A recurrence satisfied by a function asymptotic to $n^n$

Question

A function $f\colon \{0,1,\dots\}\to \mathbb{R}$ is P-recursive if it satisfies a recurrence $$ P_d(n)f(n+d)+P_{d-1}(n)f(n+d-1)+\cdots+P_0(n)f(n)=0,\ n\geq 0, $$ where each $P_i(n)\in \mathbb{R}[n]$ and $P_d(n)\neq 0$. It is known that $f(n)=n^n$ is not P-recursive. On the other hand, the function $f(n)= \frac{1}{\sqrt{2}}\left(\frac e4\right)^n \frac{(2n)!}{n!}$ is $P$-recursive and satisfies $f(n)\sim n^n$. This suggests the question: is there a P-recursive function $f(n)$ satisfying $f(n)\sim n^n$ and each $P_i(n)\in \overline{\mathbb{Q}}[n]$ or even $P_i(n)\in \mathbb{Q}[n]$, where $\overline{\mathbb{Q}}$ denotes the algebraic closure of $\mathbb{Q}$.

Answer 1

I am afraid that no. Since for fixed $k$ we have $(n+k)^{n+k}\sim n^n\cdot n^ke^k$, when we divide the recurrence by $n^n$, we get several terms equivalent to polynomials in $n$ (namely, to $e^kn^kP_k(n)$ for $k=0,1,\ldots,d$), and the leading terms of these polynomials do not cancel, since $e$ is not algebraic.