EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 1, 2023, 271-285 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the study of rainbow antimagic connection number of corona product of graphs Brian Juned Septory1,3, Liliek Susilowati1,∗, Dafik2,3, Veerabhadraiah Lokesha4, Gnaneswaran Nagamani5 1 Mathematics Department, Faculty of Science and Technology, Airlangga University, Surabaya, Indonesia 2 Mathematics Education Study Program, Faculty of Teacher Training and Education, University of Jember, Jember, Indonesia 3 Pusat Unggulan Ipteks-Perguruan Tinggi, Combinatorics and Graph, CGANT, University of Jember, Jember, Indonesia 4 Mathematics Department, Vijayanagara Sri Krishnadevaraya University, Bellary, India 5 Department of Mathematic The Gandhigram Rural Institute, Tamil Nadu, India Abstract. Given that a graph G = (V,E). By an edge-antimagic vertex labeling of graph, we mean assigning labels on each vertex under the label function f : V → {1, 2, . . . , |V (G)|} such that the associated weight of an edge uv ∈ E(G), namely w(xy) = f(x) + f(y), has distinct weight. A path P in the vertex-labeled graph G is said to be a rainbow path if for every two edges xy, x′y′ ∈ E(P ) satisfies w(xy) ̸= w(x′y′). The function f is called a rainbow antimagic labeling of G if for every two vertices x and y of G, there exists a rainbow x − y path. When we assign each edge xy with the color of the edge weight w(xy), thus we say the graph G admits a rainbow antimagic coloring. The rainbow antimagic connection number of G, denoted by rac(G), is the smallest number of colors induced from all edge weight of antimagic labeling. In this paper, we will study the rac(G) of the corona product of graphs. By the corona product of graphs G and H, denoted by G⊙H, we mean a graph obtained by taking a copy of graph G and n copies of graph H, namely H1,H2, ..., Hn, then connecting vertex vi from the copy of graph G to every vertex on graph Hi, i = 1, 2, 3, . . . , n. In this paper, we show the exact value of the rainbow antimagic connection number of Tn ⊙ Sm where Tn ∈ {Pn, Sn, Sn,p, Fn,3}. 2020 Mathematics Subject Classifications: 05C15, 05C78 Key Words and Phrases: Antimagic labeling, Rainbow connection, Rainbow antimagic con- nection number, Corona product of graphs. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4520 Email addresses: liliek-s@fst.unair.ac.id (L. Susilowati), brianseptory95@gmail.com (B. J. Septory), d.dafik@unej.ac.id (Dafik), mathematics@vskub.ac.in (V. Lokesha), nagamanigru@gmail.com (G. Nagamani) https://www.ejpam.com 271 © 2023 EJPAM All rights reserved. Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 272 1. Introduction Given two any graphs G and H. The corona operation of two graphs G and H, denoted by G⊙H, is a graph obtained by taking a copy of graph G and n copies of graph H namely H1,H2, ..., Hn then connecting vertex vi from the copy of graph G to every vertex on graph Hi, i = 1, 2, 3, . . . , n, see [8, 12] for detail. The rainbow antimagic coloring defined in above abstract is a combination of the rainbow coloring and antimagic labeling concepts. Rainbow coloring was first introduced by Chartrand et al. in 2008 [7]. Suppose that G is a nontrivially connected graph and c is the edge coloring of G. A u− v path of G, if no two edges of the u− v path are the same color is called a rainbow path. The c edge coloring, if for every vertex u, v ∈ V (G) there is a rainbow path u− v is called a rainbow connection. There are a lot of results related to the rainbow connection, see [17] and [18]. Proposition 1. [21] Let G be a connected graph of size m. The rainbow connection number rc(G) = m if and only if G is a tree. Other extension of rainbow coloring study is rainbow vertex coloring introduced in [16]. Some results on rainbow vertex connection number can see in [19], [23]. Furthermore, we also have other version of rainbow coloring study, called rainbow total coloring, see [14] and [24]. The complete survey of rainbow connection number can be found in [20]. Meanwhile, graph labeling was first introduced by Wallis et al. in 2001 [25]. Hartsfield and Ringel in 1990 [13] introduced an antimagic labeling of a graph G with edges is a bijection function f : E(G) → {1, 2, ..., |E(G)|} and w(v) = Σe∈E(v)f(e), and E(v) is the set of edges incidence to v, for vertex u, v ∈ V (G), w(u) ≠ w(v). Some results of antimagic labeling have been extensively studied by Baca et al. in [2–4]. Furthermore, Dafik et al. in 2021 [10] also contributed some results on antimagic labeling. Moreover, there are some other results related to antimagic labeling, see [6] and [9]. The concept of combining the graph coloring and the graph labeling was initiated by Arumugam et al. in 2017 [1]. He defined that for a bijection f : E(G) → {1, 2, ...|E(G)|} and w(v) = Σe∈E(v)f(e), and E(v) is the set of edges incidence to v, for each v ∈ V (G). The bijection f is called local antimagic labeling if for every two adjacent vertices u, v ∈ V (G), w(u) ̸= w(v). Thus, each local antimagic labeling is a vertex coloring at G with vertex v colored with w(v). When we consider the chromatic number of the local antimagic labeling, this notion is called a local antimagic coloring. Motivated by this combination, Dafik et al. in 2021 [11] initiates to study a rainbow antimagic coloring of graph. 2. Rainbow Antimagic Coloring Based on the description above, Dafik et al. in 2021 [11] have obtained the lower bound of the rac(G) and given some relevan results too. Proposition 2. [11] For any connected graph G, rac(G) ≥ rc(G). Theorem 1. [11] Let G be any connected graph. Let rc(G) and ∆(G) be the rainbow connection number of G and the maximum degree of G, rac(G) ≥ max{rc(G),∆(G)}. Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 273 Theorem 2. [11] Let G be a connected graph of diameter diam(G) ≤ 2. Let f be any bijective function from V (G) to the set {1, 2, . . . , |V (G)|}, there exists a rainbow path u−v. Theorem 3. [11] For Tm, being any tree of order m ≥ 3, rac(Tm) = m− 1. Some initial results for rainbow antimagic coloring have been found in [5], [11], [15] and [22]. Theorem 4. [22]For any integer m ≥ 3, rac(K2,m) = m+ 1. 3. Results In this section, we will show the rainbow antimagic connection number of Tn ⊙ Sm where Tn ∈ {Pn, Sn, Sn,p, Fn,3}. Our strategy is firstly determined the rainbow antimagic connection number of K1 + Sm. Secondly, establish the lower bound of rac(Tn ⊙ Sm). Finally, we show the exact values of rac(Pn ⊙ Sm), rac(Sn ⊙ Sm), rac(Sn,p ⊙ Sm) and rac(Fn,3 ⊙ Sm). Theorem 5. For m ≥ 3, rac(K1 + Sm) = m+ 2. Proof. The graph K1+Sm is a connected graph with vertex set V (K1+Sm) = {a1}∪ {b1} ∪ {xj , 1 ≤ j ≤ m}, and the edge set E(K1 + Sm) = {a1b1} ∪ {a1xj , b1xj , 1 ≤ j ≤ m}. The cardinality of |V (K1 +Sm)| = m+2 and the cardinality of |E(K1 +Sm)| = 2nm+1. Based on this definition, the graph K1 + Sm has ∆(K1 + Sm) = m+ 1. To prove rac(K1+Sm), first we have to show the lower bound of rac(K1+Sm). Based on Theorem 1, we have rac(G) ≥ max{rc(G),∆(G)} = m+ 1. Since, the construction of vertex labeling with the function f : V (K1 + Sm) → {1, 2, . . . |V (K1 + Sm)|} is a bijective function, assigning the most possible label for a1, b1 gives w(a1b1) must be different with other edge weights. The rest of the labels are considered to be a rainbow antimagic coloring of complete bipartite graph K2,m. Refer to Theorem 4, rac(K2,m) = ∆(K2,m) + 1 = m + 1. Since ∆(K1 + Sm) = rac(K2,m), and apart from edge a1b1, the graph K1 + S1 is K2,m, it implies that rac(K1 + Sm) ≥ max{rc(G),∆(G)} = ∆(K1 + Sm). However, if rac(K1 + Sm) ≥ ∆(K1 + Sm) then there is a conflict, since we need to include the edge a1b1, thus it must be m + 1 + 1 = ∆(K1 + Sm) + 1. It concludes that rac(K1 + Sm) ≥ ∆(K1 + Sm) + 1 = m+ 2. Secondly, we have to show the upper bound of rac(K1+Sm). Define the vertex labeling f : V (K1 + Sm) → {1, 2, ...,m+ 2} as follows. f(a1) = 1, f(b1) = 2, f(xi) = i+ 2, for 1 ≤ i ≤ m. The edge weights of the above vertex labeling f can be presented as w(a1b1) = 3, Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 274 w(a1xi) = i+ 3, for 1 ≤ i ≤ m w(b1xi) = i+ 4 for 1 ≤ i ≤ m. It is easy to see that the above edge weight will induce a rainbow antimagic coloring of graph K1+Sm. By this set of edge weight, we can easily calculate that the number of color w(a1b1) is 1. The sets w(a1xi) = {4, 5, 6, 7, . . . ,m+ 3} and w(b1xi) = {5, 6, 7, . . . ,m+ 4}, thus the number of distinct colors of w(a1xi)∪w(b1xi) is m+1. It implies the edge weights of f : V (K1 + Sm) → {1, 2, ...,m + 2} induces a rainbow antimagic coloring of 1 +m+ 1 colors. Therefore rac(K1 + Sm) ≤ m+ 2. Combining the two bounds, we have the exact value of rac(K1 + Sm) = m+ 2. The next step, we need to evaluate the existence of rainbow path of K1 + Sm. Since diam(K1 + Sm) = 2, based on Theorem 2, there exists a rainbow u− v path for any two vertices u, v ∈ V (K1 + Sm). It completes the proof. Lemma 1. Let Tn ⊙ Sm be a coronation of tree with order n and star graph with m ≥ 3. The lower bound of rac(Tn ⊙ Sm) ≥ rac(Tn) +m+ 2. Proof. The graph Tn⊙Sm is the corona product of two graphs Tn and Sm. It is obtained by taking one copy of Tn and |V (Tn)| copies of Sm and joining the i-th vertex of Tn to every vertex in the i-th copy of Sm. By this definition, it implies the graph Tn ⊙ Sm contains |V (Tn)| copies of K1 + Sm, see Figure 1. Thus, to obtain the rac(Tn ⊙ Sm), we need to consider the rac(K1+Sm) and rac(Tn). Based on Theorem 5, we have rac(K1+Sm) = m+ 2. Based on Theorem 3, we have rac(Tn) = n− 1, since E(Tn) = −1, for uv, u′v′ ∈ E(Tn) has a different colors. Thus, it implies that rac(Tn ⊙ Sm) ≥ rac(Tn) +m+ 2. Theorem 6. For odd integers n ≥ 3 and m ≥ 3, rac(Pn ⊙ Sm) = n+m+ 1. Proof. The graph Pn ⊙ Sm is a connected graph with vertex set V (Pn ⊙ Sm) = {x0i, yi, 1 ≤ i ≤ n}∪{yij , 1 ≤ i ≤ n , 1 ≤ j ≤ m} and edge set E(Pn⊙Sm) = {x0ix0i+1, 1 ≤ i ≤ n − 1} ∪ {x0iyi, 1 ≤ i ≤ n} ∪ {x0iyij , yiyij , 1 ≤ i ≤ n, 1 ≤ j ≤ m}. The cardinality of |V (Pn ⊙ Sm)| = 2n+ nm and the cardinality of |E(Pn ⊙ Sm)| = 2n+ 2nm− 1. To prove the rainbow antimagic connection number of rac(Pn ⊙ Sm), first we have to show that the lower bound of rac(Pn⊙Sm). Based on Lemma 1, we have rac(Pn⊙Sm) ≥ rac(Pn) +m+ 2. Since rac(Pn) = n− 1, thus, rac(Pn ⊙ Sm) ≥ n+m+ 1. Secondly, we have to show the upper bound of rac(Pn ⊙ Sm). Define the vertex labeling f : V (Pn ⊙ Sm) → {1, 2, ..., 2n+ nm} as follows. f(x0i) = ⌊n 2 ⌋ + i, for 1 ≤ i ≤ n f(yi) = { ⌊ n 2 ⌋ + i+ n, for 1 ≤ i ≤ ⌈n2 ⌉ i− ⌈n2 ⌉ , for ⌈n2 ⌉ + 1 ≤ i ≤ n, f(yij) = { 2n+ jn− i− 1, for 1 ≤ i ≤ ⌈n2 ⌉, 1 ≤ j ≤ m 2n+ jn− i+ 4, for ⌈n2 ⌉+ 1 ≤ i ≤ n, 1 ≤ j ≤ m, Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 275 K1 + Sm T Figure 1: The illustration of graph Tn ⊙ Sm. The edge weights of the above vertex labeling f can be presented as: for 1 ≤ i ≤ n − 1, w(x0ix0i+1) = n+ 2i, and for 1 ≤ i ≤ ⌈n2 ⌉, 1 ≤ j ≤ m w(x0iyi) = 2n+ 2i− 1, w(xiyij) = 2n+ jn+ 1, w(yiyij) = 3n+ jn+ 1. and for ⌈n2 ⌉+ 1 ≤ i ≤ n, 1 ≤ j ≤ m, w(x0iyi) = 2i− 1, w(xiyij) = 3n+ jn+ 1, w(yiyij) = 2n+ jn+ 1. Based on Theorem 3, rac(Pn) = n − 1. Since E(Pn) = n − 1, for u, v ∈ E(Pn) has a different colors. Therefore, the sum of the weights of graph Pn is n − 1. Based on Theorem 5, we have rac(K1 + Sm) = m + 2. Based on the description above, we have that the distinct weight of graph (Pn ⊙ Sm) is n +m + 1. It implies the edge weights of f : V (Pn ⊙ Sm) → {1, 2, ..., 2n+ nm} induces a rainbow antimagic coloring of m+ n+ 1 colors. Thus rac(Pn ⊙ Sm) ≤ n +m + 1. Comparing the two bounds, we have the exact value of rac(Pn ⊙ Sm) = n+m+ 1. The next step, evaluate to prove the existence of a rainbow u−v path Pn⊙Sm. Based on the definition of the graph Pn⊙Sm, then the graph Pn⊙Sm contains one graph Pn and |V (Pn)| copies of K1 + Sm, so that we can evaluate the rainbow u − v path of the graph Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 276 Pn ⊙ Sm by evaluating the rainbow u− v path on the graph Pn and the graph K1 + Sm. Since diam(K1 + Sm) = 2, based on Theorem 2, there is a rainbow u − v path for every u, v ∈ V (K1+Sm). Based on Theorem 3, rac(Pn) = n−1. Since Pn has n−1 edges, there is a rainbow u − v path for every u, v ∈ V (Pn). Therefore, according to the explanation, it can be seen that there is a rainbow u− v path for every u, v ∈ V (Pn ⊙ Sm). Rainbow antimagic coloring of graph Pn ⊙ Sm can be seen in Figure 2. y1 y3 x03 x02 x01 y24y23y22y21 y34y33y32y31y14y13y12y11 y2 4 10 13 16 22 10 16 19 13 16 22 13 10 7 5 9 14 5 7 10 32 1 16 19 13 16 13 19 1916 1913 75 1815129 1613 6 17 22 118 19 Figure 2: Rainbow antimagic coloring of graph P3 ⊙ S4. Theorem 7. For odd integers n ≥ 3 and m ≥ 3, rac(Sn ⊙ Sm) = n+m+ 2. Proof. The graph Sn ⊙ Sm is a connected graph with vertex set V (Sn ⊙ Sm) = {x0} ∪ {x0i, 1 ≤ i ≤ n} ∪ {yi, 1 ≤ i ≤ n+1} ∪ {yij , 1 ≤ i ≤ n+1, 1 ≤ j ≤ m} and edge set E(Sn⊙Sm) = {x0xi, 1 ≤ i ≤ n}∪{x0yn+1}∪{x0iyi, 1 ≤ i ≤ n}∪{x0yn+1j , yn+1yn+1j , 1 ≤ j ≤ m}∪{x0iyi, yiyij , 1 ≤ i ≤ n, 1 ≤ j ≤ m}. The cardinality of |V (Sn⊙Sm)| = 2n+nm+2 and the cardinality of |E(Sn ⊙ Sm)| = 2nm+ 2n+ 2m+ 1. To prove the rainbow antimagic connection number of rac(Sn ⊙ Sm), first we have to show that the lower bound of rac(Sn⊙Sm). Based on Lemma 1, we have rac(Sn⊙Sm) ≥ rac(Sn) +m+ 2. Since rac(Sn) = n, thus, rac(Sn ⊙ Sm) ≥ n+m+ 2. Secondly, we have to show the upper bound of rac(Sn⊙Sm). Define the vertex labeling f : V (Sn ⊙ Sm) → {1, 2, . . . 2n+ nm+ 2} as follows. f(x0) = n+ 1 f(xi) =  n+ i+ 1 for 1 ≤ i ≤ n and i is odd i for 1 ≤ i ≤ n and i is even i for i = n+ 1 f(yi) =  i for 1 ≤ i ≤ n and i is odd n+ i+ 1 for 1 ≤ i ≤ n and i is even 2n+ 2 for i = n+ 1, Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 277 f(yij) = 2n+ jn− i+ j + 3, for 1 ≤ i ≤ n+ 1, 1 ≤ j ≤ m, The edge weights of the above vertex labeling f can be presented as w(x0x0i) = { 2n+ i+ 2 for 1 ≤ i ≤ n and i is odd n+ i+ 1 for 1 ≤ i ≤ n and i is even w(x0yn+1) = 3n+ 3 w(xiyi) = n+ 2i+ 1, for 1 ≤ i ≤ n. w(x0yn+1j) = 2n+ jn+ j + 3, for 1 ≤ j ≤ m w(x0iyij) = { 3n+ jn+ j + 4 for 1 ≤ i ≤ n, 1 ≤ j ≤ m and i is odd 2n+ jn+ j + 3 for 1 ≤ i ≤ n, 1 ≤ j ≤ m and i is even w(yiyij) =  2n+ jn+ j + 3 for 1 ≤ i ≤ n+ 1, 1 ≤ j ≤ m and i is odd 3n+ jn+ j + 4 for 1 ≤ i ≤ n+ 1, 1 ≤ j ≤ m and i is even 3n+ jn+ j + 4 for i = n+ 1, Based on Theorem 3, rac(Sn) = n. Since E(Sn) = n, for u, v ∈ E(Sn) has a different colors. Therefore, the sum of the weights of graph Sn is n. Based on Theorem 5, we have rac(K1 + Sm) = m + 2. Based on the description above, we have that the distinct weight of graph (Sn ⊙ Sm) is n+m+ 2. It implies the edge weights of f : V (Sn ⊙ Sm) → {1, 2, ..., 2n + nm + 2} induces a rainbow antimagic coloring of m + n + 2 colors. Thus rac(Sn ⊙ Sm) ≤ n + m + 2. Comparing the two bounds, we have the exact value of rac(Sn ⊙ Sm) = n+m+ 2. The next step, evaluate to prove the existence of a rainbow u−v path Sn⊙Sm. Based on the definition of the graph Sn⊙Sm, then the graph Sn⊙Sm contains one graph Sn and |V (Sn)| copies of K1 + Sm, so that we can evaluate the rainbow u − v path of the graph Sn ⊙ Sm by evaluating the rainbow u− v path on the graph Sn and the graph K1 + Sm. Since diam(K1 + Sm) = 2, based on Theorem 2, there is a rainbow u − v path for every u, v ∈ V (K1 + Sm). Based on Theorem 3, rac(Sn) = n. Since Sn has n edges, there is a rainbow u − v path for every u, v ∈ V (Sn). Therefore, according to the explanation, it can be seen that there is a rainbow u− v path for every u, v ∈ V (Sn ⊙ Sm). Rainbow antimagic coloring of graph Sn ⊙ Sm can be seen in Figure 3. Theorem 8. For n = 2, m ≥ 3 and odd integers p ≥ 3, rac(Sn,p ⊙ Sm) = m+ p+ 5. Proof. The graph Sn,p ⊙ Sm is a connected graph with vertex set V (Sn,p ⊙ Sm) = {x, y, b, z} ∪ {xi, x0i, 1 ≤ i ≤ 2} ∪ {yj , y0j , 1 ≤ j ≤ p} ∪ {bk, zk, 1 ≤ k ≤ m} ∪ {xik, 1 ≤ i ≤ 2, 1 ≤ k ≤ m} ∪ {yjk, 1 ≤ j ≤ p, 1 ≤ k ≤ m} and edge set E(Sn,p ⊙ Sm) = {xy, xb, yz} ∪ {xxi, xix0i, 1 ≤ i ≤ 2} ∪ {yyj , yjy0j , 1 ≤ j ≤ p} ∪ {xbk, bbk, yzk, zzk, 1 ≤ i ≤ m} ∪ {xixik, x0ixik, 1 ≤ i ≤ 2, 1 ≤ k ≤ m} ∪ {yjyjk, y0jyjk, 1 ≤ j ≤ p, 1 ≤ k ≤ m}. The cardinality of |V (Sn,p ⊙ Sm)| = 2p+4m+ pm+8 and the cardinality of |E(Sn,p ⊙ Sm)| = 2p+ 8m+ 2pm+ 7. To prove the rainbow antimagic connection number of rac(Sn,p⊙Sm), first we have to show that the lower bound of rac(Sn,p⊙Sm). Based on Lemma 1, we have rac(Sn,p⊙Sm) ≥ Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 278 y4 y2 y3 y43 y42 y41 y51 y52 y5 y53 y31 y32 y33 y21 y22 y23 y11 y12 y13 x04 x01 x02 x03 x0 y1 6 12 17 22 15 21 26 31 16 21 26 21 26 31 16 21 26 139 1613 21 26 21 26 31 1116 21 26 21 26 31 16 21 26 31 7 26 21 7 9 11 4 211611 10 5 23 18 13 3 8 24 19 14 7 2 25 20 15 1 9 Figure 3: Rainbow antimagic coloring of graph S4 ⊙ S3. rac(Sn,p) +m+ 2. Since rac(Sn,p) = p+ 3, thus, rac(Sn,p ⊙ Sm) ≥ m+ p+ 5. Secondly, we have to show the upper bound of rac(Sn,p ⊙ Sm). Define the vertex labeling f : V (Sn,p ⊙ Sm) → {1, 2, . . . 2p+ 4m+ pm+ 8} as follows. f(x) = p+ 2 f(y) = p+ 3 f(xi) = { p+ 5 for i = 1 2p+ 8 for i = 2 f(yj) = { 2j + 1 for 1 ≤ j ≤ ⌊p 2 ⌋ 2j + 5 for ⌈p2⌉+ 1 ≤ j ≤ p f(x0i) = { 1 for i = 1, p+ 4 for i = 2 f(y0j) = { p+ 2j + 5 for 1 ≤ j ≤ ⌊p 2 ⌋ 2j + 1− p for ⌈p2⌉ ≤ j ≤ p f(b) = 2p+ 6 f(z) = 2p+ 7 f(xik) = { k(p+ 4) + 2p+ 8 for , i = 1, 1 ≤ k ≤ m, k(p+ 4) + p+ 5 for , i = 2, 1 ≤ k ≤ m, f(zk) = k(p+ 4) + p+ 6 for 1 ≤ k ≤ m, f(bk) = k(p+ 4) + p+ 7 for 1 ≤ k ≤ m, Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 279 f(yjk) = { k(p+ 4)− 2j + 2p+ 8 for 1 ≤ j ≤ ⌊p 2 ⌋ , 1 ≤ k ≤ m, k(p+ 4)− 2j + 3p+ 8 for ⌈p2⌉, 1 ≤ k ≤ m. The edge weight of the above vertex labeling f can be presented as w(xy) = 2p+ 5 w(xxi) = { 2p+ 7 for i = 1 3p+ 10 for i = 2 w(yyj) = { p+ 2j + 4 for 1 ≤ j ≤ ⌊p 2 ⌋ p+ 2j + 8 for ⌈p2⌉, 1 ≤ k ≤ m w(xix0i) = { p+ 6 for i = 1 3p+ 12 for i = 2 w(xb) = 3p+ 8 w(yz) = 3p+ 10 w(yjy0j) = { p+ 4j + 6 for 1 ≤ j ≤ ⌊p 2 ⌋ 4j + 6− p for ⌈p2⌉, 1 ≤ k ≤ m w(xixik) = k(p+ 4) + 3p+ 13, for 1 ≤ ≤ 2, 1 ≤ k ≤ m w(x0ixik) = k(p+ 4) + 2p+ 9, for 1 ≤ ≤ 2, 1 ≤ k ≤ m w(xbk) = k(p+ 4) + 2p+ 9, for 1 ≤ k ≤ m w(bbk) = k(p+ 4) + 3p+ 13, for 1 ≤ k ≤ m w(yzk) = k(p+ 4) + 2p+ 9, for 1 ≤ k ≤ m w(zzk) = k(p+ 4) + 3p+ 13, for 1 ≤ k ≤ m w(yjyjk) = { k(p+ 4) + 2p+ 9 for 1 ≤ j ≤ ⌊p 2 ⌋ , 1 ≤ k ≤ m, k(p+ 4) + 3p+ 13 for ⌈p2⌉, 1 ≤ k ≤ m. w(y0jyjk) = { k(p+ 4) + 3p+ 13 for 1 ≤ j ≤ ⌊p 2 ⌋ , 1 ≤ k ≤ m, k(p+ 4) + 2p+ 9 for ⌈p2⌉, 1 ≤ k ≤ m. Based on Theorem 3, rac(Sn,p) = p + 3. Since E(Sn,p) = p + 3, for u, v ∈ E(Sn,p) has a different colors. Therefore, the sum of the weights of graph Sn,p is p+ 3. Based on Theorem 5, we have rac(K1 + Sm) = m + 2. Based on the description above, we have that the distinct weight of graph (Sn,p ⊙ Sm) is m+ p+ 5. It implies the edge weights of f : V (Sn,p ⊙ Sm) → {1, 2, ..., 2p + 4m + pm + 8} induces a rainbow antimagic coloring of m+ n+2 colors. Thus rac(Sn,p ⊙ Sm) ≤ m+ p+5. Comparing the two bounds, we have the exact value of rac(Sn,p ⊙ Sm) = m+ p+ 5. The next step, evaluate to prove the existence of a rainbow u − v path Sn,p ⊙ Sm. Based on the definition of the graph Sn,p ⊙ Sm, then the graph Sn,p ⊙ Sm contains one graph Sn,p and |V (Sn,p)| copies of K1 + Sm, so that we can evaluate the rainbow u − v path of the graph Sn,p ⊙ Sm by evaluating the rainbow u− v path on the graph Sn,p and the graph K1 + Sm. Since diam(K1 + Sm) = 2, based on Theorem 2, there is a rainbow Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 280 u−v path for every u, v ∈ V (K1+Sm). Based on Theorem 3, rac(Sn,p) = n+p+1. Since Sn,p has n+ p+1 edges, there is a rainbow u− v path for every u, v ∈ V (Sn,p). Therefore, according to the explanation, it can be seen that there is a rainbow u − v path for every u, v ∈ V (Sn,p ⊙ Sm). Rainbow antimagic coloring of graph Sn,p ⊙ Sm can be seen in Figure 4. 50 36 43 50 22 29 36 43 50 43 36 29 22 29 36 43 29 36 43 50 22 29 36 43 29 36 43 50 22 29 36 43 29 36 43 50 2229 36 43 29 36 43 1 8 21 28 35 42 5 12 7 15 22 29 36 14 17 24 31 38 6 13 16 23 30 37 3 9 11 10 2 4 19 26 33 40 20 27 34 41 18 25 32 39 13 19 11 9 15 17 9 11 13 15 17 19 21 22 29 36 43 29 36 43 50 22 29 36 43 29 x02 y x1 y1 y2 y3 y03 y02 y01 x01 b1 b2 b3 b4 b z z2 z3 x z1 z4 x11 x12 x13 x14 x21 x22 x23 x24 y11 y12 y13 y14 y21 y22 y23 y24 y31 y32 y33 y34 x2 Figure 4: Rainbow antimagic coloring of graph S2,3 ⊙ S4. Theorem 9. For odd integers n ≥ 3 and m ≤ 3, rac(Fn,3 ⊙ Sm) = 3n+m+ 1. Proof. The graph Fn,3 ⊙ Sm is a graph with V (Fn,3 ⊙ Sm) = {xi, yi, zi, x0i, y0i, z0i1 ≤ i ≤ n} ∪ {xij , yij , zij , 1 ≤ i ≤ n, 1 ≤ j ≤ m} and edge set E(Fn,3 ⊙ Sm) = {xixi+1, 1 ≤ i ≤ n−1}∪{xiyi, yizi, xix0i, yiy0i, ziz0i, 1 ≤ i ≤ n}∪{xixij , x0ixij , yiyij , y0iyij , zizij , z0izij , 1 ≤ i ≤ n, 1 ≤ j ≤ m}. The cardinality of |V (Fn,3 ⊙ Sm)| = 6n+ 3nm and the cardinality of |E(Fn,3 ⊙ Sm)| = 6n+ 6nm− 1. To prove the rainbow antimagic connection number of rac(Fn,3⊙Sm), first we have to show that the lower bound of rac(Fn,3⊙Sm). Based on Lemma 1, we have rac(Fn,3⊙Sm) ≥ rac(Fn,3) +m+ 2. Since rac(Fn,3) = 3n− 1, thus, rac(Fn,3 ⊙ Sm) ≥ 3n+m+ 1. Secondly, we have to show the upper bound of rac(Fn,3 ⊙ Sm). Define the vertex labeling f : V (Fn,3 ⊙ Sm) → {1, 2, ..., 6n+ 3nm} as follows. f(xi) = { 3i+ ⌊ n 2 ⌋ + n for 1 ≤ i ≤ n and i is odd 3i+ n+ ⌊ n 2 ⌋ − 2 for 1 ≤ i ≤ n and i is even Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 281 f(yi) = 3i+ n+ ⌊n 2 ⌋ − 1, for 1 ≤ i ≤ n f(zi) = { 3i+ n+ ⌊ n 2 ⌋ − 2 for 1 ≤ i ≤ n and i is odd 3i+ ⌊ n 2 ⌋ + n for 1 ≤ i ≤ n and i is even f(x0i) =  3i+ ⌊ n 2 ⌋ + 4n for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is odd, n ≡ 3 mod 4 3i− n− ⌈n2 ⌉ for ⌈n2 ⌉ ≤ i ≤ n and i is odd, n ≡ 3 mod 4 3i+ ⌊ n 2 ⌋ + 4n− 2 for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is even, n ≡ 3 mod 4 3i− n− ⌈n2 ⌉ − 2 for ⌈n2 ⌉ ≤ i ≤ n and i is even, n ≡ 3 mod 4 3i+ ⌊ n 2 ⌋ + 4n for 1 ≤ i ≤ ⌈n2 ⌉ and i is odd, n ≡ 1 mod 4 3i− n− ⌈n2 ⌉ for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is odd, n ≡ 1 mod 4 3i+ ⌊ n 2 ⌋ + 4n− 2 for 1 ≤ i ≤ ⌈n2 ⌉ and i is even, n ≡ 1 mod 4 3i− n− ⌈n2 ⌉ − 2 for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is even, n ≡ 1 mod 4 f(y0i) = { 3i+ 4n+ ⌊ n 2 ⌋ − 1 for 1 ≤ i ≤ ⌈n2 ⌉ 3i− ⌈n2 ⌉ − n− 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n f(z0i) =  3i+ ⌊ n 2 ⌋ + 4n− 2 for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is odd, n ≡ 3 mod 4 3i− n− ⌈n2 ⌉ − 2 for ⌈n2 ⌉ ≤ i ≤ n and i is odd, n ≡ 3 mod 4 3i+ ⌊ n 2 ⌋ + 4n− 1 for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is even, n ≡ 3 mod 4 3i− n− ⌈n2 ⌉ for ⌈n2 ⌉ ≤ i ≤ n and i is even, n ≡ 3 mod 4 3i+ ⌊ n 2 ⌋ + 4n− 2 for 1 ≤ i ≤ ⌈n2 ⌉ and i is odd, n ≡ 1 mod 4 3i− n− ⌈n2 ⌉ − 2 for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is odd, n ≡ 1 mod 4 3i+ ⌊ n 2 ⌋ + 4n− 1 for 1 ≤ i ≤ ⌈n2 ⌉ and i is even, n ≡ 1 mod 4 3i− n− ⌈n2 ⌉ for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is even, n ≡ 1 mod 4 f(xij) =  k(3n) + 5n+ 1− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is odd, n ≡ 3 mod 4 k(3n) + 7n+ 1− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉ ≤ i ≤ n and i is odd, n ≡ 3 mod 4 k(3n) + 5n+ 3− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is even, n ≡ 3 mod 4 k(3n) + 7n+ 3− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉ ≤ i ≤ n and i is even, n ≡ 3 mod 4 k(3n) + 5n+ 1− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌈n2 ⌉ and i is odd, n ≡ 1 mod 4 k(3n) + 7n+ 1− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is odd, n ≡ 1 mod 4 k(3n) + 5n+ 3− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌈n2 ⌉ and i is even, n ≡ 1 mod 4 k(3n) + 7n+ 3− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is even, n ≡ 1 mod 4 f(yij) = { k(3n) + 4n+ ⌈n2 ⌉+ 2− 3i for 1 ≤ i ≤ ⌈n2 ⌉, 1 ≤≤ m k(3n) + 7n+ ⌈n2 ⌉+ 2− 3i for ⌈n2 ⌉+ 1 ≤ i ≤ n, 1 ≤ j ≤ m f(zij) =  k(3n) + 5n+ 3− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is odd, n ≡ 3 mod 4 k(3n) + 7n+ 3− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉ ≤ i ≤ n and i is odd, n ≡ 3 mod 4 k(3n) + 5n+ 2− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is even, n ≡ 3 mod 4 k(3n) + 7n+ 1− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉ ≤ i ≤ n and i is even, n ≡ 3 mod 4 k(3n) + 5n+ 3− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌈n2 ⌉ and i is odd, n ≡ 1 mod 4 k(3n) + 7n+ 3− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is odd, n ≡ 1 mod 4 k(3n) + 5n+ 2− 3i− ⌊ n 2 ⌋ for 1 ≤ i ≤ ⌈n2 ⌉ and i is even, n ≡ 1 mod 4 k(3n) + 7n+ 1− 3i+ ⌈n2 ⌉ for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is even, n ≡ 1 mod 4 Susilowati et. al / Eur. J. Pure Appl. Math, 16 (1) (2023), 271-285 282 The edge weights of the above vertex labeling f can be presented as w(xixi+1) = 3n+ 6i, for 1 ≤ i ≤ n− 1 w(xiyi) = { 6i+ 2n+ 2( ⌊ n 2 ⌋ )− 1 for 1 ≤ i ≤ n and i is odd 6i+ 2n+ 2( ⌊ n 2 ⌋ )− 4 for 1 ≤ i ≤ n and i is even w(yizi) = { 6i+ 2n+ 2( ⌊ n 2 ⌋ )− 3 for 1 ≤ i ≤ n and i is odd 6i+ 2n+ 2( ⌊ n 2 ⌋ )− 2 for 1 ≤ i ≤ n and i is even w(xix0i) =  6i+ 2( ⌊ n 2 ⌋ ) + 5n for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is odd, n ≡ 3 mod 4 6i− 1 for ⌈n2 ⌉ ≤ i ≤ n and i is odd, n ≡ 3 mod 4 6i+ 2( ⌊ n 2 ⌋ ) + 5n− 4 for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is even, n ≡ 3 mod 4 6i− 5 for ⌈n2 ⌉ ≤ i ≤ n and i is even, n ≡ 3 mod 4 6i+ 2( ⌊ n 2 ⌋ ) + 5n for 1 ≤ i ≤ ⌈n2 ⌉ and i is odd, n ≡ 1 mod 4 6i− 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is odd, n ≡ 1 mod 4 6i+ 2( ⌊ n 2 ⌋ ) + 5n− 4 for 1 ≤ i ≤ ⌈n2 ⌉ and i is even, n ≡ 1 mod 4 6i− 5 for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is even, n ≡ 1 mod 4 w(yiy0i) = { 6i+ 5n+ ⌊ n 2 ⌋ − 2 for 1 ≤ i ≤ ⌈n2 ⌉ 6i− 3 for ⌈n2 ⌉+ 1 ≤ i ≤ n w(ziz0i) =  6i+ 2( ⌊ n 2 ⌋ ) + 5n− 4 for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is odd, n ≡ 3 mod 4 6i− 5 for ⌈n2 ⌉ ≤ i ≤ n and i is odd, n ≡ 3 mod 4 6i+ 2( ⌊ n 2 ⌋ ) + 5n− 1 for 1 ≤ i ≤ ⌊ n 2 ⌋ and i is even, n ≡ 3 mod 4 6i− 3 for ⌈n2 ⌉ ≤ i ≤ n and i is even, n ≡ 3 mod 4 6i+ 2( ⌊ n 2 ⌋ ) + 5n− 4 for 1 ≤ i ≤ ⌈n2 ⌉ and i is odd, n ≡ 1 mod 4 6i− 5 for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is odd, n ≡ 1 mod 4 6i+ 2( ⌊ n 2 ⌋ ) + 5n− 1 for 1 ≤ i ≤ ⌈n2 ⌉ and i is even, n ≡ 1 mod 4 6i− 3 for ⌈n2 ⌉+ 1 ≤ i ≤ n and i is even, n ≡ 1 mod 4 w(xixij) =  k(3n) + 6n+ 1 for 1 ≤ i ≤ ⌈n2 ⌉ and n ≡ 3 mod 4 k(3n) + 9n+ 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n and n ≡ 3 mod 4 k(3n) + 6n+ 1 for 1 ≤ i ≤ ⌊ n 2 ⌋ and n ≡ 1 mod 4 k(3n) + 9n+ 1 for ⌈n2 ⌉ ≤ i ≤ n and n ≡ 1 mod 4 w(x0ixij) =  k(3n) + 9n+ 1 for 1 ≤ i ≤ ⌈n2 ⌉ and n ≡ 3 mod 4 k(3n) + 6n+ 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n and n ≡ 3 mod 4 k(3n) + 9n+ 1 for 1 ≤ i ≤ ⌊ n 2 ⌋ and n ≡ 1 mod 4 k(3n) + 6n+ 1 for ⌈n2 ⌉ ≤ i ≤ n and n ≡ 1 mod 4 w(yiyij) = { k(3n) + 6n+ 1 for 1 ≤ i ≤ ⌈n2 ⌉, 1 ≤≤ m k(3n) + 9n+ 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n, 1 ≤ j ≤ m w(zizij) =  k(3n) + 6n+ 1 for 1 ≤ i ≤ ⌊ n 2 ⌋ and n ≡ 3 mod 4 k(3n) + 9n+ 1 for ⌈n2 ⌉ ≤ i ≤ n and n ≡ 3 mod 4 k(3n) + 6n+ 1 for 1 ≤ i ≤ ⌈n2 ⌉ and n ≡ 1 mod 4 k(3n) + 9n+ 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n and n ≡ 1 mod 4 REFERENCES 283 w(z0izij) =  k(3n) + 9n+ 1 for 1 ≤ i ≤ ⌊ n 2 ⌋ and n ≡ 3 mod 4 k(3n) + 6n+ 1 for ⌈n2 ⌉ ≤ i ≤ n and n ≡ 3 mod 4 k(3n) + 9n+ 1 for 1 ≤ i ≤ ⌈n2 ⌉ and n ≡ 1 mod 4 k(3n) + 6n+ 1 for ⌈n2 ⌉+ 1 ≤ i ≤ n and n ≡ 1 mod 4 Based on Theorem 3, rac(Fn,3) = 3n − 1. Since E(Fn,3) = 3n − 1, for u, v ∈ E(Fn,3) has a different colors. Therefore, the sum of the weights of graph Fn,3 is 3n − 1. Based on Theorem 5, we have rac(K1 + Sm) = m+ 2. Based on the description above, we have that the distinct weight of graph (Fn,3 ⊙Sm) is 3n+m+1. It implies the edge weights of f : V (Fn,3⊙Sm) → {1, 2, ..., 6n+3nm} induces a rainbow antimagic coloring of 3n+m+1 colors. Thus rac(Fn,3⊙Sm) ≤ 3n+m+1. Comparing the two bounds, we have the exact value of rac(Fn,3 ⊙ Sm) = 3n+m+ 1. The next step, evaluate to prove the existence of a rainbow u − v path Fn,3 ⊙ Sm. Based on the definition of the graph Fn,3 ⊙ Sm, then the graph Fn,3 ⊙ Sm contains one graph Fn,3 and |V (Fn,3)| copies of K1 + Sm, so that we can evaluate the rainbow u − v path of the graph Fn,3 ⊙ Sm by evaluating the rainbow u− v path on the graph Fn,3 and the graph K1 + Sm. Since diam(K1 + Sm) = 2, based on Theorem 2, there is a rainbow u− v path for every u, v ∈ V (K1 + Sm). Based on Theorem 3, rac(Fn,3) = 3n− 1. Since Fn,3 has 3n− 1 edges, there is a rainbow u− v path for every u, v ∈ V (Fn,3). Therefore, according to the explanation, it can be seen that there is a rainbow u − v path for every u, v ∈ V (Fn,3 ⊙ Sm). 4. Concluding Remarks We have studied the rainbow antimagic coloring of the corona product on a graph with a star graph. Based on the results we have the exact value of the rainbow antimagic connection number of graph Tn ⊙ Sm where Tn is path Pn, star Sn, double star Sn,p and fire craker Fn,3. However, if Tn is not a tree graph, it is still difficult to determine the exact value of the rainbow antimagic connection number. Therefore, this study raises an open problem. Determine the exact value of the rainbow antimagic connection number of graph G⊙Sm where G is not tree. 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