Timeline for No intermediate denominators growth for holonomic functions?
Current License: CC BY-SA 4.0
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| Mar 6, 2020 at 18:32 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
edited tags; edited title
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| Mar 6, 2020 at 17:32 | history | edited | Vesselin Dimitrov | CC BY-SA 4.0 |
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| Mar 6, 2020 at 6:55 | comment | added | user6976 | I did not write that the entries of matrices should be numbers. If $p_i$ are polynomials, the entries should be probably rational functions. | |
| Mar 6, 2020 at 6:40 | comment | added | Vesselin Dimitrov | @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. | |
| Mar 6, 2020 at 3:47 | comment | added | user6976 | 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. | |
| Mar 6, 2020 at 3:27 | comment | added | Vesselin Dimitrov | @MarkSapir: For $k=1$ it shouldn't be too hard to prove that this dichotomy is indeed true. It is also easy when $\sum_{i=0}^{k-1} \deg{p_i} \in \{0,1\}$. | |
| Mar 6, 2020 at 3:18 | comment | added | user6976 | Did you try $k=1,2$? | |
| Mar 6, 2020 at 1:52 | history | asked | Vesselin Dimitrov | CC BY-SA 4.0 |