Determinant of a certain Toeplitz matrix

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 $$ {bf N}_{,n} (x) = left( {matrix{ x & {1 - x} & 1 & 1 & cdots & 1 cr {1 - x} & {2left( {x - 1} right)} & { - x} & 0 & cdots & 0 cr 1 & { - x} & {2left( {x - 1} right)} & { - x} & ddots & 0 cr 1 & 0 & { - x} & {2left( {x - 1} right)} & ddots & 0 cr vdots & vdots & ddots & ddots & ddots & { - x} cr 1 & 0 & 0 & cdots & { - x} & {2left( {x - 1} right)} cr } } right) $$ and developing the determinat along e.g. the last column it seems that it might be possible to develop a recursive relation for it.

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.