Let $M=(\alpha_{k,n})_{(k,n)\in \mathbb{N}}=\begin{pmatrix} \alpha_{0,0}& & & (0) \\ \alpha_{1,0}&\alpha_{1,1} & & \\ \alpha_{2,0}&\alpha_{2,1} &\alpha_{2,2} & \\ \vdots &\vdots &\vdots & \ddots \end{pmatrix} $ be an infinite matrix with coefficients in $\mathbb{R}$ such that all the diagonal elements are non zero, and $\beta\in\mathbb{R}$.
We define the sequence $(u_n)_{n\in\mathbb{N}}$ such that for all $n\in \mathbb{N}$, we have :
\begin{equation}
\sum_{k=0}^n \alpha_{k,n} u_k = \beta
\end{equation}
I want to find a formula for $u_n$ depending only on the $\alpha_{k,n}$ and $\beta$. I tried to find a pattern :
\begin{equation}
u_0=\frac{1}{\alpha_{0,0}}\beta
\end{equation}
\begin{equation}
u_1=\frac{\alpha_{0,0}-\alpha_{1,0}}{\alpha_{0,0}\alpha_{1,1}}\beta
\end{equation}
\begin{equation}
u_2=\frac{\alpha_{1,1}\alpha_{0,0}-\alpha_{2,1}\alpha_{0,0}+\alpha_{2,1}\alpha_{1,0}-\alpha_{2,0}\alpha_{1,1}}{\alpha_{0,0}\alpha_{1,1}\alpha_{2,2}} \beta
\end{equation}
\begin{equation}
u_3=\frac{\alpha_{2,2}\alpha_{1,1}\alpha_{0,0}-\alpha_{3,2}\alpha_{1,1}\alpha_{0,0}+\alpha_{3,2}\alpha_{2,1}\alpha_{0,0}-\alpha_{3,2}\alpha_{2,1}\alpha_{1,0}+\alpha_{3,2}\alpha_{2,0}\alpha_{1,1}-\alpha_{3,1}\alpha_{2,2}\alpha_{0,0}+\alpha_{3,1}\alpha_{2,2}\alpha_{1,0}-\alpha_{3,0}\alpha_{2,2}\alpha_{1,1}}{\alpha_{3,3}\alpha_{2,2}\alpha_{1,1}\alpha_{0,0}}\beta
\end{equation}
At this rate, it seems the number of terms in the numerator keeps growing exponentially, estimating $2^n$ terms for the $u_n$ with a product of $n$ terms in each additive term.
I feel that there is a pattern, but I can't really figure it out.
If my estimation is right, I'm searching a set of $2^n$ elements that has something with combining couples $(x,y)$ where $0 \leq x \leq n$ and $ 0 \leq y \leq x$.
For the alterning sign, I'm thinking of something analogous to the parity of a permutation.
These are my guesses for now, any help will be welcome.
2026-03-28 20:38:34.1774730314