Growth of recurrences with mixed multifold convolutions

TL;DR

This paper develops a method to accurately compute the growth rate of complex recurrence sequences that generalize classical combinatorial numbers, involving mixed convolution and maximum operations.

Contribution

It introduces a novel approach to determine the asymptotic growth rate of generalized recurrence sequences with mixed convolution and maximum terms, providing bounds with converging ratios.

Findings

Established bounds for the growth rate   the ratio converges to 1 as n increases.

Provided explicit inequalities to approximate the growth rate to any desired precision.

Demonstrated the method's applicability to sequences generalizing Catalan and Schr46der numbers.

Abstract

Generalizing some popular sequences like Catalan's number, Schr\"oder's number, etc, we consider the sequence $s_n$ with $s_0=1$ and for $n\ge 1$, \begin{multline*} s_n=\sum_{x_1+\dots+x_{\ell_1}=n-1} \kappa_1 s_{x_1}\dots s_{x_{\ell_1}} + \dots +\sum_{x_1+\dots+x_{\ell_{t'}}=n-1} \kappa_{t'} s_{x_1}\dots s_{x_{\ell_{t'}}}+\\ \max_{x_1+\dots+x_{\ell_{t'+1}}=n-1} \kappa_{t'+1} s_{x_1}\dots s_{x_{\ell_{t'+1}}} + \dots + \max_{x_1+\dots+x_{\ell_t}=n-1} \kappa_t s_{x_1}\dots s_{x_{\ell_t}}, \end{multline*} where $x_i$ are nonnegative integers, $\ell_1,\dots,\ell_t$ are positive integers, and $\kappa_1,\dots,\kappa_t$ are positive reals. We show that it is possible to compute the growth rate $\lambda$ of $s_n$ to any precision. In particular, for every $n\ge 2$, \[ \sqrt[n]{\frac{\kappa^*}{\mathcal L(n-1) s_1} s_n} \le \lambda \le \sqrt[n]{3^{18\log 3 + 2\log\frac{s_1\mathcal L^2}{\kappa^*}} n^{3\log n + 12\log 3 + \log\frac{s_1\mathcal L^2}{\kappa^*}} s_n}, \]where $\mathcal L=\max_i \ell_i$ and $\kappa^*=\kappa_i$ for some $i$ with $\ell_i\ge 2$, and the logarithm has the base $\frac{\mathcal L+1}{\mathcal L}$. The constants in the inequalities are not very well optimized and serve mostly as a proof of concept with the ratio of the upper bound and the lower bound converging to $1$ as $n$ goes to infinity.