I would like to simplify this block
\begin{equation}
\sum_{k=2}^{m-1}\binom{m}{k}S(k+1,3)S_{3}(m-k+3,1+3) \qquad \qquad (1)
\end{equation}
where, $$ n\in \mathbb {N} $$ and
\begin{equation}
S(k+1,3)=\frac{1}{2}(3^{k}-2^{k+1}+1) \qquad \qquad (2)
\end{equation}
also, we can conclude that
\begin{equation}
S_{3}(m-k+3,1+3)=(1+3)^{m-k}-3^{m-k} \qquad \qquad (3)
\end{equation}
I am trying to get a simple general form for (1), and this example explained it
\begin{equation}
\sum_{k=2}^{4-1}\binom{m}{k}S(k+1,3)S_{3}(m-k+3,1+3)=\binom{4}{2}S(3,3)S_{3}(5,4)+\binom{4}{3}S(4,3)S_{3}(4,4)\\
\nonumber
\end{equation}
\begin{equation}
=42+24=66 \qquad \qquad (4)
\nonumber
\end{equation}
Indeed I am trying by three methods
Firstly; is to simplify $$ \sum_{k=2}^{m-1}\binom{m}{k}S(k+1,3)(1+3)^{m-k}-3^{m-k}$$
Secondly; is to simplify $$ \sum_{k=2}^{m-1}\binom{m}{k}S_{3}(m-k+3,1+3)\frac{1}{2}(3^{k}-2^{k+1}+1) $$
Thirdly; is to verify the following identity in (1) $$ S_{r}(m+r,i+r)=\sum_{k}\binom{m}{k}S_{p}(m-k+p,i+p)(r-p)^{k} $$ Indeed, I am not get any exactly results that is verify (4), i need some source, books and articles about r-Stirling number and multiply r-Stirling number. Thanks for any explanation.