No label defined (Q6574111)

Let \( a \) and \( b \) be positive integers. Define a polynomial sequence \( \{F_i(x)\}_{i\in \mathbb{N}} \) given by the recurrence relation: \( F_0(x)=0 \), \( F_1(x)=1 \), and \( F_{i+2}(x)=axF_{i+1}(x)+bF_i(x) \) for all \( i\ge 0 \). The generating function of such a polynomial sequence is given by:\N\begin{align*}\Nf(x,t)=\sum_{i=0}^{\infty}F_i(x)t^{i}=\frac{t}{1-axt-bt^2}.\N\end{align*}\NFurthermore, define a companion polynomial sequence \( \{L_i(x)\}_{i\in \mathbb{N}} \) given by the recurrence relation: \( L_0(x)=2 \), \( L_1(x)=ax \), and \( L_{i+2}(x)=axL_{i+1}(x)+bL_i(x) \) for all \( i\ge 0 \). The generating function of such a polynomial sequence is given by:\N\begin{align*}\Nl(x,t)=\sum_{i=0}^{\infty}L_i(x)t^{i}=\frac{2-axt}{1-axt-bt^2}.\N\end{align*}\NIn the paper under review, the author proves the following theorems, which are the main results in the paper. \N\NTheorem 1. Suppose that \( b \) divides \( a \), and let \( q(x)\in \mathbb{Q}(x) \) be a rational function over \( \mathbb{Q} \). For the generating function \( f(x,t) \), \( f(x,q(x))\in \mathbb{Z}[x] \) if and only if\N\begin{align*}\Nq(x)\in \left\{\frac{F_i(x)}{F_{i+1}(x)}\right\}_{i\in \mathbb{N}} \quad \text{or}\quad q(x)\in \left\{-\frac{F_{i+1}(x)}{bF_{i}(x)}\right\}_{i\in \mathbb{N}^+}.\N\end{align*}\NTheorem 2. Suppose that \( b \) divides \( a \), and let \( q(x)\in \mathbb{Q}(x) \) be a rational function over \( \mathbb{Q} \). For the generating function \( l(x,t) \), \( l(x,q(x))\in \mathbb{Z}[x] \) if and only if\N\begin{align*}\Nq(x)\in \left\{\frac{F_i(x)}{F_{i+1}(x)},\frac{L_i(x)}{L_{i+1}(x)},-\frac{L_{i+1}(x)}{bL_{i}(x)}\right\}_{i\in \mathbb{N}} \quad \text{or}\quad q(x)\in \left\{-\frac{F_{i+1}(x)}{bF_{i}(x)}\right\}_{i\in \mathbb{N}^+}.\N\end{align*}\NAs an application of Theorem 1 and Theorem 2, for a square-free \( d\in \mathbb{N} \), the author verifies the results are of the same form as the above for the generating function of the sequence satisfying the recurrence relation \( F_{i+2}(\sqrt{d})=a\sqrt{d}F_{i+1}(\sqrt{d})+bF_i(\sqrt{d}) \) with the initial values \( (F_0(\sqrt{d}), F_(\sqrt{d}))=(0,1) \).