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](http://www2.math.su.se/~per/files.php?file=Determinant.Recurrence.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 $\Delta\_{n,n} = t-x\_n$.



Find recurrence in Pascal-like triangle of polynomials
I found an explicit way to express all these recurrences.


Put $t \mapsto -t$ in this question,
[Constructing new polynomials by product of roots](https://mathoverflow.net/questions/88853/constructing-new-polynomials-by-product-of-roots)
and you have the characteristic polynomial for the recurrence.


This is actually a type of plethysm.


