t}\right)x^n= \frac {1+3x^2 }{1-xt+x^2}= \left (1+3x^2 \biggl\lvert 1+x^2 \right )\left(\frac {1}{1-xt} \right )$

then, by writing differently:

$\displaystyle T \left(1+3x^2 \biggl\lvert 1+x^2\right )= T \left(1+3x^2 \biggl\lvert 1\right ) T \left( \frac {1}{2}\biggl\lvert \frac {1}{2}+\frac {1}{2}x^2\right ) T \left( 2\biggl\lvert 2 \right )$

and:

$\displaystyle T \left(1+3x^2 \biggl\lvert 1+x^2\right )= T \left(1+3x^2 \biggl\lvert 1\right ) T \left( 1\biggl\lvert \sqrt{2} \right ) T \left( \frac {1}{3}\biggl\lvert \frac {1}{3}+\frac {2}{3}x^2\right ) T \left( 3\biggl\lvert \frac {3}{\sqrt{2}} \right )$

and taking into account the Riordan matrix of Fermat polynomials:

$\displaystyle T \left( \frac {1}{3}\biggl\lvert \frac {1}{3}+\frac {1}{3}x^2\right )$

we obtain finally:

$B_n \left({x}\right) = \dfrac 1 {\left({\sqrt 2}\right)^n} F_n \left({\dfrac {2 \sqrt 2 x} 3}\right) + \dfrac 3 {\left({\sqrt 2}\right)^{n-2}} F_{n-2} \left({\dfrac {2 \sqrt 2 x} 3}\right): \quad n = 0, 1, 2, \ldots$

$\blacksquare$


Also see


Sources

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