Some Algebraic Properties of Sierpi\'nski-Type Graphs

TL;DR

This paper explores algebraic properties of Sierpiński graphs and their generalizations, including spectra, Cayley graph characterization, and introduces new non-Cayley vertex-transitive graphs and conjectures on Laplacian spectra.

Contribution

It provides the spectrum and Cayley graph characterization for regular generalized Sierpiński graphs, and introduces new non-Cayley vertex-transitive graphs.

Findings

Spectrum of regular generalized Sierpiński graphs determined

Characterization of Cayley graphs within this family

New non-Cayley vertex-transitive graphs and non-Cayley numbers

Abstract

This paper deals with some of the algebraic properties of Sierpi\'nski graphs and a family of regular generalized Sierpi\'nski graphs. For the family of regular generalized Sierpi\'nski graphs, we obtain their spectrum and characterize those graphs that are Cayley graphs. As a by-product, a new family of non-Cayley vertex-transitive graphs, and consequently, a new set of non-Cayley numbers are introduced. We also obtain the Laplacian spectrum of Sierpi\'nski graphs in some particular cases, and make a conjecture on the general case.

Equations

$$X(\Gamma {)}^{\top }X(\Gamma )=2I+A(\mathcal{L}(\Gamma )),$$

$$X(\Gamma )X(\Gamma {)}^{\top }=kI+A(\Gamma ).$$

$$\left|\begin{array}[]{cc}A&B\\ C&D\end{array}\right|=|A|\cdot\left|D-CA^{-1}B\right|,$$

$$|xI-M{M}^{\top }|=x^{p-q}|xI-M^{\top }M|.$$

$$f(x)=x^2+(2-k)x-k,$$

$$P_n(x)={\left(x(x+2)\right)}^{k^{n-2}\left(\left(\genfrac{}{}{0ex}{}{k}{2}\right)-1\right)}{P}_{n-1}(f(x)),n\ge 2,$$

$$YY^{\top}=\left[\begin{array}[]{cc}kI_{p}&X\\ X^{\top}&2I_{q}\end{array}\right],$$

$$\left|xI-Y{Y}^{\top }\right|$$

$$=(x-k{)}^p\left|(x-2)I_q-\frac{1}{x-k}{X}^{\top }X\right|$$

$$=(x-k{)}^{p-q}\left|(x-2)(x-k)I_q-X^{\top }X\right|$$

$$=(x-k{)}^{p-q}((x-2)(x-k){)}^{q-p}\left|(x-2)(x-k)I_p-X{X}^{\top }\right|(by\text{eqref}eq:MMT)$$

$$=(x-2{)}^{q-p}\left|((x-2)(x-k)-k)I_p-A({\Gamma }_{n-1})\right|(by\text{eqref}eq:xxt)$$

$$=(x-2{)}^{q-p}{P}_{n-1}((x-2)(x-k)-k).$$

$$P_n(x)=\left|(x+2)I_{2q}-Y^{\top }Y\right|=(x+2{)}^{q-p}\left|(x+2)I_{p+q}-Y{Y}^{\top }\right|.$$

$$P_n(x)=(x(x+2){)}^{q-p}{P}_{n-1}(x(x+2-k)-k),$$

$$P_2(x)=(x(x+2){)}^{\left(\genfrac{}{}{0ex}{}{k}{2}\right)-1}(f(x)-k)(f(x)+1{)}^k,$$

$$f(x)-k=(x+2)(x-k),$$

$$P_{n-1}(x)=(x-k){\left(f^{n-2}(x)+1\right)}^k\prod _{j=0}^{n-3}{\left(f^j(x)\right)}^{m_{n-3-j}}{\left(f^j(x)+2\right)}^{1+m_{n-3-j}},$$

$$P_{n-1}(f(x))=(f(x)-k){\left(f^{n-1}(x)+1\right)}^k\prod _{j=1}^{n-2}{\left(f^j(x)\right)}^{m_{n-2-j}}{\left(f^j(x)+2\right)}^{1+m_{n-2-j}}.$$

$$\frac{1}{2}(k-2)\pm \frac{1}{2}\sqrt{k(k+2)\pm 2\sqrt{k(k+2)\pm 2\sqrt{\cdots \pm 2\sqrt{k^2+4}}}}$$

$$\frac{1}{2}(k-2)\pm \frac{1}{2}\sqrt{k(k+2)\pm 2\sqrt{k(k+2)\pm 2\sqrt{\cdots \pm 2\sqrt{k^2-4}}}},$$

$$-1,k-1,\frac{1}{2}(k-2)\pm \frac{1}{2}\sqrt{k^2+4k},\frac{1}{2}(k-2)\pm \frac{1}{2}\sqrt{k(k+2)\pm 2\sqrt{k^2+4k}},\dots ,$$

$$\frac{1}{2}(k-2)\pm \frac{1}{2}\sqrt{k(k+2)\pm 2\sqrt{k(k+2)\pm 2\sqrt{\cdots \pm 2\sqrt{k^2+4k}}}},$$

$$C:=\sum _{i=1}^k\sum _{j=1}^k(E_{ij}\otimes E_{ji}),$$

$$C(M\otimes N)C=N\otimes M.$$

$$\left\{0^{[1]},k^{\left[\left(\genfrac{}{}{0ex}{}{k}{2}\right)\right]},(k+2{)}^{\left[\left(\genfrac{}{}{0ex}{}{k-1}{2}\right)\right]},{\left(\frac{1}{2}(k+2)\pm \frac{1}{2}\sqrt{k^2+4}\right)}^{[k-1]}\right\}.$$

$$\dim ({\mathcal{E}}_1(B)\cap {\mathcal{E}}_{k+1}(Q))$$

$$\dim ({\mathcal{E}}_{-1}(B)\cap {\mathcal{E}}_{k+1}(Q))$$

$$Q=I_k\otimes ((k+1)I_k-J_k),$$

$$Q^2-(k+2)Q+(k+1)I=O.$$

$$BQB=((k+1)I_k-J_k)\otimes I_k.$$

$$BQBQ=QBQB.$$

$$(L^2-(k$$

$$=((Q-B{)}^2-(k+2)(Q-B)+kI)(QB-BQ)$$

Full text

Keywords: Sierpiński Graph, Spectrum, Laplacian, Cayley graph, Non-Cayley number

AMS Mathematics Subject Classification (2010): 05C50, 05C25, 05C75

\par\parfootnotetext: E-mail Addresses: farrokhi@iasbs.ac.ir, m.farrokhi.d.g@gmail.com (M. Farrokhi D. G.), e_ghorbani@ipm.ir (E. Ghorbani), maimani@ipm.ir (H. R. Maimani), farhad.rahimi@sru.ac.ir (F. Rahimi Mahid)

1 Introduction

Sierpiński-type graphs show up in a wide range of areas; for instance, physics, dynamical systems, probability, and topology, to name a few. The Sierpiński gasket graphs form one of the most significant families of such graphs that are obtained by a finite number of iterations that give the Sierpiński gasket in the limit. Several more families of Sierpiński-type graphs have been introduced and studied in the literature (see Barrière, Comellas, and Dalfó [1] and Hinz, Klavžar, and Zemljič [6]). In this paper, we deal with two families of them, as described below.

For positive integers $n,k$, the Sierpiński 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

• •

$u_{i}=v_{i}$ for $i=1,\ldots,t-1$,

• •

$u_{t}\neq v_{t}$,

• •

$u_{j}=v_{t}$ and $v_{j}=u_{t}$ for $j=t+1,\ldots,n$.

For instance, the graphs $S(3,3)$ and $S(2,4)$ are depicted in Figure 1.

Sierpinśki graphs $S(n,k)$ were introduced in Klavžar and Milutinović [8]. 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 extreme vertices.

In addition to $S(n,k)$, we consider a ‘regularization’ of them as another family of Sierpiński-type graphs. The graphs $S^{++}(n,k)$, introduced in Klavžar and Mohar [9], are defined as follows. The graph $S^{++}(1,k)$ is the complete graph $K_{k+1}$. For $n\geq 2$, $S^{++}(n,k)$ 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 2 for an illustration of $S^{++}(3,3)$.

Many properties of Sierpiński-type graphs, including those of $S(n,k)$ and $S^{++}(n,k)$, have been studied in the literature, for a survey see Hinz, Klavžar, and Zemljič [6]. 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 2, 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 3, 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 4. 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 5, and we discuss its distribution.

2 Spectrum of $S^{++}(n,k)$

Let $\Gamma$ be a simple graph with vertex set $V(\Gamma)=\{v_{1},\ldots,v_{n}\}$ and edge set $E(\Gamma)$. Its 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 spectrum of $\Gamma$.

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

We first recall some basic facts.

The 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 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

[TABLE]

and, moreover, if $\Gamma$ is $k$-regular, then

[TABLE]

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

Lemma 1**.**

The graph $S^{++}(n+1,k)$ is isomorphic to ${\cal L}({\cal S}(S^{++}(n,k)))$.

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:

• (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 $r\leq n-2$ and $u\neq v$;

• (2)

${\bf u}=(u_{1},\ldots,u_{n-1},u)$, ${\bf v}=(u_{1},\ldots,u_{n-1},v)$ with $u\neq v$;

• (3)

${\bf u}=(0,u,\ldots,u)$, ${\bf v}=(u,u,\ldots,u)$.

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:

• (i)

If $e$ is of type (1) or (2), then $\psi(e_{\bf u})=({\bf u},v)$ and $\psi(e_{\bf v})=({\bf v},u)$;

• (ii)

If $e$ is of type (3), then $\psi(e_{\bf u})=({\bf u},u)$ and $\psi(e_{\bf v})=({\bf v},u)$.

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 $e^{\prime}$ 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 $\psi(e^{\prime})=({\bf x},z)$ for some $y\neq z$. Then, it is clear that $\psi(e)$ and $\psi(e^{\prime})$ 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 $\psi(e^{\prime})$ 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. ∎

We recall that if $A$ is a non-singular square matrix, then

[TABLE]

where $|\cdot|$ denotes the determinant of a matrix. Also, recall that if $M$ is a $p\times q$ matrix, then

[TABLE]

(Note that (4) might not be valid if $p\leq q$ 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

[TABLE]

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.

Theorem 2**.**

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

[TABLE]

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:

• (i)

$k$* with multiplicity $1$,*

• (ii)

the zeros of $f^{n-1}(x)+1$ each with multiplicity $k$,

• (iii)

the zeros of $f^{j}(x)$ each with multiplicity $k^{n-2-j}\left({k\choose 2}-1\right)$ for $j=0,1,\ldots,n-2$,

• (iv)

the zeros of $f^{j}(x)+2$ each with multiplicity $k^{n-2-j}\left({k\choose 2}-1\right)+1$ for $j=0,1,\ldots,n-2$.

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 1, $\Gamma_{n}$ is isomorphic to ${\cal L}({\cal S}(\Gamma_{n-1}))$. It follows that

[TABLE]

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 (3),

[TABLE]

On the other hand, by (1) and (2), we have

[TABLE]

Now, from (7) it follows that

[TABLE]

implying (6).

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 (6), we conclude that

[TABLE]

and since

[TABLE]

the assertion follows for $n=2$. Now assume that $n\geq 3$ and the assertion holds for $n-1$. So, we have

[TABLE]

in which $m_{i}=k^{i}\left({k\choose 2}-1\right)$. It follows that

[TABLE]

This, together with (6) and (8), implies the result. ∎

Remark 3**.**

It is straightforward to see that the zeros of $f^{j}(x)$ and $f^{j}(x)+2$ for $j=1$ are $\frac{1}{2}(k-2\pm\sqrt{k^{2}+4})$ and $\frac{1}{2}(k-2\pm\sqrt{k^{2}-4}),$ respectively, and for $j\geq 2$ are of the form

[TABLE]

and

[TABLE]

respectively, each of them consisting of $j$ nested radicals in iterative forms. Moreover, the zeros of $f^{n-1}(x)+1$ are

[TABLE]

where the last one consists of $n-2$ nested radicals.

3 Laplacian spectrum of $S(n,k)$

For a graph $\Gamma$, the matrix $L(\Gamma)=D(\Gamma)-A(\Gamma)$ is the 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 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 7 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 [math], except the $(i,j)$ entry that is $1$. Consider the $k^{2}\times k^{2}$ matrix

[TABLE]

where ‘$\otimes$’ denotes the Kronecker product. The matrix $C$ is called the commutation matrix. The main property of the commutation matrix (see Magnus and Neudecker [10]) is that it commutes the Kronecker product: for any $k\times k$ matrices $M,N$,

[TABLE]

Note that each row and each column of $C$ corresponds with a pair $(i,j)$ for $1\leq i,j\leq k$. 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 $1\leq i,j\leq k$.

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 $S(2,k)$.

Theorem 4**.**

The Laplacian spectrum of $S(2,k)$ is the following:

[TABLE]

Proof.

First, note that the graph $S(2,k)$ consists of $k$ copies of $K_{k}$ together with a matching $M$ of size $k\choose 2$; exactly one edge for each pair of copies of $K_{k}$. Let $L$ denote the Laplacian matrix of $S(2,k)$, and $L^{\prime}$ 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 $B=I-L^{\prime}$. 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 $Q$:

[TABLE]

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 $k\choose 2$ and ${k\choose 2}-k$, respectively.

We also have

[TABLE]

and from the eigenvalues of $Q$,

[TABLE]

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 $i\neq j$. It turns out that $B$ is the commutation matrix, and thus

[TABLE]

The right sides of (9) and (11) commute, and so

[TABLE]

Next, we see that

[TABLE]

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 $QB-BQ$:

[TABLE]

To show (12), suppose $P$ is a $k\times k$ matrix whose first column is $\frac{1}{\sqrt{k}}(1,\ldots,1)^{\top}$ and that $PP^{\top}=I_{k}$. Then, $J_{k}=P(kE_{11})P^{\top}$, and so

[TABLE]

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 $2,3,\ldots,k,k+1,2k+1,3k+1,\ldots,(k-1)k+1$, (12) 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 [math] eigenvalue. Thus, we have obtained so far ${k\choose 2}+{k\choose 2}-k+2(k-1)+1=k^{2}-1$ 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. ∎

Based on empirical evidence, we put forward the following conjecture.

Conjecture 5**.**

For $n,k\geq 2$, the Laplacian spectrum of $S(n,k)$ consists of the following eigenvalues:

• (i)

[math] with multiplicity $1$.

• (ii)

The zeros of $f^{j}(k-x)$, each with multiplicity $\frac{1}{2}(k^{n-j}-2k^{n-j-1}+k)$ for $j=0,1,\ldots,n-1$, where $f$ is given in (5).

• (iii)

The zeros of $f^{j}(k-x)+2$, each with multiplicity $\frac{1}{2}(k^{n-j-1}-1)(k-2)$ for $j=0,1,\ldots,n-2$.

Remark 6**.**

The zeros of $f^{j}(k-x)$ and $f^{j}(k-x)+2$ for $j=1$ are $\frac{1}{2}(k+2\pm\sqrt{k^{2}+4})$ and $\frac{1}{2}(k+2\pm\sqrt{k^{2}-4}),$ respectively, and for $j\geq 2$ are of the form

[TABLE]

and

[TABLE]

respectively, each of them consisting of $j$ nested radicals.

Remark 7**.**

The graph $S(n,2)$ is the path graph on $2^{n}$ vertices. Proposition 8 below shows that Conjecture 5 holds for $S(n,2)$. In Grigorchuk and Šunić [5], 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 [5], 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 5. In summary, Conjecture 5 holds for $n=2$ and for $k=2,3$.

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 5. For the sake of completeness, we give its complete argument here.

Proposition 8**.**

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 $g(x)=x^{2}-2$.

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 [18]). From the identities given in Cvetković, Doob, and Sachs [2, p. 220], 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 [18, p. 98]), we get

[TABLE]

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. ∎

4 What $S^{++}(n,k)$ are Cayley graphs?

Recall that a graph $\Gamma$ is 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 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 [12]. It is expected to continue to be this way for larger orders. In fact, it is conjectured in Praeger, Li, and Niemeyer [15] 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.

Proposition 9**.**

The graph $S^{++}(n,k)$ is vertex-transitive if and only if either $n\leq 2$ or $k\leq 2$.

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 $2^{n-1}\cdot 3$ vertices, which are all vertex-transitive graphs. By Lemma 1, $S^{++}(2,k)$ 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 $e^{\prime}=\{i^{\prime},\{i^{\prime},j^{\prime}\}\}$, the automorphism induced by a permutation $\sigma$ that $\sigma(i)=i^{\prime}$ and $\sigma(j)=j^{\prime}$, maps $e$ to $e^{\prime}$. 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 $n\geq 3$ and $k\geq 3$. 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. ∎

From Proposition 9, it follows that the graphs $S^{++}(n,k)$ for $n\geq 3$ and $k\geq 3$ 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 $k\geq 3$. This is our goal in the rest of this section.

Definition 1**.**

Let $\Gamma$ be a graph and $\Delta$ a subgraph of $\Gamma$. We say that $\Gamma$ is strongly $\Delta$-partitioned if:

• (i)

The vertex set of $\Gamma$ is partitioned by the vertex sets of copies $\Delta_{0},\ldots,\Delta_{k}$ of $\Delta$.

• (ii)

Apart from $\Delta_{0},\ldots,\Delta_{k}$, the graph $\Gamma$ contains no further copies of $\Delta$.

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 $n\geq 2$ and $k\geq 3$. Note that this is not the case for $k=2$ because $S^{++}(n,2)$, that is a cycle with $3\cdot 2^{n-1}$ 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.

Proposition 10**.**

Let $n\geq 2$ and $k\geq 3$. The graph $S^{++}(n,k)$ is strongly $S(n-1,k)$-partitioned.

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 $k\geq 3$, 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 $n\geq 3$. 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 $\Delta\neq\Gamma_{i}$ 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 $\Gamma^{\prime}:=\Gamma[V(\Gamma_{s})\cup V(\Gamma_{t})]$. However, $uv$ is a bridge for $\Gamma^{\prime}$ and thus a bridge for $\Delta$, a contradiction. ∎

The following lemma reveals the structure of strongly $\Delta$-partitioned Cayley graphs.

Lemma 11**.**

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 $\Gamma$.

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. ∎

Definition 2**.**

Let $\Gamma$ be a strongly $\Delta$-partitioned graph. We say that $\Gamma$ has 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 $\mathcal{SP}_{c}(\Delta)$.

Remark 12**.**

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.

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 Frobenius group with 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 [17, 8.5.5]). The other concepts we use in the following are standard and can be found in Robinson [17].

Theorem 13**.**

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 $N\cong\mathbb{Z}_{p}^{m}$ and Frobenius complement $H$, $C=C^{\prime}\cup\{c\}$ with $\Delta\cong\mathrm{Cay}(H,C^{\prime})$ and $c^{2}=1$, and either

• (i)

$c\in N$* and $H=\langle C^{\prime}\rangle$, or*

• (ii)

$c=h^{n}$* for some $h\in H\setminus\{1\}$ and $n\in N\setminus\{1\}$, and $H=\langle C^{\prime},h\rangle$.*

Conversely, if $\Delta$ satisfies the above conditions, then $\mathrm{Cay}(G,C)\in\mathcal{SP}_{1}(\Delta)$.

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 11, $H$ is a subgroup of $G$. Let $C^{\prime}$ be the neighborhood of $1$ in $\Gamma[H]$. Since $\Gamma$ is $(d+1)$-regular, besides the elements of $C^{\prime}$, the vertex $1$ has exactly one other neighbor, say $c\in G\setminus H$. So $C=C^{\prime}\cup\{c\}$. Since $H$ is a subgroup of $G$, $C^{\prime-1}\subseteq H$, which implies that $C^{\prime-1}=C^{\prime}$. 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,

[TABLE]

from which it follows that $H\cap H^{c}=\{1\}$. Now, a simple verification shows that $Hch\cap Hch^{\prime}=\emptyset$ for all distinct elements $h,h^{\prime}\in H$. 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

[TABLE]

where the unions are disjoint. As a result, every element $g\in G\setminus H$ can be written as $g=hch^{\prime}$ for some $h,h^{\prime}\in H$, from which it follows that

[TABLE]

Hence, $G$ is a Frobenius group with complement $H$. Let $N$ be the Frobenius kernel of $G$. By [17, 10.5.1(i)], $N$ 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 [17, 1.5.6(iii)]). 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 [17, 1.5.6(ii)]). If $N\neq N_{0}$, then $N_{0}H$ is a Frobenius group for $N_{0}H$ is a subgroup of $G$ and $H\cap H^{g}=1$ for all $g\in N_{0}H\setminus H$. 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 [17, Exercises 8.5(6)]. Thus $N=N_{0}$ so that $k+1=|N|=p^{m}$ is a prime power for some $m\geq 1$. Note that $N$ is a minimal normal subgroup of $G$ for if $N$ contains a nontrivial normal subgroup $N_{0}$ of $G$ properly, then $N_{0}H$ would be a Frobenius group which leads us to the same contradiction as above. If $c\in N$, then since $G\subseteq N\langle C^{\prime}\rangle$ it follows that $H=\langle C^{\prime}\rangle$. Now assume that $c\notin N$. Then $c^{n}\in H\setminus\{1\}$ for some $n\in N\setminus\{1\}$. As $G\subseteq N\langle C^{\prime},c^{n}\rangle$ it follows that $H=\langle C^{\prime},c^{n}\rangle$, as required. The converse is straightforward. ∎

We are now in a position to conclude the main result of this section.

Theorem 14**.**

The graph $S^{++}(n,k)$ is a Cayley graph if and only if either

• (i)

$n=1$,

• (ii)

$k\leq 2$, or

• (iii)

$n=2$* and $k+1=p^{m}$ is a prime power.*

Furthermore, in the case (iii), we have

[TABLE]

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 $c\in G\setminus H$.

Proof.

By Proposition 9, $S^{++}(n,k)$ for $n\geq 3$ and $k\geq 3$ 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 $k\geq 3$.

First, we show that $S^{++}(2,q-1)$ are Cayley graphs for all prime powers $q$. Let $\mathbb{F}_{q}$ denote the finite field with $q$ elements. Then $G:=\mathbb{F}^{*}_{q}\times\mathbb{F}_{q}$ together with the multiplication

[TABLE]

forms a group known as one dimensional affine group. We show that $S^{++}(2,q-1)\cong\mathrm{Cay}(G,C)$, where $C=\left\{(x,0):1\neq x\in\mathbb{F}^{*}_{q}\right\}\cup\{(-1,-1)\}$. To this end, let $H:=\{(x,0):x\in\mathbb{F}^{*}_{q}\}$ 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 $h^{\prime}g(hg)^{-1}=h^{\prime}h^{-1}\in C$ for all distinct elements $hg$ and $h^{\prime}g$ 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 $\mathbb{F}^{*}_{q}$ and $\mathbb{F}_{q}$, respectively, it follows that every right coset of $H$ has a representative of the form $(1,b)$ for some $b\in\mathbb{F}_{q}$. Let $Hg$ and $Hg^{\prime}$ be distinct right cosets of $H$ with $g=(1,a)$ and $g^{\prime}=(1,a^{\prime})$. Then an element $hg$ of $Hg$ is adjacent to an element $h^{\prime}g^{\prime}$ of $Hg^{\prime}$ if and only if $h^{\prime}g^{\prime}g^{-1}h^{-1}=(h^{\prime}g^{\prime})(hg)^{-1}\in C$ or equivalently $(h^{\prime}g^{\prime})(hg)^{-1}=(-1,-1)$ as $g^{\prime}g^{-1}\notin H$. A simple verification shows that this equation has a unique solution for $(h,h^{\prime})$ so that there is a unique edge between any two right cosets of $H$. Indeed, $h=(x,0)$ and $h^{\prime}=(x^{\prime},0)$ satisfy the equation if and only if $-x^{\prime}=x=(a^{\prime}-a)^{-1}$. 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 10, $\Gamma$ 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 11 that $H$ is a subgroup of $G$. By Theorem 13, $k+1=p^{m}$ is a prime power, $G=N\rtimes H$ is a Frobenius group with Frobenius kernel $N$ and Frobenius complement $H$ such that $N\cong\mathbb{Z}_{p}^{m}$ is a minimal normal subgroup of $G$, $C=C^{\prime}\cup\{c\}$, $C^{\prime-1}=C^{\prime}\subseteq H$, $c^{2}=1$, and either

• (a)

$c\in N$ and $H=\langle C^{\prime}\rangle$, or

• (b)

$c=h^{n}$ for some $h\in H\setminus\{1\}$ and $n\in N\setminus\{1\}$, and $H=\langle C^{\prime},h\rangle$.

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. ∎

As a generalization of Theorem 13, we pose the following problem.

Problem 15**.**

Let $\Delta$ be a regular graph. Classify all Cayley graphs in $\mathcal{SP}_{c}(\Delta)$ for $c\geq 2$.

5 New non-Cayley numbers

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 Cayley number if all vertex-transitive graphs of order $n$ are Cayley graphs. Marušič [11] 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 [13] and Iranmanesh and Praeger [7] 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.

Theorem 16** (McKay and Praeger [13]).**

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 $n=12$.

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.

Theorem 17** (Dobson and Spiga [3]).**

There exists an infinite set of primes every finite product of its distinct elements is a Cayley number.

As a consequence of Theorem 14, 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.

Theorem 18**.**

Let $k$ be any positive integer such that $k(k+1)$ is square-free, and $k+1$ is not a prime. Then, $k(k+1)\in\mathcal{NC}$.

As mentioned in Dobson and Spiga [3], 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 18 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 18.

Theorem 19**.**

The density of the set

[TABLE]

is about $0.3226$

Proof.

Let $f\in\mathbb{Z}[t]$ be a primitive polynomial (that is, the greatest common divisor of its coefficients is 1) without multiple roots such that its image on $\mathbb{N}$ 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

[TABLE]

where $\varrho(d)$ denotes the number of roots of $f$ in $\mathbb{Z}_{d}$. Ricci [16] (see also Pappalardi [14]) proved that

[TABLE]

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

[TABLE]

where $C_{\text{Feller-Tornier}}$ is the Feller-Tornier constant (see Finch [4, §2.4.1]). Since primes have zero density in the set of all positive integers, the result follows. ∎

To date, all the numbers $n$ whose membership in $\mathcal{NC}$ is known are determined based on the results of [7, 12, 13]. Using a computer search, we see that the list of the numbers whose membership in $\mathcal{NC}$ are not yet determined begins with

[TABLE]

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:

[TABLE]

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.