Diameter of io-decomposable Riordan graphs of the Bell type

TL;DR

This paper investigates the diameter of io-decomposable Riordan graphs of the Bell type, providing counterexamples, proofs for specific cases, and proposing new conjectures to advance understanding of their properties.

Contribution

It refutes an existing conjecture with a counterexample, proves the conjecture for certain graph sizes, and introduces a new conjecture for these graph classes.

Findings

Counterexample disproves the first conjecture.

First conjecture holds for particular graph sizes.

Second conjecture is valid for some special graphs.

Abstract

Recently, in the paper \cite{CJKM1} we suggested the two conjectures about the diameter of io-decomposable Riordan graphs of the Bell type. In this paper, we give a counterexample for the first conjecture. Then we prove that the first conjecture is true for the graphs of some particular size and propose a new conjecture. Finally, we show that the second conjecture is true for some special io-decomposable Riordan graphs.

Equations

$$\displaystyle{\cal A}(G)\equiv(zg,f)_{n}+(zg,f)_{n}^{T},\;\textrm{i.e.}\;r_{i,j}=r_{j,i}\equiv\left\{\begin{array}[]{ll}[z^{i-2}]gf^{j-1}&\text{if $i>j$}\\ 0&\text{if $i=j$}\end{array}\right.$$

$$C(z)=\frac{1-\sqrt{1-4z}}{2z}=\sum _{n\ge 0}\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)z^n=1+z+2z^2+5z^3+14z^4+\cdots .$$

$$2=\mathrm{diam}(P{G}_n)\le \mathrm{diam}(G_n)\le \mathrm{diam}(C{G}_n)$$

$${\ell }_{i+1,j+1}=a_0{\ell }_{i,j}+a_1{\ell }_{i,j+1}+\cdots +a_{i-j}{\ell }_{i,i}.$$

$$f=zA(f),\mathrm{or}\mathrm{equivalently}A=z/\bar{f}$$

$$G_n^r(g,f)=G_n(g(\bar{f})\cdot (\bar{f}{)}^{\prime }\cdot (z/\bar{f}{)}^{n-1},\bar{f})$$

$$G_n^r=G_n((zA(z){)}^{\prime }\cdot A^{n-2}(z),z/A(z)).$$

$$G_n^r=G_n(A^{\prime }(z)\cdot A^{n-2}(z),z/A(z)).$$

$$g(\bar{f})=z/\bar{f}=A(z)\mathrm{and}(\bar{f}{)}^{\prime }={\left(\frac{z}{A(z)}\right)}^{\prime }\equiv \frac{A(z)+z{A}^{\prime }(z)}{A^2(z)}=(zA(z){)}^{\prime }{A}^{-2}(z),$$

$$g(\bar{f})\cdot (\bar{f}{)}^{\prime }\cdot (z/\bar{f}{)}^{n-1}\equiv (zA(z){)}^{\prime }\cdot A^{n-2}(z)\mathrm{and}\bar{f}=z/A(z).$$

$$\left(\genfrac{}{}{0ex}{}{n}{m}\right)\equiv \prod _i\left(\genfrac{}{}{0ex}{}{n_i}{m_i}\right)(\mathrm{mod}p).$$

$$\mathrm{diam}(C{G}_{2^k})=k.$$

$$A^{\prime }(z)\cdot A^{2^k-2}(z)$$

$$C{G}_{2^k}^r=G_{2^k}((1-z{)}^{-2^k},z+z^2).$$

$$\left(\genfrac{}{}{0ex}{}{2^k+j-1}{2^k-1}\right)\equiv 1\text{for j=0}\mathrm{and}\left(\genfrac{}{}{0ex}{}{2^k+j-1}{2^k-1}\right)\equiv 0\text{for 1≤j≤2k−1}$$

$$m_i=\max \{j\in V(C{G}_{2^k}^r)|[z^{j-2}](z+z^2{)}^{i-1}\equiv 1\}=2i.$$

$$2^0\to 2^1\to \cdots \to 2^{k-1}\to j\mathrm{where}j\in \{2^{k-1}+1,\dots ,2^k\}$$

$$d_{C{G}_{2^k}^r}(i,j)\le k-1\text{if i,j∈V1 or i,j∈V2.}$$

$$\displaystyle\mathcal{A}(CG_{4}^{r})=\left(\begin{array}[]{cccc}0&1&0&0\\ 1&0&1&1\\ 0&1&0&1\\ 0&1&1&0\\ \end{array}\right),$$

$$d_{C{G}_{2^k}^r}(i,j)$$

$$\le k-1\text{(by induction)}$$

$$\displaystyle\mathcal{A}(CG_{8}^{r})=\left(\begin{array}[]{cccccccc}0&1&0&0&0&0&0&0\\ 1&0&1&1&0&0&0&0\\ 0&1&0&1&0&1&0&0\\ 0&1&1&0&1&1&1&1\\ 0&0&0&1&0&1&0&0\\ 0&0&1&1&1&0&1&1\\ 0&0&0&1&0&1&0&1\\ 0&0&0&1&0&1&1&0\end{array}\right).$$

$$\mathrm{diam}(C{G}_{2^k-1})=k-1.$$

$$\displaystyle{\rm diam}(G_{n})\leq\left\{\begin{array}[]{ll}s+2&\text{if $m=1$;}\\ s+3&\text{otherwise.}\end{array}\right.$$

$$d_{G_n}(i,j)$$

$$\le d_{⟨V_1⟩}(i,2^k+1)+d_{⟨V_2⟩}(2^k+1,2^k+2^m+1)+d_{⟨V_2⟩}(2^k+2^m+1,j)$$

$$\le 3.$$

$$d_{G_n}(i,j)$$

$$\displaystyle{\rm diam}(CG_{1+2^{m}+2^{k}})=\left\{\begin{array}[]{ll}2&\text{if $m=1$;}\\ 3&\text{otherwise.}\end{array}\right.$$

$$C{G}_{1+2^m+2^k}^r=G_{1+2^m+2^k}((1-z{)}^{-1-2^m-2^k},z+z^2).$$

$$\displaystyle c_{2^{m}+2^{k},j}=r_{2^{m}+2^{k}+2-j,2}\equiv\left\{\begin{array}[]{ll}1&\text{if $j=2^{k}+2^{m}+1$;}\\ 0&\text{if $j=2^{k}+2^{m}$;}\\ \text{$[z^{2^{k}+2^{m}-j}]z(1-z)^{-2^{k}-2^{m}}$}&\text{otherwise.}\end{array}\right.$$

$$[z^{2^k+2^m-j}]z(1-z{)}^{-2^k-2^m}=(\genfrac{}{}{0ex}{}{2^{k+1}+2^{m+1}-j-2}{2^k+2^m-1}),$$

Full text

AMS classifications: 05C75, 05A15

Key words: Riordan graph; io-decomposable Riordan graph; diameter; Catalan graph.

1 Introduction

Let $\kappa[[z]]$ be the ring of formal power series in the variable $z$ over an integral domain $\kappa$. A Riordan matrix [13] $L=[\ell_{n,k}]_{n,k\geq 0}$ is defined by a pair of formal power series $(g,f)\in\kappa[[z]]\times\kappa[[z]]$ with $f(0)=0$ such that $[z^{n}]gf^{k}=\ell_{n,k}$ for $k\geq 0$ where $[z^{n}]$ is the coefficient extraction operator. Usually, the Riordan matrix is denoted by $L=(g,f)$ and its leading principal submatrix of order $n$ is denoted by $(g,f)_{n}$. Since $f(0)=0$, every Riordan matrix $(g,f)$ is an infinite lower triangular matrix. Most studies on the Riordan matrices were related to combinatorics [6, 9, 11, 14, etc.] or algebraic structures [1, 2, 3, 7, etc.].

Throughout this paper, we write $a\equiv b$ for $a\equiv b\;({\rm mod\;2})$.

Recently, we in [4, 5] introduced a Riordan graph by using the notion of the Riordan matrix modulo 2 as follows.

Definition 1.1**.**

A simple labelled graph $G$ with $n$ vertices is a Riordan graph of order $n$ if the adjacency matrix ${\cal A}(G)=[r_{i,j}]_{1\leq i,j\leq n}$ of $G$ is an $n\times n$ symmetric $(0,1)$-matrix given by

[TABLE]

for some Riordan matrix $(g,f)$ over $\mathbb{Z}$. We denote such $G$ by $G_{n}(g,f)$. In particular, the Riordan graph $G_{n}(g,f)$ is called proper if $[z^{0}]g=[z^{1}]f=1$.**

For example, consider the Catalan graph $CG_{n}=G_{n}(C(z),zC(z))$ where $C(z)$ is the generating function for the Catalan numbers, i.e.

[TABLE]

When $n=6$ we have Figure 1.

In [4], we studied the structural properties of families of Riordan graphs obtained from infinite Riordan graphs, which include a fundamental decomposition theorem and certain conditions on Riordan graphs to have an Eulerian trail/cycle or a Hamiltonian cycle. A Riordan graph $G_{n}(g,f)$ is called Bell type if $f=zg$. Moreover, we studied the following Riordan graphs of special Bell type.

Definition 1.2**.**

[4*]** *Let $G_{n}=G_{n}(g,f)$ be a proper Riordan graph with the odd and even vertex sets $V_{o}=\{i\in V(G_{n})\;|\;i\equiv 1\}$ and $V_{e}=\{i\in V(G_{n})\;|\;i\equiv 0\}$, respectively. The graph $G_{n}$ is said to be io-decomposable if $\left<V_{o}\right>\cong G_{\lceil n/2\rceil}(g,f)$ and $\left<V_{e}\right>$ is a null graph.

A vertex in a graph $G$ is universal if it is adjacent to all other vertices in $G$. The distance between two vertices $u,v$ in a graph $G$ is the number of edges in a shortest path between $u$ and $v$. The diameter of $G$ is the maximum distance between all pairs of vertices, and it is denoted by diam$(G)$.

We found several properties of an io-decomposable Riordan graph of the Bell type as follows.

Lemma 1.3**.**

[4]** Let $G_{n}=G_{n}(g,zg)$ be an io-decomposable Riordan graph of the Bell type. Then we have the following.

• (i)

If $n=2^{k}+1$ for $k\geq 0$, then $G_{n}$ and $G_{n+1}$ have at least one universal vertex, namely $2^{k}+1$.

• (ii)

$G_{n}$* is a $(\lceil\log_{2}n\rceil+1)$-partite graph.*

• (iii)

The chromatic number and the clique number of $G_{n}$ are $\lceil\log_{2}n\rceil+1$.

• (iv)

The diameter of $G_{n}$ is bounded by $\mathrm{diam}(G_{n})\leq\lfloor{\rm log}_{2}n\rfloor$. In particular, if $n=2^{k}+2$ or $2^{k+1}+1$, for $k\geq 1$, then $\mathrm{diam}(G_{n})=2$.

• (v)

If $2^{k}+1<n<2^{k+1}$ then $\mathrm{diam}(G_{n})\leq\lfloor{\rm log}_{2}(n-2^{k})\rfloor+1.$

A graph is called weakly perfect if its chromatic number equals its clique number. By (iii) of Lemma 1.3, every io-decomposable Riordan graph of the Bell type is weakly perfect. It is known [10] that almost all $K_{k}$-free graphs are $(k-1)$-partite for $k\geq 3$. By (ii) and (iii) of Lemma 1.3, every io-decomposable Riordan graph $G_{n}(g,zg)$ of the Bell type is $K_{\lceil\log_{2}n\rceil+2}$-free and ($\lceil\log_{2}n\rceil+1$)-partite for $n\geq 2$. Thus the io-decomposable Riordan graph of the Bell type is very interesting object in Riordan graph theory.

It is known [4] that the Pascal graph $PG_{n}=G_{n}(1/(1-z),z/(1-z))$ and the Catalan graph $CG_{n}=G_{n}(C(z),zC(z))$ are the io-decomposable Riordan graphs of the Bell type. The following two conjectures introduced in [4] show significance of the Pascal graph $PG_{n}$ and the Catalan graph $CG_{n}$.

Conjecture 1**.**

[4*]** *Let $G_{n}$ be an io-decomposable Riordan graph of the Bell type. Then

[TABLE]

for $n\geq 4$. Moreover, $PG_{n}$ is the only graph in the class of io-decomposable graphs of the Bell type whose diameter is $2$ for all $n\geq 4$.**

Conjecture 2**.**

[4]** We have that diam$(CG_{2^{k}})=k$ and there are no io-decomposable Riordan graphs $G_{2^{k}}\not\cong CG_{2^{k}}$ of the Bell type satisfying diam$(G_{2^{k}})=k$ for all $k\geq 1$.

We note that ${\rm diam}(PG_{n})=1$ if $n=2,3$ and ${\rm diam}(PG_{n})=2$ if $n\geq 4$ since the vertex 1 is adjacent to all other vertices, $PG_{n}\cong K_{n}$ if $n=2,3$ and $PG_{n}\not\cong K_{n}$ if $n\geq 4$.

In this paper, we first give a counterexample of the upper bound in Conjecture 1. Then we prove that the upper bound in Conjecture 1 is true for the graph of some particular size and we propose a new conjecture for an upper bound of the diameter of an io-decomposable Riordan graph of the Bell type. Finally, we show that Conjecture 2 is true for some special io-decomposable Riordan graphs.

2 Upper bound of Conjecture 1

It is known [12] that an infinite lower triangular matrix $L=[\ell_{i,j}]_{i,j\geq 0}$ with $\ell_{0,0}\neq 0$ is a proper Riordan matrix if and only if there is a unique sequence $(a_{0},a_{1},\ldots)$ with $a_{0}\neq 0$ such that, for $i\geq j\geq 0$,

[TABLE]

The sequence $(a_{i})_{i\geq 0}$ is called the $A$-sequence of the Riordan array. Also, if $L=(g,f)$ then

[TABLE]

where $A=\sum_{i\geq 0}a_{i}z^{i}$ is the generating function for the $A$-sequence of $(g,f)$. In particular, if $L$ is a binary Riordan matrix $L\equiv(g,f)$ with $f^{\prime}(0)=1$ then the sequence is called the binary $A$-sequence $(1,a_{1},a_{2},\ldots)$ where $a_{k}\in\{0,1\}$.

We in [4] characterized the io-decomposable Riordan graph $G_{n}(g,zg)$ of the Bell type, see the following lemma.

Lemma 2.1**.**

[4]** Let $G_{n}=G_{n}(g,zg)$ be a Riordan graph of the Bell type. Then $G_{n}$ is io-decomposable if and only if the binary $A$-sequence of $(g,zg)$ is $(1,1,a_{2},a_{2},a_{4},a_{4},\ldots)$ where $a_{2j}\in\{0,1\}$ for all $j\geq 1$.

Let $G_{n}$ be the io-decomposable Riordan graph of the Bell type with its binary $A$-sequence generating function $A(z)=\sum_{i=0}^{15}z^{i}$. By Lemma 2.1, the Riordan graph $G_{n}$ is io-decomposable. By using the sage, we compare the diameters between $CG_{n}$ and $G_{n}$ up to degree $n=100$. Then we obtain the following 13 counterexamples for the upper bound of Conjecture 1.

[TABLE]

If for a Riordan graph $G_{n}(g,f)$ with $[z^{1}]f=1$, the relabelling is done by reversing the vertices in $[n]$, that is, by replacing a label $i$ by $n+1-i$ for each $i\in[n]$, then the resulting graph will always be a Riordan graph given by the following lemma. We denote the reverse relabelling of $G_{n}$ by $G_{n}^{r}$.

Lemma 2.2**.**

[4]** The reverse relabelling of a Riordan graph $G_{n}(g,f)$ with $f^{\prime}(0)=1$ is the Riordan graph

[TABLE]

where $\bar{f}$ is the compositional inverse of $f$.

Now, we prove that the upper bound in Conjecture 1 is true if $n=2^{k}$, $n=2^{k}-1$ or $n=1+2^{m}+2^{k}$ where $k\geq 2$ and $1\leq m<k$.

Lemma 2.3**.**

Let $G_{n}=G_{n}(g,zg)$ be a proper Riordan graph and $A(z)$ be the generating function for its binary $A$-sequence. Then the reverse relabelling of $G_{n}$ is the Riordan graph given by

[TABLE]

In particular, if $G_{n}$ is an io-decomposable Riordan graph of the Bell type then the reverse relabelling of $G_{n}$ is the Riordan graph given by

[TABLE]

Proof.

Let $f=zg$ and $\bar{f}$ be the compositional inverse of $f$. Since (2) leads to

[TABLE]

we obtain

[TABLE]

Thus, by Lemma 2.2, we obtain the desired result. In particular, if $G_{n}$ is an io-decomposable Riordan graph of the Bell type then it follows from Lemma 2.1 that $(zA(z))^{\prime}\equiv A^{\prime}(z)$. Hence the proof follows. ∎

If the base $p$ (a prime) expansion of $n$ and $m$ is $n=n_{0}+n_{1}p+n_{2}p^{2}+\cdots$ and $m=m_{0}+m_{1}p+m_{2}p^{2}+\cdots$ respectively then

[TABLE]

This is called the Lucas’s theorem.

Let $G_{n}$ be an io-decomposable Riordan graph of the Bell type. Since diam$(G_{n})\leq k$ if $n=2^{k}$ with $k\geq 0$ by (iv) of Lemma 1.3, the following theorem shows that the upper bound of Conjecture 1 is true if $n=2^{k}$ for $k\geq 1$. We denote the distance between two vertices $u,v$ in a graph $G$ by $d_{G}(u,v)$.

Theorem 2.4**.**

For an integer $k\geq 1$, we obtain

[TABLE]

Proof.

First we show that $CG_{2^{k}}^{r}=G_{2^{k}}(1,z+z^{2})$ for $k\geq 1$. It follows from (1) and (2) that the generating function for $A$-sequence of $(C,zC)$ is ${1\over 1-z}$. By Lemma 2.3, we obtain

[TABLE]

so that by Lemma 2.2 the reverse relabelling of the Catalan graph is

[TABLE]

Since $(1-z)^{-2^{k}}=\sum_{j\geq 0}{2^{k}+j-1\choose 2^{k}-1}z^{j}$, by Lucas’s theorem we obtain

[TABLE]

which imply $G_{2^{k}}((1-z)^{-2^{k}},z+z^{2})=G_{2^{k}}(1,z+z^{2})$. Thus, by (3), $CG_{2^{k}}^{r}=G_{2^{k}}(1,z+z^{2})$. Let $m_{i}={\rm max}\{j\in V(CG_{2^{k}}^{r})\;|\;ij\in E(CG_{2^{k}}^{r})\}.$ For each $i\in\{1,2,\ldots,2^{k-1}\}$, we obtain

[TABLE]

Since $m_{1}<m_{2}<\cdots<m_{2^{k-1}}$, a unique shortest path from 1 to $2^{k}$ in $CG_{2^{k}}^{r}$ is $2^{0}\rightarrow 2^{1}\rightarrow\cdots\rightarrow 2^{k-1}\rightarrow 2^{k}$ so that $d_{CG_{2^{k}}^{r}}(1,2^{k})=k$. Hence, by (iv) of Lemma 1.3, we obtain the desired result. ∎

By Lemma 2.1, the following lemma is obtained from [4, Theorem 3.7] when $\ell=0$.

Lemma 2.5**.**

Let $G_{n}$ be an io-decomposable Riordan graph of the Bell type and $G=\lim_{n\rightarrow\infty}G_{n}$. For each $s\geq 0$, $G$ has the following fractal properties:

• (i)

$G_{2^{s}+1}=\left<\{1,2,\ldots,2^{s}+1\}\right>\cong\left<\{\alpha 2^{s}+1,\alpha 2^{s}+2,\ldots,(\alpha+1)2^{s}+1\}\right>$;

• (ii)

$G_{2^{s}}=\left<\{1,2,\ldots,2^{s}\}\right>\cong\left<\{\alpha 2^{s}+1,\alpha 2^{s}+2,\ldots,(\alpha+1)2^{s}\}\right>$**

where $\alpha\geq 1$.

We can ask that how many vertex pairs $(u,v)$ can have the maximal distance $k$ in $CG_{2^{k}}$. By using Lemma 2.5, the answer is given by the following theorem.

Theorem 2.6**.**

Let $k\geq 1$ be an integer. There exist exactly $2^{k-1}$ vertex pairs $(i,2^{k})$ with $i\in\{1,\ldots,2^{k-1}\}$ such that $d_{CG_{2^{k}}}(i,2^{k})=k$ is the maximal distance in $CG_{2^{k}}$.

Proof.

Since $CG_{2^{k}}^{r}$ is the reverse relabelling of $CG_{2^{k}}$, this theorem is equivalent to the following:

• •

$d_{CG_{2^{k}}^{r}}(i,j)=k$ if $i=1$ and $j\in\{2^{k-1}+1,\ldots,2^{k}\}$;

• •

$d_{CG_{2^{k}}^{r}}(i,j)\leq k-1$ otherwise.

Since by (i) of Lemma 1.3 the vertex $2^{k-1}+1$ is adjacent to all vertices $1,\ldots,2^{k-1}$ in $CG_{2^{k}}$, the vertex $2^{k-1}$ is adjacent to all vertices $2^{k-1}+1,\ldots,2^{k}$ in $CG_{2^{k}}^{r}$. So by (4) the shortest path from 1 to $j$ in $CG_{2^{k}}^{r}$ is

[TABLE]

and thus $d_{CG_{2^{k}}^{r}}(i,j)=k$ if $i=1$ and $j\in\{2^{k-1}+1,\ldots,2^{k}\}$.

Let $V_{1}=\{i\in V(CG_{2^{k}}^{r})\;|\;1\leq i\leq 2^{k-1}\}$ and $V_{2}=\{j\in V(CG_{2^{k}}^{r})\;|\;2^{k-1}<j\leq 2^{k}\}$. Since $\left<V_{1}\right>\cong\left<V_{2}\right>\cong CG_{2^{k-1}}^{r}$ by Lemma 2.5 and $G_{2^{k-1}}\cong CG_{2^{k-1}}$, it follows from Theorem 2.4 that

[TABLE]

Now it is enough to show that $d_{CG_{2^{k}}^{r}}(i,j)<k$ if $i\in V_{1}\backslash\{1\}$ and $j\in V_{2}$ for $k\geq 2$. We prove this by induction on $k\geq 2$. Let $k=2$. Since the adjacency matrix of $CG_{4}^{r}$ is given by

[TABLE]

we see that $d_{CG_{4}^{r}}(2,3)=d_{CG_{4}^{r}}(2,4)=1<2$. Thus it holds for $k=2$. Let $k\geq 3$. Since $\left<V_{1}\right>\cong CG_{2^{k-1}}^{r}$ and the vertex $2^{k-1}$ is adjacent to all vertices $j\in V_{2}$ in $CG_{2^{k}}^{r}$, we obtain

[TABLE]

where $i\in V_{1}\backslash\{1\}$ and $j\in V_{2}$. Hence the proof follows. ∎

Example 2.7**.**

Let us consider the Catalan graph $CG_{8}=G_{8}(C(z),zC(z))$ of order 8. Since its reverse relabeling is $CG_{8}^{r}=G_{8}(1,z+z^{2})$, we obtain Figure 2 from the adjacency matrix

[TABLE]

Thus we can see that the four vertex pairs $(1,5),(1,6),(1,7)$ and $(1,8)$ in $CG_{8}^{r}$ have maximal distance 3 i.e., the four vertex pairs $(8,4),(8,3),(8,2)$ and $(8,1)$ in $CG_{8}$ have the maximal distance 3.

Let $G_{n}$ be an io-decomposable Riordan graph of the Bell type. Since it follows from (iv) of Lemma 1.3 that diam$(G_{n})\leq k-1$ if $n=2^{k}-1$, the following corollary shows that the upper bound of Conjecture 1 is true if $n=2^{k}-1$ for $k\geq 1$.

Corollary 2.8**.**

For an integer $k\geq 1$, we obtain

[TABLE]

Proof.

By Theorem 2.6, we obtain diam$(CG_{2^{k}-1})\leq k-1$. It follows from Lemma 2.3 that one can show $CG_{2^{k}-1}^{r}=G_{n}(1+z,z+z^{2})$. By using the similar proof in Theorem 2.4, we can show that $2^{1}-1\rightarrow 2^{2}-1\rightarrow\cdots\rightarrow 2^{k}-1$ is the shortest path from 1 to $2^{k}-1$ in $CG_{2^{k}-1}^{r}$, i.e. $d_{CG_{2^{k}-1}^{r}}(1,2^{k}-1)=k-1$. Since it follows from Theorem 2.6 that diam$(CG_{2^{k}-1})\leq k-1$, we obtain diam$(CG_{2^{k}-1})=k-1$. Hence the proof follows. ∎

The following lemma is useful to obtain Theorem 2.10 and Conjecture 3.

Lemma 2.9**.**

Let $n=1+2^{m}+\sum_{j=0}^{s}2^{k+j}$ be an integer with $k>m\geq 1$. If $G_{n}$ be the io-decomposable Riordan graph of the Bell type, then we obtain

[TABLE]

Proof.

We prove this by induction on $s\geq 0$. Let $s=0$, i.e. $n=1+2^{m}+2^{k}$. If $m=1$ then it follows from (v) of Lemma 1.3 that ${\rm diam}(G_{2^{k}+3})=2$. For $k>m\geq 2$, let $V_{1}=\{i\in V(G_{n})\;|\;1\leq i\leq 2^{k}+1\}$ and $V_{2}=\{j\in V(G_{n})\;|\;2^{k}+1\leq j\leq n\}$. Since $\left<V_{1}\right>\cong G_{2^{k}+1}$ and $\left<V_{2}\right>\cong G_{2^{m}+1}$ by Lemma 2.5, it follows from (iv) of Lemma 1.3 that ${\rm diam}(\left<V_{1}\right>)={\rm diam}(\left<V_{2}\right>)=2$. Let $i\in V_{1}\backslash\{2^{k}+1\}$ and $j\in V_{2}\backslash\{2^{k}+1\}$. Now it is enough to show that $d(i,j)\leq 3$. Since the vertices $2^{k}+1$ and $2^{k}+2^{m}+1$ are the universal vertices in $\left<V_{1}\right>$ and $\left<V_{2}\right>$ respectively, we obtain

[TABLE]

Thus the theorem holds for $s=0$.

Let $s\geq 1$, i.e. $n=1+2^{m}+\sum_{j=0}^{s}2^{k+j}$. For $k>m\geq 1$, let $W_{1}=\{i\in V(G_{n})\;|\;1\leq i\leq 2^{k+s}+1\}$ and $W_{2}=\{j\in V(G_{n})\;|\;2^{k+s}+1\leq j\leq n\}$. Since by Lemma 2.5 we obtain $\left<W_{1}\right>\cong G_{2^{k+s}+1}$ and $\left<W_{2}\right>\cong G_{n-2^{k+s}}$ , by (iv) of Lemma 1.3 we obtain ${\rm diam}(\left<W_{1}\right>)=2$ and by induction we obtain ${\rm diam}(\left<W_{2}\right>)\leq s+1$ if $m=1$ or ${\rm diam}(\left<W_{2}\right>)\leq s+2$ if $k>m>1$. Let $i\in W_{1}\backslash\{2^{k+s}+1\}$ and $j\in W_{2}\backslash\{2^{k+s}+1\}$. Now it is enough to show that $d_{G_{n}}(i,j)\leq s+2$ if $m=1$ or $d_{G_{n}}(i,j)\leq s+3$ if $k>m>1$. Since the vertices $2^{k+1}+1$ are the universal vertices in $\left<W_{1}\right>$, we obtain

[TABLE]

Hence, by induction, we obtain the desired result. ∎

From Lemma 2.9, the following theorem shows that the upper bound of Conjecture 1 is true if $n=1+2^{m}+2^{k}$ for $k>m\geq 1$.

Theorem 2.10**.**

Let $k$ and $m$ be integers with $k>m\geq 1$. Then

[TABLE]

Proof.

Since by Lemma 2.9 we obtain ${\rm diam}(CG_{2^{k}+3})=2$, it is enough to show that diam$(CG_{1+2^{m}+2^{k}})=3$ for $k>m>1$. Now let $k$ and $m$ be integers with $k>m>1$. By Lemma 2.2, the reverse relabelling of the Catalan graph $CG_{2^{k}+2^{m}+1}$ is

[TABLE]

Let $\mathcal{A}(CG_{1+2^{m}+2^{k}})=[c_{i,j}]$ and $\mathcal{A}(CG_{1+2^{m}+2^{k}}^{r})=[r_{i,j}]$. By (5), we obtain

[TABLE]

Since

[TABLE]

by Lucas’s theorem we obtain for $j=1,\ldots,2^{k}+2^{m}-1$

[TABLE]

By (9) and (2), the set $N(2^{k}+2^{m})$ of neighbors of the vertex $2^{k}+2^{m}$ in $CG_{2^{k}+2^{m}+1}$ is

[TABLE]

It is known [8] that $[z^{n}]C(z)\equiv 1$ if and only if $n=2^{k}-1$ for $k\geq 1$. It implies

[TABLE]

Thus the set $N(1)$ of neighbors of the vertex $1$ in $CG_{2^{k}+2^{m}+1}$ is

[TABLE]

Since

[TABLE]

the distance between vertices $1$ and $2^{k}+2^{m}$ in $CG_{1+2^{m}+2^{k}}$ is at least 3 so that by Lemma 2.9 we obtain diam$(CG_{1+2^{m}+2^{k}})=3$. Hence the proof follows. ∎

We end this section with the following conjecture.

Conjecture 3**.**

Let $n=1+2^{m}+\sum_{j=0}^{s}2^{k+j}$ be an integer with $k>m\geq 1$ and $s\geq 1$. Then

[TABLE]

Remark 2.11**.**

If Conjecture 3 is true, then by Lemma 2.9 the upper bound of Conjecture 1 is true if $n=1+2^{m}+\sum_{j=0}^{s}2^{k+j}$ for $k>m\geq 1$ and $s\geq 1$. By using the sage, we have checked that Conjecture 3 is true for $n\leq 2^{8}$.

3 Conjecture 2

In this section, we show that Conjecture 2 is true for some special io-decomposable Riordan graphs of the Bell type.

Lemma 3.1**.**

Let $G_{n}=G_{n}(g,zg)$ be an io-decomposable Riordan graph. If there exists $k\geq 2$ such that diam$(G_{2^{k}})=s$ then diam$(G_{2^{k+m}})\leq s+m$ for all $m\geq 1$.

Proof.

Let $V_{1}=\{i\in V(G_{n})\;|\;1\leq i\leq 2^{k+m-1}+1\}$ and $V_{2}=\{j\in V(G_{n})\;|\;2^{k+m-1}+1\leq j\leq 2^{k+m}\}$ be the vertex subsets of $V(G_{2^{k+m}})$. Since $\left<V_{1}\right>\cong G_{2^{k+m-1}+1}$ has a universal vertex $2^{k+m-1}+1$ and by Lemma 2.5 we obtain $\left<V_{2}\right>\cong G_{2^{k+m-1}}$, we obtain

[TABLE]

Let diam$(G_{2^{k}})=s$. Applying for $m=1$ in (13), we obtain diam$(G_{2^{k+1}})\leq s+1$. Applying again for $m=2$ in (13), we obtain diam$(G_{2^{k+2}})\leq s+2$. By repeating this process, we obtain the desired result. ∎

Let ${\cal B}(g,f)$ denote a binary Riordan matrix, i.e. ${\cal B}(g,f)\equiv(g,f)$. We note that a Riordan matrix $[b_{i,j}]_{i,j\geq 0}$ is of the Bell type given by ${\cal B}(g,zg)$ with $g(0)=1$ if and only if, for $i\geq j\geq 0$,

[TABLE]

where $(1,a_{1},\ldots)$ is the binary $A$-sequence of ${\cal B}(g,zg)$. Let $G_{n}=G_{n}(g,zg)$ and $\mathcal{A}(G_{n})=[r_{i,j}]_{1\leq i,j\leq n}$ where $g(0)=1$. Since $r_{i,j}=b_{i-2,j-1}$ for $i>j\geq 1$, by (14) we need the finite term $(1,a_{1},\ldots,a_{n-2})$ of the binary $A$-sequence to determine $\mathcal{A}(G_{n})$.

Theorem 3.2**.**

Let $G_{2^{k}}=G_{2^{k}}(g,zg)$ be an io-decomposable Riordan graph. If the binary $A$-sequence of $(g,zg)$ is of the following form

[TABLE]

then for $k\geq 4$ we obtain

[TABLE]

Proof.

First we show that diam$(G_{2^{m}})=m-1$. Since the induced subgraph $H$ of $\{1,2\ldots,2^{m}-1\}$ in $G_{2^{m}}$ is $H=CG_{2^{m}-1}$ and $CG_{2^{m}-1}^{r}=G_{2^{m}-1}(1+z,z+z^{2})$, the $(2^{m}-1)$th row of $\mathcal{A}(G_{2^{m}})=[r_{i,j}]$ is given by

[TABLE]

By (14), (15) and (16), the $2^{m}$th row in $\mathcal{A}(G_{2^{m}})=[r_{i,j}]$ is given by

[TABLE]

which means the only two vertices 1 and $2^{m}-1$ are adjacent to the vertex $2^{m}$ in $G_{2^{m}}$. Let $V_{1}=\{1,\ldots,2^{m-1}+1\}$ and $V_{2}=\{2^{m-1}+1,\ldots,2^{m}-1\}$ be the vertex subsets of $V(G_{2^{m}})$. Since $\left<V_{1}\right>$ has the universal vertex $2^{m-1}+1$ and $\left<V_{2}\right>\cong CG_{2^{m-1}-1}$, if $v_{1}\in V_{1}$ and $v_{2}\in V_{2}$ then we respectively obtain $d_{G_{2^{m}}}(v_{1},2^{m})\leq 3$ and

[TABLE]

which implies diam$(G_{2^{m}})=m-1$. Hence, by Lemma 3.1, we obtain the desired result. ∎

[TABLE]

Table 1  Diameters of io-decomposable Riordan graphs of the Bell type with degree 8

[TABLE]

Table 2  Diameters of io-decomposable Riordan graphs of the Bell type with degree 16 such that the first 6 entries of its $A$-sequence are all 1s

By Lemma 3.1, using the results in Table 1 and 2 we obtain the following theorem.

Theorem 3.3**.**

For $k\geq 4$, let $G_{2^{k}}=G_{2^{k}}(g,zg)$ be an io-decomposable Riordan graph and $G_{2^{k}}\not\cong CG_{2^{k}}$. If the first 16 entries in the binary $A$-sequence of $(g,zg)$ are not all 1s then

[TABLE]