\end{matrix} \right|,$$

where

$$N_1 = \frac {N}{2},$$

$$\label{eq:18}
\begin{aligned}
L_{N_1} (\varepsilon) = \left.\begin{array}{l}\left| \begin{matrix} \varepsilon& 1 &&&&&&\\ 
1&\varepsilon&1&&&&& \\
&1&\varepsilon&1&&&& \\
&&&\bold{.}&&&& \\
&&&&\bold{.}&&& \\
&&&&&\bold{.}&& \\
&&&&&1&\varepsilon&1 \\
&&&&&&1&\varepsilon
 \end{matrix} \right|\end{array}\right\}
 \end{aligned}
  N_1$$

\
and $I_{N_1}$ being the unity matrix of order $N_1$.\
The determinant can be brought into a quasi-triangular form. For this, we need to add row number $(N_1+1)$ to the first row, row $(N_1+2)$ to the second and so on until row $N_1$ is added to row $2N_1$. Then, the first column is subtracted from column $(N_1+1)$, second from $(N_1+2)$ and analogously column $N_1$ from $2N_1$.\
As result we obtain

$$\label{eq:19}
D_N = L_{N_1} (\varepsilon + 1) L_{N_1} (\varepsilon - 1),$$

where $L_n(x)$ are well-known polynomials, expressions of which are given in \[10\] (for $g_n(x)$ there) for $n \leq 20$. It is also known that representation of $L_n(x)$ via the trigonometrical functions is \[11\]

$$\label{eq:20}
L_n (x) = \frac{sin ((n+1) \theta)}{sin \theta},$$

where $x=2cos \theta$.

Then, for obtaining the energy values $D_N (\varepsilon) = 0$ is reduced to conditions, which need to satisfy at least one of two of the following equations:

$$\label{eq:21}
L_{N_1} (\varepsilon + 1) = 0$$

or

$$\label{eq:22}
L_{N_1} (\varepsilon - 1) = 