1. Introduction

ojdm

Open Journal of Discrete Mathematics

2161-7635 2161-7643

Scientific Research Publishing

10.4236/ojdm.2025.151001

ojdm-138493

Articles

Physics Mathematics

k-Product Cordial Labeling of Path Graphs

Robinson Santrin

Sabibha

¹ Kruz Jeya

Daisy

² Pon

Jeyanthi

³ Maged Zakaria

Youssef

⁴ ⁵

aDepartment of Science and Humanities, Vins Christian College of Engineering, Nagercoil, India

aDepartment of Mathematics, Holy Cross College, Nagercoil, India

aResearch Centre, Department of Mathematics, Govindammal Aditanar College for Women, Tiruchendur, India

aDepartment of Mathematics and Statistics, College of Science, Imam Mohammad Ibn Saud Islamic University, Riyadh, Saudi Arabia

aDepartment of Mathematics, Faculty of Science, Ain Shams University, Cairo, Egypt

26 12 2024

15 01 1 29 8, August 2024 23, August 2024 23, December 2024

2014

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

In 2012, Ponraj et al. defined a concept of k-product cordial labeling as follows: Let f be a map from

V (G)

{0,1, \dots, k - 1}

where k is an integer,

1 \leq k \leq | V (G) |

. For each edge

u v

assign the label

f (u) f (v) (\mod k)

. f is called a k-product cordial labeling if

| v_{f} (i) - v_{f} (j) | \leq 1

, and

| e_{f} (i) - e_{f} (j) | \leq 1

i, j \in {0,1, \dots, k - 1}

, where

v_{f} (x)

and

e_{f} (x)

denote the number of vertices and edges respectively labeled with x (

x = 0,1, \dots, 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.

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

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 ${0,1}$ and for each edge $x y$ assign the label $| f (x) - f (y) |$ . 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 $V (G)$ to ${0,1}$ . For each edge $u v$ , assign the label $f (u) f (v)$ . Then f is called product cordial labeling if $| v_{f} (0) - v_{f} (1) | \leq 1$ and $| e_{f} (0) - e_{f} (1) | \leq 1$ where $v_{f} (i)$ and $e_{f} (i)$ denotes the number of vertices and edges respectively labeled with $i (i = 0,1)$ . 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, $K_{2} + m K_{1}$ , $K_{2, n}$ and $K_{n}^{c} + 2 K_{2}$ . 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 $k \geq 2$ 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 $⌊ x ⌋$ stands for the largest integer less than or equal to x and $⌈ x ⌉$ stands for the smallest integer greater than or equal to x.

Definition 1. An integer $p >1$ is called a prime number if 1 and p are the only divisor of p.

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

3. Main Results

In this section, we study the k-product cordial labeling of path graph $P_{n}$ where $k = 5,6,7,10,11$ 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 $V (P_{n}) = {v_{i};1 \leq i \leq n}$ and $E (P_{n}) = {(v_{i}, v_{i + 1});1 \leq i \leq n - 1}$ respectively.

Theorem 3. For $n \geq 3$ , the path $P_{n}$ is 5-product cordial.

Proof. Define $f : V (P_{n}) \to {0,1,2,3,4}$ as follows:

$f (v_{i}) = 0$ ; $1 \leq i \leq ⌊ \frac{n}{5} ⌋$ .

For $i = ⌊ \frac{n}{5} ⌋ + j;1 \leq j \leq n - ⌊ \frac{n}{5} ⌋$ ,

$f (v_{i}) = {\begin{array}{l} 4 & if j \equiv 1, 6 (\mod 8) \\ 1 & if j \equiv 2, 5 (\mod 8) \\ 2 & if j \equiv 3, 7 (\mod 8) \\ 3 & if j \equiv 4, 0 (\mod 8) \end{array}$

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

Case (i): If $n \equiv 1 (\mod 5)$ ,

$e_{f} (i) = ⌊ \frac{n}{5} ⌋$ for $0 \leq i \leq 4$ .

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,2,3,4 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 1 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,1,2,3 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 4 \end{array}$

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

Case (ii): If $n \equiv 2 (\mod 5)$ ,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,2,3 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 1,4 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,1,2,3 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 4 \end{array}$

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

Case (iii): If $n \equiv 3 (\mod 5)$ ,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,3 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 1,2,4 \end{array}$

For n is even, $e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,1,2 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 3,4 \end{array}$

For n is odd, $e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,1,3 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 2,4 \end{array}$

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

Case (iv): If $n \equiv 4 (\mod 5)$ ,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 1,2,3,4 \end{array}$

For n is even, $e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,3 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 1,2,4 \end{array}$

For n is odd, $e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{5} ⌋ & if i = 0,2 \\ ⌊ \frac{n}{5} ⌋ + 1 & if i = 1,3,4 \end{array}$

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

Case (v): If $n \equiv 0 (\mod 5)$ ,

$v_{f} (i) = \frac{n}{5}$ for $0 \leq i \leq 4$ .

For n is even, $e_{f} (i) = {\begin{array}{l} \frac{n}{5} - 1 & if i = 2 \\ \frac{n}{5} & if i = 0,1,3,4 \end{array}$

For n is odd, $e_{f} (i) = {\begin{array}{l} \frac{n}{5} - 1 & if i = 3 \\ \frac{n}{5} & if i = 0,1,2,4 \end{array}$

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

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

Figure 1 Figure 1. 5-product cordial labeling of P12.

Theorem 4. For $n \geq 3$ , the path $P_{n}$ is 7-product cordial.

Proof. Define $f : V (P_{n}) \to {0,1,2,3,4,5,6}$ as follows:

$f (v_{i}) = 0$ ; $1 \leq i \leq ⌊ \frac{n}{7} ⌋$ .

For $i = ⌊ \frac{n}{7} ⌋ + j; 1 \leq j \leq n - ⌊ \frac{n}{7} ⌋$ ,

$f (v_{i}) = {\begin{array}{l} 6 & if j \equiv 1,8 (\mod 12) \\ 1 & if j \equiv 2,7 (\mod 12) \\ 2 & if j \equiv 3,10 (\mod 12) \\ 5 & if j \equiv 4,9 (\mod 12) \\ 3 & if j \equiv 5,0 (\mod 12) \\ 4 & if j \equiv 6,11 (\mod 12) \end{array}$

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

Case (i): If $n \equiv 1 (\mod 7)$ ,

$e_{f} (i) = ⌊ \frac{n}{7} ⌋$ for $0 \leq i \leq 6$ .

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,2,3,4,5,6 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,1,2,3,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 6 \end{array}$

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

Case (ii): If $n \equiv 2 (\mod 7)$ .

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,2,3,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,6 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,1,2,3,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 6 \end{array}$

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

Case (iii): If $n \equiv 3 (\mod 7)$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,1,3,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 2,6 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,3,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,6 \end{array}$

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,2,3,4 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,5,6 \end{array}$

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

Case (iv): If $n \equiv 4 (\mod 7)$ ,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,3,4 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,5,6 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,1,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 2,3,6 \end{array}$

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

Case (v): If $n \equiv 5 (\mod 7)$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,4,5 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,3,6 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,4 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,3,5,6 \end{array}$

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,3 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,4,5,6 \end{array}$

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

Case (vi): If $n \equiv 6 (\mod 7)$ ,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,3,4,5,6 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{7} ⌋ & if i = 0,4 \\ ⌊ \frac{n}{7} ⌋ + 1 & if i = 1,2,3,5,6 \end{array}$

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

Case (vii): If $n \equiv 0 (\mod 7)$ ,

$v_{f} (i) = \frac{n}{7}$ for $0 \leq i \leq 6$ , $e_{f} (i) = {\begin{array}{l} \frac{n}{7} - 1 & if i = 4 \\ \frac{n}{7} & if i = 0,1,2,3,5,6 \end{array}$

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

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

Figure 2 Figure 2. 7-product cordial labeling of P11.

Theorem 5. For $n \geq 3$ , the path $P_{n}$ is 11-product cordial.

Proof. Define $f : V (P_{n}) \to {0,1,2, \dots,10}$ as follows:

$f (v_{i}) = 0$ ; $1 \leq i \leq ⌊ \frac{n}{11} ⌋$ .

For $i = ⌊ \frac{n}{11} ⌋ + j;1 \leq j \leq n - ⌊ \frac{n}{11} ⌋$ ,

$f (v_{i}) = {\begin{array}{l} 10 & if j \equiv 1,12 (\mod 20) \\ 1 & if j \equiv 2,11 (\mod 20) \\ 9 & if j \equiv 3,14 (\mod 20) \\ 2 & if j \equiv 4,13 (\mod 20) \\ 7 & if j \equiv 5,16 (\mod 20) \\ 4 & if j \equiv 6,15 (\mod 20) \\ 3 & if j \equiv 7,18 (\mod 20) \\ 8 & if j \equiv 8,17 (\mod 20) \\ 6 & if j \equiv 9,0 (\mod 20) \\ 5 & if j \equiv 10,19 (\mod 20) \end{array}$

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

Case (i): If $n \equiv 1 (\mod 11)$ ,

$e_{f} (i) = ⌊ \frac{n}{11} ⌋$ for $0 \leq i \leq 10$ .

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,2,3,4,5,6,7,8,9,10 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,1,2,3,4,5,6,7,8,9 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 10 \end{array}$

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

Case (ii): If $n \equiv 2 (\mod 11)$ .

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,2,3,4,5,6,7,8,9 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,10 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,1,2,3,4,5,6,7,8,9 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 10 \end{array}$

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

Case (iii): If $n \equiv 3 (\mod 11)$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,1,2,3,4,5,6,7,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 9,10 \end{array}$

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,3,4,5,6,7,8,9 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,10 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,2,3,4,5,6,7,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,9,10 \end{array}$

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

Case (iv): If $n \equiv 4 (\mod 11)$ .

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,3,4,5,6,7,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,9,10 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,1,2,3,4,5,6,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 7,9,10 \end{array}$

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

Case (v): If $n \equiv 5 (\mod 11)$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,1,2,4,5,6,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 3,7,9,10 \end{array}$

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,3,5,6,7,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,4,9,10 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,3,4,5,6,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,7,9,10 \end{array}$

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

Case (vi): If $n \equiv 6 (\mod 11)$ .

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,3,5,6,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,4,7,9,10 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,1,2,4,5,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 3,6,7,9,10 \end{array}$

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

Case (vii): If $n \equiv 7 (\mod 11)$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,2,4,5,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,3,6,7,9,10 \end{array}$

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,3,5,6 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,4,7,8,9,10 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,5,6,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,4,7,9,10 \end{array}$

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

Case (viii): If $n \equiv 8 (\mod 11)$ .

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,5,6 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,4,7,8,9,10 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,4,5,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,6,7,9,10 \end{array}$

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

Case (ix): If $n \equiv 9 (\mod 11)$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,5,8 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,4,6,7,9,10 \end{array}$

For n is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,6 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,4,5,7,8,9,10 \end{array}$

For n is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,5 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,4,6,7,8,9,10 \end{array}$

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

Case (x): If $n \equiv 10 (\mod 11)$ .

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0 \\ ⌊ \frac{n}{11} ⌋ + 1 & if 1 \leq i \leq 10 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{11} ⌋ & if i = 0,5 \\ ⌊ \frac{n}{11} ⌋ + 1 & if i = 1,2,3,4,6,7,8,9,10 \end{array}$

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

Case (xi): If $n \equiv 0 (\mod 11)$ .

$v_{f} (i) = \frac{n}{11}$ for $0 \leq i \leq 10$ .

$e_{f} (i) = {\begin{array}{l} \frac{n}{11} - 1 & if i = 5 \\ \frac{n}{11} & if i = 0,1,2,3,4,6,7,8,9,10 \end{array}$

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

An example of 11-product cordial labeling of $P_{8}$ is shown in Figure 3 .

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 $n \geq 3$ , the path $P_{n}$ is k-product cordial graph if k is prime.

Theorem 7. For $n \geq 3$ , the path $P_{n}$ is 6-product cordial.

Proof. Define $f : V (P_{n}) \to {0,1,2,3,4,5}$ .

We have the following six cases.

Case (i): If $n \equiv 1 (\mod 6)$ ,

$f (v_{i}) = {\begin{array}{l} 0 & if 1 \leq i \leq ⌊ \frac{n}{6} ⌋ \\ 3 & if ⌊ \frac{n}{6} ⌋ + 1 \leq i \leq 2 ⌊ \frac{n}{6} ⌋ \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 2 - j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋ + 1$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 2 - j}) = {\begin{array}{l} 1 & if j \equiv 1,0 (\mod 4) \\ 5 & if j \equiv 2,3 (\mod 4) \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 1 + j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 1 + j}) = {\begin{array}{l} 4 & if j \equiv 1,0 (\mod 4) \\ 2 & if j \equiv 2,3 (\mod 4) \end{array}$

Therefore,

$e_{f} (i) = ⌊ \frac{n}{6} ⌋$ for $0 \leq i \leq 5$ .

For $⌊ \frac{n}{6} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,1,2,3,4 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 5 \end{array}$

For $⌊ \frac{n}{6} ⌋$ is even,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,2,3,4,5 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 1 \end{array}$

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

Case (ii): If $n \equiv 2 (\mod 6)$ ,

$f (v_{i}) = {\begin{array}{l} 0 & for 1 \leq i \leq ⌊ \frac{n}{6} ⌋ \\ 3 & for ⌊ \frac{n}{6} ⌋ + 1 \leq i \leq 2 ⌊ \frac{n}{6} ⌋ \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 3 - j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋ + 2$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 3 - j}) = {\begin{array}{l} 1 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 2 + j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 2 + j}) = {\begin{array}{l} 4 & for j \equiv 1,0 (\mod 4) \\ 2 & for j \equiv 2,3 (\mod 4) \end{array}$

Therefore,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,2,3,4 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 1,5 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,1,2,3,4 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 5 \end{array}$

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

Case (iii): If $n \equiv 3 (\mod 6)$ .

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

We label the vertices $v_{i} (1 \leq i \leq n - 1)$ as in Case (ii), then assign 2 to $v_{n}$ .

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

We label the vertices $v_{i} (1 \leq i \leq n - 1)$ as in Case (ii), then assign 4 to $v_{n}$ .

Therefore,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,1,2,3 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 4,5 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is even,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,2,3 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 1,4,5 \end{array}$

For $⌊ \frac{n}{6} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,3,4 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 1,2,5 \end{array}$

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

Case (iv): If $n \equiv 4 (\mod 6)$ .

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

We label the vertices $v_{i} (1 \leq i \leq n - 1)$ as in Case (iii) Subcase (i), then assign 4 to $v_{n}$ .

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

We label the vertices $v_{i} (1 \leq i \leq n - 1)$ as in Case (iii) Subcase (ii), then assign 2 to $v_{n}$ .

Therefore,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,3 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 1,2,4,5 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,1,3 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 2,4,5 \end{array}$

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

Case (v): If $n \equiv 5 (\mod 6)$ ,

$f (v_{i}) = {\begin{array}{l} 0 & for 1 \leq i \leq ⌊ \frac{n}{6} ⌋ \\ 3 & for ⌊ \frac{n}{6} ⌋ + 1 \leq i \leq 2 ⌊ \frac{n}{6} ⌋ + 1 \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 4 - j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋ + 2$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 4 - j}) = {\begin{array}{l} 1 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 3 + j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋ + 2$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 3 + j}) = {\begin{array}{l} 4 & for j \equiv 1,0 (\mod 4) \\ 2 & for j \equiv 2,3 (\mod 4) \end{array}$

Therefore,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 1,2,3,4,5 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,1 \\ ⌊ \frac{n}{6} ⌋ + 1 & for i = 2,3,4,5 \end{array}$

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

Case (vi): If $n \equiv 0 (\mod 6)$ ,

$f (v_{i}) = {\begin{array}{l} 0 & if 1 \leq i \leq ⌊ \frac{n}{6} ⌋ \\ 3 & if ⌊ \frac{n}{6} ⌋ + 1 \leq i \leq 2 ⌊ \frac{n}{6} ⌋ \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + 1 - j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + 1 - j}) = {\begin{array}{l} 1 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 4 ⌊ \frac{n}{6} ⌋ + j;1 \leq j \leq 2 ⌊ \frac{n}{6} ⌋$ ,

$f (v_{4 ⌊ \frac{n}{6} ⌋ + j}) = {\begin{array}{l} 4 & for j \equiv 1,0 (\mod 4) \\ 2 & for j \equiv 2,3 (\mod 4) \end{array}$

Therefore,

$v_{f} (i) = ⌊ \frac{n}{6} ⌋$ for $0 \leq i \leq 5$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{6} ⌋ & for i = 0,2,3,4,5 \\ ⌊ \frac{n}{6} ⌋ - 1 & for i = 1 \end{array}$

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

Figure 4 shows the 6-product cordial labeling of $P_{9}$ .

Figure 4 Figure 4. 6-product cordial labeling of P9.

Theorem 8. For $n \geq 3$ , the path $P_{n}$ is 10-product cordial.

Proof. Define $f : V (P_{n}) \to {0,1,2,3,4,5,6,7,8,9}$ .

We have the following four cases.

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

$f (v_{i}) = {\begin{array}{l} 0 & if 1 \leq i \leq ⌊ \frac{n}{10} ⌋ \\ 5 & if ⌊ \frac{n}{10} ⌋ + 1 \leq i \leq 2 ⌊ \frac{n}{10} ⌋ \end{array}$

For $i = 6 ⌊ \frac{n}{10} ⌋ + 1 + t - j;1 \leq j \leq 4 ⌊ \frac{n}{10} ⌋ + t$ where $t = 1,2,3,4$ ,

$f (v_{6 ⌊ \frac{n}{10} ⌋ + 1 + t - j}) = {\begin{array}{l} 3 & for j \equiv 1,6 (\mod 8) \\ 7 & for j \equiv 2,5 (\mod 8) \\ 9 & for j \equiv 3,7 (\mod 8) \\ 1 & for j \equiv 4,0 (\mod 8) \end{array}$

For $i = 6 ⌊ \frac{n}{10} ⌋ + t + j;1 \leq j \leq 4 ⌊ \frac{n}{10} ⌋$ ,

$f (v_{6 ⌊ \frac{n}{10} ⌋ + t + j}) = {\begin{array}{l} 4 & for j \equiv 1,6 (\mod 8) \\ 6 & for j \equiv 2,5 (\mod 8) \\ 8 & for j \equiv 3,0 (\mod 8) \\ 2 & for j \equiv 4,7 (\mod 8) \end{array}$

From the above labeling we have the following subcases:

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

$e_{f} (i) = ⌊ \frac{n}{10} ⌋$ for $0 \leq i \leq 9$ .

For $⌊ \frac{n}{10} ⌋$ is even,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,1,2,4,5,6,7,8,9 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 3 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,1,2,3,4,5,6,8,9 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 7 \end{array}$

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

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,3,4,5,6,7,8,9 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1 \end{array}$

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,1,2,4,5,6,8,9 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 3,7 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,1,2,4,5,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 3,7,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is even,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,4,5,6,7,8,9 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,3 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,3,4,5,6,8,9 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,7 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,1,2,4,5,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,3,7,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is even,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,4,5,6,7,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,3,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,3,4,5,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,7,9 \end{array}$

Hence, $P_{n}$ is a 10-product cordial graph if $n = 10 ⌊ \frac{n}{10} ⌋ + t$ where

$t = 1,2,3,4$ .

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

$f (v_{i}) = {\begin{array}{l} 0 & for 1 \leq i \leq ⌊ \frac{n}{10} ⌋ \\ 5 & for ⌊ \frac{n}{10} ⌋ + 1 \leq i \leq 2 ⌊ \frac{n}{10} ⌋ + 1 \end{array}$

For $i = 6 ⌊ \frac{n}{10} ⌋ + 6 - j;1 \leq j \leq 4 ⌊ \frac{n}{10} ⌋ + 4$ ,

$f (v_{6 ⌊ \frac{n}{10} ⌋ + 6 - j}) = {\begin{array}{l} 3 & for j \equiv 1,6 (\mod 8) \\ 7 & for j \equiv 2,5 (\mod 8) \\ 9 & for j \equiv 3,7 (\mod 8) \\ 1 & for j \equiv 4,0 (\mod 8) \end{array}$

For $i = 6 ⌊ \frac{n}{10} ⌋ + 5 + j;1 \leq j \leq 4 ⌊ \frac{n}{10} ⌋$ ,

$f (v_{6 ⌊ \frac{n}{10} ⌋ + 5 + j}) = {\begin{array}{l} 4 & for j \equiv 1,6 (\mod 8) \\ 6 & for j \equiv 2,5 (\mod 8) \\ 8 & for j \equiv 3,0 (\mod 8) \\ 2 & for j \equiv 4,7 (\mod 8) \end{array}$

Therefore, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,4,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,3,5,7,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is even,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,4,6,7,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,3,5,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 0,2,3,4,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 1,5,7,9 \end{array}$

Hence, $P_{n}$ is a 10-product cordial graph if $n = 10 ⌊ \frac{n}{10} ⌋ + 5$ .

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

$f (v_{i}) = {\begin{array}{l} 0 & if 1 \leq i \leq ⌊ \frac{n}{10} ⌋ + 1 \\ 5 & if ⌊ \frac{n}{10} ⌋ + 2 \leq i \leq 2 ⌊ \frac{n}{10} ⌋ + 2 \end{array}$

For $i = 6 ⌊ \frac{n}{10} ⌋ + 7 - j;1 \leq j \leq 4 ⌊ \frac{n}{10} ⌋ + 4$ ,

$f (v_{6 ⌊ \frac{n}{10} ⌋ + 7 - j}) = {\begin{array}{l} 3 & for j \equiv 1,6 (\mod 8) \\ 7 & for j \equiv 2,5 (\mod 8) \\ 9 & for j \equiv 3,7 (\mod 8) \\ 1 & for j \equiv 4,0 (\mod 8) \end{array}$

For $i = 6 ⌊ \frac{n}{10} ⌋ + 6 + j;1 \leq j \leq 4 ⌊ \frac{n}{10} ⌋ + t - 6$ where $t = 6,7,8,9$ ,

$f (v_{6 ⌊ \frac{n}{10} ⌋ + 6 + j}) = {\begin{array}{l} 4 & for j \equiv 1,6 (\mod 8) \\ 6 & for j \equiv 2,5 (\mod 8) \\ 8 & for j \equiv 3,0 (\mod 8) \\ 2 & for j \equiv 4,7 (\mod 8) \end{array}$

From the above labeling we have the following subcases:

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2,4,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,3,5,7,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is even,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2,4,6,7,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,3,5,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2,3,4,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,5,7,9 \end{array}$

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

For $⌊ \frac{n}{10} ⌋$ is even,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,3,4,5,7,9 \end{array}$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 4,6,7,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & f o r i = 0,1,2,3,5,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2,4,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,3,5,6,7,9 \end{array}$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 3,4,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,5,7,9 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,3,4,5,6,7,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is even,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 6,7,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,3,4,5,9 \end{array}$

For $⌊ \frac{n}{10} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 3,6,8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,4,5,7,9 \end{array}$

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

For $⌊ \frac{n}{10} ⌋$ is even,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 2 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,3,4,5,6,7,8,9 \end{array}$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 6,7 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,3,4,5,8,9 \end{array}$ .

For $⌊ \frac{n}{10} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 8 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,3,4,5,6,7,9 \end{array}$ ,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 3,6 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,4,5,7,8,9 \end{array}$ .

Hence, $P_{n}$ is a 10-product cordial graph if $n = 10 ⌊ \frac{n}{10} ⌋ + t$ where

$t = 6,7,8,9$ .

Case (iv): If $n = 10 ⌊ \frac{n}{10} ⌋$ .

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

We label the vertices $v_{i} (1 \leq i \leq n - 1)$ as in Case (iii), then assign 8 to $v_{n}$ .

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

We label the vertices $v_{i} (1 \leq i \leq n - 1)$ as in Case (iii), then assign 2 to $v_{n}$ .

From this label we get,

$v_{f} (i) = ⌊ \frac{n}{10} ⌋$ for $0 \leq i \leq 9$ .

For $⌊ \frac{n}{10} ⌋$ is odd,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ - 1 & for i = 7 \\ ⌊ \frac{n}{10} ⌋ & for i = 0,1,2,3,4,5,6,8,9 \end{array}$ .

For $⌊ \frac{n}{10} ⌋$ is even,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{10} ⌋ & for i = 3 \\ ⌊ \frac{n}{10} ⌋ + 1 & for i = 0,1,2,4,5,6,7,8,9 \end{array}$ .

Hence, $P_{n}$ is a 10-product cordial graph if $n = 10 ⌊ \frac{n}{10} ⌋$ . □

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

Figure 5 Figure 5. 10-product cordial labeling of P10.

Theorem 9 For $n \geq 3$ , the path $P_{n}$ is 15-product cordial.

Proof. Define $f : V (P_{n}) \to {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14}$ .

We have the following five cases.

Case (i): If $n = 15 ⌊ \frac{n}{15} ⌋ + t$ where $1 \leq t \leq 8$ then

$f (v_{i}) = 0;1 \leq i \leq ⌊ \frac{n}{15} ⌋ .$

For $t = 1,2,3,5$ and $i = 3 ⌊ \frac{n}{15} ⌋ + 1 - j;1 \leq j \leq 2 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{3 ⌊ \frac{n}{15} ⌋ + 1 - j}) = {\begin{array}{l} 5 & for j \equiv 1,0 (\mod 4) \\ 10 & for j \equiv 2,3 (\mod 4) \end{array}$

For $t = 4,6,7,8$ and $i = 3 ⌊ \frac{n}{15} ⌋ + 1 - j;1 \leq j \leq 2 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{3 ⌊ \frac{n}{15} ⌋ + 1 - j}) = {\begin{array}{l} 10 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 1 + t - j;1 \leq j \leq 8 ⌊ \frac{n}{15} ⌋ + t$ where $1 \leq t \leq 8$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 1 + t - j}) = {\begin{array}{l} 2 & for j \equiv 1 (\mod 8) \\ 8 & for j \equiv 2 (\mod 8) \\ 11 & for j \equiv 3 (\mod 8) \\ 13 & for j \equiv 4 (\mod 8) \\ 14 & for j \equiv 5 (\mod 8) \\ 4 & for j \equiv 6 (\mod 8) \\ 1 & for j \equiv 7 (\mod 8) \\ 7 & for j \equiv 0 (\mod 8) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + t + j;1 \leq j \leq 4 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + t + j}) = {\begin{array}{l} 6 & for j \equiv 1,6 (\mod 8) \\ 9 & for j \equiv 2,5 (\mod 8) \\ 12 & for j \equiv 3,0 (\mod 8) \\ 3 & for j \equiv 4,7 (\mod 8) \end{array}$

From the above labeling we have the following subcases:

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,1,3,4,5,6,7,8,9,10,11,12,13,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 2 \end{array}$

$e_{f} (i) = ⌊ \frac{n}{15} ⌋$ ; $0 \leq i \leq 14$ .

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,1,3,4,5,6,7,9,10,11,12,13,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 2,8 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,2,3,4,5,6,7,8,9,10,11,12,13,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,1,3,4,5,6,7,9,10,12,13,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 2,8,11 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,2,3,4,5,6,7,8,9,10,11,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,13 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,1,3,4,5,6,7,9,10,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 2,8,11,13 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,2,3,4,5,6,7,9,10,11,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,13,8 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,1,3,4,5,6,7,9,10,11,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 2,8,11,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,4,5,6,7,9,10,11,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,8,13 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,1,3,5,6,7,9,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 2,4,8,11,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,4,5,6,7,9,10,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,8,11,13 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,7,9,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,8,11,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,7,9,10,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,8,11,13 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,9,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,7,8,11,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,9,10,12,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,7,8,11,13 \end{array}$

Therefore, $P_{n}$ is a 15-product cordial graph if $n = 15 ⌊ \frac{n}{15} ⌋ + t$ where

$1 \leq t \leq 8$ .

Case (ii): If $n = 15 ⌊ \frac{n}{15} ⌋ + t$ , where $9 \leq t \leq 12$ then

$f (v_{i}) = 0;1 \leq i \leq ⌊ \frac{n}{15} ⌋ .$

For $i = 3 ⌊ \frac{n}{15} ⌋ + 1 - j;1 \leq j \leq 2 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{3 ⌊ \frac{n}{15} ⌋ + 1 - j}) = {\begin{array}{l} 10 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 9 - j;1 \leq j \leq 8 ⌊ \frac{n}{15} ⌋ + 8$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 9 - j}) = {\begin{array}{l} 2 & for j \equiv 1 (\mod 8) \\ 8 & for j \equiv 2 (\mod 8) \\ 11 & for j \equiv 3 (\mod 8) \\ 13 & for j \equiv 4 (\mod 8) \\ 14 & for j \equiv 5 (\mod 8) \\ 4 & for j \equiv 6 (\mod 8) \\ 1 & for j \equiv 7 (\mod 8) \\ 7 & for j \equiv 0 (\mod 8) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 8 + j;1 \leq j \leq 4 ⌊ \frac{n}{15} ⌋ + t - 8$ where $9 \leq t \leq 12$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 8 + j}) = {\begin{array}{l} 6 & for j \equiv 1,6 (\mod 8) \\ 9 & for j \equiv 2,5 (\mod 8) \\ 12 & for j \equiv 3,0 (\mod 8) \\ 3 & for j \equiv 4,7 (\mod 8) \end{array}$

From the above labeling we have the following subcases:

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

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,9,10,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,7,8,11,12,13 \end{array}$

For $⌊ \frac{n}{15} ⌋$ is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,9,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,6,7,8,11,13,14 \end{array}$

For $⌊ \frac{n}{15} ⌋$ is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,7,8,9,11,13,14 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,6,7,8,9,11,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,6,10,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,7,8,9,11,12,13 \end{array}$

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

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,5,6,10,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,3,4,7,8,9,11,12,13 \end{array}$

For $⌊ \frac{n}{15} ⌋$ is even, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,3,5,10 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,4,6,7,8,9,11,12,13,14 \end{array}$

For $⌊ \frac{n}{15} ⌋$ is odd, $v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,5,10,12 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,3,4,6,7,8,9,11,13,14 \end{array}$

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

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,5,10 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,3,4,6,7,8,9,11,12,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 0,5,10,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 1,2,3,4,6,7,8,9,11,12,13 \end{array}$

Therefore, $P_{n}$ is a 15-product cordial graph if $n = 15 ⌊ \frac{n}{15} ⌋ + t$ where

$9 \leq t \leq 12$ .

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

$f (v_{i}) = 0;1 \leq i \leq ⌊ \frac{n}{15} ⌋ + 1.$

For $i = 3 ⌊ \frac{n}{15} ⌋ + 2 - j;1 \leq j \leq 2 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{3 ⌊ \frac{n}{15} ⌋ + 2 - j}) = {\begin{array}{l} 10 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 10 - j;1 \leq j \leq 8 ⌊ \frac{n}{15} ⌋ + 8$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 10 - j}) = {\begin{array}{l} 2 & for j \equiv 1 (\mod 8) \\ 8 & for j \equiv 2 (\mod 8) \\ 11 & for j \equiv 3 (\mod 8) \\ 13 & for j \equiv 4 (\mod 8) \\ 14 & for j \equiv 5 (\mod 8) \\ 4 & for j \equiv 6 (\mod 8) \\ 1 & for j \equiv 7 (\mod 8) \\ 7 & for j \equiv 0 (\mod 8) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 9 + j;1 \leq j \leq 4 ⌊ \frac{n}{15} ⌋ + 4$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 9 + j}) = {\begin{array}{l} 6 & for j \equiv 1,6 (\mod 8) \\ 9 & for j \equiv 2,5 (\mod 8) \\ 12 & for j \equiv 3,0 (\mod 8) \\ 3 & for j \equiv 4,7 (\mod 8) \end{array}$

Therefore,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 5,10 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 0,1,2,3,4,6,7,8,9,11,12,13,14 \end{array}$

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 5,10,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 0,1,2,3,4,6,7,8,9,11,12,13 \end{array}$

Hence, $P_{n}$ is a 15-product cordial graph if $n = 15 ⌊ \frac{n}{15} ⌋ + 13$ .

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

$f (v_{i}) = 0;1 \leq i \leq ⌊ \frac{n}{15} ⌋ + 1.$

For $i = 3 ⌊ \frac{n}{15} ⌋ + 2 - j;1 \leq j \leq 2 ⌊ \frac{n}{15} ⌋ + 1$ ,

$f (v_{3 ⌊ \frac{n}{15} ⌋ + 2 - j}) = {\begin{array}{l} 10 & for j \equiv 1,0 (\mod 4) \\ 5 & for j \equiv 2,3 (\mod 4) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 11 - j;1 \leq j \leq 8 ⌊ \frac{n}{15} ⌋ + 8$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 11 - j}) = {\begin{array}{l} 2 & for j \equiv 1 (\mod 8) \\ 8 & for j \equiv 2 (\mod 8) \\ 11 & for j \equiv 3 (\mod 8) \\ 13 & for j \equiv 4 (\mod 8) \\ 14 & for j \equiv 5 (\mod 8) \\ 4 & for j \equiv 6 (\mod 8) \\ 1 & for j \equiv 7 (\mod 8) \\ 7 & for j \equiv 0 (\mod 8) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 10 + j;1 \leq j \leq 4 ⌊ \frac{n}{15} ⌋ + 4$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 10 + j}) = {\begin{array}{l} 6 & for j \equiv 1,6 (\mod 8) \\ 9 & for j \equiv 2,5 (\mod 8) \\ 12 & for j \equiv 3,0 (\mod 8) \\ 3 & for j \equiv 4,7 (\mod 8) \end{array}$

Therefore,

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 5,14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 0,1,2,3,4,6,7,8,9,10,11,12,13 \end{array}$

For $⌊ \frac{n}{15} ⌋$ is even,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 5 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 0,1,2,3,4,6,7,8,9,10,11,12,13,14 \end{array}$

For $⌊ \frac{n}{15} ⌋$ is odd,

$v_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 10 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 0,1,2,3,4,5,6,7,8,9,11,12,13,14 \end{array}$

Hence, $P_{n}$ is a 15-product cordial graph if $n = 15 ⌊ \frac{n}{15} ⌋ + 14$ .

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

$f (v_{i}) = 0;1 \leq i \leq ⌊ \frac{n}{15} ⌋ .$

For $i = 3 ⌊ \frac{n}{15} ⌋ + 1 - j;1 \leq j \leq 2 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{3 ⌊ \frac{n}{15} ⌋ + 1 - j}) = {\begin{array}{l} 10 & if j \equiv 1,0 (\mod 4) \\ 5 & if j \equiv 2,3 (\mod 4) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + 1 - j;1 \leq j \leq 8 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + 1 - j}) = {\begin{array}{l} 2 & for j \equiv 1 (\mod 8) \\ 8 & for j \equiv 2 (\mod 8) \\ 11 & for j \equiv 3 (\mod 8) \\ 13 & for j \equiv 4 (\mod 8) \\ 14 & for j \equiv 5 (\mod 8) \\ 4 & for j \equiv 6 (\mod 8) \\ 1 & for j \equiv 7 (\mod 8) \\ 7 & for j \equiv 0 (\mod 8) \end{array}$

For $i = 11 ⌊ \frac{n}{15} ⌋ + j;1 \leq j \leq 4 ⌊ \frac{n}{15} ⌋$ ,

$f (v_{11 ⌊ \frac{n}{15} ⌋ + j}) = {\begin{array}{l} 6 & for j \equiv 1,6 (\mod 8) \\ 9 & for j \equiv 2,5 (\mod 8) \\ 12 & for j \equiv 3,0 (\mod 8) \\ 3 & for j \equiv 4,7 (\mod 8) \end{array}$

Therefore,

$v_{f} (i) = ⌊ \frac{n}{15} ⌋$ ; $0 \leq i \leq 14$ .

$e_{f} (i) = {\begin{array}{l} ⌊ \frac{n}{15} ⌋ & if i = 14 \\ ⌊ \frac{n}{15} ⌋ + 1 & if i = 0,1,2,3,4,5,6,7,8,9,10,11,12,13 \end{array}$

Hence, $P_{n}$ is a 15-product cordial graph if $n = 15 ⌊ \frac{n}{15} ⌋$ . □

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

Figure 6 Figure 6. 15-product cordial labeling of P16.

Remark 10. [12] The path $P_{n}$ is 4-product cordial if and only if $n \leq 11$ .

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

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

edge labels $i p$ and ( $i = 1,2,3, \dots, p - 1$ ), so $\sum_{i = 1}^{p - 1} e_{f} (i p) \geq 4 p - 6$ and using the pigeonhole principle, we get at least i from the set ${p,2 p, \dots, (p - 1) p}$ such that $e_{f} (i) > 2$ which contradicts that f is a p²-product cordial labeling of $P_{2 p^{2}}$ .

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

Conjecture 12. For all $n \geq 3$ , 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.

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