Determinant of a circulant matrix as Chebyshev-like recurrence It is while studying the Hückel Method of Physical Chemistry that I came across the following recurrence relation:
\begin{align*}
U_n(x)=xU_{n-1}(x)-U_{n-2}(x)+(-1)^{n-1}(4+2x)
\end{align*}
Where 
\begin{align*}
U_n(x):=\underbrace{\left|\begin{matrix} x& 1 &&&&1\\
1 & x& 1 &&& \\
& 1 & x &1&& \\
&& 1 &x&& \\
&&&& \ddots &1\\ 
1 &&&&1&x
\end{matrix}\right|}_n
\end{align*}
For the related determinant $\displaystyle D_n(x):=\underbrace{\left|\begin{matrix} x& 1 &&&&\\
1 & x& 1 &&& \\
& 1 & x &1&& \\
&& 1 &x&& \\
&&&& \ddots &1\\ 
 &&&&1&x
\end{matrix}\right|}_n$, we have the nicer recurrence relation:
\begin{align*}
D_n(x)=xD_{n-1}(x)-D_{n-2}(x)
\end{align*}
which is a Chebyshev polynomial of the first kind.
However, I would like to somehow find a closed form for $U_n(x)$. Is this at all possible?
 A: First, some clarifying remarks: The Chebyshev polynomials of the first and second kinds are traditionally written as $T_n(x)$ and $U_n(x)$ respectively. To avoid confusion with the problem as written, I will write the latter as $S_n(x)$ instead.
Observe that $U_n(x)$ as given above is only well-defined for $n\geq 3$. But we can extend this to $n\geq 1$ if we take $U_1(x)=x+2,$  $U_2(x)=\left\vert\begin{matrix}
x&2\\2&x \end{matrix}\right\vert=x^2-4$ which may be checked to be consistent with the recurrence relation and the polynomials for $n\geq 3$.
Additionally, the canonical Chebyshev polynomials of the second kind $S_n(x)$ may be checked to relate to the second set of polynomials as $S_n(x)=D_n(2x)$, i.e. the main diagonal is $2x$ instead. For that reason I will introduce $V_n(x):=U_n(2x)$ for convenience. These satisfy
$$V_n=2xV_{n-1}-V_{n-2}+(-1)^{n-1}(4+4x)\text{ for }n\geq 3,\;\; V_1=2x+2,\;\; V_2=4x^2-4.$$
Here I have initiated the convention of suppressing the explicit $x$-dependence of $V_n$.
We now are ready to proceed. To obtain $\{V_n\}$, let us define the generating function $\mathcal{V}(t)=\sum_{n=1}^\infty V_n t^n$ for these polynomials. Then
\begin{align}
\mathcal{V}(t)
&=V_1 t+V_2 t^2+\sum_{n=3}^\infty V_n t^n\\
&=V_1 t+V_2 t^2+2x\sum_{n=3}^\infty V_{n-1}t^n-\sum_{n=3}^\infty V_{n-2}t^n+(4+4x)\sum_{n=3}^\infty (-1)^{n-1}t^n\\
&=V_1 t+V_2 t^2+2xt\left(\mathcal{V}(t)-V_1 t\right)-t^2 \mathcal{V}(t)-(4+4x)\cdot\frac{t^3}{1+t}\\\\
\implies&\mathcal{V}(t)=\frac{t}{1-2xt+t^2}
\left[(1-2xt) V_1+V_2 t+\frac{(4+4x)t^2}{1+t}\right]\\
\end{align}
Simplifying the factor within brackets using the initial values yields
$$ (1-2xt) (2x+2)+(4x^2-4) t+\frac{(4+4x)t^2}{1+t}
 = (2x+2)\frac{1-t}{1+t} \\
\implies \mathcal{V}(t)=\dfrac{2x+2}{1-2xt+t^2}\dfrac{t-t^2}{1+t}
=\frac{1-t^2}{1-2x t+t^2}-\frac{2}{1+t}+1.$$
Now, suppose we had been doing the second-kind Chebyshev polynomials $S_n$, in which case the inhomogeneous term $(-1)^{n-1}(4x+4)$ isn't present. In that case, one may check that the generating function obtained is simply
$\mathcal{S}(t)=\sum\limits_{n=0}^\infty S_n t^n=\dfrac{1}{1-2xt+t^2}$ and so 
$$\mathcal{V}(t)=(1-t^2)\mathcal{S}(t)-\frac{2}{1+t}+1$$
From this we can directly identify $U_n(x)=V_n(x/2)$ for all $n$ as

\begin{align}
U_0(x) &= S_0(x/2)-2+1\\&=1-2+1=0,\\
U_1(x) &= S_1(x/2)+2 \\&= x+2,\\
U_{n\geq 2}(x)&=S_n(x/2)-S_{n-2}(x/2)-2(-1)^n,
\end{align}

which represents our final result.
