On the transcendence of growth constants associated with polynomial recursions

Let $P(x):=a_d x^d+\cdots+a_0\in\mathbb{Q}[x]$, $a_d>0$, be a polynomial of degree $d\geq 2$. Let $(x_n)$ be a sequence of integers satisfying \begin{equation*} x_{n+1}=P(x_n)\mbox{for all}\quad n=0,1,2\ldots,\quad\mbox{and} \quad x_n\to\infty\quad\mbox{as}\quad n\to\infty. \end{equation*} Set $\alpha:=\lim_{n\to\infty} x^{d^{-n}}_n$. Then, under the assumption $a_d^{1/(d-1)}\in\mathbb{Q}$, in a recent result by Dubickas \cite{dubickas}, either $\alpha$ is transcendental, or $\alpha$ can be an integer, or a quadratic Pisot unit with $\alpha^{-1}$ being its conjugate over $\mathbb{Q}$. In this paper, we study the nature of such $\alpha$ without the assumption that $a_d^{1/(d-1)}$ is in $\mathbb{Q}$, and we prove that either the number $\alpha$ is transcendental, or $\alpha^h$ is a Pisot number with $h$ being the order of the torsion subgroup of the Galois closure of the number field $\mathbb{Q}(\alpha, a_d^{-\frac{1}{d-1}})$. Other results presented in this paper investigate the solutions of the inequality $||q_1 \alpha_1^n+\cdots+q_k \alpha_k^n +\beta||<\theta^n$ in $(n,q_1,\ldots,q_k)\in \mathbb{N}\times(K^\times)^k$, considering whether $\beta$ is rational or irrational. Here, $K$ represents a number field, and $\theta\in (0,1)$. The notation $||x||$ denotes the distance between $x$ and its nearest integer in $\mathbb{Z}$.