text:
\section{Introduction}
Sierpi\'nski-type graphs show up in a wide range of areas; for instance, physics, dynamical systems, probability, and topology, to name a few.
The Sierpi\'nski gasket graphs form one of the most significant families of such graphs that are obtained by a finite number of iterations that give the Sierpi\'nski gasket in the limit. Several more families of Sierpi\'nski-type graphs have been introduced and studied in the literature (see Barri\`{e}re, Comellas, and Dalf\'o \cite{bcd} and Hinz, Klav\v{z}ar, and Zemlji\v{c} \cite{hkz}). In this paper, we deal with two families of them, as described below.

For positive integers $n,k$, the {\em Sierpi\'nski graph} $S(n, k)$ is defined with vertex set $[k]^n$, where $[k]:=\{1,\ldots,k\}$, and two different vertices $(u_1, \ldots , u_n)$ and $(v_1, \ldots , v_n)$ are adjacent if and only if there exists a $t\in[n]$ such that
\begin{itemize}
  \item $u_i=v_i$ for $i=1,\ldots,t-1$,
  \item $u_t\ne v_t$,
  \item $u_j=v_t$ and $v_j=u_t$ for $j=t+1,\ldots,n$.
\end{itemize}
For instance, the graphs $S(3,3)$ and $S(2,4)$ are depicted in Figure~\ref{fig:S}.
\begin{figure}[h!]
\centering
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\end{tikzpicture}~~~~
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(a1) edge (a4)
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(f1) edge (f4)
(f4) edge (a4)
(f1) edge (a4)
(a1) edge (f4)
(b1) edge (a2)
(b1) edge (b2)
(b2) edge (a2)
(b3) edge (a3)
(b3) edge (b4)
(e1) edge (e2)
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(e3) edge (f3)
(f3) edge (e4)
(a3) edge (b4)
(e3) edge (e4);
\end{tikzpicture}
  \caption{The Sierpi\'nski graphs $S(3,3)$ (left) and $S(2,4)$ (right)}\label{fig:S}
\end{figure}
Sierpin\'ski graphs $S(n,k)$ were introduced in Klav\v{z}ar and Milutinovi\'c \cite{kmi}.
The graph $S(n,3)$ is indeed isomorphic to the graph of the Tower
of Hanoi with $n$ disks. The graph $S(n,k)$ has $k^n-k$ vertices of degree $k$ and $k$ vertices of degree $k-1$ that are $(i,\ldots,i)$ for $i\in[k]$.
 These vertices are called the {\em extreme vertices}.

In addition to $S(n,k)$, we consider a `regularization' of them as another
 family of Sierpi\'nski-type graphs.  The graphs $S^{++}(n, k)$, introduced in Klav\v{z}ar and Mohar \cite{km}, are defined as follows. The graph
$S^{++}(1, k)$  is the complete graph $K_{k+1}$. For $n\ge2$, $S^{++}(n, k)$ is the graph obtained
from the disjoint union of $k+1$ copies of $S(n-1, k)$ in which the extreme vertices in
distinct copies of $S(n - 1, k)$ are connected as the complete graph $K_{k+1}$. See Figure~\ref{fig:S++} for an illustration of $S^{++}(3,3)$.
\begin{figure}[h!]
\centering
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 \caption{The graph $S^{++}(3,3)$}\label{fig:S++}
\end{figure}

Many properties of Sierpi\'nski-type graphs, including those of $S(n,k)$ and  $S^{++}(n,k)$, have been studied in the literature, for a survey see Hinz, Klav\v{z}ar, and Zemlji\v{c} \cite{hkz}.
In this paper, we investigate some of the algebraic properties of the two families of graphs, namely the spectrum and the property of being a Cayley graph.
More precisely, in Section~\ref{sec:S++}, we determine the spectrum of the graphs $S^{++}(n,k)$.
The Laplacian spectrum of $S(n,k)$ is already known for $k=2,3$. We establish the case $n=2$, in Section~\ref{sec:S},
and make a conjecture on the Laplacian spectrum of $S(n,k)$ in general.
We also characterize the graphs  $S^{++}(n,k)$ that are Cayley graphs in Section~\ref{sec:Cayley}. As a by-product, a new family of non-Calyley vertex-transitive graphs are obtained. From this result, we conclude a new set of square-free non-Cayley numbers in Section~\ref{sec:NC}, and we discuss its distribution.

\section{Spectrum of $S^{++}(n,k)$}\label{sec:S++}

Let $\Gamma$ be a simple graph with vertex set $V(\Gamma)=\{v_1, \ldots ,v_n \}$  and edge set $E(\Gamma)$. Its {\em adjacency matrix} $A(\Gamma)=[a_{ij}]$ is an $n \times n$ symmetric matrix with $a_{ij}=1$ if $v_i$ and $v_j$ adjacent, and $a_{ij}=0$ otherwise.
 The multi-set of the eigenvalues of $A(\Gamma)$ is called the {\em spectrum} of $\Gamma$.

In this section, we determine the spectrum of $S^{++}(n,k)$. As we will see, the recursive structure of these Sierpi\'nski-type graphs also shows up in their spectrum.

We first recall some basic facts.

The {\em incidence matrix} of a graph $\Gamma$ is a 0-1 matrix $X(\Gamma)=[x_{ve}]$, with rows indexed by the vertices and columns
indexed by the edges of $\Gamma$, where  $x_{ve}=1$ if the vertex $v$ is an endpoint of the edge $e$.
For a graph $\Gamma$, ${\cal L}(\Gamma)$ denotes the {\em line graph} of $\Gamma$, in which $V({\cal L}(\Gamma))$ corresponds with $E(\Gamma)$, and  two vertices of ${\cal L}(\Gamma)$  are adjacent if and only if they have a common vertex as edges of $\Gamma$. The subdivision graph ${\cal S}(\Gamma)$ of $\Gamma$ is the graph obtained by inserting a new vertex into every edge of $\Gamma$.
It is easy to verify that
\begin{equation}\label{eq:xtx}X(\Gamma)^\top X(\Gamma)= 2I+A({\cal L}(\Gamma)),\end{equation}
and, moreover, if $\Gamma$ is $k$-regular, then
\begin{equation}\label{eq:xxt}X(\Gamma)X(\Gamma)^\top= kI+A(\Gamma).\end{equation}

The following lemma gives a recursive relation for the graphs $S^{++}(n,k)$.

\begin{lemma}\label{lineiso}
The graph $S^{++}(n+1,k)$ is isomorphic to ${\cal L}({\cal S}(S^{++}(n,k)))$.
\end{lemma}
\begin{proof} Let $k$ be fixed.
The graph $\Gamma_n:=S^{++}(n,k)$ can be obtained by the union of $S(n,k)$ and $S(n-1,k)$ by adding a matching between the extreme vertices of the two graphs.
If we consider $\{ 0\}\times[k]^{n-1}$ as the vertex set of $S(n-1,k)$ (to make them compatible with the length $n$ of the vertices of $S(n,k)$),
then $\left(\{0\}\cup[k]\right)\times[k]^{n-1}$ is the vertex set of $\Gamma_n$.
It follows that any edge $e=\{{\bf u},{\bf v}\}$ of $\Gamma_n$ is of one of the following types:
\begin{itemize}
  \item[(1)] ${\bf u}=(u_1,\ldots,u_r,u,v,\ldots,v)$, ${\bf v}=(u_1,\ldots,u_r,v,u,\ldots,u)$  for some  $r\le n-2$ and $u\ne v$;
  \item[(2)] ${\bf u}=(u_1,\ldots,u_{n-1},u)$, ${\bf v}=(u_1,\ldots,u_{n-1},v)$ with $u\ne v$;
  \item[(3)] ${\bf u}=(0,u,\ldots,u)$, ${\bf v}=(u,u,\ldots,u)$.
\end{itemize}
 Each $e=\{{\bf u},{\bf v}\}\in E(\Gamma_n)$ is divided into two new edges $e_{\bf u}$ and $e_{\bf v}$  in ${\cal S}(\Gamma_n)$, where we assume that ${\bf u}\in e_{\bf u}$ and ${\bf v}\in e_{\bf v}$.
  We define a map $\psi:E({\cal S}(\Gamma_n))\to\left(\{0\}\cup[k]\right)\times[k]^n$
  based on the type of $e$ as follows:
\begin{itemize}
  \item[(i)] If $e$ is of type (1) or (2), then  $\psi(e_{\bf u})=({\bf u},v)$  and $\psi(e_{\bf v})=({\bf v},u)$;
  \item[(ii)] If $e$ is of type (3), then  $\psi(e_{\bf u})=({\bf u},u)$ and $\psi(e_{\bf v})=({\bf v},u)$.
\end{itemize}
It is easily seen that $\psi$ is a one-to-one map.
We show that $\psi$ is an isomorphism from ${\cal L}({\cal S}(\Gamma_n))$ to $\Gamma_{n+1}$.
Let $e$ and $e'$ be two edges that share a vertex ${\bf x}$ of ${\cal S}(\Gamma_n)$.
If ${\bf x}=(x_1,\ldots,x_n)$ is an `old' vertex of ${\cal S}(\Gamma_n)$, then $\psi(e)=({\bf x},y)$ and $\psi(e')=({\bf x},z)$ for some $y\ne z$.
Then, it is clear that $\psi(e)$ and $\psi(e')$ are adjacent in $\Gamma_{n+1}$.
If ${\bf x}$ is a `new' vertex of ${\cal S}(\Gamma_n)$, then from (i) and (ii), it is clear that $\psi(e)$ and $\psi(e')$ are adjacent in $\Gamma_{n+1}$.
This shows that $\psi$ is indeed a one-to-one homomorphism. As ${\cal L}({\cal S}(\Gamma_n))$ and $\Gamma_{n+1}$ have the same number of edges, it follows that $\psi$ is an isomorphism.
\end{proof}

We recall that if $A$ is a non-singular square matrix, then
\begin{equation}\label{eq:det}\left|\begin{array}{cc}A&B\\C&D\end{array}\right|=|A|\cdot\left|D-CA^{-1}B\right|,\end{equation}
where $|\cdot|$ denotes the determinant of a matrix.
Also, recall that if $M$ is a $p\times q$ matrix, then
\begin{equation}\label{eq:MMT}|xI-MM^\top|=x^{p-q}|xI-M^\top M|.\end{equation}
(Note that \eqref{eq:MMT} might not be valid if $p\le q$ and $x = 0$, but this has no effect in our argument since two polynomials that agree in all but finitely many points, agree everywhere.)

Let
\begin{equation}\label{eq:f}f(x)=x^2+(2-k)x-k,\end{equation}
 and let $f^j(x)$ denote the polynomial of degree $2^j$ obtained by $j$ times composition of $f$ with itself.
 As a convention, we let $f^0(x)=x$.

We now give the main result of this section.
\begin{theorem}
Let $k$  be an integer and  $P_n(x)$ denote the characteristic polynomial of the adjacency matrix of $S^{++}(n,k)$. Then, $P_n$ satisfies the recursion relation
\begin{equation}\label{recursion}
P_n(x)=\left(x(x+2)\right)^{k^{n-2}\left({k\choose 2}-1\right)}P_{n-1}(f(x)),~~n\ge2,
\end{equation}
with $P_1(x)=(x-k)(x+1)^k$.
Moreover, for $n\geq 2$, the spectrum of $S^{++}(n,k)$ consists of the following eigenvalues:
\begin{itemize}
  \item[\rm(i)] $k$ with multiplicity $1$,
   \item[\rm(ii)]  the zeros of $f^{n-1}(x)+1$ each with  multiplicity $k$,
  \item[\rm(iii)]  the zeros of $f^j(x)$ each with multiplicity $k^{n-2-j}\left({k\choose 2}-1\right)$ for $j=0,1,\ldots,n-2$,
  \item[\rm(iv)] the zeros of $f^j(x)+2$ each with multiplicity $k^{n-2-j}\left({k\choose 2}-1\right)+1$ for $j=0,1,\ldots,n-2$.
\end{itemize}
\end{theorem}

\begin{proof} Let $\Gamma_n:=S^{++}(n,k)$.
Suppose that $X$ and $Y$ are the incidence matrices of $\Gamma_{n-1}$ and ${\cal S}(\Gamma_{n-1})$, respectively.
By Lemma~\ref{lineiso}, $\Gamma_n$ is isomorphic to ${\cal L}({\cal S}(\Gamma_{n-1}))$. It follows that
$$YY^\top=\left[\begin{array}{cc}kI_p & X \\ X^\top &   2I_q\end{array}\right],$$
where the matrix is divided according to the partition of the vertices into $p=k^{n-1}+k^{n-2}$ `old' vertices of $\Gamma_{n-1}$
and $q=\frac{1}{2} (k^n+k^{n-1})$ `new' vertices (which have degree $2$) added to $\Gamma_{n-1}$ to obtain ${\cal S}(\Gamma_{n-1})$.
Therefore, from \eqref{eq:det},
\begin{align}
    \left| xI-YY^\top\right|&=\left|(x-k)I_p\right|\cdot\left|(x-2)I_q-X^\top\left((x-k)I_p\right)^{-1}X\right|\nonumber\\
    &=(x-k)^p\left|(x-2)I_q-\frac{1}{x-k}X^\top X\right|\nonumber\\
    &=(x-k)^{p-q}\left|(x-2)(x-k)I_q-X^\top X\right|\nonumber\\
    &=(x-k)^{p-q}((x-2)(x-k))^{q-p}\left|(x-2)(x-k)I_p-XX^\top\right|~~~\hbox{(by \eqref{eq:MMT})}\nonumber\\
     &=(x-2)^{q-p}\left|((x-2)(x-k)-k)I_p-A(\Gamma_{n-1})\right|~~~\hbox{(by \eqref{eq:xxt})}\nonumber\\
     &=(x-2)^{q-p}P_{n-1}((x-2)(x-k)-k).\label{eq:Pn-1}
\end{align}
On the other hand, by \eqref{eq:xtx} and \eqref{eq:xxt}, we have
$$P_n(x)=\left| (x+2)I_{2q}-Y^\top Y\right|=(x+2)^{q-p}\left| (x+2)I_{p+q}-YY^\top\right|.$$
Now, from \eqref{eq:Pn-1} it follows that
$$P_n(x)=(x(x+2))^{q-p}P_{n-1}(x(x+2-k)-k),$$
implying \eqref{recursion}.

To prove the second part of the theorem, note that as $\Gamma_1=K_{k+1}$, we have $P_1(x)=(x-k)(x+1)^k$. From \eqref{recursion}, we conclude that
$$P_2(x)=(x(x+2))^{{k\choose2}-1}(f(x)-k)(f(x)+1)^k,$$
and since
\begin{equation}\label{f-k}
f(x)-k=(x+2)(x-k),
\end{equation}
 the assertion follows for $n=2$.
Now assume that $n\ge3$ and the assertion holds for $n-1$. So, we have
$$P_{n-1}(x)=(x-k)\left(f^{n-2}(x)+1\right)^k\prod_{j=0}^{n-3}\left(f^j(x)\right)^{m_{n-3-j}}\left(f^j(x)+2\right)^{1+m_{n-3-j}},$$
in which $m_i=k^i\left({k\choose 2}-1\right)$.
It follows that
$$P_{n-1}(f(x))=(f(x)-k)\left(f^{n-1}(x)+1\right)^k\prod_{j=1}^{n-2}\left(f^j(x)\right)^{m_{n-2-j}}\left(f^j(x)+2\right)^{1+m_{n-2-j}}.$$
This, together with \eqref{recursion} and \eqref{f-k},  implies the result.
\end{proof}

\begin{remark}\rm It is straightforward to see that the zeros of $f^j(x)$ and  $f^j(x)+2$ for
	$j=1$ are $\frac12(k-2\pm\sqrt{k^{2}+4})$ and  $\frac12(k-2\pm\sqrt{k^{2}-4}),$
			 respectively, and for $j\ge2$ are of the form
$$\frac{1}{2}(k-2)\pm\frac{1}{2}\sqrt{k(k+2) \pm 2\sqrt{k(k+2) \pm2\sqrt{ \cdots \pm 2\sqrt{k^2+4}}}}$$
and
$$\frac{1}{2}(k-2)\pm\frac{1}{2}\sqrt{k(k+2) \pm 2\sqrt{k(k+2) \pm2\sqrt{ \cdots \pm 2\sqrt{k^2-4}}}},$$
respectively, each of them consisting of $j$ nested radicals in iterative forms. Moreover, the zeros of $f^{n-1}(x)+1$  are
\begin{align*}
&-1,\,k-1,\, \frac{1}{2}(k-2)\pm\frac{1}{2}\sqrt{k^2+4k},\, \frac{1}{2}(k-2)\pm\frac{1}{2}\sqrt{k(k+2) \pm 2\sqrt{k^2+4k}},\ldots, \\
&~~~\frac{1}{2}(k-2)\pm\frac{1}{2}\sqrt{k(k+2) \pm 2\sqrt{k(k+2) \pm2\sqrt{ \cdots \pm 2\sqrt{k^2+4k}}}},
\end{align*}
where the last one consists of $n-2$ nested radicals.
\end{remark}

\section{Laplacian spectrum of $S(n,k)$}\label{sec:S}

 For a graph $\Gamma$, the matrix $L(\Gamma)=D(\Gamma)-A(\Gamma)$ is the {\em Laplacian matrix} of $\Gamma$, where
$D(\Gamma)$ is the diagonal matrix of vertex degrees.
The multi-set of eigenvalues of $L(\Gamma)$ is called the {\em Laplacian spectrum} of $\Gamma$.
In this section, we deal with the Laplacian spectrum of $S(n,k)$.
This is trivial for $n=1$ or $k=1$.
For $k=2,3$, the Laplacian spectrum of $S(n,k)$ is already known (see Remark~\ref{rem:conj} below). We establish the case $n=2$,
and put forward a conjecture  explicitly describing the Laplacian spectrum of $S(n,k)$ in general.

Let $E_{ij}$ be a $k\times k$ matrix in which all entries are $0$, except the $(i,j)$ entry that is $1$.
Consider the $k^2\times k^2$ matrix
$$C:=\sum_{i=1}^k\sum_{j=1}^k(E_{ij}\otimes E_{ji}),$$
where `$\otimes$' denotes the Kronecker product. The matrix $C$ is called  the {\em commutation matrix}. The main property of the commutation matrix (see Magnus and Neudecker \cite{mn}) is that it commutes the Kronecker product:
for any $k\times k$ matrices $M,N$,
$$C(M\otimes N)C=N\otimes M.$$
Note that each row and each column of $C$ corresponds with a pair $(i,j)$ for $1\le i,j\le k$.
Moreover, $C$ is indeed a permutation matrix in which the only $1$ entry in the row $(i,j)$ is located at the column $(j,i)$ for every $1\le i,j\le k$.

For $n=1$, the Laplacian spectrum of $S(1,k)=K_k$ is $\left\{0^{[1]},\,k^{[k-1]}\right\}$, where the superscripts indicate multiplicities.
In the following theorem, we determine the Laplacian spectrum of $S(2,k)$.

\begin{theorem}
The Laplacian spectrum of $S(2,k)$ is the following:
$$\left\{0^{[1]},~ k^{\left[k\choose 2\right]},~(k+2)^{\left[k-1\choose 2\right]},~\left(\frac{1}{2}(k+2)\pm \frac{1}{2}\sqrt{k^2+4}\right)^{[k-1]}\right\}.$$
\end{theorem}
\begin{proof}
First, note that the graph $S(2,k)$ consists of $k$ copies of $K_k$ together with a matching $M$ of size $k\choose2$; exactly one edge for each pair of copies of $K_k$. Let $L$ denote the Laplacian matrix of $S(2,k)$, and $L'$ be the Laplacian matrix of the induced subgraph by the edges of $M$.
It is seen that $L=Q-B$, where $Q=L(kK_k)+I$ and $B=I-L'$. Note that $B$ is a permutation matrix with ${k\choose 2}+k$ eigenvalues $1$ and ${k\choose 2}$ eigenvalues $-1$. Observe that $Q$ has $k$ eigenvalues $1$ and $k^2-k$ eigenvalues $k+1$.
 We have the following bounds on the dimensions of the  intersections of the eigenspaces of $B$ and $Q$:
\begin{align*}
\dim({\cal E}_1(B)\cap{\cal E}_{k+1}(Q))&\geq k^2-k+{k\choose 2} +k-k^2={k\choose 2},\\
\dim({\cal E}_{-1}(B)\cap{\cal E}_{k+1}(Q))&\geq k^2-k+{k\choose 2}-k^2={k\choose 2}-k,
\end{align*}
in which ${\cal E}_\lambda$ denotes the eigenspace corresponding to the eigenvalue $\lambda$.
For ${\bf x}\in{\cal E}_1(B)\cap{\cal E}_{k+1}(Q)$, we have $L{\bf x}=k{\bf x}$ and for ${\bf x}\in{\cal E}_{-1}(B)\cap{\cal E}_{k+1}(Q)$, $L{\bf x}=(k+2){\bf x}$.
This means that $L$ has eigenvalues $k$ and $k+2$ with multiplicities at least $k\choose2$ and ${k\choose2}-k$, respectively.

We also have
\begin{equation}\label{eq:Q}Q=I_k\otimes ((k+1)I_k-J_k),\end{equation}
and from the eigenvalues of $Q$,
\begin{equation}\label{eq:Q^2}Q^2-(k+2)Q+(k+1)I=O.\end{equation}
Coming back to $B$, for each of the extreme vertices $(1,1),\ldots,(k,k)$ of $S(2,k)$, there is a $1$ on all the entries of the diagonal of $B$.
The off-diagonal $1$'s correspond with the edges of $M$. By the definition of $S(2,k)$, the edges of $M$ connect the vertices $(i,j)$ and $(j,i)$ for $i\ne j$.
It turns out that $B$ is the commutation matrix, and thus
\begin{equation}\label{eq:BQB}BQB=((k+1)I_k-J_k)\otimes I_k.\end{equation}
The right sides of \eqref{eq:Q} and \eqref{eq:BQB} commute, and so
$$BQBQ=QBQB.$$
Next, we see that
\begin{align*}
 (L^2-(k&+2)L+kI)(QB-BQ)\\&=((Q-B)^2-(k+2)(Q-B)+kI)(QB-BQ)\\
 &=\left(Q^2-(k+2)Q+kI+B^2+(k+2)B-QB-BQ\right)(QB-BQ)\\
  &=\left((k+2)B-QB-BQ\right)(QB-BQ)~~~~\hbox{(by \eqref{eq:Q^2} and since $B^2=I$)}\\
 &=Q^2-(k+2)Q-B\left(Q^2-(k+2)Q\right)B-(QB)^2+(BQ)^2\\
 &=(BQ)^2-(QB)^2\\
 &=O.
\end{align*}

The above equality shows that every vector in the column space of $QB-BQ$ is an eigenvector for $L$ with eigenvalues $\lambda$, where $\lambda^2-(k+2)\lambda+k=0$. To obtain the multiplicity of such $\lambda$, we compute the rank of $QB-BQ$:
\begin{align}
  {\rm rank}(QB-BQ)&={\rm rank}(Q-BQB)\nonumber\\
  &={\rm rank}(I_k\otimes ((k+1)I_k-J_k)-((k+1)I_k-J_k)\otimes I_k)\nonumber\\
  &={\rm rank}(J_k\otimes I_k-I_k\otimes J_k)\nonumber\\
  &=2k-2.\label{eq:2k-2}
\end{align}
To show \eqref{eq:2k-2}, suppose  $P$ is a $k\times k$ matrix whose first column is $\frac1{\sqrt k}(1,\ldots,1)^{\top}$ and that $PP^{\top}=I_k$. Then, $J_k = P(kE_{11})P^{\top}$, and so
$$J_k\otimes I_k-I_k\otimes J_k=(P\otimes P)\big((kE_{11}\otimes I_k)-(I_k\otimes kE_{11})\big)(P^{\top}\otimes P^{\top}).$$
Since $(kE_{11}\otimes I_k)-(I_k \otimes kE_{11})$
 is a diagonal matrix having precisely $2k-2$ non-zero entries in
the columns $2,3,\ldots,k,k+1,2k+1,3k+1,\ldots,(k-1)k+1$,  \eqref{eq:2k-2} follows.

As  $x^2-(k+2)x+k$ is an irreducible polynomial, each of its roots is an  eigenvalue of $L$ with multiplicity at least $k-1$.
The matrix $L$ has a $0$ eigenvalue. Thus, we have obtained so far ${k\choose2}+{k\choose2}-k+2(k-1)+1=k^2-1$ eigenvalues of $L$.
As the sum of the eigenvalues of $L$ is twice the number of edges of $S(2,k)$, it follows that the remaining eigenvalue is $k+2$. So the proof is complete.
\end{proof}

Based on empirical evidence, we put forward the following conjecture.
\begin{conj}\label{conj}\rm
For $n,k\geq 2$, the Laplacian spectrum of $S(n,k)$ consists of the following eigenvalues:
\begin{itemize}
  \item[\rm(i)] $0$ with multiplicity $1$.
    \item[\rm(ii)]  The zeros of $f^j(k-x)$, each with multiplicity $\frac12(k^{n-j}-2k^{n-j-1}+k)$ for $j=0,1,\ldots,n-1$, where $f$ is given in \eqref{eq:f}.
  \item[\rm(iii)] The zeros of $f^j(k-x)+2$, each with multiplicity $\frac12(k^{n-j-1}-1)(k-2)$ for $j=0,1,\ldots,n-2$.
\end{itemize}
\end{conj}
\begin{remark}\rm The zeros of $f^j(k-x)$ and  $f^j(k-x)+2$
	 for	$j=1$ are $\frac12(k+2\pm\sqrt{k^{2}+4})$ and  $\frac12(k+2\pm\sqrt{k^{2}-4}),$
	respectively, and for $j\ge2$ are of the form
$$\frac{1}{2}(k+2)\pm\frac{1}{2}\sqrt{k(k+2) \pm 2\sqrt{k(k+2) \pm2\sqrt{ \cdots \pm 2\sqrt{k^2+4}}}}$$
and
$$\frac{1}{2}(k+2)\pm\frac{1}{2}\sqrt{k(k+2) \pm 2\sqrt{k(k+2) \pm2\sqrt{ \cdots \pm 2\sqrt{k^2-4}}}},$$
respectively, each of them consisting of $j$ nested radicals.
\end{remark}
\begin{remark}\label{rem:conj}\rm
The graph $S(n,2)$ is the path graph on $2^n$ vertices. Proposition~\ref{prop:path} below shows that Conjecture~\ref{conj} holds for $S(n,2)$.
In Grigorchuk and \v{S}uni\'c \cite{gs}, the spectrum of the Schreier graph $\Gamma_n$ was determined.
The graph $\Gamma_n$ is, in fact, the graph obtained from $S(n,3)$ by adding a loop on each extreme vertex.
By the way the adjacency matrix of $A(\Gamma_n)$ is defined in \cite{gs}, for each loop a $1$ entry on the diagonal is considered, so that each row and column of $A(\Gamma_n)$ has constant sum $3$. It is then observed that the Laplacian spectrum of $S(n,3)$ can be deduced from the spectrum of $\Gamma_n$, which agrees with Conjecture~\ref{conj}.
In summary, Conjecture~\ref{conj} holds for $n=2$ and for $k=2,3$.
\end{remark}

It is known in the literature that the characteristic polynomial of the Laplacian matrix of the paths can be expressed in terms of the
Chebyshev polynomials. From this fact, for paths with $2^n$ vertices, we obtain the iterated form according to Conjecture~\ref{conj}. For the sake of completeness, we give its complete argument here.
\begin{proposition} \label{prop:path}
The characteristic polynomial of the Laplacian matrix of the path graph on $2^n$ vertices is equal to
$x\prod_{j=0}^{n-1}g^j(2-x)$, where $g(x)=x^2-2$.
\end{proposition}
\begin{proof}  Let $\phi_m$ be the characteristic polynomial of the Laplacian matrix of the path graph on $m$ vertices.
Let $T_m$ and $U_m$ be  the Chebyshev polynomials of degree $m$ of the first and the
second kind, respectively. Then $T_m$ is the only polynomial satisfying
$T_m(\cos\theta)=\cos m\theta$ and $U_m(x) =\sin((m+1)\arccos x)/\sin(\arccos x)$ (Snyder \cite {s}).
From the identities given in Cvetkovi\'c, Doob, and Sachs \cite[p.~220]{cds}, it follows that
$\phi_m(x)=xU_{m-1}(x/2-1)$. By successive use of the identity
$U_{2k-1}(x)=2T_k(x)U_{k-1}(x)$ (see \cite[p.~98]{s}), we get
$$U_{2^n-1}(x)=2^{n-1}T_{2^{n-1}}(x)T_{2^{n-2}}(x)\cdots T_2(x)U_1(x).$$
Note that $U_1(x)=2x$ and $T_2(x)=2x^2-1$. It is seen that $2T_2(x/2-1)=x^2-4x+2=g(2-x)$.
This, together with the identity $T_{2k}(x)=T_2(T_k(x))$, implies that
$2T_{2^j}(x/2-1)=g^j(2-x)$. The proof is now complete.
\end{proof}

\section{What $S^{++}(n,k)$ are Cayley graphs?}\label{sec:Cayley}

Recall that a graph $\Gamma$ is  {\em vertex-transitive} if for any two vertices $u,v$  of $\Gamma$, there exists an automorphism $\sigma$ of $\Gamma$ such that
$\sigma(u)=v$. Let $G$ be a group and $C \subset G$ such that $1\not\in C$ and $c\in C$ implies that $c^{-1}\in C$.
The {\em Cayley graph} $\mathrm{Cay}(G,C)$ with the group $G$ and the `connection set' $C$ is the
graph with vertex set  $G$ in which vertex $u$ is connected to $v$ if and only
if $vu^{-1} \in C$.

It is known that any Cayley graph is vertex-transitive.
 In the other way around, at least for small orders, it seems that the great majority of vertex-transitive graphs are Cayley graphs, see McKay and Praeger \cite{bdm-cep1}. It is expected to continue to be this way for larger orders. In fact, it is conjectured in Praeger, Li, and Niemeyer \cite{cep-chl-acn} that  most vertex-transitive graphs are Cayley graphs.
   In this section, we first determine what $S^{++}(n,k)$ are vertex-transitive and, then, classify $S^{++}(n,k)$ that are Cayley graphs.

\begin{proposition}\label{prop:transitive}
The graph $S^{++}(n,k)$ is vertex-transitive if and only if either $n\le2$ or $k\le2$.
\end{proposition}
\begin{proof}
We have $S^{++}(n,1)\cong K_2$, $S^{++}(1,k)\cong K_{k+1}$, and $S^{++}(n,2)$ is the cycle graph on $2^{n-1}\cdot3$ vertices, which are all vertex-transitive graphs. By Lemma \ref{lineiso}, $S^{++}(2,k)$ is isomorphic to ${\cal L}({\cal S}(K_{k+1}))$.
In the graph ${\cal S}(K_{k+1})$, the `new' vertices  are in one-to-one correspondence with $2$-subsets of $[k+1]$.
It is then easy to see that that any permutation of $[k+1]$ induces an automorphism of  ${\cal S}(K_{k+1})$. Now,
for a given pair of edges of ${\cal S}(K_{k+1})$ which can be represented as $e=\{i,\{i,j\}\}$ and $e'=\{i',\{i',j'\}\}$,
the automorphism induced by a permutation $\sigma$ that $\sigma(i)=i'$ and $\sigma(j)=j'$, maps $e$ to $e'$.
It follows that   ${\cal S}(K_{k+1})$ is edge-transitive, and so ${\cal L}({\cal S}(K_{k+1}))\cong S^{++}(2,k)$ is vertex-transitive.  Hence, assume that $n\geq3$ and $k\ge3$.
  We observe that an extreme vertex of a copy $\Delta$ of $S(n-1,k)$ in $\Gamma=S^{++}(n,k)$ cannot be mapped to a non-extreme vertex of  $\Delta$ by any automorphism of $\Gamma$.
    To be more precise, let ${\bf u}=(1,\ldots,1,1)$ and ${\bf v}=(1,\ldots,1,2)$.
    It can be seen that ${\bf u}$ is a cut vertex for the induced subgraph by the vertices at distance at most $3$ form ${\bf u}$,
    while ${\bf v}$ is not a cut vertex for the induced subgraph by the vertices at distance at most $3$ form ${\bf v}$.
    It follows that ${\bf u}$ cannot be mapped to ${\bf v}$ by any automorphism of $\Gamma$, and thus $\Gamma$ is not vertex-transitive.
\end{proof}

From Proposition~\ref{prop:transitive}, it follows that the graphs $S^{++}(n,k)$ for $n\ge3$ and $k\ge3$ cannot be Cayley graphs.
The graphs $S^{++}(n,1)$, $S^{++}(n,2)$, and $S^{++}(1,k)$ are all Cayley graphs. It remains to characterize what $S^{++}(2,k)$ are Cayley graphs for $k\ge3$.
This is our goal in the rest of this section.

\begin{definition}\em
Let $\Gamma$ be a graph and $\Delta$ a subgraph of $\Gamma$.
We say that $\Gamma$ is  {\em strongly $\Delta$-partitioned} if:
\begin{itemize}
\item[\rm(i)] The vertex set of $\Gamma$ is partitioned by the vertex sets of copies $\Delta_0,\ldots,\Delta_k$ of $\Delta$.
\item[\rm(ii)] Apart from $\Delta_0,\ldots,\Delta_k$, the graph $\Gamma$ contains no further copies of $\Delta$.
\end{itemize}
\end{definition}

By the way $S^{++}(n,k)$ is defined, it is constructed based on $k+1$ copies of $S(n-1,k)$.
The following proposition gives a structural property of $S^{++}(n,k)$
that it is indeed strongly $S(n-1,k)$-partitioned for $n\ge2$ and $k\ge3$.
Note that this is not the case for $k=2$ because $S^{++}(n,2)$, that is a cycle with $3\cdot2^{n-1}$ vertices, contains more than three copies of $S(n-1,2)$, which is a path on  $2^{n-1}$ vertices. Although we only need the case $n=2$ of the proposition, we state it in its full generality because it could be of independent interest.

\begin{proposition}\label{vertex partition}
Let $n\ge2$ and $k\ge3$. The graph $S^{++}(n,k)$ is strongly $S(n-1,k)$-partitioned.
\end{proposition}
\begin{proof}
Let $\Gamma:=S^{++}(n,k)$ and $\Gamma_0,\ldots,\Gamma_k$ be the $k+1$ copies of $S(n-1,k)$ used to construct $\Gamma$ by its definition.
Clearly, $V(\Gamma_0),\ldots,V(\Gamma_k)$ is a partition of $V(\Gamma)$.
We show that $\Gamma$ contains no more copies of $S(n-1,k)$. Let $\Delta$ be a subgraph of $\Gamma$ isomorphic to $S(n-1,k)$.

First, assume that $n=2$. Let $u\in V(\Delta)\cap V(\Gamma_t)$ for some $t$, with $0\leq t\leq k$. Since $u$ has at most one neighbor in $V(\Gamma)\setminus V(\Gamma_t)$ and $k\geq3$, there  exists another vertex $v\in V(\Delta)\cap V(\Gamma_t)$ adjacent to $u$. Now, if $w$ is any vertex of $\Delta$, then since $w$ is adjacent to two vertices $u$ and $v$ of $\Gamma_t$ it must belong to $V(\Gamma_t)$. Hence $V(\Delta)\subseteq V(\Gamma_t)$ and, consequently, $\Delta=\Gamma_t$.

Now, let $n\geq3$. Note that $S(n-1,k)$ is connected and has no bridges since every edge of $S(n-1,k)$ lies on a cycle (which can be seen by induction on $n$). If $\Delta\ne\Gamma_i$ for $i=0,\ldots,k$, then $\Delta$ shares its vertices with at least two $\Gamma_s$ and $\Gamma_t$.
By the definition, exactly one extreme vertex, say $u$, of $\Gamma_s$ is adjacent to exactly one extreme vertex, say $v$, of $\Gamma_t$.
Because of the connectivity, $\Delta$ must contain the edge $uv$. Note that
for any vertex $w$ outside $\Gamma_s$ and $\Gamma_t$, the distance between $w$ and either $u$ or $v$ is greater than the diameter of $S(n-1,k)$, and so
$w\not\in V(\Delta)$. It follows that $\Delta$ is a subgraph of $\Gamma':=\Gamma[V(\Gamma_s)\cup V(\Gamma_t)]$.
However, $uv$ is a bridge for $\Gamma'$ and thus a bridge for $\Delta$, a contradiction.
\end{proof}

The following lemma reveals the structure of strongly $\Delta$-partitioned Cayley graphs.
\begin{lemma}\label{Cayley partitioned graph: induced subgraph=coset}
Let $\Gamma$ be a Cayley graph with  a subgraph $\Delta$ such that $\Gamma$ is strongly $\Delta$-partitioned. Then, the vertex sets of the copies of $\Delta$ are all the right cosets of a subgroup of the underlying group of $\Gamma$.
\end{lemma}
\begin{proof}
Let $\Gamma$ be a Cayley graph on a group $G$, and $X\subseteq G$ be such that $1\in X$ and $\Gamma[X]$, the subgraph of $\Gamma$ induced by $X$, is isomorphic to $\Delta$. Since for any $x\in X$, $\Gamma[Xx^{-1}]$ is isomorphic to $\Delta$ and $1\in Xx^{-1}$, from the hypothesis of the lemma, it follows that $Xx^{-1}=X$. Thus $XX^{-1}=X$ and, hence, $X$ is a subgroup of $G$. As for any $g\in G$, $\Gamma[Xg]$ is isomorphic to $\Delta$ and the sets $Xg$ cover all elements of $G$, it follows that every induced subgraph of $\Gamma$ isomorphic to $\Delta$ is a right coset of $X$, as required.
\end{proof}

\begin{definition}\em
Let $\Gamma$ be a strongly $\Delta$-partitioned graph. We say that $\Gamma$ has {\em connection constant} $c$ if there are exactly $c$ edges between any two copies of $\Delta$ in $\Gamma$. We denote the set of all strongly $\Delta$-partitioned graphs with connection constant $c$ by $\mathcal{SP}_c(\Delta)$.
\end{definition}

\begin{remark}\rm  The family $\mathcal{SP}_1(K_d)$ contains an only regular graph. However, this is not the case in any regular graph $\Delta$.
 If $\Gamma\in\mathcal{SP}_1(\Delta)$ is a regular graph with $\Delta$ being a $d$-regular graph on $k$ vertices, then $\Gamma$ necessarily contains $k+1$ copies of $\Delta$ and thus $\Gamma$ is $(d+1)$-regular with $k(k+1)$ vertices. For instance, in the case in which $\Delta$ is $C_4$, the cycle on $4$ vertices, $\Gamma$ is a cubic graph on $20$ vertices.
By a computer search, we found all the regular graphs in $\mathcal{SP}_1(C_4)$. It turned out that there are seven non-isomorphic such graphs, among which only one is a Cayley graph.
\end{remark}

Here we recall some notions from group theory that will be used in what follows. Let $G$ be a finite group and $H$ be a nontrivial proper subgroup of $G$. The conjugate of $H$ by an element $g$ of $G$ is defined as $H^g=\{h^g:h\in G\}$, where $h^g:=g^{-1}hg$ denotes the conjugate of $h$ by $g$. The group $G$ is called a \textit{Frobenius group} with \textit{Frobenius complement} $H$ if $H\cap H^g=\{1\}$ for all $g\in G\setminus H$. A celebrated theorem of Frobenius states that  $N:=G\setminus\bigcup_{g\in G}(H\setminus\{1\})^g$ is a normal subgroup of $G$, called the Frobenius kernel of $G$, satisfying $G=NH$ and $N\cap H=\{1\}$, that is, $G=N\rtimes H$ is a semidirect product of $N$ by $H$ (see \cite[8.5.5]{djsr}). The other concepts we use in the following are standard and can be found in Robinson \cite{djsr}.
\begin{theorem}\label{Cayley pseudo-Sierpinski graph with c=1}
Suppose that $\Gamma$ and $\Delta$ are two regular graphs and $\Gamma\in\mathcal{SP}_1(\Delta)$.
If $\Gamma$ is a Cayley graph $\mathrm{Cay}(G,C)$, then $|\Delta|+1=p^m$ is a prime power, $G=N\rtimes H$ is a Frobenius group with minimal normal Frobenius kernel $N\cong\Bbb{Z}_p^m$ and Frobenius complement $H$, $C=C'\cup\{c\}$ with $\Delta\cong\mathrm{Cay}(H,C')$ and $c^2=1$, and either
\begin{itemize}
\item[\rm(i)]$c\in N$ and $H=\gen{C'}$, or
\item[\rm(ii)]$c=h^n$ for some $h\in H\setminus\{1\}$ and $n\in N\setminus\{1\}$, and $H=\gen{C',h}$.
\end{itemize}
Conversely, if $\Delta$ satisfies the above conditions, then $\mathrm{Cay}(G,C)\in\mathcal{SP}_1(\Delta)$.
\end{theorem}
\begin{proof}
Let $\Gamma=\mathrm{Cay}(G,C)$, and $\Delta_0,\Delta_1,\ldots,\Delta_k$ be the copies of $\Delta$ in $\Gamma$.
As there is exactly one edge between any two copies of $\Delta$, it is observed that $|\Delta|=k$ and $\Gamma$ is $(d+1)$-regular if $\Delta$ is $d$-regular. Let $H:=V(\Delta_0)$ and assume, without loss of generality, that $1\in H$. By Lemma \ref{Cayley partitioned graph: induced subgraph=coset}, $H$ is a subgroup of $G$. Let $C'$ be the neighborhood of $1$ in $\Gamma[H]$. Since $\Gamma$ is $(d+1)$-regular, besides the elements of $C'$, the vertex $1$ has exactly one other neighbor, say $c\in G\setminus H$. So $C=C'\cup\{c\}$. Since $H$ is a subgroup of $G$, $C'^{-1}\subseteq H$, which implies that $C'^{-1}=C'$. Thus, $c=c^{-1}$ is an involution. Clearly, $Hc\neq H$ so that $\Gamma[Hc]=\Delta_i$ for some $1\leq i\leq k$. On the other hand,
\[1=|E(\Delta_0,\Delta_i)|=|\{\{h,ch\}:h\in H\cap H^c\}|=|H\cap H^c|,\]
from which it follows that $H\cap H^c=\{1\}$. Now, a simple verification shows that $Hch\cap Hch'=\emptyset$ for all distinct elements $h,h'\in H$. Since $\Gamma[H]=\Delta_0$ and $\Gamma[Hch]$ ($h\in H$) are equal to $\Delta_1,\ldots,\Delta_k$ in some order, we must have
\[G=H\cup\bigcup_{h\in H}Hch,\]
where the unions are disjoint. As a result, every element $g\in G\setminus H$ can be written as $g=hch'$ for some $h,h'\in H$, from which it follows that
\[H\cap H^g=(H^{{h'}^{-1}}\cap (H^h)^c)^{h'}=(H\cap H^c)^{h'}=\{1\}^{h'}=\{1\}.\]
Hence, $G$ is a Frobenius group with complement $H$. Let $N$ be the Frobenius kernel of $G$. By \cite[10.5.1(i)]{djsr}, $N$ is nilpotent. Let $N_0$ be a nontrivial characteristic subgroup of $N$ with minimum order. Then $N_0$ is a normal subgroup of $G$ (see \cite[1.5.6(iii)]{djsr}). Note that $N_0$ is an elementary Abelian $p$-group for $N_0$ is nilpotent and the subgroup of $N_0$ generated by central elements of a given prime order $p$ dividing $|Z(N_0)|$ is a characteristic subgroup of $N_0$ and hence of $N$ (see \cite[1.5.6(ii)]{djsr}). If $N\neq N_0$, then $N_0H$ is a Frobenius group for $N_0H$ is a subgroup of $G$ and $H\cap H^g=1$ for all $g\in N_0H\setminus H$. Moreover, as a proper subgroup of $N$, $|N_0|\leq|N|/2\leq(k+1)/2$ and hence $|N_0|-1$ is not divisible by $|H|=k$ contradicting \cite[Exercises 8.5(6)]{djsr}. Thus $N=N_0$ so that $k+1=|N|=p^m$ is a prime power for some $m\geq1$. Note that $N$ is a minimal normal subgroup of $G$ for if $N$ contains a nontrivial normal subgroup $N_0$ of $G$ properly, then $N_0H$ would be a Frobenius group which leads us to the same contradiction as above. If $c\in N$, then since $G\subseteq N\gen{C'}$ it follows that $H=\gen{C'}$. Now assume that $c\notin N$. Then $c^n\in H\setminus\{1\}$ for some $n\in N\setminus\{1\}$. As $G\subseteq N\gen{C',c^n}$ it follows that $H=\gen{C',c^n}$, as required. The converse is straightforward.
\end{proof}

We are now in a position to conclude the main result of this section.
\begin{theorem}\label{Cayley S++(n,k)}
The graph $S^{++}(n,k)$  is a Cayley graph if and only if either
\begin{itemize}
\item[\rm(i)] $n=1$,
\item[\rm(ii)] $k\le2$, or
\item[\rm(iii)] $n=2$ and $k+1=p^m$ is a prime power.
\end{itemize}
Furthermore, in the case {\rm(iii)}, we have
\[S^{++}(n,k)\cong\mathrm{Cay}(G,(H\setminus\{1\})\cup\{c\}),\]
for every Frobenius group $G$ with complement $H$ of order $p^m-1$, elementary Abelian minimal normal Frobenius kernel of order $p^m$, and involution $c\in G\setminus H$.
\end{theorem}
\begin{proof}
By Proposition~\ref{prop:transitive}, $S^{++}(n,k)$ for $n\ge3$ and $k\ge3$ is not a Cayley graph.
As mentioned above, $S^{++}(n,1)$, $S^{++}(n,2)$, and $S^{++}(1,k)$ are all Cayley graphs.
So, we may assume that $n=2$ and $k\ge3$.

First, we show that $S^{++}(2,q-1)$ are Cayley graphs for all prime powers $q$. Let $\Bbb{F}_q$ denote the finite field with $q$ elements. Then $G:=\Bbb{F}^*_q\times\Bbb{F}_q$ together with the multiplication
\[(x,a)\cdot(y,b)=(xy,xb+a),\]
forms a group known as {\em one dimensional affine group}. We show that $S^{++}(2,q-1)\cong\mathrm{Cay}(G,C)$, where $C=\left\{(x,0): 1\ne x\in\Bbb{F}^*_q\right\}\cup\{(-1,-1)\}$. To this end, let $H:=\{(x,0): x\in\Bbb{F}^*_q\}$ be a subgroup of $G$ of order $q-1$. Then $H$ has $q$ right cosets each of which induces a complete subgraph in $\mathrm{Cay}(G,C)$ for $h'g(hg)^{-1}=h'h^{-1}\in C$ for all distinct elements $hg$ and $h'g$ of a right coset $Hg$ of $H$. Since $(x,0)(1,ax^{-1})=(x,a)$ covers all elements of $G$ when $x$ and $a$ ranges over $\Bbb{F}^*_q$ and $\Bbb{F}_q$, respectively, it follows that every right coset of $H$ has a representative of the form $(1,b)$ for some $b\in \Bbb{F}_q$. Let $Hg$ and $Hg'$ be distinct right cosets of $H$ with $g=(1,a)$ and $g'=(1,a')$. Then an element $hg$ of $Hg$ is adjacent to an element $h'g'$ of $Hg'$ if and only if $h'g'g^{-1}h^{-1}=(h'g')(hg)^{-1}\in C$ or equivalently $(h'g')(hg)^{-1}=(-1,-1)$ as $g'g^{-1}\notin H$. A simple verification shows that this equation has a unique solution for $(h,h')$ so that there is a unique edge between any two right cosets of $H$. Indeed, $h=(x,0)$ and $h'=(x',0)$ satisfy the equation if and only if $-x'=x=(a'-a)^{-1}$. Hence, from the definition, it follows that $S^{++}(2,q-1)\cong\mathrm{Cay}(G,C)$.

Now, assume that $\Gamma:=S^{++}(2,k)\cong\mathrm{Cay}(G,C)$ be a presentation of $S^{++}(2,k)$ as a Cayley graph. By Proposition~\ref{vertex partition}, $\Gamma$ is strongly $\Gamma_0$-partitioned for some complete subgraph $\Gamma_0$ of $\Gamma$ of order $k$. Let $H:=V(\Gamma_0)$ and assume that $1\in H$. We know from Lemma \ref{Cayley partitioned graph: induced subgraph=coset} that $H$ is a subgroup of $G$.
By Theorem~\ref{Cayley pseudo-Sierpinski graph with c=1}, $k+1=p^m$ is a prime power, $G=N\rtimes H$ is a Frobenius group with Frobenius kernel $N$ and Frobenius complement $H$ such that $N\cong \Bbb{Z}_p^m$ is a minimal normal subgroup of $G$, $C=C'\cup\{c\}$, $C'^{-1}=C'\subseteq H$, $c^2=1$, and either
\begin{itemize}
\item[\rm(a)]$c\in N$ and $H=\gen{C'}$, or
\item[\rm(b)]$c=h^n$ for some $h\in H\setminus\{1\}$ and $n\in N\setminus\{1\}$, and $H=\gen{C',h}$.
\end{itemize}
Since $\Gamma_0$ is a complete graph, we must have $H\setminus\{1\}\subseteq C$. Then, (a) and (b) together are equivalent to say that $c\in G\setminus H$. The proof is now complete.
\end{proof}

As a generalization of Theorem~\ref{Cayley pseudo-Sierpinski graph with c=1}, we pose the following problem.

\begin{problem}\rm
Let $\Delta$ be a regular graph. Classify all Cayley graphs in $\mathcal{SP}_c(\Delta)$ for $c\ge2$.
\end{problem}

\section{New non-Cayley numbers}\label{sec:NC}

In this final section, we give a partial answer to a famous rather old open problem in algebraic graph theory. A positive integer $n$ is called a \textit{Cayley number} if all vertex-transitive graphs of order $n$ are Cayley graphs. Maru\v{s}i\v{c} \cite{dm} in 1983 posed the problem of characterizing the set $\mathcal{NC}$ of all non-Cayley numbers. Since disjoint unions of copies of vertex-transitive (non-Cayley) graphs are again vertex-transitive (non-Cayley) graphs, it follows that every multiple of a non-Cayley number is again a non-Cayley number.
Hence the problem of determining $\mathcal{NC}$ reduces to finding `minimal' non-Cayley numbers. It is well-known that all primes are Cayley numbers. Following a series of papers by various authors, McKay and Praeger \cite{bdm-cep2} and Iranmanesh and Praeger \cite{mai-cep} provided necessary and sufficient conditions under which the product of two and three distinct primes is a Cayley number, respectively. In the same paper, McKay and Praeger established the following remarkable result determining all non-square-free Cayley numbers.
\begin{theorem}[McKay and Praeger \cite{bdm-cep2}]
Let $n$ be a positive integer that is divisible by the square of a prime $p$. Then $n\in\mathcal{NC}$ unless $n=p^2$, $n=p^3$, or $n=12$.
\end{theorem}

It follows that, for determining $\mathcal{NC}$, it is enough to consider only square-free positive integers. While the problem is yet open for the products of at least four distinct primes, there are partial results worth to mention here.
\begin{theorem}[Dobson and Spiga \cite{td-ps}]
There exists an infinite set of primes every finite product of its distinct elements is a Cayley number.
\end{theorem}

As a consequence of Theorem~\ref{Cayley S++(n,k)}, the graph $S^{++}(2,k)$ that has $k(k+1)$ vertices is not a Cayley graph if $k+1$ is not a prime power.
Therefore, we obtain a new infinite class of square-free non-Cayley numbers as follows.
\begin{theorem}\label{non-Cyaley numbers}
Let $k$ be any positive integer such that $k(k+1)$ is square-free, and $k+1$ is not a prime. Then, $k(k+1)\in\mathcal{NC}$.
\end{theorem}

As mentioned in Dobson and Spiga \cite{td-ps}, it is straightforward by making use of the group-theoretic and the number-theoretic results already available in the literature to prove that Cayley numbers have density zero in the natural numbers, and hence the density
of non-Cayley numbers is $1$. In the light of this  fact, one might wonder about the distribution of the numbers $k$ satisfying the conditions of Theorem~\ref{non-Cyaley numbers} in the set of positive integers.
The following theorem shows that for large enough $N$, more than one third of positive integers less than or equal to $N$  satisfies the conditions of
Theorem~\ref{non-Cyaley numbers}.
\begin{theorem}
The density of the set
\[\{k :  k(k+1)~\hbox{is square-free, and $k+1$ is not a prime}\}\]
is about $0.3226$
\end{theorem}
\begin{proof}
Let $f\in\Bbb{Z}[t]$ be a primitive polynomial (that is, the greatest common divisor of its coefficients is 1) without multiple roots such that its image on $\Bbb{N}$ has $k$-free greatest common divisor. Recall that a number that is not divisible by any proper $k$-th power is called $k$-free. Let $\mathcal{S}_f^k(x)$ denote the number of all positive integers $n\leq x$ such that $f(n)$ is $k$-free, and consider
\[\delta_{f,k}:=\prod_{p\ \text{prime}}\left(1-\frac{\varrho(p^k)}{p^k}\right),\]
where $\varrho(d)$ denotes the number of roots of $f$ in $\Bbb{Z}_d$. Ricci \cite{gr} (see also Pappalardi \cite{fp}) proved that
\[\mathcal{S}_f^k(x)\sim\delta_{f,k}x\]
provided that $\deg f\leq k$. Clearly, the function $f(t):=t(t+1)$ satisfies the above requirements of Ricci's theorem for $k=2$. Also, it is obvious that $\varrho(p^2)=2$ for all primes $p$. Thus, by Ricci's theorem, the density of all positive integers $k$, for which $k(k+1)$ is square-free, in the set of all positive integers, is equal to
\[\delta_{f,2}=\prod_{p\ \text{prime}}\left(1-\frac{2}{p^2}\right)=2C_{\text{Feller-Tornier}}-1\approx0.3226340989,\]
where $C_{\text{Feller-Tornier}}$ is the Feller-Tornier constant (see Finch \cite[\S 2.4.1]{srf}). Since primes have zero density in the set of all positive integers, the result follows.
\end{proof}

To date, all the numbers $n$ whose membership in $\mathcal{NC}$ is known are determined based on the results of \cite{mai-cep, bdm-cep1, bdm-cep2}.
Using a computer search, we see that the list of the numbers whose membership in $\mathcal{NC}$ are not yet determined begins with
$$9982,\ 12958,\ 18998,\ 19646,\ 20398,\ 21574,\ 24662,\ 25438,\ 25606,\ \ldots.$$
A simple computation reveals that among the numbers leass than or equal to $10^8$, there are $2763$ square-free integers of the form $k(k+1)$, with $k+1$ not a prime of which the following eight integers are new non-Cayley numbers:
$$1386506,\ 2668322,\ 15503906,\ 23985506,\ 38359442,\ 74261306,\ 89898842,\ 95912642.$$

\section*{Acknowledgments}
The authors would like to thank anonymous referees for constructive comments which led to improvement of the presentation of the paper. The second author carried this work during a Humboldt Research Fellowship at the University of Hamburg. He thanks the Alexander von Humboldt-Stiftung for financial support.