Let $c,n>0$ be positive integers. Consider the following sum of products of Catalan type numbers, $0\leq m\leq n$. \begin{align*} r(m)=&\sum_{k=0}^m \frac{c}{k+c}\binom{2k+c-1}{k}\cdot\frac{1}{n+1-k}\binom{2(n-k)}{n-k}\\ =&\sum_{k=0}^m \left(\binom{2k+c-1}{k}-\binom{2k+c-1}{k-1}\right)\left(\binom{2(n-k)}{n-k}-\binom{2(n-k)}{n-k-1}\right). \end{align*} It is known that $r(0)=C_n$ is the $n-$th Catalan number, \begin{equation*} r(n)=\frac{c+1}{n+c+1}\binom{2n+c}{n}=\binom{2n+c}{n}-\binom{2n+c}{n-1}, \end{equation*} and \begin{equation*} \sum_{m=0}^n r(m)=\binom{2n+c}{n}. \end{equation*} My question is whether it is possible to simplify the intermediate terms $r(m)$ for $0<m<n$?
For $n=1,2,3$: \begin{align*} n=1:&\; r(0)=1,\; r(1)=c+1;\\ n=2:&\; r(0)=2,\; r(1)=c+2;\; r(2)=\frac{(c+1)(c+4)}{2};\\ n=3:&\; r(0)=5,\; r(1)=2c+5;\; r(2)=\frac{(c+2)(c+5)}{2},\; r(3)=\frac{(c+1)(c+5)(c+6)}{6}. \end{align*}