**1.** Let us write the connection matrix \({\mathbf{J}}\) of the 1D Ising model as a sum of a one-step shifting matrix \({\mathbf{T}}\) and the transposed matrix \({{{\mathbf{T}}}^{ + }}\):

$${\mathbf{J}} = \left( {\begin{array}{*{20}{c}} 0&1&0&1 \\ 1&0& \ddots &0 \\ 0& \ddots & \ddots &1 \\ 1&0&1&0 \end{array}} \right) = {\mathbf{T}} + {{{\mathbf{T}}}^{ + }},\,\,{\text{where}}\,\,{\mathbf{T}} = \left( {\begin{array}{*{20}{c}} 0&0&0&1 \\ 1&0& \ddots &0 \\ 0& \ddots & \ddots &0 \\ 0&0&1&0 \end{array}} \right).$$

(A1)

The interaction constant is equal to one: \({{J}_{{ii + 1}}} = 1\).

Let \({{{\mathbf{e}}}_{i}}\) be the \(i\)-th unit vector in the space \({{{\mathbf{R}}}^{{\mathbf{N}}}}\): \(\left( {{{{\mathbf{e}}}_{i}},{{{\mathbf{e}}}_{j}}} \right) = {{\delta }_{{ij}}}\), where \({{\delta }_{{ij}}}\) is the Kronecker symbol and \(i = 1,...,N\). The matrix \({\mathbf{T}}\) transforms the \(i\)-th unit vector in the \((i + 1)\)-th unit vector: \({\mathbf{T}}{{{\mathbf{e}}}_{i}} = {{{\mathbf{e}}}_{{i + 1}}}\). In addition we suppose that \({\mathbf{T}}{{{\mathbf{e}}}_{N}} = {{{\mathbf{e}}}_{1}}\). By \(F({\mathbf{s}})\) we denote the value of the quadratic form: \(F({\mathbf{s}}) = \left( {{\mathbf{Ts}},{\mathbf{s}}} \right)\). Then the energy of the state \({\mathbf{s}}\) is equal to:

$$E({\mathbf{s}}) = - \frac{{\left( {{\mathbf{Js}},{\mathbf{s}}} \right)}}{{2N}} = - \frac{{F({\mathbf{s}})}}{N}.$$

(A2)

In what follows it is more convenient to use not the energy \(E({\mathbf{s}})\) but the function \(F({\mathbf{s}})\).

We choose the ground state \({{{\mathbf{s}}}_{0}} = (1,1,...,1) = \sum\nolimits_{i = 1}^N {{{{\mathbf{e}}}_{i}}} \) as an initial configuration and by \({{{\mathbf{s}}}_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}\) we denote the configurations from the *n*-vicinity of \({{{\mathbf{s}}}_{0}}\) supposing that spins “–1” are at the positions \({{i}_{1}}\), \({{i}_{2}}\), ..., and \({{i}_{n}}\):

$${{{\mathbf{s}}}_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}} = {{{\mathbf{s}}}_{0}} - 2\left( {{{{\mathbf{e}}}_{{{{i}_{1}}}}} + {{{\mathbf{e}}}_{{{{i}_{2}}}}}... + {{{\mathbf{e}}}_{{{{i}_{n}}}}}} \right).$$

It is easy to see that

$$F({{{\mathbf{s}}}_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}) = N - 4 \cdot n + 4 \cdot \left( {{{{\mathbf{e}}}_{{{{i}_{1}} + 1}}} + {{{\mathbf{e}}}_{{{{i}_{2}} + 1}}} + ... + {{{\mathbf{e}}}_{{{{i}_{n}} + 1}}},{{{\mathbf{e}}}_{{{{i}_{1}}}}} + {{{\mathbf{e}}}_{{{{i}_{2}}}}} + ... + {{{\mathbf{e}}}_{{{{i}_{n}}}}}} \right).$$

(A3)

By

$${{\Delta }_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}} = \left( {{{{\mathbf{e}}}_{{{{i}_{1}} + 1}}} + {{{\mathbf{e}}}_{{{{i}_{2}} + 1}}} + ... + {{{\mathbf{e}}}_{{{{i}_{n}} + 1}}},{{{\mathbf{e}}}_{{{{i}_{1}}}}} + {{{\mathbf{e}}}_{{{{i}_{2}}}}} + ... + {{{\mathbf{e}}}_{{{{i}_{n}}}}}} \right),$$

(A4)

we denote the scalar product in

eq. (A3).

The set of different \({{\Delta }_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}\) defines the energy spectrum in the *n*-vicinity. It is not difficult to understand the construction of these scalar products. At first, suppose that the indices \({{i}_{j}}\) are in series: \({{i}_{2}} = {{i}_{1}} + 1,\;{{i}_{3}} = {{i}_{2}} + 1,...\), and \({{i}_{n}} = {{i}_{{n - 1}}} + 1\). It is easy to see that in this case the value of \({{\Delta }_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}\) is equal to \(n - 1\). Next, let the set of indices falls into two isolated subgroups. Inside each of the group the indices are in series so that the first group is \(i,i + 1,..,i + a\) and the second group is \(k,k + 1,..,k + b\). Evidently, \(a + b + 2 = n\) but there is a gap between the value of the index \(i + a\) and the index \(k\): \(k > i + a + 1\). It can easily be checked that independent of the structures of the groups the value of \({{\Delta }_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}\) is equal to \(n - 2\). Similarly, when the set of the indices \({{i}_{1}},{{i}_{2}},..,{{i}_{n}}\) falls into three isolated subgroups inside which the indices are in series the value of the scalar product \({{\Delta }_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}\) is equal to \(n - 3\) and it is independent of the structures of the groups, and so on. When the set of indices \({{i}_{1}},{{i}_{2}},..,{{i}_{n}}\) falls into \(k\) isolated subgroups (inside each of which the indices are in series) the value of \({{\Delta }_{{{{i}_{1}}{{i}_{2}}..{{i}_{n}}}}}\) is equal to \(n - k\). When \(n \leqslant {N \mathord{\left/ {\vphantom {N 2}} \right. \kern-0em} 2}\) the maximal possible number of such isolated subgroups is equal to \({{k}_{{\max }}} = n\). Using equations (A2), (A3) and (A4), we obtain that the energies of the states from the *n*-vicinity take exactly \(n\) different values:

$$E(k) = - \left( {1 - {{4k} \mathord{\left/ {\vphantom {{4k} N}} \right. \kern-0em} N}} \right),\,\,\,\,k = 1,2,...n,\,\,{\text{where}}\,\,n = 1,2,..,{N \mathord{\left/ {\vphantom {N 2}} \right. \kern-0em} 2}.$$

(A5)

**2.** Now let us find the degeneracies of the energy levels \(E(k)\) in the *n*-vicinity. By \(D(n,k)\) we denote the number of the state in the *n*-vicinity whose energy is equal to \(E(k)\).

The 0-vicinity consists of the configuration \({{{\mathbf{s}}}_{0}}\) itself, \(F({{{\mathbf{s}}}_{0}}) = \left( {{{{\mathbf{s}}}_{0}},{{{\mathbf{s}}}_{0}}} \right) = N\) and we can write: \({{E}_{0}} = - 1\) and \(D(0,0) = 1\).

The 1-vicinity of \({{{\mathbf{s}}}_{0}}\) consists of \(N\) configurations that differ from \({{{\mathbf{s}}}_{0}}\) by different value of one spin only. Consequently, for all \(N\) configurations from the 1-vicinity the energies are equal to the same value \(E(1)\) and \(D(1,1) = N\).

It is easy to see that \(C_{N}^{2}\) configurations from the 2-vicinity of \({{{\mathbf{s}}}_{0}}\) falls into two subgroups. First, there are \(N\) configurations where two “–1” spins are in series. Let \(E(1)\) be the energy of the states from this subgroup. Second, all other configurations belong to the second subgroup. The energy of these states is equal to \(E(2)\). Consequently,

$$D(2,1) = N,\,\,\,\,D(2,2) = \frac{{N(N - 3)}}{2}.$$

It is not difficult to show that in the 3-vicinity there are exactly \(N\) configurations that are characterized by the energy \(E(1)\), \(N(N - 4)\) configurations that are characterized by the energy \(E(2)\), and the energy of all other configurations is equal to \(E(3)\). Then we obtain:

$$D(3,1) = N,\,\,\,\,D(3,2) = N(N - 4),\,\,\,\,D(3,3) = \frac{{N(N - 4)(N - 5)}}{{3!}}.$$

From now on, the calculations are more cumbersome but with the aid of a straightforward analysis for the 4-vicinity and 5-vicinity we obtain the following results:

$$D(4,1) = N,\,\,\,\,D(4,2) = \frac{3}{2}N(N - 5),\,\,\,\,D(4,3) = \frac{{N(N - 5)(N - 6)}}{2},$$

and

$$D(4,4) = \frac{{N(N - 5)(N - 6)(N - 7)}}{{4!}};$$

$$D(5,1) = N,\,\,\,\,D(5,2) = 2N(N - 6),\,\,\,\,D(5,3) = N(N - 6)(N - 7),\,\,\,\,D(5,4) = \frac{{N(N - 6)(N - 7)(N - 8)}}{{3!}},$$

and

$$D(5,5) = \frac{{N(N - 6)(N - 7)(N - 8)(N - 9)}}{{5!}}$$

respectively.

An analyses of the expressions for the values of \(D(n,k)\) obtained for the five vicinities of the ground state allowed us to write a combinatorial formula that we confirmed by computer simulations for different values of \(N,n\) and \(k\):

$$D(n,k) = NC_{{N - n - 1}}^{{k - 1}} \cdot {{C_{n}^{k}} \mathord{\left/ {\vphantom {{C_{n}^{k}} n}} \right. \kern-0em} n} = \frac{{Nk}}{{(N - n)n}}C_{{N - n}}^{k}C_{n}^{k},\,\,\,\,\begin{array}{*{20}{c}} {n = 1,2,...{N \mathord{\left/ {\vphantom {N 2}} \right. \kern-0em} 2};} \\ {k = 1,2,...n.} \end{array}$$

(A6)