Some New Graph Interpretations of Padovan Numbers

Abstract

Padovan numbers and Perrin numbers belong to the family of numbers of the Fibonacci type and they are well described in the literature. In this paper, by studying independent (1,2)-dominating sets in paths and cycles, we obtain new binomial formulas for Padovan and Perrin numbers. As a consequence of graph interpretation, we propose a new dependence between Padovan and Perrin numbers. By studying special independent (1,2)-dominating sets in a composition of two graphs, we define Padovan polynomials of graphs. By the fact that every independent (1,2)-dominating set includes the set of leaves as a subset, in some cases a symmetric structure of the independent (1,2)-dominating set can be used.

1. Introduction

The Fibonacci sequence ($F_n$) is defined by the second-order linear recurrence relation $F_n=F_{n-1}+F_{n-2}$, for $n\ge 2$, with initial conditions $F_0=F_1=1$. This recurrence first appears in the book Liber Abaci in 1202 as a solution of the famous rabbit problem [1]. This book wrote by Leonardo Pisano, better known as Fibonacci, played an important role in European mathematics as a compendium of arithmetical and algebraic knowledge of those times. However, it is worth mentioning that the Fibonacci sequence was discovered earlier by Indian mathematicians, Pingala and Virachanka [2]. For centuries, the Fibonacci sequence was intensively studied by mathematics enthusiasts because of its interesting appearance in nature and its connections with the golden ratio. There are many sources that look at the properties of this ratio in nature and art, especially the work of Leonardo da Vinci and, in the 20th century, the religious art of Salvatore Dali.

Actually, there is a huge interest of modern science in the application of the golden section and Fibonacci numbers. They play a significant role in theoretical physics, appearing in models of quasicrystals and quantum field theories, where they help describe the arrangement of particles and energy states. In informatics, Fibonacci numbers are used in algorithms for efficient data structures and searching techniques, such as Fibonacci heaps and the Fibonacci search method. Their unique properties also make them useful in coding theory and cryptography, enhancing data security and compression. Other applications can be found in, for example, [3].

Numbers defined recursively by the linear homogeneous recurrence relation with constant coefficients are also named as numbers of the Fibonacci type; a long list of well-known sequences of the Fibonacci type can be found in [4]. A Padovan sequence ($Pv\left(n\right)$) is a sequence of the Fibonacci type defined by $Pv\left(n\right)=Pv(n-2)+Pv(n-3)$ for $n\ge 3$ with $Pv\left(0\right)=Pv\left(1\right)=Pv\left(2\right)=1$. The origin of this sequence has many similarities, with the plastic number being the unique real solution of the equation $x^3-x-1=0$. A so-called cyclic version of the Padovan sequence is the Perrin sequence ($Pr\left(n\right)$) defined by the same recurrence as the Padovan sequence with the initial conditions $Pr\left(0\right)=3$, $Pr\left(1\right)=0$, and $Pr\left(2\right)=2$. The Padovan sequence has many interpretations (see for details the survey given in the Online Encyclopedia of Integer Sequences, which is an online database of integer sequences; see [5]). Among them, graph interpretations play an important role in studying Padovan sequences (see for example [6]). We need to recall the necessary graph definitions and notation. Let G be an undirected, connected simple graph with the vertex set $V\left(G\right)$ and the edge set $E\left(G\right)$. By $P_n$, $n\ge 2$ and $C_n$, $n\ge 3$, we mean a path and a cycle, respectively.

A subset $S\subseteq V\left(G\right)$ is an independent set of a graph G if no two vertices from S are adjacent. Moreover, every subset containing exactly one vertex and the empty set are independent. The cardinality of the largest independent set is the independence number of a graph G denoted by $\alpha \left(G\right)$. The cardinality of the smallest maximal independent set is the lower independence number of a graph G denoted by $i\left(G\right)$. A subset $D\subseteq V\left(G\right)$ is a dominating set of G if every vertex is either in D or has a neighbor in D. A subset which is independent and dominating is named as the independent dominating set or a kernel. The classical concept of independent dominating sets was introduced by Neuman and Morgernstern in the context of game theory [7]. There are some variants and generalizations of independent dominating sets (for example see [8,9]). In the context of this paper, it is necessary to mention the multiple domination and the generalization of domination in the distance sense. For independent multiple domination, interesting results were given by Nagy [10,11]. A special case of independent multiple-dominating sets was considered, among others, by Bednarz and Paja [12,13].

Many types of dominating sets can be found in the book [14]. Some of them can be studied with independence as an additional restriction. One of the interesting generalizations of the classical domination is the concept of (1,2)-dominating sets introduced by Hedetniemi et al. in [15]. A subset $S\subseteq V\left(G\right)$ is a (1,2)-dominating set of G if, for every vertex $v\in V\left(G\right)\setminus S$, there are distinct vertices $u,w\in S$, such that $uv\in E\left(G\right)$ and $d_G(v,w)\le 2$. From the above, it immediately follows that a (1,2)-dominating set has at least two vertices. In particular, the vertex set is also (1,2)-dominating. If a (1,2)-dominating set S is independent, then we simultaneously say that S is independent (1,2)-dominating (shorthand: (1,2)-IDS).

The (1,2)-IDS of a graph G was considered and next studied, among others, by Hedetniem et al [15], Michalski et al. [16,17,18], and Raczek [19,20] with respect to the existing problems or parameters of $(1,2)$ domination. A graph does not always have a (1,2)-IDS. For example, in a complete graph, there is not a (1,2)-IDS. Moreover, a complete characterization of graphs with (1,2)-IDS is not given, only some sufficient conditions are known.

The following has been proved:

Theorem 1 

([15]). Every connected graph G having at least two non-adjacent vertices and no triangles has a (1,2)-IDS of cardinality $\alpha \left(G\right)$.

Theorem 2 

([16]). If the set S is the (1,2)-IDS of a graph G, then S is a maximal (1,2)-IDS including the set of leaves of G.

Theorem 3 

([16]). If S is a maximal independent set of a graph G including the set of leaves of G, then S is the (1,2)-IDS of graph G.

If we assume that a graph is non-complete and without triangles, then it has a (1,2)-IDS, and as a result, we can count all (1,2)-IDS. An investigation of the literature shows that the problem of counting of independent sets or independent dominating sets is intensively studied, and it is also interesting for applications, among others, in combinatorial chemistry. From the pure mathematical point of view, counting independent sets was initiated by Prodinger and Tichy in [21]. In that short paper, they observed that the total number of independent sets in a path is equal to the Fibonacci number, and this result gave an impetus for counting independent sets. In the next few decades, a number of results were obtained. The survey wrote by Gutman and Wagner collected and classified results related to the number of independent sets in graphs. The bibliography of this paper contains 128 papers; most of them were published at the beginning of the XX century (see [22]).

2. Combinatorial Definitions of Padovan and Perrin Numbers and Their Graph Interpretations

Recent works on integer sequences involving new combinatorial approaches were reviewed in the literature, introducing new visualizations of their therms in a combinatorial manner. The Online Encyclopedia of Integer Sequences collects and classifies interpretations of Fibonacci-type sequences (see [5]). Some interesting combinatorial interpretations of Padovan numbers and their generalizations were obtained recently by Vieira et al. [23,24]. In this section, we give a visualization of Padovan numbers being the total number of special subsets of the set of n integers.

Let $X=\{1,2,\cdots ,n\}$, $n\ne 3$ and let $Y\subset X$ such that

(i)

$\left|Y\right|=p$ for fixed $p\ge 2$;

(ii)

if $i,j\in Y$, then $|i-j|\ge 2$;

(iii)

if $t\notin Y$, then there are $q,r\in Y$, $q\ne r$ such that $|t-q|=1$ and $|t-r|\le 2$.

For a fixed $p\ge 2$, let $\mathcal{Y}(n,p)$ be the family of all p-element subsets Y satisfying the conditions (i)–(iii) and $\mathcal{Y}\left(n\right)={\displaystyle \bigcup _{p\ge 2}}\mathcal{Y}(n,p)$. The cardinality of the family $\mathcal{Y}(n,p)$ will be denoted by $i_{1,2}(n,p)$ and the cardinality of $\mathcal{Y}\left(n\right)$ by ${\mathcal{I}}_{1,2}\left(n\right)$; so, ${\mathcal{I}}_{1,2}\left(n\right)={\displaystyle \sum _{p\ge 2}}{i}_{1,2}(n,p)$.

To determine numbers $i_{1,2}(n,p)$ and consequently ${\mathcal{I}}_{1,2}\left(n\right)$, firstly, we must prove the necessary lemmas.

Lemma 1. 

Let $n\ge 3$ and $p\ge 2$ be integers. Then, $\{1,n\}\subseteq Y$ for every $Y\subseteq \mathcal{Y}(n,p)$.

Proof. 

Let Y be an arbitrary subset of $\mathcal{Y}(n,p)$ and suppose that $1\notin Y$ or $n\notin Y$. Without a loss of generality, suppose that $n\notin Y$. Then, by (iii), we have $\{n-1,n-2\}\subseteq Y$, a contradiction with (ii). □

Lemma 2. 

Let $n\ge 3$ and $p\ge 2$ be integers. If $\mathcal{Y}(n,p)\ne \varnothing$, then $2p-1\le n\le 3p-2$.

Proof. 

Let $Y\subseteq \mathcal{Y}(n,p)$. Then, by Lemma 1, we have $\{1,n\}\subseteq Y$, and by the definition of the subset Y, consecutive inner integers from Y are at the distance 2 or 3. Then, $n=2p-1$ if the minimum distance equals 2 in each case or $n=3p-2$ if the maximum distance equals 3 in each case. Consequently, $2p-1\le n\le 3p-2$, which ends the proof. □

Theorem 4. 

Let $p\ge 2$ and $2p-1\le n\le 3p-2$ be fixed integers. Then, $i_{1,2}(n,p)=\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p+1}}\right)$.

Proof. 

Let $p\ge 2$ and $2p-1\le n\le 3p-2$ be fixed integers. Let Y be an arbitrary subset of the family $\mathcal{Y}(n,p)$ satisfying conditions (i)–(iii). Our aim is to calculate the number of all subsets $Y\subseteq \mathcal{Y}(n,p)$. Let us observe that instead of the subset Y, we can consider a binary n-tuple $\alpha =({\alpha }_1,\cdots ,{\alpha }_n)$ such that ${\alpha }_i$, $i\in \{1,\cdots ,n\}$ satisfies the following conditions:

(a)

${\alpha }_i=\left\{\begin{array}{cc}1\hfill & \mathrm{if}i\in Y,\hfill \\ 0\hfill & \mathrm{otherwise}\hfill \end{array}\right.$;

(b)

${\displaystyle \sum _{i=1}^n}{\alpha }_i=p$;

(c)

if ${\alpha }_i={\alpha }_j=1$ then $|i-j|\ge 2$;

(d)

if ${\alpha }_t=0$ then there are ${\alpha }_r={\alpha }_q=1$ such that $|t-r|=1$ and $|t-q|\le 2$.

To calculate the number of all n-tuples $\alpha$, we start with a p-tuple $\beta =({\beta }_1,\cdots ,{\beta }_p)$, $p\ge 2$, where ${\beta }_s=1$ for each $s\in \{1,\cdots ,p\}$. Next, we construct a $(2p-1)$-tuple ${\beta }^{\prime }=({\beta }_1,{\beta }_1^{\prime },{\beta }_2,{\beta }_2^{\prime },\cdots ,{\beta }_{p-1},{\beta }_{p-1}^{\prime },{\beta }_p)$, where ${\beta }_r^{\prime }=0$ for each $r\in \{1,\cdots ,p-1\}$. For building the n-tuple $\alpha$, it suffices to extend ${\beta }^{\prime }$ by $n-(2p-1)$ zeros in such a way that we can put at most one zero between each three words ${\beta }_r,{\beta }_r^{\prime },{\beta }_{r+1}$ for $r\in \{1,\cdots ,p-1\}$. We can do this in $C_{p-1}^{n-(2p-1)}=\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p+1}}\right)$, which ends the proof. □

Corollary 1. 

Let $p\ge 2$ and $2p-1\le n\le 3p-2$ be fixed integers. Then, $I_{1,2}(n,p)={\displaystyle \sum _{p\ge 2}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p+1}}\right)$.

We show that numbers $i_{1,2}(n,p)$ and $I_{1,2}\left(n\right)$ have a graph interpretation. To study the number of all (1,2)-IDS, let us observe that every (1,2)-IDS in graph G includes the set of leaves as a subset and every (1,2)-IDS in a graph is a maximal independent set. Firstly, the concept of counting maximal independent sets including the set of leaves in graphs, in particular in trees, was studied by Włoch in [25]. It was proved that the largest number of maximal independent sets including the set of leaves is realized by the $(n-3)$rd Padovan number and the extremal tree achieving this maximum value is a graph $P_n$, $n\ge 3$.

In [16], it was observed that the number of all (1,2)-IDS in $P_n$, $n\ge 3$ is equal to $Pv(n-3)$ and the number of (1,2)-IDS in $C_n$, $n\ge 3$ is equal to $Pr\left(n\right)$. However, the investigation of the literature shows that the problem of counting of (1,2)-IDS has not yet been sufficiently studied. Only some initial results are included in the paper in [16], even though the problem of counting independent sets or their variants in graphs is intensively studied in that paper.

Based on the above, we deduce that the number $i_{1,2}(n,p)$ is equal to the number of all p-elements (1,2)-IDS in $P_n$, $n\ge 3$, and consequently, the number $I_{1,2}\left(n\right)$ is equal to the total number (1,2)-IDS in $P_n$, $n\ge 3$.

From the above graph interpretation and by Theorem 4, we obtain a new binomial formula for Padovan numbers.

Theorem 5. 

Let $p\ge 2$ and $n\ge 3$ be integers. Then, $Pv\left(n\right)={\displaystyle \sum _{p\ge 2}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p+4}}\right)$.

The number $i_{1,2}(n,p)$ can also be given by a recurrence relation. Through inspection of p-elements (1,2)-IDS in $P_n$, we have

$$i_{1,2}(n,p)=i_{1,2}(n-2,p-1)+i_{1,2}(n-3,p-1),forn\ge 6,p\ge 3$$

(1)

with the initial conditions $i_{1,2}(n,2)=1$ for $n\in \{3,4\}$ and $i_{1,2}(5,3)=1$.

Let $X=\{1,2,\cdots ,n\}$, $n\ge 4$, and let $Y^{*}$ be a p-element, $p\ge 2$, subset of X such that $Y^{*}$ does not contain either two consecutive integers or both 1 and n simultaneously. Moreover, if $t\notin Y^{*}$, then there is $q,r\in Y^{*}$ such that $|t-q|\in \{1,n-1\}$ and $|t-r|\in \{1,2,n-1,n-2\}$. For a fixed $p\ge 2$, let ${\mathcal{Y}}^{*}(n,p)$ be a family of all p-element subsets $Y^{*}$ and ${\mathcal{Y}}^{*}\left(n\right)={\displaystyle \bigcup _{p\ge 2}}{\mathcal{Y}}^{*}(n,p)$. The cardinality of the family ${\mathcal{Y}}^{*}(n,p)$ will be denoted by $i_{1,2}^{*}(n,p)$ and the cardinality of ${\mathcal{Y}}^{*}\left(n\right)$ by ${\mathcal{I}}_{1,2}^{*}\left(n\right)$. Then, ${\mathcal{I}}_{1,2}^{*}\left(n\right)={\displaystyle \sum _{p\ge 2}}{i}_{1,2}^{*}(n,p)$.

Theorem 6. 

Let $p\ge 2$ and $2p\le n\le 3p$ be integers. Then,

$$i_{1,2}^{*}(n,p)={\displaystyle \frac{n}{n-2p}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p-1}}\right).$$

Proof. 

Let $p\ge 2$ and $2p\le n\le 3p$ be fixed integers. Analogous to in the proof of Theorem 4, instead of the subset $Y^{*}$, we can consider a binary n-tuple $t=(t_1,\cdots ,t_n)$ such that $t_i$, $i\in \{1,\cdots ,n\}$ satisfies the following conditions:

(e)

$t_i=\left\{\begin{array}{cc}1\hfill & \mathrm{if}i\in Y^{*},\hfill \\ 0\hfill & \mathrm{otherwise};\hfill \end{array}\right.$

(f)

${\displaystyle \sum _{i=1}^n}{t}_i=p$;

(g)

if $t_i=t_j=1$ then $2\le |i-j|\le n-2$;

(h)

if $t_i=0$ then there are $t_r=t_q=1$ such that $|i-r|\in \{1,2,n-1,n-2\}$ and $|i-q|\in \{1,n-1\}$.

To construct an n-tuple, we start with a p-tuple $b=(b_1,\cdots ,b_p)$, $p\ge 2$, where $b_s=1$ for each $s\in \{1,\cdots ,p\}$. Next, we construct either a $2p$-tuple $b^{\prime }=(b_1^{\prime },b_1,\cdots ,b_p^{\prime },b_p)$ or $b^{\prime \prime }=(b_1,b_1^{\prime \prime },\cdots ,b_p,b_p^{\prime \prime })$ such that $b_s^{\prime }=b_s^{\prime \prime }=0$ for $s\in \{1,\cdots ,p\}$. In each case, $b_1^{\prime }\ne b_p$ and $b_1\ne b_p^{\prime \prime }$. Next, to build the n-tuple t, we have to add at most one zero between words $b_i^{\prime },b_i,b_{i+1}^{\prime }$ and $b_i,b_i^{\prime \prime },b_{i+1}$, $i\in \{1,\cdots ,p-1\}$ in each case of $2p$ tuple $b^{\prime }$ and $b^{\prime \prime }$. Clearly, in each case, we can do this in $C_p^{n-2p}=\left({\displaystyle \genfrac{}{}{0pt}{}{p}{n-2p}}\right)$ ways and because we have two $2p$-tuples $b^{\prime }$ and $b^{\prime \prime }$, so there are $2\left({\displaystyle \genfrac{}{}{0pt}{}{p}{n-2p}}\right)$ possibilities in this case. Assume now that in the n-tuple, t holds $t_1=t_n=0$. Then, $t_2=t_{n-1}=1$; otherwise, the condition (h) does not hold. Then, it suffices to consider an $(n-2)$-tuple satisfying conditions (a)–(d) in the proof of Theorem 4. Consequently, we have $\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p-1}}\right)$ possibilities in this case. Hence, from the above cases, $i_{1,2}^{*}(n,p)=2\left({\displaystyle \genfrac{}{}{0pt}{}{p}{n-2p}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p-1}}\right)$. Applying the well-known formula $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)={\displaystyle \frac{n}{k}}\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)$, we have $i_{1,2}^{*}(n,p)={\displaystyle \frac{n}{n-2p}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p-1}}\right)$, which ends the proof. □

Corollary 2. 

Let $p\ge 2$ be a fixed integer and $2p\le n\le 3p$. Then,

$${\mathcal{I}}_{1,2}^{*}(n,p)=\sum _{p\ge 2}\left({\displaystyle \frac{n}{n-2p}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p-1}}\right)\right).$$

In the graph interpretation, $i_{1,2}^{*}(n,p)$ is equal to the number of all p-elements (1,2)-IDS of $C_n$, and consequently, the number ${\mathcal{I}}_{1,2}^{*}\left(n\right)$ is equal to the total number of (1,2)-IDS in a graph $C_n$, $n\ge 4$; so, ${\mathcal{I}}_{1,2}^{*}\left(n\right)=Pr\left(n\right)$. Using the above graph interpretation, we can find relations between $i_{1,2}(n,p)$ and $i_{1,2}^{*}(n,p)$, and consequently, we find a new identity between Padovan and Perrin numbers.

Theorem 7. 

Let $n\ge 8$, $p\ge 3$ be integers. Then,

$i_{1,2}^{*}(n,p)=2i_{1,2}(n-3,p-1)+5i_{1,2}(n-4,p-1)+3i_{1,2}(n-5,p-1)$.

Proof. 

To prove this theorem, it suffices to calculate the number of (1,2)-IDS in a graph $C_n$, $n\ge 8$. Let $V\left(C_n\right)=\{x_1,\cdots ,x_n\}$ with the numbering of vertices in the natural fashion. Without a loss of generality, let us suppose that S is an arbitrary (1,2)-IDS of $C_n$ and let us consider the following cases.

• $x_1\in S$
  Then, $\{x_n,x_2\}\cap S=\varnothing$ and the set S has the form $S=\left\{x_1\right\}\cup S^{*}$, where $S^{*}$ is a (1,2)-IDS of a special subpath of $C_n$. We have the following possibilities.

  1.1.

  $\{x_{n-1},x_3\}\subset S^{*}$. Then, $S^{*}$ is a (1,2)-IDS of $P_{n-3}$ and there exists $i_{1,2}(n-3,p-1)$ such (1,2)-IDS.

  1.2.

  Either $\{x_{n-1},x_4\}\subset S^{*}$ or $\{x_{n-2},x_3\}\subset S^{*}$. Then, $S^{*}$ is an arbitrary (1,2)-IDS of $P_{n-4}$, and by a symmetry of possibilities, there are $2i_{1,2}(n-4,p-1)$ such sets.

  1.3.

  $\{x_{n-2},x_4\}\subset S^{*}$. In this case, $\{x_2,x_3,x_n,x_{n-1}\}\cap S=\varnothing$. Hence, $S^{*}$ is an (1,2)-IDS of $P_{n-5}$, so there exists $i_{1,2}(n-5,p-1)$ such sets.

• $x_1\notin S$.
  Then, the set $S=S^{*}$, where $S^{*}$ is a (1,2)-IDS of a special subpath of $C_n$.

  2.1.

  $\{x_n,x_2\}\subset S^{*}$. Then, $S^{*}$ is an arbitrary (1,2)-IDS of $P_{n-1}$ and there are $i_{1,2}(n-1,p)$ such sets.

  2.2.

  Either $\{x_n,x_3\}\subset S^{*}$ or $\{x_{n-1},x_2\}\subset S^{*}$. Then, $S^{*}$ is an arbitrary (1,2)-IDS of $P_{n-2}$, so there are $2i_{1,2}(n-2,p)$ sets S by symmetry of this case. Summing up the above possibilities, we have $i_{1,2}^{*}(n,p)=i_{1,2}(n-1,p)+2i_{1,2}(n-2,p)+i_{1,2}(n-3,p-1)+2i_{1,2}(n-4,p-1)+i_{1,2}(n-5,p-1)$. Using (1), we can transform the right-hand side of the previous equation to $i_{1,2}^{*}(n,p)=2i_{1,2}(n-3,p-1)+5i_{1,2}(n-4,p-1)+3i_{1,2}(n-5,p-1)$, which ends the proof.

□

Corollary 3. 

Let $n\ge 8$ be an integer. Then,

$$Pr\left(n\right)=2Pv(n-6)+5Pv(n-7)+3Pv(n-8).$$

3. Padovan Polynomials of a Graph

A natural method of generalization of Fibonacci-type sequences is polynomials (see for example [26,27]). The classical Fibonacci polynomials are defined by the recurrence relation $f_n\left(x\right)=x{f}_{n-1}\left(x\right)+f_{n-2}\left(x\right)$, for $n\ge 2$, with the initial conditions $f_0\left(x\right)=1$, and $f_1\left(x\right)=x$. Clearly, if $x=1$, then $f_n\left(1\right)=F_n$. Fibonacci polynomials were introduced by the Belgian mathematician E. C. Catalan in 1883 and studied, among others, by the German mathematician E. Jacobsthal. Throughout history, different properties and generalizations of Fibonacci polynomials have been described. In particular, relationships between Fibonacci polynomials and Pascal’s triangle were given, Binet’s type of formulas and generating function and roots were determined (see for a description [26,27]). Another type of Fibonacci polynomial was defined using the graph method.

In [28], G. Hopkins and W. Stanton introduced the Fibonacci polynomial of a graph which gives the total number of independent sets in the composition of two graphs. We recall this definition.

For two graphs G and H, a composition is a graph $G\left[H\right]$ such that $V\left(G\right[H\left]\right)=V\left(G\right)\times V\left(H\right)$ and $(x_i,y_p)(x_j,y_q)\in E\left(G\left[H\right]\right)$ if and only if $x_i{x}_j\in E\left(G\right)$ or $i=j$ and $y_p{y}_q\in E\left(H\right)$. A composition of two graphs is called a lexicographic product.

A generalized Fibonacci polynomial of a graph with respect to the number of distance independent sets was introduced by I. Włoch [29] and in [26].

In this paper, we consider Padovan polynomials belonging to the family of Fibonacci-type polynomials.

Padovan polynomials $Pv(n,x)$ are defined by the recurrence relation of the form $Pv(n,x)=xPv(n-2,x)+Pv(n-3,x)$ for $n\ge 3$ with the initial conditions $Pv(0,x)=Pv(1,x)=Pv(2,x)=1$.

The sequence of Padovan polynomials has the form $1,1,1,x+1,x+1,x^2+x+1,x^2+2x+1,x^3+x^2+2x+1,...$ for a few initial values of n. If $x=1$, then $Pv(n,1)=Pv\left(n\right)$; consequently, we obtain the Padovan sequence $1,1,1,2,2,3,4,5,\cdots$.

In this paper, we introduce the Padovan polynomial of a graph which gives the total number of special (1,2)-IDS in a composition of two graphs. Let ${\sigma }_{max}\left(G\right)$ denote the number of all maximal independent sets of G. Let $S\subset V\left(G\right[H\left]\right)$ be a (1,2)-IDS of $G\left[H\right]$. Then, by the definition of $G\left[H\right]$, the set ${\pi }_G\left(S\right)$ is a maximal independent set of G.

Let S be a $(1,2)$-IDS of $G\left[H\right]$ such that ${\pi }_G\left(S\right)$ is a $(1,2)$-IDS of G and let ${\mathcal{I}}_{{\pi }_G(1,2)}\left(G\left[H\right]\right)$ be the number of all such subsets S.

For a fixed integer $x\ge 2$, let H be a graph such that $i\left(H\right)\ge 2$ and ${\sigma }_{max}\left(H\right)=x$, for the integer $x\ge 2$. Let G be a graph with $(1,2)$-IDS.

The Padovan polynomial $Pv(G,x)$ of a graph G is defined by $Pv(G,x)={\mathcal{I}}_{{\pi }_G(1,2)}\left(G\left[H\right]\right)$.

Theorem 8. 

Let H be a graph with $i\left(H\right)\ge 2$ and ${\sigma }_{max}\left(H\right)=x$ for integer $x\ge 2$. For a connected graph G on order n, $n\ge 3$, with $(1,2)$-IDS, $Pv(G,x)={\displaystyle \sum _{p\ge 2}}{i}_{1,2}(G,n,p)x^p$.

Proof. 

Let G and H be as in the statement of the theorem. To determine the Padovan polynomial $Pv(G,x)$, it suffices to show that the number of all $(1,2)$-IDS of $G\left[H\right]$ such that the projection into G gives an $(1,2)$-IDS is equal to ${\displaystyle \sum _{p\ge 2}}{i}_{1,2}(G,n,p)x^p$. By the definition of $G\left[H\right]$, we deduce that to obtain an $(1,2)$-IDS of $G\left[H\right]$, say S, such that ${\pi }_G\left(S\right)$ is a $(1,2)$-IDS, we first have to choose a $(1,2)$-IDS of G. Clearly, every $(1,2)$-IDS has at least two vertices; so, let us consider a p-element $(1,2)$-IDS of G, $p\ge 2$. A p-element $(1,2)$-IDS can be chosen in $i_{1,2}(G,n,p)$ ways. Next, in each of the chosen copies of H, we have to choose a maximal independent set among x vertices. By assuming that every maximal independent set of H has at least two vertices and G is connected, this guarantees that every vertex of H is (1,2)-dominated. Because ${\sigma }_{max}\left(H\right)=x$, in each copy of H, a maximal $(1,2)$-IDS can be chosen in x ways; so, we have $i_{1,2}(G,n,p)x^p$$(1,2)$-IDS that a projection on G is a $(1,2)$-IDS. Finally, $Pv(G,x)={\displaystyle \sum _{p\ge 2}}{i}_{1,2}(G,n,p)x^p$, which ends the proof. □

Using results given in Theorem 5 and Theorem 6, respectively, we obtain Padovan polynomials of paths and cycles.

Theorem 9. 

Let $p\ge 2$. Then, for $n\ge 3$, we have $Pv(P_n,x)={\displaystyle \sum _{p\ge 2}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p+1}}\right)x^p$ and for $2p\le n\le 3p$ we have $Pv(C_n,x)={\displaystyle \sum _{p\ge 2}}\left({\displaystyle \frac{n}{n-2p}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p-1}}\right)\right)x^p$.

As an illustration of Theorem 9, let us consider a composition $P_7\left[C_n\right]$ for $n\ge 4$. Because $i\left(C_n\right)\ge 2$ and ${\sigma }_{max}\left(C_n\right)=x$, where $x\ge 2$, $Pv\left(P_7\left[C_n\right]\right)={\displaystyle \sum _{p\ge 2}}\left({\displaystyle \genfrac{}{}{0pt}{}{p-1}{n-2p+1}}\right)x^p=x^3+x^4$.

In particular, $Pv\left(P_5\left[C_4\right]\right)=24$.

4. Conclusions

In this paper, we have obtained new binomial formulas for Padovan and Perrin numbers by studying independent (1,2)-dominating sets in paths and cycles. This approach provides a new interpretation of these Fibonacci-type numbers and discovery of a new dependency between Padovan and Perrin numbers, extending the existing knowledge of their relationships. Moreover, by analyzing special independent (1,2)-dominating sets in the composition of two graphs, we introduced the concept of the Padovan polynomial for graphs.

The results presented in this paper contribute to the ongoing study of Fibonacci-type sequences and their applications, particularly in graph theory, and open new directions for further investigation into the symmetrical aspects of these sets.