q > 1$. Thus we can conclude that $\alpha$ cannot be rational, unless $q=1$ so that $\alpha$ is an integer. \end{proof} Now the proof of Theorem~\ref{thmrecseq} is straightforward. \begin{proof}[Proof of Theorem~\ref{thmrecseq}] As derived in Section~\ref{secasymp}, if an integer sequence satisfies a recursion of the form $x_{n+1} = P(x_n)$ for some polynomial $P$ of degree $d > 1$ with rational coefficients, and $x_n \to \infty$ as $n \to \infty$, then an asymptotic formula of the form $$x_n = A \alpha^{d^n} + B + O(\alpha^{-d^n})$$ holds. If $\alpha$ is rational, but not an integer, then we have an immediate contradiction to Theorem~\ref{thmirrcrit}. \end{proof} \section{Further generalizations} Let us remark that Theorem~\ref{thmirrcrit} can be extended to number fields: \begin{theorem} Let $L$ be a number field, and let $\mathfrak a_L$ be its maximal order. Assume that the sequence $(G_n)_{n\geq 0}$ attains values in $\mathcal O_L$ infinitely often, and that it satisfies an asymptotic formula of the form $$G_n=A \alpha^n+B+O(|\alpha|^{-\epsilon n}),$$ where $|\alpha| > 1$, $A$ and $B$ are algebraic numbers with $A \neq 0$, $\epsilon > 0$, and the constant implied by the $O$-term does not depend on $n$. Then the number $\alpha$ is either an algebraic integer in $\mathfrak a_L$, or $\alpha\not\in L$. \end{theorem} The proof of this theorem is similar to the proof of Theorem~\ref{thmirrcrit}. In particular, let $K$ be the normal closure of $L(A,B)$, and assume that $\alpha=p/q$ with $p\in \mathfrak a_L$ and $q\in \Z$ with $q>1$. Then we consider the prime ideal factorizations $$(p)=\mathfrak p_1^{n_1} \cdots \mathfrak p_k^{n_k} \quad \text{and} \quad (q)=\mathfrak q_1^{m_1} \cdots \mathfrak q_{\ell}^{m_\ell}$$ in $K$. We can construct the same linear forms as in the proof of Theorem~\ref{thmirrcrit} and use the subspace theorem to get a contradiction. It is also possible to consider a higher-dimensional variant of Theorem~\ref{thmrecseq}. Let $f_1,\ldots,f_N \in \Z[X_1,\ldots,X_N]$ be polynomials of degree $d > 1$. Then we can consider a sequence $(\mathbf x_n)_{n \geq 0}$ with $\mathbf x_n=\left(x^{(1)}_n,\ldots,x^{(N)}_n\right)\in \Z^N$ for all $n\geq 0$ satisfying the polynomial recursion $$\mathbf x_{n+1}=f(\mathbf x_n)=(f_1(\mathbf x_n),\ldots,f_N(\mathbf x_n)).$$ With this notation at hand we pose the following problem: \begin{problem} Assume that $\max \left\{x^{(1)}_n,\ldots,x^{(N)}_n\right\} \rightarrow \infty$ as $n\rightarrow \infty$, and let $$\alpha=\lim_{n\rightarrow \infty} \big( \max\left\{x^{(1)}_n,\ldots,x^{(N)}_n\right\} \big)^{d^{-n}}.$$ Is $\alpha$ necessarily irrational or an integer? \end{problem} \section{Acknowledgment} This paper resulted from an Erasmus+ staff exchange. Financial support from Erasmus+ EU funds is gratefully acknowledged. \bibliographystyle{jis} \bibliography{wagner4} \bigskip \hrule \bigskip \noindent 2020 {\it Mathematics Subject Classification}: Primary 11J72; Secondary 11B37. \noindent \emph{Keywords: } irrationality, polynomial recursion, growth constant. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000058}, \seqnum{A003095}, and \seqnum{A076949}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received August 12 2020; revised versions received August 13 2020; December 31 2020; January 1 2021. Published in {\it Journal of Integer Sequences}, January 2 2021. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}