Consider an infinite, upper-triangular Toepliz-matrix, with first row $x_1,x_2,\dots,x_n.$
Then there is a sequence of determinants obtained from the $m \times m-$ sub-matrices with upper left corner at first row, and $k^{th}$ column (thus $x_k$ is in the upper left corner of the sub-matrix). Thus, for each $n$ and $k$ with $1\leq k\leq n,$ we have a sequence of determinants. By a paper of H. Zakrajsek and M. Petrovsek, there is a recurrence for such sequence of determinants, with length $\binom{n-1}{k-1}$.
Let $\Delta_{n,k}(t)$ be the characteristic polynomial for that recurrence.
We may then arrange $\Delta_{n,k}(t)$ in a triangle, such that the degree (in $t$) of the polynomial in row $n$, column $k$ is $\binom{n-1}{k-1}$.
This triangle has several nice properties:
The polynomials in row $n$ is in the variables $t,x_1,\dots,x_n$.
$\Delta_{n,k}(t) = \phi(\Delta_{n,n-k+1}(t))$ where $\phi$ sends $x_i$ to $x_{n-i+1}.$
and most importantly, $\deg_t(\Delta_{n,k}) = \deg_t(\Delta_{n-1,k}\Delta_{n-1,k-1})$ which hints of a Pascal-like recurrence, but with some sort of multiplication instead of addition.
Question: Is there a nice way to describe this non-commutative multiplication, or some general rule, that gives $\Delta_{n,k}$ in terms of $\Delta_{n-1,k}$ and $\Delta_{n-1,k-1}$?
I have computed the first 6 rows in the triangle, see this pdf. Each colored row consists of the following: The framed number indicates the row in the triangle. The three following polynomials are $\Delta_{n-1,k-1},$ $\Delta_{n-1,k}$ and $\Delta_{n,k}$. What is the rule to obtain the bottom polynomial from the two ones above?
So far I have: $\Delta_{n,k} = t^b \Delta_{n-1,k} + x_n P(t,x_1,\dots,x_n)$ for $b = \binom{n-1}{k-1} - \binom{n-2}{k-1},$ and $P$ is some polynomial. Thus, the problem is to figure out how to introduce the variable $x_n.$
Also, easy to see that $\Delta_{n,1} = t-x_1$ and $\Delta_{n,n} = t-x_n$.
