I am currently playing around with recursive formulars. The following problem is not taken froma book or course.
Denote by $R_k$ a matrix of dimension $k \times k$ and by $r_k$ a vector of dimension $1 \times k$. Furthermore we are given a lower unittriangular matrix $R_2$ and the recursive formula $R_{k+1} = \begin{pmatrix} R_k & 0 \\ r_kR_k & 1 \end{pmatrix}$. It is easy to compute the matrix $R_3, R_4, R_5...$ one after another, but does there exist a closed formula such that given an integer $n$ one can compute the matrix $R_n$ without computing $R_3, ..., R_{n-1}$?
I already tried computing the sequence and find an explicit description but the therms become very fast very complex: Denote by $r_k^i$ the i-th entry of the vector $r_k$:
$R_2=\begin{pmatrix} 1 & 0 \\ r_1^1 & 1 \end{pmatrix}$
$R_3 =\begin{pmatrix} 1 & 0 & 0 \\ r_1^1 & 1 & 0 \\ r_2^1+r_2^2r_1^1 & r_2^2 & 1 \end{pmatrix}$
$R_4 =\begin{pmatrix} 1 & 0 & 0 & 0 \\ r_1^1 & 1 & 0 & 0 \\ r_2^1+r_2^2r_1^1 & r_2^2 & 1 & 0\\ r_3^1+r_1^1r_3^2+r_3^3r_2^1+r_3^3r_2^2r_1^1 & r_3^2+r_3^3r_2^2 & r_3^3 & 1 \end{pmatrix}$
So far I only see the pattern at the subdiagonals.
Thank you in advance for your help.
WLOG, $R_0=(1)$. Then
$$R_1=\begin{pmatrix}1&0\\r&1\end{pmatrix}$$
$$R_2=\begin{pmatrix}1&0&0\\r&1&0\\s+rs'&s'&1\end{pmatrix}$$ $$R_3=\begin{pmatrix}1&0&0&0\\r&1&0&0\\s+rs'&s'&1&0\\t+rt'+(s+rs')t''&t_1+s't''&t''&1\end{pmatrix}$$
does not seem to lead us anywhere.