How to find the asymptotics of a linear two-dimensional recurrence relation Let $d$ be a positive number.
There is a two dimensional recurrence relation as follow:
$$R(n,m) = R(n-1,m-1) + R(n,m-d)$$
where
$R(0,m) = 1$ and $R(n,0) = R(n,1) = \cdots = R(n, d-1) = 1$ for all $n,m>0$.
How to analyze the asymptotics of $R(n, kn)$ for fixed $k$?
It is easy to see that
$$R(n, kn) = O\left( c_{k,d}^{n} \cdot (n+k+d)^{O(1)} \right)$$
Is there a way (or an algorithm) to find $c_{k,d}$ given $k$ and $d$?
PS: I have calcuated the bivariate generating function of $R(\cdot, \cdot)$:
\begin{align}
f(x,y) 
&= \frac{1 - xy - y^{d} + xy^{d}}{(1 - x)(1 - y)(1 - xy - y^{d})} \\
&= \frac{1}{(1 - x)(1 - y)} + \frac{xy^{d}}{(1 - x)(1 - y)(1 - xy - y^{d})} \\
\end{align}

This is to complement Blanco's answer by showing that
\begin{equation*}
    R(n,kn)=\exp\{(C_{k,d}+o(1))\,n\} \tag{0}\label{0}
\end{equation*}
(as $n\to\infty$), where $k\ge1$ and $d\ge1$ (are fixed),
\begin{equation*}
    C_{k,d}:=\frac k{1+y_{k,d}\,d}\,\big(\ln(1+y_{k,d})+y_{k,d}\,\ln(1+1/y_{k,d})\big),
\end{equation*}
\begin{equation*}
    y_{k,d}:=\max\Big(x_d,\frac{k-1}d\Big), 
\end{equation*}
and $x_d$ is the unique positive root of the equation
\begin{equation*}
    x_d(1+x_d)^{d-1}=1. \tag{0.5}\label{0.5}
\end{equation*}
In view of Blanco's answer, it is enough to show that
\begin{equation*}
    B(n):=B(n,kn)=\exp\{(C_{k,d}+o(1))\,n\}, \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
    B(n):=\max\{c_{a,b}\colon (a,b)\in E_{n,k,d}\},
\end{equation*}
\begin{equation*}
    E_{n,k,d}:=\{(a,b)\colon 0\le a\le n-1,b\ge0,bd+a\le kn-d,\ a,b \text{ are integers}\},
\end{equation*}
\begin{equation*}
    c_{a,b}:=\binom{a+b}b=\binom{a+b}a. 
\end{equation*}
Note that $c_{a,b}$ is increasing in $a$ and in $b$.
Note also that $c_{a,b}=(a+b)^{O(1)}=n^{O(1)}$ for $(a,b)\in E_{n,k,d}$ if $a=O(1)$ or $b=O(1)$.
So, it remains to consider the case when $a\to\infty$ and $b\to\infty$. Then, by Stirling's formula,
\begin{equation*}
\begin{aligned}
    \ln c_{a,b}\sim a\ln(1+b/a)+b\ln(1+a/b). 
\end{aligned}
\tag{2}\label{2}
\end{equation*}
Also,
\begin{equation*}
\begin{aligned}
    \frac{c_{a+d,b-1}}{c_{a,b}}&=b\frac{(a+d+1)\cdots(a+d+b-1)}{(a+1)\cdots(a+b)} \\ 
    &= b\frac{(a+b+1)\cdots(a+b+d-1)}{(a+1)\cdots(a+d)} \\ 
    &\sim b\frac{(a+b)^{d-1}}{a^d} =\frac ba\Big(1+\frac ba\Big)^{d-1}. 
\end{aligned}
\end{equation*}
So, for a fixed value of $bd+a$, the maximum of $c_{a,b}$ occurs when $b/a\to x_d$ (recall \eqref{0.5}).
Let now $(a,b)\in E_{n,k,d}$ be a maximizer of $c_{a,b}$ (such that $a\to\infty$ and $b\to\infty$). Then, since $c_{a,b}$ is increasing in $a$ and in $b$, we have
\begin{equation*}
    bd+a\sim kn. 
\end{equation*}
The conditions $b/a\to x_d$ and $bd+a\sim kn$ imply
\begin{equation*}
a\sim k\frac1{1+x_d\,d}\,n, \tag{3}\label{3}
\end{equation*}
and the latter condition is compatible with condition $a\le n-1$ only if
\begin{equation*}
    k\frac1{1+x_d\,d}\le1. \tag{4}\label{4} 
\end{equation*}
If this is the case, then
\begin{equation*}
b\sim k\frac{x_d}{1+x_d\,d}\,n, 
\end{equation*}
so that, by \eqref{2},
\begin{equation*}
\begin{aligned}
    \frac{\ln c_{a,b}}n\to \frac k{1+x_d\,d}\,(\ln(1+x_d)+x_d\,\ln(1+1/x_d)). 
\end{aligned}
\tag{5}\label{5}
\end{equation*}
Also, \eqref{4} implies $x_d\ge(k-1)/d$, so that $y_{k,d}=x_d$.
So, we have proved \eqref{1}, and thus \eqref{0} -- in the case when condition (3) is compatible with condition $a\le n-1$.
Otherwise, we have $a=n-1\sim n$ and still $bd+a\sim kn$, whence $b\sim(k-1)n/d$. So, by \eqref{2}, here
\begin{equation*}
\begin{aligned}
    \frac{\ln c_{a,b}}n\to \ln\Big(1+\frac{k-1}d\Big)+\frac{k-1}d\,\ln\Big(1+\frac d{k-1}\Big). 
\end{aligned}
\tag{6}\label{6}
\end{equation*}
Also, in this "incompatibility" case, we have $k\frac1{1+x_d\,d}\ge1$ -- cf. \eqref{4}. So, here $x_d\le(k-1)/d$ and hence $y_{k,d}=\frac{k-1}d$.
So, we have proved \eqref{1}, and thus \eqref{0} -- in the case when condition (3) is incompatible with condition $a\le n-1$.
Thus, in either case, \eqref{0} is proved. $\quad\Box$
