Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 1 https://internationalpubls.com On (α, β)-Total Edge Irregularity Strength of Some Graphs P. Pandiaraj1, a, K. Muthugurupackiam2*, S. Anbalagan3, R. Gurusamy4, I. Muthuselvam5 1 Part time Research scholar, Rajah Serfoji Government College (Autonomous), Affiliated to Bharathidasan University,Tiruchirappalli, Tamil Nadu, India. 2 Department of Mathematics, Government Arts and Science College, Srivilliputtur – 626 125, Tamil Nadu, India. 3 Department of Mathematics, Raja serfoji Government College (Autonomous), Thanjavur – 613 005, Tamil Nadu, India. 4 Department of Mathematics, Mepco Schlenk Engineering College, Sivakasi - 626 005, Tamil Nadu, India. 5 Department of Mathematics, Kalasalingam Academy of Research and Education, Krishnankoil – 626 126, Tamil Nadu,India. a Department of Mathematics, Kamaraj College of Engineering and Technology, Virudhunagar - 625 701, Tamil Nadu, India Email: 1pandiaraj0@gmail.com, 2gurupackiam@yahoo.com, 3sms.anbu18@gmail.com, 4sahama2010@gmail.com, 5selvam08mm11@gmail.com Article History: Received: 11-11-2024 Revised:24-12-2024 Accepted:09-01-2025 Abstract: Let 𝑉(𝐺), and 𝐸(𝐺) having cardinality 𝑙 and 𝑚 respectively, symbolize the vertex set, and edge set of a simple graph 𝐺 together with a total ℏ– labeling ℓ: ℳ → {1, 2, 3, … , ℏ } where ℳ = 𝑉(𝐺) ∪ 𝐸(𝐺). Let ℂ = {𝛼, 𝛼 + 𝛽, 𝛼 + 2𝛽, … , 𝛼 + (𝑚 − 1)𝛽}. If an invertible function exists, say ℱ: 𝐸(𝐺) → ℂ specified by ℱ(𝒾𝒿) = ℓ(𝒾) + ℓ(𝒿) + ℓ(𝒾𝒿) for every edge 𝒾𝒿 belongs to 𝐸(𝐺) with 𝛼 ≥ 3, 𝛽 ≥ 2, then ℓ is termed as (𝛼, 𝛽)– total edge irregular labeling. Also, the value ℱ(𝒾𝒿) is said to be the edge weight of 𝒾𝒿. The least ℏ for which 𝐺 admits (𝛼, 𝛽) – edge irregular ℏ is indicated by (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺), known as (𝛼, 𝛽) −Total Edge irregularity Strength of the graph 𝐺. The precise value of (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺) for standard graph families and Mycielskian path graphs is explored in this work. In addition, an open problem of (3, 2) – 𝑡𝑒𝑠(𝑃𝑛 𝑘) and (3, 2) – 𝑡𝑒𝑠(𝑇) are solved partially. Keywords: (𝛼, 𝛽)- Irregular labeling, Edge irregular labeling, Irregular labeling, Irregularity strength, Total edge irregular labeling. 1. Introduction This paper considers finite simple undirected graphs. Under certain conditions, graph labeling refers to assigning numerical values to graph components like vertices, edges, or both. These specifications described by means of some evaluating function values (weights). Chartrand et al. [1] initially proposed irregular graph labeling as follows: Let 𝐺 = (𝑉, 𝐸) be a connected graph of order at least 3 having edge ℏ– labeling ℊ ∶ 𝐸 → {1, 2, . . . , ℏ} and 𝓌(𝑣) = ∑ ℊ(𝔢)𝓋∈𝔢 , where 𝓌(𝑣) denotes vertex 𝓋 weight. When the vertex weights for all vertices differ, this is termed as irregular labeling. The irregularity strength 𝑠(𝐺) is the least natural number ℏ for which 𝐺 reveals irregular graph labeling. Baca et al. [2] reshaped irregular labeling as total vertex irregular labeling and total edge irregular labeling. The weight of an edge 𝑒 = 𝓈𝓉 ∈ 𝐸 is defined as 𝓌(𝑒) = 𝒻(𝓈) + 𝒻(𝓉) + 𝒻(𝑒) where 𝒻 ∶ 𝔒 → 𝔔 is a function with 𝔒 = 𝑉 ∪ 𝐸 and 𝔔 = {𝜏 ∶ 1 ≤ 𝜏 ≤ 𝓂 and 𝜏𝜖ℕ}. If every one of the edge weights are distinct, 𝒻 is referred to as a Total Edge Irregular labeling. The Total Edge irregularity Strength 𝑡𝑒𝑠(𝐺) is the least positive integer 𝓂 for which total edge irregular labeling 𝒻 ∶ 𝔒 → 𝔔 exists. For a centralized uniform theta graphs, Riyan [3] computed the total edge irregularity Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 2 https://internationalpubls.com strength (TES). Susanti Y, et al. [4] predicted asymmetric graphs and symmetric graphs 𝑡𝑒𝑠(𝐺). Rosyida I, et al. [5] determined the exact value of 𝑡𝑒𝑠(𝐺) for n-uniform cactus chain graphs. Muthu Guru Packiam [6] introduced the concept of (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺) as follows: Consider a simple graph 𝐺 = (𝑉, 𝐸) on 𝑙 vertices and 𝑚 edges together with a total ℎ – labeling 𝓅 or 𝜌: 𝑉(𝐺) ∪ 𝐸(𝐺) → {1, 2, 3, … , ℎ}. Then 𝜌 is called (𝛼, 𝛽)–total edge irregular labeling if there exists a one-to-one correspondence, say 𝜓: 𝐸(𝐺) → {𝛼, 𝛼 + 𝛽, 𝛼 + 2𝛽, … + 𝛼 + (𝑚 − 1)𝛽} defined by 𝜓(𝑢𝑣) = 𝜌(𝑢) + 𝜌(𝑣) + 𝜌(𝑢𝑣) for all 𝑢𝑣 ∈ 𝐸(𝐺), where 𝛼 ≥ 3, 𝛽 ≥ 2. Also, the value 𝜓(𝑢𝑣) is said to be the edge weight of 𝑢𝑣. The least ℎ for which 𝐺 admits (𝛼, 𝛽) – edge irregular ℎ is indicated by (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺), known as (𝛼, 𝛽) −Total Edge irregularity Strength of the graph 𝐺. Additionally, they presented upper and lower limitations for the parameter and established the evaluations of (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺) for some graph families. The purpose of this paper is to focus on the study of (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺). Subdivision of a graph G that results from dividing each edge by a vertex [7]. In a graph G, attaching two pendant vertices to each all vertices of G is denoted by G ∘ S2. Middle graph of a graph 𝐺 = (𝕍, 𝔼) denoted by 𝑀(𝐺) has vertex set 𝕍 ∪ 𝔼 and a pair of vertices are adjacent if and only if they are incident with one other or adjacent edges. Any two adjacent vertices of 𝐺 are adjacent in the middle graph of 𝐺 is named as Total graph 𝑇(𝐺) of a graph 𝐺. The vertices of path graph 𝑃 𝑛 are 𝓋1, 𝓋2, … , 𝓋𝑛. Graph 𝑃 𝑛 3 can be constructed from 𝑃 𝑛 by connecting every 𝓋𝑖 𝑡ℎ vertex to 𝓋𝑖+2 𝑡ℎ vertex and every 𝓋𝑖 𝑡ℎvertex to the 𝓋𝑖+3 𝑡ℎ vertex by an edge. If all unordered pairs of 𝑛 vertices are adjacent in 𝐺 is termed as complete graph [8] denoted by 𝐾𝑛. If a vertex set is separated like 𝕌 and 𝕍 with |𝕌| = 𝑎 and |𝕍|= 𝑏 such that every vertex of 𝕌 is adjacent to every vertex of 𝕍 is referred as 𝐾𝑎,𝑏- Complete bipartite graph. In case 𝑎 = 1 𝑜𝑟 b = 1 in 𝐾𝑎,𝑏 is named as star graph 𝐾1,𝑏 𝑜𝑟 𝐾𝑎,1. The central vertex of two copies 𝐾1,𝓃 is joined by an edge is 𝐵(𝓃, 𝓃) - Bistar graph. Mycielskian [9] µ(𝐺) of a graph 𝐺 = (𝕍, 𝔼) has vertex set 𝕍 ∪ 𝕍 ′ ∪ {𝑢}, where 𝕍 ′ = {𝓋𝑖′ ∶ 𝓋𝑖 ∈ 𝕍 } and edge set 𝔼 ∪ {𝓋𝑖𝓋𝑗′ ∶ 𝓋𝑖𝓋𝑗 ∈ 𝔼} ∪ {𝓋𝑖′𝑢 ∶ 𝓋𝑖′ ∈ 𝕍 ′}. Proposition 1 [6]: Let 𝐺 be a graph with 𝑞 edges and 𝑝 vertices. Then ⌈ 𝛼+(𝑞−1)𝛽 3 ⌉ ≤ (𝛼, 𝛽) − 𝑡𝑒𝑠(𝐺) ≤ 𝛼 − 2 + (𝑞 − 1)𝛽 for any two natural number 𝛼 ≥ 3 and 𝛽 ≥ 2. Proposition 2 [10]: (3, 2) − 𝑡𝑒𝑠(𝑃 𝑛) = ⌈ 2𝑛−1 3 ⌉ for path 𝑃𝑛. 2. Results and Discussion Theorem 1: (3, 2)– 𝑡𝑒𝑠(𝑆(𝑃 𝑛 ∘ 𝑆2)) = 4𝑛 − 1. Proof. Consider the index sets 𝕀 = {1, 2, … , 𝑛} and 𝒥 = {1, 2, … , 𝑛 − 1}. Let the vertex set of 𝑆(𝑃 𝑛 ∘ 𝑆2) = ⋃ 𝕍𝜎 6 𝜎=1 where 𝑉1 = {𝑢𝔦}, 𝑉2 = {𝑣𝔦}, 𝑉3 = {𝑤𝔦}, 𝑉4 = {𝑥𝒿}, 𝑉5 = {𝑦𝔦} and 𝑉6 = {𝑧𝔦} and edge set of (𝑆(𝑃 𝑛 ∘ 𝑆2)) = ⋃ 𝔼𝜑 6 𝜑=1 where 𝔼1 = {𝑢𝔦𝑥𝔦 }, 𝔼2 = {𝑥𝒿𝑢𝒿+1 }, 𝔼3 = {𝑢𝔦𝑦𝔦 }, 𝔼4 = {𝑦𝔦𝑣𝔦 }, 𝔼5 = {𝑢𝔦𝑧𝔦} and 𝔼6 = {𝑧𝔦𝑤𝔦 } for every 𝔦 𝜖 𝕀 and 𝒿 𝜖 𝒥 . Total labeling 𝜌: ℑ → ℵ where ℑ = 𝑉(𝑆(𝑃 𝑛 ∘ 𝑆2)) ∪ 𝐸(𝑆(𝑃 𝑛 ∘ 𝑆2)) and ℵ = {1, 2, … ,4𝑛 − 1} is given below: 𝜌(𝑢1) = 3, 𝜌(𝑢𝑖) = 4𝑖 − 3, 𝜌(𝑣𝑖) = 4𝑖 − 1, 𝑖 ≠ 1. 𝜌(𝑣1) = 1, 𝜌(𝑤𝑖) = 4𝑖 − 1, 𝜌(𝑦𝑖) = 4𝑖 − 3, 𝜌(𝑧𝑖) = 4𝑖 − 1, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 3 https://internationalpubls.com 𝜌(𝑥𝒿) = 4𝒿 + 1, 𝜌(𝑢1𝑥1) = 3, 𝜌(𝑢𝔦𝑥𝔦) = 1 + 4𝔦, 𝔦 ≠ 1. 𝜌(𝑥𝔦𝑢𝔦+1) = −1 + 4𝔦, 𝜌(𝑢𝔦𝑦𝔦) = 𝜌(𝑦𝔦𝑣𝔦) = −3 + 4𝑖, 𝜌(𝑢1𝑧1) = 1, 𝜌(𝑢𝑖𝑧𝑖) = 4𝑖 − 1, 𝜌(𝑧𝑖𝑤𝑖) = −1 + 4𝑖, 𝑖 ≠ 1. Induced edge weight function 𝜓: 𝐸(𝑆(𝑃 𝑛 ∘ 𝑆2)) → {3, 5, 7, … ,12𝑛 − 3} is given by 𝜓(𝑢𝔦𝑥𝔦) = −1 + 12𝔦, 𝜓(𝑥𝔦𝑢𝔦+1) = 1 + 12𝔦, 𝜓(𝑢𝑖𝑦𝑖) = 12𝑖 − 9, 𝜓(𝑦𝑖𝑣𝑖) = 12𝑖 − 7, 𝑖 ≠ 1 𝜓(𝑢1𝑦1) = 5, 𝜓(𝑦1𝑣1) = 3, 𝜓(𝑢𝑖𝑧𝑖) = 12𝑖 − 5, 𝜓(𝑧𝑖𝑤𝑖) = 12𝑖 − 3, Therefore, the induced edge weights of 𝑆(𝑃 𝑛 ∘ 𝑆2) differs by 2 in the arithmetic progression. Thus, (3, 2)– 𝑡𝑒𝑠(𝑆(𝑃 𝑛 ∘ 𝑆2)) ≤ −1 + 4𝑛. Proposition 1 confirms that(3, 2)– 𝑡𝑒𝑠(𝑆(𝑃 𝑛 ∘ 𝑆2)) ≥ −1 + 4𝑛, this concludes the proof. Example 1: (3, 2) total edge irregular labeling for 𝑆(𝑃 5 ∘ 𝑆2) is shown in Fig 1. Figure 1. (3, 2)– 𝑡𝑒𝑠(𝑆(𝑃 5 ∘ 𝑆2)) = 19. Remark 1: When 𝑛 = 1, 𝑃 1 ∘ 𝑆2 ≅ 𝑃 5 and hence it follows from the proposition 2. Requisition Theorem 2: (3, 2)– 𝑡𝑒𝑠(𝑀(𝑃 𝑛)) = 2𝑛 − 2. Proof. Consider the index sets 𝕀 = {1, 2, … , 𝑛} and 𝒥 = {1, 2, … , 𝑛 − 1}. Let 𝑉(𝑀(𝑃 𝑛)) = 𝑉1 ∪ 𝑉2 where 𝑉1 = {𝑢𝒾} and 𝑉2 = {𝑣𝒿} and let 𝐸(𝑀(𝑃 𝑛)) = {𝑣ℓ𝑣ℓ+1} ∪ {𝑢𝓀𝑣𝓀 } ∪ {𝑣𝓀𝑢𝓀+1 } denotes the vertex set and edge set of 𝑀(𝑃 𝑛) for all 𝑖 𝜖 𝕀, 𝒿 , 𝓀 𝜖 𝒥 and ℓ𝜖 𝕀 − {𝑛 − 1, 𝑛}. Total labeling 𝜌: 𝒜 → ℬ where 𝒜 = 𝑉(𝑀(𝑃 𝑛)) ∪ 𝐸(𝑀(𝑃 𝑛)) and ℬ = 𝕀 ⋃{𝑛 + 1, 𝑛 + 2, … ,2𝑛 − 2} is prescribed by: 𝜌(𝑣1) = 1, 𝜌(𝑣2) = 3, 𝜌(𝑣𝑖) = 2𝑖, 𝑖 = 3,4,5, … , 𝑛 𝜌(𝑢1) = 1, 𝜌(𝑢2) = 3,𝜌(𝑢𝔦) = −2 + 2𝔦, 3 ≤ 𝔦 ≤ 𝑛 𝜌(𝑣1𝑣2) = 3, 𝜌(𝑣2𝑣3) = 4, 𝜌(𝑣𝑖𝑣𝑖+1) = −1 + 2𝑖, 𝑖 = 3,4,5, … , 𝑛 − 2 𝜌(𝑢𝒿𝑣𝒿) = −1 + 2𝒿 , 𝜌(𝑣1𝑢2) = 1, 𝜌(𝑣2𝑢3) = 4, 𝜌(𝑣𝔦𝑢𝔦+1) = −1 + 2𝔦, 𝔦 = 3,4,5, … , 𝑛 − 1 Induced edge weight function 𝜓: 𝐸(𝑀(𝑃 𝑛)) → {3, 5, 7, … , 6𝑛 − 7} is given by Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 4 https://internationalpubls.com 𝜓(𝑣ℓ𝑣ℓ+1) = 1 + 6ℓ, ℓ𝜖 𝕀 − {𝑛 − 1, 𝑛} 𝜓(𝑢𝒿𝑣𝒿) = −3 + 6𝒿, 𝜓(𝑣𝒿𝑢𝒿+1) = −1 + 6𝒿 Therefore, the induced edge weights of 𝑆(𝑀(𝑃 𝑛)) differs by 2 in the arithmetic progression. Hence, (3, 2)– 𝑡𝑒𝑠(𝑀(𝑃 𝑛)) ≤ −2 + 2𝑛. Proposition 1 reveals that(3, 2)– 𝑡𝑒𝑠(𝑀(𝑃 𝑛)) ≥ −2 + 2𝑛. Hence the result follows. Example 2: (3, 2) − 𝑡𝑒𝑠(𝑀(𝑃8)is shown in Fig 2. Figure 2. (3, 2)– 𝑡𝑒𝑠(𝑀(𝑃8)) = 14. Theorem 3: (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃 𝑛)) = ⌈ 8𝑛 3 ⌉ − 3 for every positive integer 𝑛 at least 3. Proof. Consider the index sets 𝕀 = {1, 2, … , 𝑛} and 𝒥 = {1, 2, … , 𝑛 − 1}. Let 𝑉(𝑇(𝑃 𝑛)) = 𝔸1 ∪ 𝕍2 where 𝔸1 = {𝔞𝒾} and 𝕍2 = {𝑣𝒿} and 𝐸(𝑇(𝑃 𝑛)) = {𝔞𝒿𝔞𝒿+1 } ∪ {𝑣𝓉𝑣𝓉+1 } ∪ {𝔞𝒿𝑣𝒿} ∪ {𝑣𝒿𝔞𝒿+1} denotes the vertex set and edge set of 𝑇(𝑃 𝑛) for all 𝑖 𝜖 𝕀, 𝒿 𝜖 𝒥 and 𝓉 𝜖 𝒥 − {𝑛 − 1}. Total labeling 𝓅: 𝒞 → 𝒟 where 𝒞 = 𝑉(𝑇(𝑃 𝑛)) ∪ 𝐸(𝑇(𝑃 𝑛)) and 𝒟 = {1, 2, … , ⌈ 8𝑛 3 ⌉ − 3} is represented by: Case i Suppose 𝑛 ≡ 0 (𝑚𝑜𝑑 3) 𝓅(𝑣1) = 1, 𝜌(𝑣2) = 5, 𝓅(𝑣3𝔦) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1) = 3 + 8𝔦, 𝓅(𝑣3𝔦+2) = 5 + 8𝔦, 𝓅( 𝔞1) = 1, 𝓅(𝔞2) = 3, 𝓅( 𝔞3𝔦) = −3 + 8𝔦, 𝓅( 𝔞3𝔦+1) = 8𝔦 + 1, 𝓅( 𝔞3𝔦+2) = 3 + 8𝔦, 𝓅(𝑣1𝑣2) = 3, 𝓅(𝑣3𝔦−1𝑣3𝔦) = 8𝔦 − 3, 𝓅(𝑣3𝔦𝑣3𝔦+1) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1𝑣3𝔦+2) = 8𝔦 + 1, 𝓅( 𝔞1 𝔞2) = 1, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 5 https://internationalpubls.com 𝓅( 𝔞3𝔦−1 𝔞3𝔦) = −3 + 8𝔦, 𝓅( 𝔞3𝔦 𝔞3𝔦+1) = −1 + 8𝔦, 𝓅( 𝔞3𝔦+1 𝔞3𝔦+2) = 8𝔦 + 1, 𝓅( 𝔞1𝑣1) = 1, 𝓅( 𝔞3𝔦𝑣3𝔦) = 𝓅( 𝔞3𝔦+1 𝑣3𝔦+1) = −1 + 8𝔦, 𝓅( 𝔞3𝔦−1 𝑣3𝔦−1) = −5 + 8𝔦, 𝓅(𝑣1 𝔞2) = 3, 𝓅(𝑣3𝔦−1 𝔞3𝔦) = −3 + 8𝔦, 𝓅(𝑣3𝔦 𝔞3𝔦+1) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1 𝔞3𝔦+2) = 8𝔦 + 1, Case ii Let 𝑛 ≡ 1 (𝑚𝑜𝑑 3) 𝓅(𝑣1) = 1, 𝓅(𝑣3𝔦−1) = 8𝔦 − 3, 𝓅(𝑣3𝔦) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1) = 3 + 8𝔦, 𝓅( 𝔞1) = 1, 𝓅( 𝔞2) = 3, 𝓅( 𝔞𝑛) = 8 3 (𝑛 − 1) 𝓅( 𝔞3𝔦) = −3 + 8𝔦, 𝓅( 𝔞3𝔦+1) = 8𝔦 + 1, 𝓅( 𝔞3𝔦+2) = 3 + 8𝔦, 𝓅(𝑣1𝑣2) = 3, 𝓅(𝑣3𝔦−1𝑣3𝔦) = −3 + 8𝔦, 𝓅(𝑣3𝔦𝑣3𝔦+1) = −1 + 8𝔦, 𝓅(𝑣3𝔦+1𝑣3𝔦+2) = 1 + 8𝔦, 𝓅( 𝔞1 𝔞2) = 1, 𝓅 ( 𝔞𝑛−1 𝔞𝑛) = 8 3 (𝑛 − 1) 𝓅( 𝔞3𝔦−1 𝔞3𝔦) = −3 + 8𝔦, 𝓅( 𝔞3𝔦 𝔞3𝔦+1) = 8𝔦 − 1, 𝓅( 𝔞3𝔦+1 𝔞3𝔦+2) = 1 + 8𝔦, 𝓅(𝑣1 𝔞2) = 3, 𝓅(𝑣3𝔦−1 𝔞3𝔦) = −3 + 8𝔦, 𝓅(𝑣𝑛−1 𝔞𝑛) = 8 3 (𝑛 − 1), 𝓅(𝑣3𝔦 𝔞3𝔦+1) = 8𝔦 − 1, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 6 https://internationalpubls.com 𝓅(𝑣3𝔦+1 𝔞3𝔦+2) = 1 + 8𝔦, 𝓅( 𝔞1𝑣1) = 1 𝓅( 𝔞3𝔦−1 𝑣3𝔦−1) = −5 + 8𝔦, 𝓅( 𝔞3𝔦 𝑣3𝔦) = −1 + 8𝔦, 𝓅( 𝔞3𝔦+1 𝑣3𝔦+1) = −1 + 8𝔦, Case iii Consider 𝑛 ≡ 2 (𝑚𝑜𝑑 3) 𝓅(𝑣1) = 1, 𝓅(𝑣3𝔦−1) = −3 + 8𝔦, 𝓅(𝑣3𝔦) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1) = 3 + 8𝔦, 𝓅( 𝔞1) = 1 𝓅( 𝔞3𝔦−1) = −5 + 8𝔦, 𝓅( 𝔞3𝔦) = −3 + 8𝔦, 𝓅( 𝔞3𝔦+1) = 8𝔦 + 1, 𝓅(𝑣1𝑣2) = 3, 𝓅(𝑣3𝔦−1 𝑣3𝔦) = −3 + 8𝔦, 𝓅(𝑣3𝔦 𝑣3𝔦+1) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1 𝑣3𝔦+2) = 8𝔦 + 1, 𝓅( 𝔞1 𝔞2) = 1 𝓅( 𝔞3𝔦−1 𝔞3𝔦) = −3 + 8𝔦, 𝓅( 𝔞3𝔦 𝔞3𝔦+1) = −1 + 8𝔦, 𝓅( 𝔞3𝔦+1 𝔞3𝔦+2) = 8𝔦 + 1, 𝓅( 𝔞1 𝑣1) = 1 𝓅( 𝔞3𝔦−1 𝑣3𝔦−1) = −5 + 8𝔦, 𝓅( 𝔞3𝔦 𝑣3𝔦) = 𝓅( 𝔞3𝔦+1 𝑣3𝔦+1) = 8𝔦 − 1, 𝓅(𝑣1 𝔞2) = 3, 𝓅(𝑣3𝔦−1 𝔞3𝔦) = −3 + 8𝔦, 𝓅(𝑣3𝔦 𝔞3𝔦+1) = 8𝔦 − 1, 𝓅(𝑣3𝔦+1 𝔞3𝔦+2) = 8𝔦 + 1 . Induced edge weight function 𝜓: 𝐸(𝑇(𝑃 𝑛)) → {3, 5, 7, … ,8𝑛 − 9} is given by 𝜓(𝔞𝒿𝔞𝒿+1) = −3 + 8𝒿, 𝜓(𝑣𝓉𝑣𝓉+1) = 1 + 8𝓉, 𝜓(𝔞𝒿𝑣𝒿) = −5 + 8𝒿, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 7 https://internationalpubls.com 𝜓(𝑣𝒿𝔞𝒿+1) = 8𝒿 − 1, Therefore, the induced edge weights of 𝑇(𝑃 𝑛)generates an arithmetic progression and it differs by 2 and hence (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃 𝑛)) ≤ ⌈ 8𝑛 3 ⌉ − 3. From Proposition 1, we get (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃 𝑛)) ≥ ⌈ 8𝑛 3 ⌉ − 3 and hence the result follows. Example 3: (3, 2) − 𝑡𝑒𝑠(𝑇(𝑃9)) is given in Fig 3. Figure 3. (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃9)) = 21. Remark 2: When 𝑛 = 2, 𝑇(𝑃 2) ≅ 𝐶3 and hence (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃2)) = 3. (3, 2) total edge irregular labeling for 𝑇(𝑃9) is shown in Fig 4. Figure 4. (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃2)) = 3. Theorem 4: (3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 ∘ 𝐾2) = ⌈ 8(𝑛+1) 3 ⌉ − 3. Proof. Consider the index sets 𝕀 = {1, 2, … , 𝑛} and 𝒥 = {1, 2, … , 𝑛 − 1}. Let 𝑉(𝑃 𝑛 ∘ 𝐾2) = ⋃ 𝑉𝑛 3 𝑛=1 where 𝑉1 = {𝔰𝑖}, 𝑉2 = {𝑣𝑖} and 𝑉3 = {𝔱𝑖} and let 𝐸(𝑃 𝑛 ∘ 𝐾2) = {𝔰𝒿𝔰𝒿+1 } ∪ {𝔰𝑖𝑣𝑖} ∪ {𝑣𝑖𝔱𝑖 } ∪ {𝔰𝑖𝔱𝑖 } denotes the vertex set and edge of 𝑃 𝑛 ∘ 𝐾2 respectively for all 𝑖 𝜖 𝕀, 𝒿 𝜖 𝒥. Total labeling 𝓅: ℒ → ℳwhere ℒ = 𝑉(𝑃 𝑛 ∘ 𝐾2) ∪ 𝐸(𝑃 𝑛 ∘ 𝐾2) and ℳ = {1, 2, … , ⌈ 8(𝑛+1) 3 ⌉ − 3} is described by: Case i Consider 𝑛 ≡ 0 (𝑚𝑜𝑑 3) Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 8 https://internationalpubls.com 𝓅( 𝔰𝑛) = 8𝑛 3 , 𝓅( 𝑣𝑛) = 8𝑛 3 − 3, 𝓅( 𝔱𝑛) = 8𝑛 3 𝓅( 𝔰3𝔦) = 1 + 8𝔦, 𝓅( 𝔰3𝔦−1) = −3 + 8𝔦, 𝓅( 𝔰3𝑖−2) = 8𝑖 − 5, 𝓅( 𝑣3𝔦) = −3 + 8𝔦, 𝓅( 𝑣3𝑖−1) = 8𝑖 − 5, 𝓅( 𝑣3𝑖−2) = 8𝑖 − 7, 𝓅( 𝔱1) = 1 𝓅( 𝔱3𝑖) = 8𝑖 + 1, 𝓅( 𝔱3𝔦−1) = −3 + 8𝔦, 𝓅( 𝔱3𝑖−2) = 8𝑖 − 5, 𝑖 ≠ 1 𝓅( 𝔰3𝔦−2 𝔰3𝔦−1) = −7 + 8𝔦, 𝓅( 𝔰3𝑖−1 𝔰3𝑖) = 8𝑖 − 5, 𝓅( 𝔰3𝑖 𝔰3𝑖+1) = 8𝑖 − 3, 𝓅( 𝔰1𝑣1) = 1 𝓅( 𝔰3𝑖−1𝑣3𝑖−1) = 8𝑖 − 5, 𝓅( 𝔰3𝑖𝑣3𝑖) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−2𝑣3𝑖−2) = 8𝑖 − 9, 𝑖 ≠ 1 𝓅(𝑣3𝔦−2𝔱3𝔦−2) = −7 + 8𝔦, 𝓅(𝑣3𝔦−1𝔱3𝔦−1) = −3 + 8𝔦, 𝓅(𝑣3𝑖𝔱3𝑖) = 8𝑖 − 1, 𝓅( 𝔰1𝔱1) = 3 𝓅( 𝔰3𝑖−1𝔱3𝑖−1) = 8𝑖 − 3, 𝓅( 𝔰3𝑖𝔱3𝑖) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−2𝔱3𝑖−2) = 8𝑖 − 7, 𝑖 ≠ 1 𝓅( 𝔰𝑛−1 𝔰𝑛) = 8𝑛 3 − 4, 𝓅( 𝔰𝑛 𝑣𝑛) = 8𝑛 3 − 2, 𝓅( 𝔰𝑛 𝔱𝑛) = 8𝑛 3 − 1, 𝓅( 𝑣𝑛 𝔱𝑛) = 8𝑛 3 Case ii If 𝑛 − 1 ≡ 0 (𝑚𝑜𝑑 3) 𝓅(𝑣3𝔦) = −3 + 8𝔦, 𝓅(𝑣3𝑖−1) = 8𝑖 − 5, 𝓅(𝑣3𝑖−2) = 8𝑖 − 7, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 9 https://internationalpubls.com 𝓅( 𝔰3𝔦) = 1 + 8𝔦, 𝓅( 𝔰3𝔦−1) = −3 + 8𝔦, 𝓅( 𝔰3𝔦−2) = 8𝑖 − 5, 𝓅( 𝔱1) = 1 𝓅( 𝔱3𝑖) = 8𝑖 + 1, 𝓅( 𝔱3𝔦−1) = −3 + 8𝔦, 𝓅( 𝔱3𝔦−2) = −5 + 8𝔦, 𝑖 ≠ 1 𝓅( 𝔰3𝑖−1 𝔰3𝑖) = 8𝑖 − 5, 𝓅( 𝔰3𝔦−2 𝔰3𝔦−1) = −7 + 8𝔦, 𝓅( 𝔰3𝑖 𝔰3𝑖+1) = 8𝑖 − 3, 𝓅( 𝔰1𝑣1) = 1 𝓅( 𝔰3𝑖−1 𝑣3𝑖−1) = 8𝑖 − 5, 𝓅( 𝔰3𝑖 𝑣3𝑖) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−2 𝑣3𝑖−2) = 8𝑖 − 9, 𝓅(𝑣3𝔦−2 𝔱3𝔦−2) = −7 + 8𝔦, 𝓅(𝑣3𝔦−1 𝔱3𝔦−1) = −3 + 8𝔦, 𝓅(𝑣3𝑖 𝔱3𝑖) = 8𝑖 − 1, 𝓅(𝔰1𝔱1) = 3 𝓅(𝔰3𝑖−1 𝔱3𝑖−1) = 8𝑖 − 3, 𝓅( 𝔰3𝑖 𝔱3𝑖) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−2 𝔱3𝑖−2) = 8𝑖 − 7, 𝑖 ≠ 1 Case iii If 𝑛 ≡ 2 (𝑚𝑜𝑑 3) 𝓅( 𝔰3𝑖−2) = 8𝑖 − 5, 𝓅( 𝔰3𝔦−1) = −3 + 8𝔦, 𝓅( 𝔰3𝔦) = 1 + 8𝔦, 𝓅(𝑣3𝔦−2) = −7 + 8𝔦, 𝓅(𝑣3𝑖−1) = 8𝑖 − 5, 𝓅(𝑣3𝔦) = −3 + 8𝔦, 𝓅( 𝔱1) = 1 𝓅( 𝔱3𝑖) = 8𝑖 + 1, 𝓅( 𝔱3𝔦−1) = −3 + 8𝔦, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 10 https://internationalpubls.com 𝓅( 𝔱3𝔦−2) = −5 + 8𝔦, 𝑖 ≠ 1 𝓅( 𝔰3𝑖 𝔰3𝑖+1) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−1 𝔰3𝑖) = 8𝑖 − 5, 𝓅( 𝔰3𝔦−2 𝔰3𝔦−1) = −7 + 8𝔦, 𝓅( 𝔰1𝑣1) = 1 𝓅( 𝔰3𝑖−1𝑣3𝑖−1) = 8𝑖 − 5, 𝓅( 𝔰3𝑖 𝑣3𝑖) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−2 𝑣3𝑖−2) = 8𝑖 − 9, 𝑖 ≠ 1 𝓅(𝑣3𝔦−2 𝔱3𝔦−2) = −7 + 8𝔦, 𝓅(𝑣3𝔦−1 𝔱3𝔦−1) = −3 + 8𝔦, 𝓅(𝑣3𝑖 𝔱3𝑖) = 8𝑖 − 1, 𝓅( 𝔰1 𝔱1) = 3 𝓅( 𝔰3𝑖−1 𝔱3𝑖−1) = 8𝑖 − 3, 𝓅( 𝔰3𝑖 𝔱3𝑖) = 8𝑖 − 3, 𝓅( 𝔰3𝑖−2 𝔱3𝑖−2) = 8𝑖 − 7, 𝑖 ≠ 1 Induced weight function for edge 𝜓: 𝐸(𝑃𝑛 ∘ 𝐾2) → {3, 5, 7, … , 8𝑛 − 1} is given by 𝜓(𝔰𝒿𝔰𝒿+1) = 1 + 8𝒿, 𝜓(𝔰1𝑣1) = 5 𝜓(𝔰𝔦𝑣𝔦) = −5 + 8𝔦, 𝑖 ≠ 1 𝜓(𝔰𝔦𝔱𝔦) = −1 + 8𝔦, 𝜓(𝑣1𝔱1) = 3 𝜓(𝑣𝔦𝔱𝔦) = −3 + 8𝔦, 𝑖 ≠ 1 Therefore, the induced edge weights of 𝑃 𝑛 ∘ 𝐾2differs by 2 in the arithmetic progression. Thus, (3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 ∘ 𝐾2) ≤ ⌈ 8(𝑛+1) 3 ⌉ − 3. Proposition 1 exhibits that(3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 ∘ 𝐾2) ≥ ⌈ 8(𝑛+1) 3 ⌉ − 3. Example 4: (3, 2) − 𝑡𝑒𝑠(𝑃 8 ∘ 𝐾2)is shown in Fig 5. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 11 https://internationalpubls.com Figure 5. (3, 2)– 𝑡𝑒𝑠(𝑃 8 ∘ 𝐾2) = 21. Theorem 5: (3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 3) = 2𝑛 − 3 for every positive integer 𝑛 more than 3. Proof. Consider the index set 𝕀 = {1, 2, … , 𝑛}. Let 𝑉(𝑃 𝑛 3) = {𝑣𝑖 } and let 𝐸(𝑃 𝑛 3) = ⋃ 𝐸𝑛 3 𝑛=1 where 𝐸1 = {𝑣𝑖𝑣𝑖+1 }, 𝐸2 = {𝑣𝑖𝑣𝑖+2} and 𝐸3 = {𝑣𝑖𝑣𝑖+3 } denotes the vertex set and edge set of 𝑃 𝑛 3 for all 𝑖 𝜖 𝕀. Total labeling 𝜌: 𝒫 → 𝒬 where 𝒫 = 𝑉(𝑃 𝑛 3) ∪ 𝐸(𝑃 𝑛 3) and 𝒬 = {1, 2, … ,2𝑛 − 3} is predicted by: 𝜌(𝑣1) = 1, 𝑓(𝑣2) = 1, 𝜌(𝑣𝔦) = 2𝔦 − 3, 𝑖 ≠ 1, 2. 𝜌(𝑣𝔦𝑣𝔦+1) = { 1 𝔦 = 1 −1 + 2𝔦 𝔦 ≠ 1 𝜌(𝑣𝔦𝑣𝔦+2) = { 1 𝑖 = 1 1 + 2𝔦 𝔦 ≠ 1 𝜌(𝑣𝔦𝑣𝔦+3) = { 3 𝔦 = 1 2𝑖 + 3 𝔦 ≠ 1 Induced edge weight function 𝜓: 𝐸(𝑃 𝑛 3) → {3, 5, 7, … ,6𝑛 − 11} is given by 𝜓(𝑣1𝑣2) = 3 𝜓(𝑣𝔦𝑣𝔦+1) = −5 + 6𝔦, 𝔦 ≠ 1 𝜓(𝑣𝔦𝑣𝔦+2) = −1 + 6𝔦, 𝜓(𝑣𝔦𝑣𝔦+3) = 3 + 6𝔦, Therefore, induced edge weights of 𝑃 𝑛 3differs by 2 in arithmetic progression. Thus, (3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 3) ≤ −3 + 2𝑛. By Proposition 1, we get (3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 3) ≥ −3 + 2𝑛. This concludes the proof. Example 5: (3, 2) − 𝑡𝑒𝑠(𝑃 7 3) is shown in Fig 6. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 12 https://internationalpubls.com Figure 6. (3, 2)– 𝑡𝑒𝑠(𝑃 7 3) = 11. Theorem 6: If 𝑇 is a tree other than star on 𝑝 vertices, then (3, 2)– 𝑡𝑒𝑠(𝑇) = ⌈ 2𝑝−1 3 ⌉. Proof. Step i: Arrange the given tree from left to right by considering one pendant as starting node. Step ii: Label the edges of tree from left to right by completing lower branch and then upper branch using the induced edge weights {3, 5, 7, … ,2𝑝 − 1}. Step iii: Each pair of vertices and its edge should be labeled by at most odd numbers which are less than or equal to either the corresponding induced edge weights or ⌈ 2𝑝−1 3 ⌉ with non-decreasing order. Step iv: From the above labeling, (3, 2)– 𝑡𝑒𝑠(𝑇) ≤ ⌈ 2𝑝−1 3 ⌉. Proposition 1exhibits that(3, 2)– 𝑡𝑒𝑠(𝑇) ≥ ⌈ 2𝑝−1 3 ⌉ and hence the result follows. Example 6: (3, 2) total edge irregular labeling for tree with 10 vertices is shown in Fig 7. Figure 7. (3, 2)– 𝑡𝑒𝑠(𝑇) = 7 Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 13 https://internationalpubls.com Theorem 7: (3, 2)– 𝑡𝑒𝑠(𝐾𝑛) = 𝑛2 − 5𝑛 + 9 for 𝑛 ≥ 4. Proof. Consider the index set 𝕀 = {1, 2, … , 𝑛}. Let 𝑉(𝐾𝑛) = {𝑢𝑖} and let 𝐸(𝐾𝑛) = {𝑢𝔦𝑢𝔦+1 } ∪ {𝑢𝔦𝑢𝑗/ 1 ≤ 𝔦 < 𝑗 ≤ 𝑛} denotes the vertex set and edge set of 𝐾𝑛 for all 𝑖 𝜖 𝕀. Total labeling 𝜌: 𝒫 → 𝒬 where 𝒫 = 𝑉(𝐾𝑛) ∪ 𝐸(𝐾𝑛) and 𝒬 = {1, 2, … , 𝑛2 − 5𝑛 + 9} is given below: 𝜌(𝑢1) = 1 𝜌(𝑢𝔦) = −3 + 2𝔦, 𝔦 ≠ 1 𝜌(𝑢1𝑢𝔦) = 1, 𝜌(𝑢𝔦𝑢𝑗) = 2𝑛𝔦 − 2𝑛 − (𝔦2 + 3𝔦 − 7), 𝔦 ≠ 1, 𝑛 − 1 , 𝔦 + 1 ≤ 𝑗 ≤ 𝑛, 𝜌(𝑢𝑛−1𝑢𝑛) = 𝑛2 − 5𝑛 + 9 Induced weight function for edge 𝜓: 𝐸(𝐾𝑛) → {3, 5, 7, … ,1 − 𝑛 + 𝑛2} is given by 𝜌(𝑢1𝑢𝔦) = −1 + 2𝔦, 𝜌(𝑢𝑛−1𝑢𝑛) = 1 − 𝑛 + 𝑛2 𝜌(𝑢𝔦𝑢𝑗) = 2𝑛𝔦 − 2𝑛 − (3 + 𝔦 − 𝔦2) − 2 − 2𝔦 + 2𝑗, 𝔦 ≠ 1, 𝑛 − 1 , 𝔦 + 1 ≤ 𝑗 ≤ 𝑛, The induced edge weights of 𝐾𝑛 differs by 2 in arithmetic progression. Hence, (3, 2)– 𝑡𝑒𝑠(𝐾𝑛) ≤ 9 − 5𝑛+𝑛2. Further it is not possible to obtain (3, 2)– total edge irregular labelling by assigning label fewer than 𝑛2 − 5𝑛 + 9, hence (3, 2)– 𝑡𝑒𝑠(𝐾𝑛) = 𝑛2 − 5𝑛 + 9. Example 7: (3, 2) − 𝑡𝑒𝑠(𝐾6) is given in Fig 8. Figure 8. (3, 2)– 𝑡𝑒𝑠(𝐾6) = 15. Remark 2: When 𝑛 = 3, (3, 2)– 𝑡𝑒𝑠(𝐾3) = 3, total edge irregular labelling for 𝐾3 is shown in Fig 9. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 14 https://internationalpubls.com Figure 9. (3, 2)– 𝑡𝑒𝑠(𝐾3) = 3. Corollary 1. If a simple connected graph 𝐺 has 𝑝 vertices and 𝑞 edges, then ⌈ 2𝑞+1 3 ⌉ ≤ (3, 2) − 𝑡𝑒𝑠(𝐺) ≤ 9 − 5𝑝 + 𝑝2. Theorem 8: (3, 2)– 𝑡𝑒𝑠(𝜇(𝑃𝑛)) = ⌊ 8𝑛 3 ⌋ − 1 for 𝑛 ≥ 3. Proof. Consider the index set 𝕀 = {1, 2, … , 𝑛}. The vertex set and edge set of 𝑃𝑛 are 𝒱1and ℰ1 respectively. Let 𝑉(𝜇(𝑃𝑛)) = ⋃ 𝒱𝜑 3 𝜑=1 and 𝐸(𝜇(𝑃𝑛)) = ⋃ ℰℴ 4 ℴ=1 denotes the vertex set and edge set of 𝜇(𝑃𝑛) where 𝒱1 = {𝔵𝑖}, 𝒱2 = {𝔵𝑖′ }, 𝒱3 = {𝑤}, ℰ1 = {𝔵𝔦𝔵𝔦+1 }, ℰ2 = {𝔵𝔦′𝔵𝔦+1}, ℰ3 = {𝔵𝔦𝔵𝔦+1′} and ℰ4 = {𝔵𝔦′𝑤} for all 𝑖 𝜖 𝕀. Total labeling 𝓅: 𝒳 → 𝒴 where 𝒳 = 𝑉(𝜇(𝑃𝑛)) ∪ 𝐸(𝜇(𝑃𝑛)) and 𝒴 = {1, 2, … , ⌊ 8𝑛 3 ⌋ − 1} is given below: Case i Suppose 𝑛 ≡ 0 (𝑚𝑜𝑑 3) 𝓅( 𝔵1) = 8𝑛 3 − 1, 𝓅( 𝔵3𝔦) = 1 + 2𝔦, 𝓅( 𝔵3𝔦−1) = 2𝔦 − 1, 𝓅( 𝔵3𝔦−2) = 2𝔦 − 1, 𝓅(𝔵′3𝔦) = 2𝑛 + 2𝑖 − 1, 𝓅( 𝔵′3𝔦−1) = 2𝑛 + 2𝔦 − 3, 𝓅( 𝔵′3𝔦−2) = 2𝑛 + 2𝔦 − 3, 𝓅( 𝔵3𝔦−1 𝔵3𝔦) = 𝓅( 𝔵3𝔦−2 𝔵3𝔦−1) = 2𝔦 − 1, 𝓅( 𝔵3𝔦 𝔵3𝔦+1) = 2𝔦 − 1, 𝓅( 𝔵3𝔦−2 𝔵′3𝔦−1) = −7 + 8𝔦, 𝓅( 𝔵3𝔦−1 𝔵′3𝔦−2) = 𝓅( 𝔵3𝔦−1 𝔵′3𝔦) = 8𝔦 − 5, 𝓅( 𝔵3𝔦 𝔵′3𝔦+1) = −3 + 8𝔦, 𝓅( 𝔵3𝔦 𝔵′3𝔦−1) = 8𝔦 − 3, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 15 https://internationalpubls.com 𝓅( 𝔵3𝔦+1 𝔵′3𝔦) = −1 + 8𝔦, 𝓅( 𝔵′3𝔦−1𝑤) = 𝓅( 𝔵′3𝔦𝑤) = 4𝑛 3 + 1 + 4(𝔦 − 1). 𝓅( 𝔵′3𝔦−2𝑤) = 4𝑛 3 − 1 + 4(𝔦 − 1), Case ii Consider 𝑛 ≡ 1 (𝑚𝑜𝑑 3) 𝓅( 𝔵1) = 8 3 (𝑛 − 1) + 1, 𝓅(𝔵3𝔦) = 2𝔦 + 1, 𝓅( 𝔵3𝔦−1) = −1 + 2𝔦, 𝓅( 𝔵3𝔦−2) = 2𝔦 − 1, 𝓅( 𝔵′ 3𝔦) = 2(𝑛 + 𝔦) − 1 , 𝓅( 𝔵′3𝔦−1) = 2(𝑛 + 𝔦) − 3, 𝓅( 𝔵′3𝔦−2) = 2𝑛 + 2𝔦 − 3, 𝓅( 𝔵3𝔦−2 𝔵3𝔦−1) = 𝓅( 𝔵3𝔦−1 𝔵3𝔦) = 𝓅( 𝔵3𝔦 𝔵3𝔦+1) = 2𝔦 − 1, 𝓅( 𝔵3𝔦−2 𝔵′3𝔦−1) = 8𝔦 − 7, 𝓅( 𝔵3𝑖−1 𝔵′3𝑖) = 8𝑖 − 5, 𝓅( 𝔵3𝑖 𝔵′3𝑖+1) = 8𝑖 − 3, 𝓅(𝔵3𝑖−1 𝔵′3𝑖−2) = 8𝑖 − 5, 𝓅( 𝔵3𝑖 𝔵′3𝑖−1) = 8𝑖 − 3, 𝓅( 𝔵3𝑖+1 𝔵′3𝑖) = 8𝑖 − 1, 𝓅( 𝔵′3𝔦−1𝑤) = 𝓅( 𝔵′3𝔦𝑤) = 4(𝑛 − 1) 3 + 4𝔦 − 1. 𝓅( 𝔵′3𝔦−2𝑤) = 4𝑛+12𝔦−7 3 , Case iii Suppose 𝑛 ≡ 2 (𝑚𝑜𝑑 3) 𝓅( 𝔵1) = 8𝑛−4 3 , 𝓅( 𝔵3𝔦) = 2𝔦 + 1, 𝓅( 𝔵3𝔦−1) = 𝜌( 𝔵3𝔦−2) = 2𝔦 − 1, 𝓅( 𝔵′3𝔦) = 2𝔦 − 1 + 2𝑛, 𝓅( 𝔵′3𝔦−1) = 𝓅( 𝔵′3𝔦−2) = −3 + 2𝔦 + 2𝑛, 𝓅( 𝔵3𝔦−2 𝔵3𝔦−1) = −1 + 2𝔦, 𝓅( 𝔵3𝔦−1 𝔵3𝔦) = 𝓅( 𝔵3𝔦 𝔵3𝔦+1) = 2𝔦 − 1, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 16 https://internationalpubls.com 𝓅( 𝔵3𝔦−2 𝔵′3𝔦−1) = 8𝔦 − 7, 𝓅( 𝔵3𝔦 𝔵′3𝔦−1) = 8𝔦 − 3, 𝓅( 𝔵3𝔦−1 𝔵′3𝔦) = 8𝔦 − 5, 𝓅( 𝔵3𝔦+1 𝔵′3𝔦) = −1 + 8𝔦, 𝓅( 𝔵3𝔦 𝔵′3𝔦+1) = 8𝔦 − 3, 𝓅( 𝔵3𝔦−1 𝔵′3𝔦−2) = 8𝔦 − 5, 𝓅( 𝔵′3𝔦 𝑤) = 4𝑛 + 12𝔦 − 8 3 , 𝓅( 𝔵′3𝔦−1 𝑤) = 4𝑛 + 12𝔦 − 8 3 , 𝓅( 𝔵′3𝔦−2 𝑤) = 4𝑛 + 12𝔦 − 14 3 , Induced weight function for edge𝜓: 𝐸(𝜇(𝑃𝑛)) → {3, 5, 7, … , −5 + 8𝑛} is given by 𝜓(𝔵𝔦𝔵𝔦+1) = 2𝔦 + 1, 𝔦 ≠ 1 𝜓(𝔵𝔦𝔵′𝔦+1) = −3 + 2𝑛 + 4𝔦, 𝔦 ≠ 1 𝜓(𝔵𝔦+1𝔵′𝔦) = −1 + 4𝔦 + 2𝑛, 𝔦 ≠ 1 𝜓(𝔵′𝔦𝑤) = −5 + 6𝑛 + 2𝔦. Therefore, induced weights in edges of 𝜇(𝑃𝑛)generates an arithmetic progression and it differed by 2. Hence(3, 2)– 𝑡𝑒𝑠(𝜇(𝑃𝑛)) ≤ ⌊ 8𝑛 3 ⌋ − 1. From Proposition 1, we get (3, 2)– 𝑡𝑒𝑠(𝜇(𝑃𝑛)) ≥ −1 + ⌊ 8𝑛 3 ⌋ and it concludes the result. Example 8: (3, 2) total edge irregular labeling of 𝜇(𝑃6)is given in Fig 10. Figure 10. (3, 2)– 𝑡𝑒𝑠(𝜇(𝑃6)) = 15. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 9s (2025) 17 https://internationalpubls.com Remark 3: When 𝑛 = 2, 𝜇(𝑃2) ≅ 𝐶5, and (3, 2)– 𝑡𝑒𝑠(𝐶 5) = 4, total edge irregular labelling for 𝐶 5 is shown in Fig 11. Figure 11. (3, 2)– 𝑡𝑒𝑠(𝐶5) = 4. 3. Conclusion The exact value of (3, 2) – 𝑡𝑒𝑠(𝐺) is presented in this article namely: (3, 2)– 𝑡𝑒𝑠(𝑆(𝑃 𝑛 ∘ 𝑆2)) = 4𝑛 − 1, (3, 2)– 𝑡𝑒𝑠(𝑀(𝑃 𝑛)) = 2𝑛 − 2, (3, 2)– 𝑡𝑒𝑠(𝐾𝑛) = 𝑛2 − 5𝑛 + 9 for 𝑛 ≥ 4, (3, 2)– 𝑡𝑒𝑠(𝑇(𝑃 𝑛)) = ⌈ 8𝑛 3 ⌉ − 3 for 𝑛 ≥ 3, (3, 2)– 𝑡𝑒𝑠(𝑃 𝑛 ∘ 𝐾2) = ⌈ 8(𝑛+1) 3 ⌉ − 3, and (3, 2)– 𝑡𝑒𝑠(𝜇(𝑃𝑛)) = ⌊ 8𝑛 3 ⌋ − 1 for 𝑛 ≥ 3. 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