0.$$

In view of , these equations lead to

$$\label{eq:23}
sin[(N_1 + 1) \theta_+] = 0$$

and

$$\label{eq:24}
sin[(N_1 + 1) \theta_-] = 0,$$

where the $\theta_{\pm}$ are determined from

$$\label{eq:25}
2cos \theta_{\pm} = \varepsilon \pm 1 .$$

The solutions of and are $$\label{eq:26}
\theta_{\pm} = \frac{n_\pm \pi}{N_1 + 1},$$

where $n_{\pm}$ = 1,2,…,$N_1$.

Then, from we get the formula for the energy values of a closed cluster with $2N_1$ atoms, which describes the nanoribbon

$$\label{eq:27}
\varepsilon_{n_\pm} = 2cos \frac{n_\pm \pi}{N_1 + 1} \pm 1 .$$

Taking into account that $n_+$ and $n_-$ both may possess $N_1$ values and due to the $\pm 1$ part we see from , that $\varepsilon_{n_\pm}$ possesses $2N_1=N$ values, as this has to be the case for a cluster with $N$ atoms.

If the ribbon is transformed to an infinite layer $(N_1 \rightarrow \infty)$, then we get the boundary energy spectrum values from , which are $\varepsilon_{max} = 3$ at $n_+=1$ and $\varepsilon_{min} = - 3$ at $n_-=N_1$. These values also coincide with the result in section\[sec:Ham\].

Using equation one can easily find the values of $N_1$, under which the solutions $\varepsilon_{n \pm}=0$ exist, that is the ribbon is metallic. The condition for this is 