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My question concerns holonomic sequences of rational numbers, meaning assignments $a(n) : \mathbb{N} \to \mathbb{Q}$ fulfilling a linear recurrent relation of the form $$ a(n+k) = \sum_{i=0}^{k-1} p_i(n)a(n+i), \quad n = 1, 2, \ldots, $$ where $p_i(x) \in \mathbb{Q}(x)$ are rational functions. In other words, the generating function $f(X) := \sum_{n} a(n)X^n \in \mathbb{Q}[[X]]$ and fulfills a linear ODE with polynomial coefficients.

Define $D_n$ to be the least common denominator of the rational numbers $a(1), a(2), \ldots, a(n)$. That is the minimum positive integer such that all $a(i) \in \frac{1}{D_n}\mathbb{Z}$ for $i \leq n$. A generalization (which I have no clue how to establish) of the prime number theorem would be that $\lim_{n \to \infty} \frac{1}{n}\log{D_n}$ converges (possibly to $+\infty$).

Question. If $D_n$ is unbounded, does it grow at least exponentially in $n$?

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  • $\begingroup$ Did you try $k=1,2$? $\endgroup$
    – user6976
    Mar 6, 2020 at 3:18
  • $\begingroup$ @MarkSapir: For $k=1$ it shouldn't be too hard to prove that this dichotomy is indeed true. It is also easy when $\sum_{i=0}^{k-1} \deg{p_i} \in \{0,1\}$. $\endgroup$ Mar 6, 2020 at 3:27
  • $\begingroup$ How about $k=2$? It may help to rewrite the recurrence relation in a matrix form and reduce the problem to a problem about eigenvalues of powers of a matrix. $\endgroup$
    – user6976
    Mar 6, 2020 at 3:47
  • $\begingroup$ @MarkSapir: But the $p_i(n)$ need not be constant. What you write applies to the $p_i \equiv \mathrm{const}$ case, i.e. when $f(X) \in \mathbb{Q}(X)$ is rational. Example: if $a(n) = 1/n$, or $f(X) = -\log(1-X)$, we have $D_n = [1,\ldots,n] = e^{n + o(n)}$, by the prime number theorem. I have no idea how to prove the $k=2$ case; rather trying to get a sense if the statement looks plausible in general. $\endgroup$ Mar 6, 2020 at 6:40
  • $\begingroup$ I did not write that the entries of matrices should be numbers. If $p_i$ are polynomials, the entries should be probably rational functions. $\endgroup$
    – user6976
    Mar 6, 2020 at 6:55

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