What is the general $n \times n$ form of the divisibility matrix? Background + Motivation
I had the following idea of using digits as vectors. Let us have say I have a three digit number $a$ in the basis $\lambda$:
$$ a = a_0 + a_1 \lambda + a_2 \lambda^2$$ 
Now, we want to convert $\lambda$ coefficients to $\lambda+1$ coefficients:  
$$ a = a_0 -a_1 +a_2 + (a_1 - 2a_2) (\lambda +1) + a_2(\lambda +1)^2 $$
We note if $a_0 -a_1 +a_2$ is divisible by $\lambda+1$ then so is $a$. And since we are only interested checking if the number is divisible, we introduce the divisibility matrix (for $3$ digits):
$$
(\lambda+1) |a   \implies (\lambda+ 1)|
\begin{pmatrix}
    1       & 0 & 0  \\
\end{pmatrix}
\begin{pmatrix}
    1       & -1 & 1  \\
    0       & 1 & -2 \\
    0       & 0 & 1
\end{pmatrix}
\begin{pmatrix}
    a_0 \\
    a_1 \\
    a_2
\end{pmatrix} 
$$ 
Similarly  if the $a$ is divisible by $\lambda+2$
$$
(\lambda+2) | a \implies (\lambda+ 2) |
\begin{pmatrix}
    1       & 0 & 0  \\
\end{pmatrix}
\begin{pmatrix}
    1       & -1 & 1  \\
    0       & 1 & -2 \\
    0       & 0 & 1
\end{pmatrix}^2
\begin{pmatrix}
    a_0 \\
    a_1 \\
    a_2
\end{pmatrix}
$$ 
Example
Let $\lambda = 10$ and $a=121$
Then we verify $1-2+1 = 0$ which is indeed divisible by $10 + 1 = 11$
Question
We have only done this for a $3$ digit number. What is general form of the divisibility matrix for a $n$ digit number?

We make use of the following: 
$$a = a_0 + a_1 \lambda + a_2 \lambda^2 + \dots $$
Or with $\lambda + 1$ coefficients:
$$ a = b_0 + b_1 (\lambda +1) + b_2 (\lambda +1)^2 + \dots$$
We define $\lambda +1 = \beta $ and combine the above $2$ equations:
$$ a_0 + a_1(\beta -1) + a_2 (\beta -1)^2 + \dots = b_0 + b_1 \beta + b_2\beta^2 +\dots $$
Setting $\beta = 0$:
$$b_0 = a_0 -a_1 +a_2 -a_3 + \dots = \sum_{i=0}^\infty (-1)^i a_i $$
Differentiating and setting $\beta = 0$ again:
$$ b_1 = a_1 - 2 a_2 + 3 a_3 - 4 a_4 + \dots = \sum_{i=1}^\infty   a_i ( -1)^{i+1} i$$
Differentiating and setting $\beta = 0$ again:
$$  b_2 =  \frac{2!}{2! 0!} a_2 - \frac{3!}{1!2!} a_3 + \frac{4!}{2! 2!} a_4 - \frac{5!}{3! 2!} a_4 +  \dots = \sum_{i=2}^\infty a_i(-1)^{i} \text{ }{ }^i C_2  $$
Hence, in general:
$$ b_k = \sum_{i=k}^\infty a_i (-1)^{i-k} \text{ }{ }^i C_k  $$
with $k \neq 0$
Now, we can construct an $n \times n$ divisibility matrix:
$$D_{jk} = \begin{cases} 
      0 & j < k \\
      (-1)^k & j=0 \\
      (-1)^{j-k} \text{ }{ }^j C_k & \text{else}
   \end{cases}
$$
To write it some terms explicitly:
$$ D = 
\begin{pmatrix}
    1       & -1 & 1 & -1 & 1 &\dots  \\
    0       & 1 & -2  & 3 & -4 & \dots \\
    0       & 0 & 1 & -3 & 6 & \dots \\
    \vdots \\
    0       & 0 &  \dots &  & &1   &
\end{pmatrix}$$
Interestingly each column is related to the binomial tree.
