What is the inverse of a triangular matrix whose nonzero elements are binomial coefficients? What is the closed-form solution to a recursive relation?

Question

Let \begin{equation*} \begin{split} M_m &=\begin{pmatrix} -\binom{1}{0} & \binom{2}{0} &-\binom{3}{0} &\dotsm & (-1)^{m-1}\binom{m-1}{0} & (-1)^m\binom{m}{0}\\ 0 & \binom{2}{1} &-\binom{3}{1} &\dotsm & (-1)^{m-1}\binom{m-1}{1} & (-1)^m\binom{m}{1}\\ 0 & 0 &-\binom{3}{2} &\dotsm & (-1)^{m-1}\binom{m-1}{2} & (-1)^m\binom{m}{2}\\ \vdots & \vdots &\vdots &\ddots & \vdots & \vdots\\ 0 & 0 & 0 &\dotsm & (-1)^{m-1}\binom{m-1}{m-2} & (-1)^m\binom{m}{m-2}\\ 0 & 0 & 0 &\dotsm & 0 & (-1)^m\binom{m}{m-1} \end{pmatrix}_{m\times m}\\ &=(M_{i,j})_{m\times m}, \end{split} \end{equation*} where \begin{equation*} M_{i,j}= \begin{cases} (-1)^{j}\dbinom{j}{i-1}, & 1\le i\le j\le m;\\ 0, & 1\le j<i\le m. \end{cases} \end{equation*} For $m=5$, by the famous software Mathematica, we obtain $$ \begin{pmatrix} -1 & 1 & -1 & 1 & -1 \\ 0 & 2 & -3 & 4 & -5 \\ 0 & 0 & -3 & 6 & -10 \\ 0 & 0 & 0 & 4 & -10 \\ 0 & 0 & 0 & 0 & -5 \\ \end{pmatrix}^{-1} = \begin{pmatrix} -1 & \frac{1}{2} & -\frac{1}{6} & 0 & \frac{1}{30} \\ 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} & 0 \\ 0 & 0 & -\frac{1}{3} & \frac{1}{2} & -\frac{1}{3} \\ 0 & 0 & 0 & \frac{1}{4} & -\frac{1}{2} \\ 0 & 0 & 0 & 0 & -\frac{1}{5} \\ \end{pmatrix}. $$ What is the inverse of the triangular matrix $M_m$ for $m\in\mathbb{N}=\{1,2,\dotsc\}$?

The matrix $M_m$ comes from the recursive relation \begin{equation}\label{beta(m+1minus1)} \sum_{k=j+1}^{m}(-1)^{k}\binom{k}{j}\beta_{m+1,k} =(-1)^{j+1} \beta_{m,j}, \quad 0\le j\le m-1, \end{equation} where the first few $\beta_{m,j}$ are \begin{align*} \beta_{1,0}&=1, & & & & & &\\ \beta_{2,0}&=\frac{5}{3}, & \beta_{2,1}&=1, & & & &\\ \beta_{3,0}&=\frac{11}{5}, & \beta_{3,1}&=\frac{13}{6}, & \beta_{3,2}&=\frac{1}{2}, & &\\ \beta_{4,0}&=\frac{93}{35}, & \beta_{4,1}&=\frac{101}{30}, & \beta_{4,2}&=\frac{4}{3}, & \beta_{4,3}&=\frac{1}{6}. \end{align*} We can also derive \begin{align*}%\label{beta(m+1)m-form} \beta_{m,m-1}&=\frac{1}{(m-1)!}, \quad m\ge1,\\ \beta_{m,m-2}&=\frac{3m+4}{6(m-2)!}, \quad m\ge2,\\ \beta_{m,m-3}&=\frac{15 m^2+35 m+24}{120(m-3)!}, \quad m\ge3,\\ \beta_{m,m-4}&=\frac{105 m^3+315 m^2+364 m+176}{5040(m-4)!}, \quad m\ge4. \end{align*} We guess that \begin{equation*} \beta_{m,m-k}=\frac{1}{(2k-1)!(m-k)!}\sum_{\ell=0}^{k-1}\theta_{k,\ell} m^\ell, \quad m\ge k, \end{equation*} where $\theta_{k,\ell}$ is a sequence of positive integers.

What is the explicit or closed-form expression of the sequence $\theta_{k,\ell}$ for $0\le\ell\le n-1$? What is the explicit or closed-form expression of the sequence $\beta_{m,j}$ for $0\le j\le m-1$?

Answer 1

This may answer the question, the sequence is the only implicit thing. I consider $Q=\begin{pmatrix}Q_{i,j}\end{pmatrix}_{n\times n}$ for $n\ge 3$, where $$Q_{i,j}= \begin{cases} \dbinom{j}{i-1}, & 1\le i\le j\le n;\\ 0, & 1\le j<i\le n. \end{cases} $$ This is the same as the one defined in the question up to multiplying it by a $\pm 1$ diagonal matrix. Define sequence $u_k$ by $u_1=1$ and $$ u_k=-\sum_{i=0}^{k-2}\dfrac{u_{i+1}}{(k-i)!} $$ for every $k>1$. The $n\times n $ matrix $Q^{-1}=\begin{pmatrix}q_{i,j}\end{pmatrix}_{n\times n}$ is given by $$ \begin{cases} q_{i,j}=0,&1\le j<i\le n;\\ q_{i,i}=\dfrac{1}{i},&1\le j=i\le n;\\ q_{i,i+1}=-\dfrac{1}{2},&2\le j=i+1\le n;\\ q_{i,i+k}=u_{k+1}(i+1)\cdots(i+k-1),&k+1\le i+k\le n, k\ge 2. \end{cases} $$ The proof is formal and should be direct, also the sequence $u_k$ has some zero entries.

Answer 2

Using the answer of @Toni, the sequence $u_k$ can be related to the Bernoulli numbers $B_k$ for $k\geq 0$, \begin{align}\tag{1} u_{k+1}= \frac{B_{k}}{k!}. \end{align} After some algebra, the inverse of $M_m$ is given by \begin{align}\tag{2} M_m^{-1} = \left[ \, (-1)^i \binom{j}{i} \frac{B_{j-i}}{j} \, \right]_{i,j=1}^{m} \end{align} for $m>0$.