Determinant of a certain Toeplitz matrix 
Compute the following determinant
$$\begin{vmatrix} x & 1 & 2 & 3 & \cdots & n-1 & n\\ 1 & x & 1 & 2 & \cdots & n-2 & n-1\\ 2 & 1 & x & 1 & \cdots & n-3 & n-2\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ n-1 & n-2 & n-3 & n-4 & \cdots & x & 1\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}$$


I tried the following. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. I got:
\begin{vmatrix} x-1 & 1-x & 1 & 1 & \cdots & 1 & 1\\ -1 & x-1 & 1-x & 1 & \cdots & 1 & 1\\ -1 & -1 & x-1 & 1-x & \cdots & 1 & 1\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ -1 & -1 & -1 & -1 & \cdots & x-1 & 1-x\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}
I did the same thing with the columns. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. And I got:
\begin{vmatrix} 2x-2 & -x & 0 & 1 & \cdots & 0 & 1\\ -x & 2x-2 & -x & 1 & \cdots & 0 & 1\\ -2 & -x & 2x-2 & 1-x & \cdots & 0 & 1\\ 1 & 1 & 1-x & x & \cdots & -1 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ -2 & -2 & -2 & -1 & \cdots & 2x-2 & 1-x\\ 1 & 1 & 1 & n-3 & \cdots & 1-x &x \end{vmatrix}
I hope I didn’t make a mistake somewhere. With this part I don't know what to do next. I don't know if I'm doing it right. Thank you in advance !
 A: Here are the first few determinants with the help of WA:
$$
\begin{array}{rl}
n & \text{determinant}
\\ 1 & x^2 - 1
\\ 2 & x^3 - 6 x + 4
\\ 3 & x^4 - 20 x^2 + 32 x - 12
\\ 4 & x^5 - 50 x^3 + 140 x^2 - 120 x + 32
\\ 5 & x^6 - 105 x^4 + 448 x^3 - 648 x^2 + 384 x - 80
\\ 6 & x^7 - 196 x^5 + 1176 x^4 - 2520 x^3 + 2464 x^2 - 1120 x + 192
\\ 7 & x^8 - 336 x^6 + 2688 x^5 - 7920 x^4 + 11264 x^3 - 8320 x^2 + 3072 x - 448
\end{array}
$$
There are some patterns for the coefficients but I don't see a complete pattern:

*

*The polynomial has degree is $n+1$


*The coefficient of $x^{n+1}$ is $1$


*The coefficient of $x^{n}$ is $0$


*The coefficient of $x^{n-1}$  is $-$A002415$(n+1)$


*The independent term is $(-1)^n$A001787$(n)$
OEIS doesn't have the sequence of coefficients of $x^{n-2}$ nor of $x$.
I don't expect a nice closed form in monomial form. A recurrence is more probable.
A: Partial answer
Let's define the following $[0,n] \times [0,n]$ matrices (diagonal, subdiagonal, and all-ones Lower triangular)
$$
\eqalign{
  & {\bf I}_{\,n}  = \left( {\matrix{
   1 & {} & {} & {}  \cr 
   {} & 1 & {} & {}  \cr 
   {} & {} &  \ddots  & {}  \cr 
   {} & {} & {} & 1  \cr 
 } } \right)\quad {\bf E}_{\,n}  = \left( {\matrix{
   {} & {} & {} & {}  \cr 
   1 & {} & {} & {}  \cr 
   {} &  \ddots  & {} & {}  \cr 
   {} & {} & 1 & {}  \cr 
 } } \right)  \cr 
  & {\bf S}_{\,n}  = \left( {{\bf I}_{\,n}  - {\bf E}_{\,n} } \right)^{\, - \,{\bf 1}}
  = \left( {\matrix{
   1 & {} & {} & {}  \cr 
   1 & 1 & {} & {}  \cr 
    \vdots  &  \vdots  &  \ddots  & {}  \cr 
   1 & 1 & 1 & 1  \cr 
 } } \right) \cr} 
$$
where the zeros have been omitted to make more clear the ones-pattern,
and let's denote with an over-bar the transpose.
It is then easy to see that our matrix reads as
$$
{\bf M}_{\,n} (x) = x\,{\bf I}_{\,n}  + {\bf S}_{\,n} ^{\,{\bf 2}}  - {\bf S}_{\,n}
  + \overline {{\bf S}_{\,n} } ^{\,{\bf 2}}  - \overline {{\bf S}_{\,n} } 
$$
Since the determinant of ${\bf S}_{\,n}$ is unitary, we can multiply by its inverse
to get a simpler matrix
$$
\eqalign{
  & {\bf N}_{\,n} (x) = {\bf S}_{\,n} ^{\, - {\bf 1}} \,{\bf M}_{\,n} (x)\;\overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}  =   \cr 
  &  = x\,{\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}
  + {\bf S}_{\,n} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}  - \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}
  + {\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} }  - {\bf S}_{\,n} ^{\, - {\bf 1}}  =   \cr 
  &  = x\,{\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}
  + \left( {{\bf S}_{\,n} \overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}
  + {\bf S}_{\,n} ^{\, - {\bf 1}} \overline {{\bf S}_{\,n} } } \right)
 - \left( {\overline {{\bf S}_{\,n} } ^{\, - {\bf 1}}  + {\bf S}_{\,n} ^{\, - {\bf 1}} } \right) \cr} 
$$
Now ${\bf N}_{\,n} (x) $ has the following structure
$$
{\bf N}_{\,n} (x) = \left( {\matrix{
   x & {1 - x} & 1 & 1 &  \cdots  & 1  \cr 
   {1 - x} & {2\left( {x - 1} \right)} & { - x} & 0 &  \cdots  & 0  \cr 
   1 & { - x} & {2\left( {x - 1} \right)} & { - x} &  \ddots  & 0  \cr 
   1 & 0 & { - x} & {2\left( {x - 1} \right)} &  \ddots  & 0  \cr 
    \vdots  &  \vdots  &  \ddots  &  \ddots  &  \ddots  & { - x}  \cr 
   1 & 0 & 0 &  \cdots  & { - x} & {2\left( {x - 1} \right)}  \cr 
 } } \right)
$$
and developing the determinat along e.g. the last column it seems that it might be possible to develop
a recursive relation for it.
