Robinson Santrin Sabibha¹, Kruz Jeya Daisy², Pon Jeyanthi³, Maged Zakaria Youssef^4,5
1Department of Science and Humanities, Vins Christian College of Engineering, Nagercoil, India.
²Department of Mathematics, Holy Cross College, Nagercoil, India.
³Research Centre, Department of Mathematics, Govindammal Aditanar College for Women, Tiruchendur, India.
⁴Department of Mathematics and Statistics, College of Science, Imam Mohammad Ibn Saud Islamic University, Riyadh, Saudi Arabia.
⁵Department of Mathematics, Faculty of Science, Ain Shams University, Cairo, Egypt.
DOI: 10.4236/ojdm.2025.151001   PDF    HTML   XML   56 Downloads   286 Views   Citations

Abstract

In 2012, Ponraj et al. defined a concept of k-product cordial labeling as follows: Let f be a map from $V\left(G\right)$ to $\left\{\mathrm{0,1,}\cdots \mathrm{,}k-1\right\}$ where k is an integer, $1\le k\le \left|V\left(G\right)\right|$ . For each edge $uv$ assign the label $f\left(u\right)f\left(v\right)\left(\mathrm{mod}k\right)$ . f is called a k-product cordial labeling if $\left|v_f\left(i\right)-v_f\left(j\right)\right|\le 1$ , and $\left|e_f\left(i\right)-e_f\left(j\right)\right|\le 1$ , $i\mathrm{,}j\in \left\{\mathrm{0,1,}\cdots \mathrm{,}k-1\right\}$ , where $v_f\left(x\right)$ and $e_f\left(x\right)$ denote the number of vertices and edges respectively labeled with x ($x=\mathrm{0,1,}\cdots \mathrm{,}k-1$ ). Motivated by this concept, we further studied and established that several families of graphs admit k-product cordial labeling. In this paper, we show that the path graphs $P_n$ admit k-product cordial labeling.

Keywords

Cordial Labeling, Product Cordial Labeling, k-Product Cordial Labeling, Path Graph

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1. Introduction

All graphs considered here are simple, finite, connected and undirected. We follow the basic notations and terminology of graph theory as in [1]. While studying graph theory, one that has gained a lot of popularity during the last 60 years is the concept of labelings of graphs due to its wide range of applications. Labeling is a function that allocates the elements of a graph to real numbers, usually positive integers. In 1967, Rosa [2] published a pioneering paper on graph labeling problems. Thereafter, several authors have studied many types of graph labeling techniques. Gallian in his survey [3] beautifully classified them. Cordial labeling is one such labelings defined by Cahit [4] as follows: Let f be a function from the vertices of G to $\left\{\mathrm{0,1}\right\}$ and for each edge $xy$ assign the label $\left|f\left(x\right)-f\left(y\right)\right|$ . f is called a cordial labeling of G if the number of vertices labeled 0 and the number of vertices labeled 1 differ by at most 1, and the number of edges labeled 0 and the number of edges labeled 1 differ at most by 1. This concept was extended and defined product cordial labeling by Sundaram et al. [5] as follows: Let f be a function from $V\left(G\right)$ to $\left\{\mathrm{0,1}\right\}$ . For each edge $uv$ , assign the label $f\left(u\right)f\left(v\right)$ . Then f is called product cordial labeling if $\left|v_f\left(0\right)-v_f\left(1\right)\right|\le 1$ and $\left|e_f\left(0\right)-e_f\left(1\right)\right|\le 1$ where $v_f\left(i\right)$ and $e_f\left(i\right)$ denotes the number of vertices and edges respectively labeled with $i\left(i=\mathrm{0,1}\right)$ . Some of the researchers have shown interest in this topic and published their results. An interested reader can refer to [6]-[11].

Ponraj et al. [12] further extended the concept of product cordial labeling and defined a new labeling called k-product cordial labeling. The same authors [13] established that the 4-product cordial behaviour of graphs such as subdivision star, wheel, $K_2+m{K}_1$ , $K_{\mathrm{2,}n}$ and $K_n^c+2K_2$ . For further results on 3-product and 4-product cordial labeling one can refer to [3]. Inspired by the concept of k-product cordial labeling and the results in [12], we further studied and showed that several families of graphs admit k-product cordial labeling in our published papers [14]-[21]. In [22], it is proved that $P_n$ is 3-product cordial for every positive integer n. In this paper, we made an attempt to characterize the values of positive integer $k\ge 2$ for which the path graph $P_n$ is k-product cordial for every positive integer n.

2. Terminology and Definitions

We use the following terminology and definitions to prove our main results.

The pigeonhole principle is, if m pigeons occupy n pigeonholes and $m>n$ , then at least one pigeonhole has two or more pigeons roosting in it [23].

Let x be any real number. Then $\lfloor x\rfloor$ stands for the largest integer less than or equal to x and $\lceil x\rceil$ stands for the smallest integer greater than or equal to x.

Definition 1. An integer $p\mathrm{<mo>></mo>1}$ is called a prime number if 1 and p are the only divisor of p.

Definition 2. Let f be a map from $V\left(G\right)$ to $\left\{\mathrm{0,1,}\cdots \mathrm{,}k-1\right\}$ where k is an integer, $1\le k\le \left|V\left(G\right)\right|$ . For each edge $uv$ assign the label $f\left(u\right)f\left(v\right)\left(\mathrm{mod}k\right)$ . f is called a k-product cordial labeling if $\left|v_f\left(i\right)-v_f\left(j\right)\right|\le 1$ , and $\left|e_f\left(i\right)-e_f\left(j\right)\right|\le 1$ , $i\mathrm{,}j\in \left\{\mathrm{0,1,}\cdots \mathrm{,}k-1\right\}$ , where $v_f\left(x\right)$ and $e_f\left(x\right)$ denote the number of vertices and edges respectively labeled with x ($x=\mathrm{0,1,}\cdots \mathrm{,}k-1$ ).

3. Main Results

In this section, we study the k-product cordial labeling of path graph $P_n$ where $k=\mathrm{5,6,7,10,11}$ and 15. Also, we show that not all paths are $p^2$ -product cordial when p is odd prime. In all the results, we consider the vertex and edge set of $P_n$ be $V\left(P_n\right)=\left\{v_i\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le n\right\}$ and $E\left(P_n\right)=\left\{\left(v_i\mathrm{,}{v}_{i+1}\right)\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le n-1\right\}$ respectively.

Theorem 3. For $n\ge 3$ , the path $P_n$ is 5-product cordial.

Proof. Define $f\mathrm{<mo>:</mo>}V\left(P_n\right)\to \left\{\mathrm{0,1,2,3,4}\right\}$ as follows:

$f\left(v_i\right)=0$ ; $1\le i\le \lfloor \frac{n}{5}\rfloor$ .

For $i=\lfloor \frac{n}{5}\rfloor +j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le n-\lfloor \frac{n}{5}\rfloor$ ,

$f\left(v_i\right)=\{\begin{array}{ll}4\hfill & \text{if}j\equiv 1,6\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{if}j\equiv 2,5\left(\mathrm{mod}8\right)\hfill \\ 2\hfill & \text{if}j\equiv 3,7\left(\mathrm{mod}8\right)\hfill \\ 3\hfill & \text{if}j\equiv 4,0\left(\mathrm{mod}8\right)\hfill \end{array}$

From the above labeling pattern, we have the following cases.

Case (i): If $n\equiv 1\left(\mathrm{mod}5\right)$ ,

$e_f\left(i\right)=\lfloor \frac{n}{5}\rfloor$ for $0\le i\le 4$ .

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=1\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=4\hfill \end{array}$

Hence, $P_n$ is a 5-product cordial graph if $n\equiv 1\left(\mathrm{mod}5\right)$ .

Case (ii): If $n\equiv 2\left(\mathrm{mod}5\right)$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,2,3}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{1,4}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=4\hfill \end{array}$

Hence, $P_n$ is a 5-product cordial graph if $n\equiv 2\left(\mathrm{mod}5\right)$ .

Case (iii): If $n\equiv 3\left(\mathrm{mod}5\right)$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,3}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4}\hfill \end{array}$

For n is even, $e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,1,2}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{3,4}\hfill \end{array}$

For n is odd, $e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,1,3}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{2,4}\hfill \end{array}$

Hence, $P_n$ is a 5-product cordial graph if $n\equiv 3\left(\mathrm{mod}5\right)$ .

Case (iv): If $n\equiv 4\left(\mathrm{mod}5\right)$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=0\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4}\hfill \end{array}$

For n is even, $e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,3}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4}\hfill \end{array}$

For n is odd, $e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{5}\rfloor \hfill & \text{if}i=\mathrm{0,2}\hfill \\ \lfloor \frac{n}{5}\rfloor +1\hfill & \text{if}i=\mathrm{1,3,4}\hfill \end{array}$

Hence, $P_n$ is a 5-product cordial graph if $n\equiv 4\left(\mathrm{mod}5\right)$ .

Case (v): If $n\equiv 0\left(\mathrm{mod}5\right)$ ,

$v_f\left(i\right)=\frac{n}{5}$ for $0\le i\le 4$ .

For n is even, $e_f\left(i\right)=\{\begin{array}{ll}\frac{n}{5}-1\hfill & \text{if}i=2\hfill \\ \frac{n}{5}\hfill & \text{if}i=\mathrm{0,1,3,4}\hfill \end{array}$

For n is odd, $e_f\left(i\right)=\{\begin{array}{ll}\frac{n}{5}-1\hfill & \text{if}i=3\hfill \\ \frac{n}{5}\hfill & \text{if}i=\mathrm{0,1,2,4}\hfill \end{array}$

Hence, $P_n$ is a 5-product cordial graph if $n\equiv 0\left(\mathrm{mod}5\right)$ . □

An example of 5-product cordial labeling of $P_{12}$ is shown in Figure 1.

Figure 1. 5-product cordial labeling of P₁₂.

Theorem 4. For $n\ge 3$ , the path $P_n$ is 7-product cordial.

Proof. Define $f\mathrm{<mo>:</mo>}V\left(P_n\right)\to \left\{\mathrm{0,1,2,3,4,5,6}\right\}$ as follows:

$f\left(v_i\right)=0$ ; $1\le i\le \lfloor \frac{n}{7}\rfloor$ .

For $i=\lfloor \frac{n}{7}\rfloor +j\mathrm{<mi>\; </mi>};\mathrm{<mi>\; </mi>1}\le j\le n-\lfloor \frac{n}{7}\rfloor$ ,

$f\left(v_i\right)=\{\begin{array}{ll}6\hfill & \text{if}j\equiv \mathrm{1,8}\left(\mathrm{mod}12\right)\hfill \\ 1\hfill & \text{if}j\equiv \mathrm{2,7}\left(\mathrm{mod}12\right)\hfill \\ 2\hfill & \text{if}j\equiv \mathrm{3,10}\left(\mathrm{mod}12\right)\hfill \\ 5\hfill & \text{if}j\equiv \mathrm{4,9}\left(\mathrm{mod}12\right)\hfill \\ 3\hfill & \text{if}j\equiv \mathrm{5,0}\left(\mathrm{mod}12\right)\hfill \\ 4\hfill & \text{if}j\equiv \mathrm{6,11}\left(\mathrm{mod}12\right)\hfill \end{array}$

From the above labeling pattern, we have the following cases.

Case (i): If $n\equiv 1\left(\mathrm{mod}7\right)$ ,

$e_f\left(i\right)=\lfloor \frac{n}{7}\rfloor$ for $0\le i\le 6$ .

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=1\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=6\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 1\left(\mathrm{mod}7\right)$ .

Case (ii): If $n\equiv 2\left(\mathrm{mod}7\right)$ .

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,6}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=6\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 2\left(\mathrm{mod}7\right)$ .

Case (iii): If $n\equiv 3\left(\mathrm{mod}7\right)$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{2,6}\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,3,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,6}\hfill \end{array}$

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,5,6}\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 3\left(\mathrm{mod}7\right)$ .

Case (iv): If $n\equiv 4\left(\mathrm{mod}7\right)$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,3,4}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,5,6}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,1,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{2,3,6}\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 4\left(\mathrm{mod}7\right)$ .

Case (v): If $n\equiv 5\left(\mathrm{mod}7\right)$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,4,5}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,6}\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,4}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,5,6}\hfill \end{array}$

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,3}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,5,6}\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 5\left(\mathrm{mod}7\right)$ .

Case (vi): If $n\equiv 6\left(\mathrm{mod}7\right)$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=0\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,5,6}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{7}\rfloor \hfill & \text{if}i=\mathrm{0,4}\hfill \\ \lfloor \frac{n}{7}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,5,6}\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 6\left(\mathrm{mod}7\right)$ .

Case (vii): If $n\equiv 0\left(\mathrm{mod}7\right)$ ,

$v_f\left(i\right)=\frac{n}{7}$ for $0\le i\le 6$ , $e_f\left(i\right)=\{\begin{array}{ll}\frac{n}{7}-1\hfill & \text{if}i=4\hfill \\ \frac{n}{7}\hfill & \text{if}i=\mathrm{0,1,2,3,5,6}\hfill \end{array}$

Hence, $P_n$ is a 7-product cordial graph if $n\equiv 0\left(\mathrm{mod}7\right)$ . □

An example of 7-product cordial labeling of $P_{11}$ is shown in Figure 2.

Figure 2. 7-product cordial labeling of P₁₁.

Theorem 5. For $n\ge 3$ , the path $P_n$ is 11-product cordial.

Proof. Define $f\mathrm{<mo>:</mo>}V\left(P_n\right)\to \left\{\mathrm{0,1,2,}\cdots \mathrm{,10}\right\}$ as follows:

$f\left(v_i\right)=0$ ; $1\le i\le \lfloor \frac{n}{11}\rfloor$ .

For $i=\lfloor \frac{n}{11}\rfloor +j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le n-\lfloor \frac{n}{11}\rfloor$ ,

$f\left(v_i\right)=\{\begin{array}{ll}10\hfill & \text{if}j\equiv \mathrm{1,12}\left(\mathrm{mod}20\right)\hfill \\ 1\hfill & \text{if}j\equiv \mathrm{2,11}\left(\mathrm{mod}20\right)\hfill \\ 9\hfill & \text{if}j\equiv \mathrm{3,14}\left(\mathrm{mod}20\right)\hfill \\ 2\hfill & \text{if}j\equiv \mathrm{4,13}\left(\mathrm{mod}20\right)\hfill \\ 7\hfill & \text{if}j\equiv \mathrm{5,16}\left(\mathrm{mod}20\right)\hfill \\ 4\hfill & \text{if}j\equiv \mathrm{6,15}\left(\mathrm{mod}20\right)\hfill \\ 3\hfill & \text{if}j\equiv \mathrm{7,18}\left(\mathrm{mod}20\right)\hfill \\ 8\hfill & \text{if}j\equiv \mathrm{8,17}\left(\mathrm{mod}20\right)\hfill \\ 6\hfill & \text{if}j\equiv \mathrm{9,0}\left(\mathrm{mod}20\right)\hfill \\ 5\hfill & \text{if}j\equiv \mathrm{10,19}\left(\mathrm{mod}20\right)\hfill \end{array}$

From the above labeling pattern, we have the following cases.

Case (i): If $n\equiv 1\left(\mathrm{mod}11\right)$ ,

$e_f\left(i\right)=\lfloor \frac{n}{11}\rfloor$ for $0\le i\le 10$ .

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6,7,8,9,10}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=1\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=10\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 1\left(\mathrm{mod}11\right)$ .

Case (ii): If $n\equiv 2\left(\mathrm{mod}11\right)$ .

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,10}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=10\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 2\left(\mathrm{mod}11\right)$ .

Case (iii): If $n\equiv 3\left(\mathrm{mod}11\right)$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3,4,5,6,7,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{9,10}\hfill \end{array}$

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,3,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,10}\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6,7,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 3\left(\mathrm{mod}11\right)$ .

Case (iv): If $n\equiv 4\left(\mathrm{mod}11\right)$ .

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,3,4,5,6,7,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,9,10}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,3,4,5,6,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{7,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 4\left(\mathrm{mod}11\right)$ .

Case (v): If $n\equiv 5\left(\mathrm{mod}11\right)$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,4,5,6,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{3,7,9,10}\hfill \end{array}$

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,7,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,9,10}\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,3,4,5,6,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,7,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 5\left(\mathrm{mod}11\right)$ .

Case (vi): If $n\equiv 6\left(\mathrm{mod}11\right)$ .

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,9,10}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,1,2,4,5,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{3,6,7,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 6\left(\mathrm{mod}11\right)$ .

Case (vii): If $n\equiv 7\left(\mathrm{mod}11\right)$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,2,4,5,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,3,6,7,9,10}\hfill \end{array}$

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,8,9,10}\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,5,6,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,7,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 7\left(\mathrm{mod}11\right)$ .

Case (viii): If $n\equiv 8\left(\mathrm{mod}11\right)$ .

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,5,6}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,7,8,9,10}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,4,5,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,6,7,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 8\left(\mathrm{mod}11\right)$ .

Case (ix): If $n\equiv 9\left(\mathrm{mod}11\right)$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,5,8}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,6,7,9,10}\hfill \end{array}$

For n is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,6}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,5,7,8,9,10}\hfill \end{array}$

For n is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,5}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,6,7,8,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 9\left(\mathrm{mod}11\right)$ .

Case (x): If $n\equiv 10\left(\mathrm{mod}11\right)$ .

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=0\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}1\le i\le 10\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{11}\rfloor \hfill & \text{if}i=\mathrm{0,5}\hfill \\ \lfloor \frac{n}{11}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,6,7,8,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 10\left(\mathrm{mod}11\right)$ .

Case (xi): If $n\equiv 0\left(\mathrm{mod}11\right)$ .

$v_f\left(i\right)=\frac{n}{11}$ for $0\le i\le 10$ .

$e_f\left(i\right)=\{\begin{array}{ll}\frac{n}{11}-1\hfill & \text{if}i=5\hfill \\ \frac{n}{11}\hfill & \text{if}i=\mathrm{0,1,2,3,4,6,7,8,9,10}\hfill \end{array}$

Hence, $P_n$ is a 11-product cordial graph if $n\equiv 0\left(\mathrm{mod}11\right)$ . □

An example of 11-product cordial labeling of $P_8$ is shown in Figure 3.

Figure 3. 11-product cordial labeling of P₈.

As a consequence of Theorems 3, 4 and 5 we propose Conjecture 6.

Conjecture 6. For all $n\ge 3$ , the path $P_n$ is k-product cordial graph if k is prime.

Theorem 7. For $n\ge 3$ , the path $P_n$ is 6-product cordial.

Proof. Define $f\mathrm{<mo>:</mo>}V\left(P_n\right)\to \left\{\mathrm{0,1,2,3,4,5}\right\}$ .

We have the following six cases.

Case (i): If $n\equiv 1\left(\mathrm{mod}6\right)$ ,

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{if}1\le i\le \lfloor \frac{n}{6}\rfloor \hfill \\ 3\hfill & \text{if}\lfloor \frac{n}{6}\rfloor +1\le i\le 2\lfloor \frac{n}{6}\rfloor \hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +2-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor +1$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +2-j}\right)=\{\begin{array}{ll}1\hfill & \text{if}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{if}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +1+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +1+j}\right)=\{\begin{array}{ll}4\hfill & \text{if}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 2\hfill & \text{if}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

Therefore,

$e_f\left(i\right)=\lfloor \frac{n}{6}\rfloor$ for $0\le i\le 5$ .

For $\lfloor \frac{n}{6}\rfloor$ is odd,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,3,4}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=5\hfill \end{array}$

For $\lfloor \frac{n}{6}\rfloor$ is even,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4,5}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=1\hfill \end{array}$

Hence, $P_n$ is a 6-product cordial graph if $n\equiv 1\left(\mathrm{mod}6\right)$ .

Case (ii): If $n\equiv 2\left(\mathrm{mod}6\right)$ ,

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{for}1\le i\le \lfloor \frac{n}{6}\rfloor \hfill \\ 3\hfill & \text{for}\lfloor \frac{n}{6}\rfloor +1\le i\le 2\lfloor \frac{n}{6}\rfloor \hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +3-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor +2$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +3-j}\right)=\{\begin{array}{ll}1\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +2+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +2+j}\right)=\{\begin{array}{ll}4\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 2\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

Therefore,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{1,5}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,3,4}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=5\hfill \end{array}$

Hence, $P_n$ is a 6-product cordial graph if $n\equiv 2\left(\mathrm{mod}6\right)$ .

Case (iii): If $n\equiv 3\left(\mathrm{mod}6\right)$ .

Subcase (i): If $\lfloor \frac{n}{6}\rfloor$ is odd.

We label the vertices $v_i\left(1\le i\le n-1\right)$ as in Case (ii), then assign 2 to $v_n$ .

Subcase (ii): If $\lfloor \frac{n}{6}\rfloor$ is even.

We label the vertices $v_i\left(1\le i\le n-1\right)$ as in Case (ii), then assign 4 to $v_n$ .

Therefore,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,3}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{4,5}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is even,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,2,3}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{1,4,5}\hfill \end{array}$

For $\lfloor \frac{n}{6}\rfloor$ is odd,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,3,4}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{1,2,5}\hfill \end{array}$

Hence, $P_n$ is a 6-product cordial graph if $n\equiv 3\left(\mathrm{mod}6\right)$ .

Case (iv): If $n\equiv 4\left(\mathrm{mod}6\right)$ .

Subcase (i): If $\lfloor \frac{n}{6}\rfloor$ is odd.

We label the vertices $v_i\left(1\le i\le n-1\right)$ as in Case (iii) Subcase (i), then assign 4 to $v_n$ .

Subcase (ii): If $\lfloor \frac{n}{6}\rfloor$ is even.

We label the vertices $v_i\left(1\le i\le n-1\right)$ as in Case (iii) Subcase (ii), then assign 2 to $v_n$ .

Therefore,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,3}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{1,2,4,5}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,1,3}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{2,4,5}\hfill \end{array}$

Hence, $P_n$ is a 6-product cordial graph if $n\equiv 4\left(\mathrm{mod}6\right)$ .

Case (v): If $n\equiv 5\left(\mathrm{mod}6\right)$ ,

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{for}1\le i\le \lfloor \frac{n}{6}\rfloor \hfill \\ 3\hfill & \text{for}\lfloor \frac{n}{6}\rfloor +1\le i\le 2\lfloor \frac{n}{6}\rfloor +1\hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +4-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor +2$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +4-j}\right)=\{\begin{array}{ll}1\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +3+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor +2$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +3+j}\right)=\{\begin{array}{ll}4\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 2\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

Therefore,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=0\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{1,2,3,4,5}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,1}\hfill \\ \lfloor \frac{n}{6}\rfloor +1\hfill & \text{for}i=\mathrm{2,3,4,5}\hfill \end{array}$

Hence, $P_n$ is a 6-product cordial graph if $n\equiv 5\left(\mathrm{mod}6\right)$ .

Case (vi): If $n\equiv 0\left(\mathrm{mod}6\right)$ ,

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{if}1\le i\le \lfloor \frac{n}{6}\rfloor \hfill \\ 3\hfill & \text{if}\lfloor \frac{n}{6}\rfloor +1\le i\le 2\lfloor \frac{n}{6}\rfloor \hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +1-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +1-j}\right)=\{\begin{array}{ll}1\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=4\lfloor \frac{n}{6}\rfloor +j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{6}\rfloor$ ,

$f\left(v_{4\lfloor \frac{n}{6}\rfloor +j}\right)=\{\begin{array}{ll}4\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 2\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

Therefore,

$v_f\left(i\right)=\lfloor \frac{n}{6}\rfloor$ for $0\le i\le 5$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{6}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4,5}\hfill \\ \lfloor \frac{n}{6}\rfloor -1\hfill & \text{for}i=1\hfill \end{array}$

Hence, $P_n$ is a 6-product cordial graph if $n\equiv 0\left(\mathrm{mod}6\right)$ . □

Figure 4 shows the 6-product cordial labeling of $P_9$ .

Figure 4. 6-product cordial labeling of P₉.

Theorem 8. For $n\ge 3$ , the path $P_n$ is 10-product cordial.

Proof. Define $f\mathrm{<mo>:</mo>}V\left(P_n\right)\to \left\{\mathrm{0,1,2,3,4,5,6,7,8,9}\right\}$ .

We have the following four cases.

Case (i): If $n=10\lfloor \frac{n}{10}\rfloor +t$ where $t=\mathrm{1,2,3,4}$ then

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{if}1\le i\le \lfloor \frac{n}{10}\rfloor \hfill \\ 5\hfill & \text{if}\lfloor \frac{n}{10}\rfloor +1\le i\le 2\lfloor \frac{n}{10}\rfloor \hfill \end{array}$

For $i=6\lfloor \frac{n}{10}\rfloor +1+t-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{10}\rfloor +t$ where $t=\mathrm{1,2,3,4}$ ,

$f\left(v_{6\lfloor \frac{n}{10}\rfloor +1+t-j}\right)=\{\begin{array}{ll}3\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{3,7}\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv \mathrm{4,0}\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=6\lfloor \frac{n}{10}\rfloor +t+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{10}\rfloor$ ,

$f\left(v_{6\lfloor \frac{n}{10}\rfloor +t+j}\right)=\{\begin{array}{ll}4\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 6\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 2\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

From the above labeling we have the following subcases:

Subcase (i): If $t=1$ ,

$e_f\left(i\right)=\lfloor \frac{n}{10}\rfloor$ for $0\le i\le 9$ .

For $\lfloor \frac{n}{10}\rfloor$ is even,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=3\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,3,4,5,6,8,9}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=7\hfill \end{array}$

Subcase (ii): If $t=2$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=1\hfill \end{array}$

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,4,5,6,8,9}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{3,7}\hfill \end{array}$

Subcase (iii): If $t=3$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,4,5,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{3,7,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is even,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,4,5,6,7,8,9}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,3}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4,5,6,8,9}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,7}\hfill \end{array}$

Subcase (iv): If $t=4$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,4,5,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,3,7,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is even,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,4,5,6,7,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,3,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4,5,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,7,9}\hfill \end{array}$

Hence, $P_n$ is a 10-product cordial graph if $n=10\lfloor \frac{n}{10}\rfloor +t$ where

$t=\mathrm{1,2,3,4}$ .

Case (ii): If $n=10\lfloor \frac{n}{10}\rfloor +5$ ,

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{for}1\le i\le \lfloor \frac{n}{10}\rfloor \hfill \\ 5\hfill & \text{for}\lfloor \frac{n}{10}\rfloor +1\le i\le 2\lfloor \frac{n}{10}\rfloor +1\hfill \end{array}$

For $i=6\lfloor \frac{n}{10}\rfloor +6-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{10}\rfloor +4$ ,

$f\left(v_{6\lfloor \frac{n}{10}\rfloor +6-j}\right)=\{\begin{array}{ll}3\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{3,7}\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv \mathrm{4,0}\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=6\lfloor \frac{n}{10}\rfloor +5+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{10}\rfloor$ ,

$f\left(v_{6\lfloor \frac{n}{10}\rfloor +5+j}\right)=\{\begin{array}{ll}4\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 6\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 2\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

Therefore, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,4,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,3,5,7,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is even,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,4,6,7,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,3,5,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,2,3,4,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{1,5,7,9}\hfill \end{array}$

Hence, $P_n$ is a 10-product cordial graph if $n=10\lfloor \frac{n}{10}\rfloor +5$ .

Case (iii): If $n=10\lfloor \frac{n}{10}\rfloor +t$ where $t=\mathrm{6,7,8,9}$ then

$f\left(v_i\right)=\{\begin{array}{ll}0\hfill & \text{if}1\le i\le \lfloor \frac{n}{10}\rfloor +1\hfill \\ 5\hfill & \text{if}\lfloor \frac{n}{10}\rfloor +2\le i\le 2\lfloor \frac{n}{10}\rfloor +2\hfill \end{array}$

For $i=6\lfloor \frac{n}{10}\rfloor +7-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{10}\rfloor +4$ ,

$f\left(v_{6\lfloor \frac{n}{10}\rfloor +7-j}\right)=\{\begin{array}{ll}3\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{3,7}\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv \mathrm{4,0}\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=6\lfloor \frac{n}{10}\rfloor +6+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{10}\rfloor +t-6$ where $t=\mathrm{6,7,8,9}$ ,

$f\left(v_{6\lfloor \frac{n}{10}\rfloor +6+j}\right)=\{\begin{array}{ll}4\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 6\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 2\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

From the above labeling we have the following subcases:

Subcase (i): If $t=6$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{2,4,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,3,5,7,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is even,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{2,4,6,7,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,3,5,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{2,3,4,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,5,7,9}\hfill \end{array}$

Subcase (ii): If $t=7$ .

For $\lfloor \frac{n}{10}\rfloor$ is even,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{2,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,3,4,5,7,9}\hfill \end{array}$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{4,6,7,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{f}o\text{r}i=\mathrm{0,1,2,3,5,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{2,4,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,3,5,6,7,9}\hfill \end{array}$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{3,4,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,5,7,9}\hfill \end{array}$

Subcase (iii): If $t=8$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{2,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,3,4,5,6,7,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is even,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{6,7,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,3,4,5,9}\hfill \end{array}$

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{3,6,8}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,4,5,7,9}\hfill \end{array}$

Subcase (iv): If $t=9$ .

For $\lfloor \frac{n}{10}\rfloor$ is even,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=2\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,3,4,5,6,7,8,9}\hfill \end{array}$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{6,7}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,3,4,5,8,9}\hfill \end{array}$ .

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=8\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,3,4,5,6,7,9}\hfill \end{array}$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{3,6}\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,4,5,7,8,9}\hfill \end{array}$ .

Hence, $P_n$ is a 10-product cordial graph if $n=10\lfloor \frac{n}{10}\rfloor +t$ where

$t=\mathrm{6,7,8,9}$ .

Case (iv): If $n=10\lfloor \frac{n}{10}\rfloor$ .

Subcase (i): If $\lfloor \frac{n}{10}\rfloor$ is even.

We label the vertices $v_i\left(1\le i\le n-1\right)$ as in Case (iii), then assign 8 to $v_n$ .

Subcase (ii): If $\lfloor \frac{n}{10}\rfloor$ is odd.

We label the vertices $v_i\left(1\le i\le n-1\right)$ as in Case (iii), then assign 2 to $v_n$ .

From this label we get,

$v_f\left(i\right)=\lfloor \frac{n}{10}\rfloor$ for $0\le i\le 9$ .

For $\lfloor \frac{n}{10}\rfloor$ is odd,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor -1\hfill & \text{for}i=7\hfill \\ \lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=\mathrm{0,1,2,3,4,5,6,8,9}\hfill \end{array}$ .

For $\lfloor \frac{n}{10}\rfloor$ is even,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{10}\rfloor \hfill & \text{for}i=3\hfill \\ \lfloor \frac{n}{10}\rfloor +1\hfill & \text{for}i=\mathrm{0,1,2,4,5,6,7,8,9}\hfill \end{array}$ .

Hence, $P_n$ is a 10-product cordial graph if $n=10\lfloor \frac{n}{10}\rfloor$ . □

Figure 5 shows the 10-product cordial labeling of $P_{10}$ .

Figure 5. 10-product cordial labeling of P₁₀.

Theorem 9 For $n\ge 3$ , the path $P_n$ is 15-product cordial.

Proof. Define $f\mathrm{<mo>:</mo>}V\left(P_n\right)\to \left\{\mathrm{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}\right\}$ .

We have the following five cases.

Case (i): If $n=15\lfloor \frac{n}{15}\rfloor +t$ where $1\le t\le 8$ then

$f\left(v_i\right)=\mathrm{0<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le \lfloor \frac{n}{15}\rfloor \mathrm{.}$

For $t=\mathrm{1,2,3,5}$ and $i=3\lfloor \frac{n}{15}\rfloor +1-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{3\lfloor \frac{n}{15}\rfloor +1-j}\right)=\{\begin{array}{ll}5\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 10\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $t=\mathrm{4,6,7,8}$ and $i=3\lfloor \frac{n}{15}\rfloor +1-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{3\lfloor \frac{n}{15}\rfloor +1-j}\right)=\{\begin{array}{ll}10\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +1+t-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 8\lfloor \frac{n}{15}\rfloor +t$ where $1\le t\le 8$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +1+t-j}\right)=\{\begin{array}{ll}2\hfill & \text{for}j\equiv 1\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv 2\left(\mathrm{mod}8\right)\hfill \\ 11\hfill & \text{for}j\equiv 3\left(\mathrm{mod}8\right)\hfill \\ 13\hfill & \text{for}j\equiv 4\left(\mathrm{mod}8\right)\hfill \\ 14\hfill & \text{for}j\equiv 5\left(\mathrm{mod}8\right)\hfill \\ 4\hfill & \text{for}j\equiv 6\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv 7\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv 0\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +t+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +t+j}\right)=\{\begin{array}{ll}6\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 12\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 3\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

From the above labeling we have the following subcases:

Subcase (i): If $t=1$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,4,5,6,7,8,9,10,11,12,13,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=2\hfill \end{array}$

$e_f\left(i\right)=\lfloor \frac{n}{15}\rfloor$ ; $0\le i\le 14$ .

Subcase (ii): If $t=2$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,4,5,6,7,9,10,11,12,13,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{2,8}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6,7,8,9,10,11,12,13,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=1\hfill \end{array}$

Subcase (iii): If $t=3$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,4,5,6,7,9,10,12,13,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{2,8,11}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6,7,8,9,10,11,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,13}\hfill \end{array}$

Subcase (iv): If $t=4$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,4,5,6,7,9,10,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{2,8,11,13}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,2,3,4,5,6,7,9,10,11,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,13,8}\hfill \end{array}$

Subcase (v): If $t=5$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,4,5,6,7,9,10,11,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{2,8,11,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,4,5,6,7,9,10,11,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,8,13}\hfill \end{array}$

Subcase (vi): If $t=6$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,1,3,5,6,7,9,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{2,4,8,11,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,4,5,6,7,9,10,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,8,11,13}\hfill \end{array}$

Subcase (vii): If $t=7$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,7,9,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,8,11,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,7,9,10,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,8,11,13}\hfill \end{array}$

Subcase (viii): If $t=8$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,9,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,8,11,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,9,10,12,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,8,11,13}\hfill \end{array}$

Therefore, $P_n$ is a 15-product cordial graph if $n=15\lfloor \frac{n}{15}\rfloor +t$ where

$1\le t\le 8$ .

Case (ii): If $n=15\lfloor \frac{n}{15}\rfloor +t$ , where $9\le t\le 12$ then

$f\left(v_i\right)=\mathrm{0<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le \lfloor \frac{n}{15}\rfloor \mathrm{.}$

For $i=3\lfloor \frac{n}{15}\rfloor +1-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{3\lfloor \frac{n}{15}\rfloor +1-j}\right)=\{\begin{array}{ll}10\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +9-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 8\lfloor \frac{n}{15}\rfloor +8$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +9-j}\right)=\{\begin{array}{ll}2\hfill & \text{for}j\equiv 1\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv 2\left(\mathrm{mod}8\right)\hfill \\ 11\hfill & \text{for}j\equiv 3\left(\mathrm{mod}8\right)\hfill \\ 13\hfill & \text{for}j\equiv 4\left(\mathrm{mod}8\right)\hfill \\ 14\hfill & \text{for}j\equiv 5\left(\mathrm{mod}8\right)\hfill \\ 4\hfill & \text{for}j\equiv 6\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv 7\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv 0\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +8+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{15}\rfloor +t-8$ where $9\le t\le 12$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +8+j}\right)=\{\begin{array}{ll}6\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 12\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 3\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

From the above labeling we have the following subcases:

Subcase (i): If $t=9$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,9,10,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,8,11,12,13}\hfill \end{array}$

For $\lfloor \frac{n}{15}\rfloor$ is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,9,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,6,7,8,11,13,14}\hfill \end{array}$

For $\lfloor \frac{n}{15}\rfloor$ is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,8,9,11,13,14}\hfill \end{array}$

Subcase (ii): If $t=10$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,6,7,8,9,11,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,6,10,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,7,8,9,11,12,13}\hfill \end{array}$

Subcase (iii): If $t=11$ ,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,5,6,10,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,7,8,9,11,12,13}\hfill \end{array}$

For $\lfloor \frac{n}{15}\rfloor$ is even, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,3,5,10}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,4,6,7,8,9,11,12,13,14}\hfill \end{array}$

For $\lfloor \frac{n}{15}\rfloor$ is odd, $v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,5,10,12}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,6,7,8,9,11,13,14}\hfill \end{array}$

Subcase (iv): If $t=12$ ,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,5,10}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,6,7,8,9,11,12,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{0,5,10,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{1,2,3,4,6,7,8,9,11,12,13}\hfill \end{array}$

Therefore, $P_n$ is a 15-product cordial graph if $n=15\lfloor \frac{n}{15}\rfloor +t$ where

$9\le t\le 12$ .

Case (iii): If $n=15\lfloor \frac{n}{15}\rfloor +13$ , then

$f\left(v_i\right)=\mathrm{0<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le \lfloor \frac{n}{15}\rfloor +1.$

For $i=3\lfloor \frac{n}{15}\rfloor +2-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{3\lfloor \frac{n}{15}\rfloor +2-j}\right)=\{\begin{array}{ll}10\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +10-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 8\lfloor \frac{n}{15}\rfloor +8$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +10-j}\right)=\{\begin{array}{ll}2\hfill & \text{for}j\equiv 1\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv 2\left(\mathrm{mod}8\right)\hfill \\ 11\hfill & \text{for}j\equiv 3\left(\mathrm{mod}8\right)\hfill \\ 13\hfill & \text{for}j\equiv 4\left(\mathrm{mod}8\right)\hfill \\ 14\hfill & \text{for}j\equiv 5\left(\mathrm{mod}8\right)\hfill \\ 4\hfill & \text{for}j\equiv 6\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv 7\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv 0\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +9+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{15}\rfloor +4$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +9+j}\right)=\{\begin{array}{ll}6\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 12\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 3\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

Therefore,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{5,10}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{0,1,2,3,4,6,7,8,9,11,12,13,14}\hfill \end{array}$

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{5,10,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{0,1,2,3,4,6,7,8,9,11,12,13}\hfill \end{array}$

Hence, $P_n$ is a 15-product cordial graph if $n=15\lfloor \frac{n}{15}\rfloor +13$ .

Case (iv): If $n=15\lfloor \frac{n}{15}\rfloor +14$ , then

$f\left(v_i\right)=\mathrm{0<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le \lfloor \frac{n}{15}\rfloor +1.$

For $i=3\lfloor \frac{n}{15}\rfloor +2-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{15}\rfloor +1$ ,

$f\left(v_{3\lfloor \frac{n}{15}\rfloor +2-j}\right)=\{\begin{array}{ll}10\hfill & \text{for}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{for}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +11-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 8\lfloor \frac{n}{15}\rfloor +8$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +11-j}\right)=\{\begin{array}{ll}2\hfill & \text{for}j\equiv 1\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv 2\left(\mathrm{mod}8\right)\hfill \\ 11\hfill & \text{for}j\equiv 3\left(\mathrm{mod}8\right)\hfill \\ 13\hfill & \text{for}j\equiv 4\left(\mathrm{mod}8\right)\hfill \\ 14\hfill & \text{for}j\equiv 5\left(\mathrm{mod}8\right)\hfill \\ 4\hfill & \text{for}j\equiv 6\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv 7\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv 0\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +10+j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{15}\rfloor +4$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +10+j}\right)=\{\begin{array}{ll}6\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 12\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 3\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

Therefore,

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=\mathrm{5,14}\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{0,1,2,3,4,6,7,8,9,10,11,12,13}\hfill \end{array}$

For $\lfloor \frac{n}{15}\rfloor$ is even,

$v_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=5\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{0,1,2,3,4,6,7,8,9,10,11,12,13,14}\hfill \end{array}$

For $\lfloor \frac{n}{15}\rfloor$ is odd,

$\mathrm{<mi>\; </mi>}{v}_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=10\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{0,1,2,3,4,5,6,7,8,9,11,12,13,14}\hfill \end{array}$

Hence, $P_n$ is a 15-product cordial graph if $n=15\lfloor \frac{n}{15}\rfloor +14$ .

Case (v): If $n=15\lfloor \frac{n}{15}\rfloor$ , then

$f\left(v_i\right)=\mathrm{0<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le i\le \lfloor \frac{n}{15}\rfloor \mathrm{.}$

For $i=3\lfloor \frac{n}{15}\rfloor +1-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 2\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{3\lfloor \frac{n}{15}\rfloor +1-j}\right)=\{\begin{array}{ll}10\hfill & \text{if}j\equiv \mathrm{1,0}\left(\mathrm{mod}4\right)\hfill \\ 5\hfill & \text{if}j\equiv \mathrm{2,3}\left(\mathrm{mod}4\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +1-j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 8\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +1-j}\right)=\{\begin{array}{ll}2\hfill & \text{for}j\equiv 1\left(\mathrm{mod}8\right)\hfill \\ 8\hfill & \text{for}j\equiv 2\left(\mathrm{mod}8\right)\hfill \\ 11\hfill & \text{for}j\equiv 3\left(\mathrm{mod}8\right)\hfill \\ 13\hfill & \text{for}j\equiv 4\left(\mathrm{mod}8\right)\hfill \\ 14\hfill & \text{for}j\equiv 5\left(\mathrm{mod}8\right)\hfill \\ 4\hfill & \text{for}j\equiv 6\left(\mathrm{mod}8\right)\hfill \\ 1\hfill & \text{for}j\equiv 7\left(\mathrm{mod}8\right)\hfill \\ 7\hfill & \text{for}j\equiv 0\left(\mathrm{mod}8\right)\hfill \end{array}$

For $i=11\lfloor \frac{n}{15}\rfloor +j\mathrm{<mi>\; </mi><mo>;</mo><mi>\; </mi>1}\le j\le 4\lfloor \frac{n}{15}\rfloor$ ,

$f\left(v_{11\lfloor \frac{n}{15}\rfloor +j}\right)=\{\begin{array}{ll}6\hfill & \text{for}j\equiv \mathrm{1,6}\left(\mathrm{mod}8\right)\hfill \\ 9\hfill & \text{for}j\equiv \mathrm{2,5}\left(\mathrm{mod}8\right)\hfill \\ 12\hfill & \text{for}j\equiv \mathrm{3,0}\left(\mathrm{mod}8\right)\hfill \\ 3\hfill & \text{for}j\equiv \mathrm{4,7}\left(\mathrm{mod}8\right)\hfill \end{array}$

Therefore,

$v_f\left(i\right)=\lfloor \frac{n}{15}\rfloor$ ; $0\le i\le 14$ .

$e_f\left(i\right)=\{\begin{array}{ll}\lfloor \frac{n}{15}\rfloor \hfill & \text{if}i=14\hfill \\ \lfloor \frac{n}{15}\rfloor +1\hfill & \text{if}i=\mathrm{0,1,2,3,4,5,6,7,8,9,10,11,12,13}\hfill \end{array}$

Hence, $P_n$ is a 15-product cordial graph if $n=15\lfloor \frac{n}{15}\rfloor$ . □

Figure 6 shows the 15-product cordial labeling of $P_{16}$ .

Figure 6. 15-product cordial labeling of P₁₆.

Remark 10. [12] The path $P_n$ is 4-product cordial if and only if $n\le 11$ .

Lemma 11. The path $P_{2p^2}$ does not admit p²-product cordial labeling for every odd prime p.

Proof. Suppose that f is a p²-product cordial labeling of $P_{2p^2}$ . Then $v_f\left(i\right)=2$ for ($i=\mathrm{0,1,2,3,}\cdots \mathrm{,}{p}^2-1$ ) and $e_f\left(i\right)=2$ or 1 for ($i=\mathrm{0,1,2,3,}\cdots \mathrm{,}{p}^2-1$ ). Obviously, $v_f\left(0\right)=2$ and 0 must be assigned consecutively at the beginning or end of the path or beginning and end of the path together. Otherwise $e_f\left(0\right)>2$ , which is not possible. Thus, $e_f\left(0\right)=2$ . Now $v_f\left(ip\right)=2$ for ($i=\mathrm{1,2,3,}\cdots \mathrm{,}p-1$ ) and each vertex labels $ip$ must be labeled inconsecutively, otherwise $e_f\left(0\right)>2$ , which is not possible. These vertex labels with the other vertex labels produce the

edge labels $ip$ and ($i=\mathrm{1,2,3,}\cdots \mathrm{,}p-1$ ), so ${\displaystyle \sum _{i=1}^{p-1}{e}_f\left(ip\right)}\ge 4p-6$ and using the pigeonhole principle, we get at least i from the set $\left\{p\mathrm{,2}p\mathrm{,}\cdots \mathrm{,}\left(p-1\right)p\right\}$ such that $e_f\left(i\right)>2$ which contradicts that f is a p²-product cordial labeling of $P_{2p^2}$ .

As a consequence of Theorems 7, 8, 9, Remark 10 and Lemma 11 we propose the following conjecture:

Conjecture 12. For all $n\ge 3$ , the path $P_n$ is k-product cordial graph if k is the product of two distinct prime numbers.

Acknowledgements

We would like to thank the referees for their valuable suggestions to improve the quality of the paper.