On the rainbow antimagic coloring of vertex amalgamation of graphs

The purpose of this study is to develop rainbow antimagic coloring. This study is a combination of two notions, namely antimagic and rainbow concept. If every vertex of graph G is labeled with the antimagic labels and then edge weight of antimagic labels are used to assign a rainbow coloring. The minimum number of colors for a rainbow path to exist with the condition satisfying the edge weights w(x) ≠ w(y) for any two vertices x and y is the definition of the rainbow antimagic connection number rac(G). In this study, we use connected graphs and simple graphs in obtaining the rainbow antimagic connection number. This paper will explain the rainbow antimagic coloring on some graphs and get their formula of the rainbow antimagic connection number. We have obtained rac(G) where G is vertex amalgamation of graphs, namely path, star, broom, paw, fan, and triangular book graph.


Introduction
Over time, the topics in graph theory are developing and generating new ideas, for example rainbow antimagic coloring. The combination of two notions, namely rainbow connection and antimagic labeling is called rainbow antimagic coloring (RAC).
Hartsfield and Ringel [13] define that graph G is said to be antimagic labeling if every edges are labeled with the integers {1, 2, . . . , k} so that vertex weight of two vertices is different, that is, all vertex on graph G do not have the same vertex weight. The dense graph is proven to be antimagic according to Alon et al [2]. The antimagic labeling of trees has been discovered by Liang et al [20]. More study of antimagic labeling some graphs in [5,15,21,23].
Chartrand, et al. [6] were the first to introduce the concept of rainbow connection. In the rainbow connection concept, there are adjacent edges that may have the same color where the color uses the integers {1, 2, . . . , k}. A rainbow path in graph G can be formed if there are no adjacent edges that have the same color. A graph G is said to be rainbow connected if any two of its vertices are connected by a rainbow path. Rainbow connection can be defined as the edge coloring that makes the graph G rainbow connected. Li et al [19] have discovered the exact value of 2-connected graph and chordal graph. The rc and src of fan and sun graph was invented by Sy et al [25]. Yandera et al [27] obtain the rainbow connection of Amal t (P m,2 ). More study of rainbow connection some graphs in [7,10,11,12,18].
A graph is said to be rainbow antimagic coloring if every vertex of graph G is labeled with the antimagic labels and then edge weight of antimagic labels are used to assign a rainbow connection such that there is a rainbow path with all different edge weight [24]. The minimum number of colors for a rainbow path to exist with the conditional edge weights w(x) = w(y) for any two vertices x and y is the definition of the rainbow antimagic connection number and denoted by rac(G). Dafik et al. [8] obtained rac(G) where G is simple graph namely path, star, and cycle graph. Intan K et al. [16] found the rainbow antimagic connection number of releated wheel graph. Sulistiono et al [24] made a theorem to prove the lower bound of rac(G). More study on rainbow antimagic coloring can be seen at [1,4,9,22].
We investigated about rac(G) where G is vertex amalgamation of path, star, broom, paw, fan, and triangular book graph. Lee, et al. [17] define that the graph G is operated by vertex amalgamation if there is v 0 ∈ V (G) as a fixed center point, then graph G is multiplied by n with v 0 as fixed center point and denoted by Amal(G, v 0 , n) with n ≥ 2. The study of amalgamation product in [3,12,14,15,26]. For illustration of vertex amalgamation is provided in Figure 2.

Previous Results
Some previous results related to the concept of rainbow antimagic coloring will be presented in this chapter.

Results and Discussion
In this chapter, several theorems of rainbow antimagic coloring on vertex amalgamation products that have been obtained include: Proof. Let H 1 be a vertex amalgamation product of path with vertex set V ( We need to prove the correctness of the formula rac(H 1 ) is mn − m with m, n ≥ 2, it is necessary to prove it by using the lower and upper bounds, namely rac(H 1 ) ≥ mn − m and rac(H 1 ) ≤ mn − m.
First, we prove the correctness of the lower bound of rac(H 1 ) is rac(H 1 ) ≥ mn − m. We know that H 1 graph is classified as a tree graph. Based on Proposition 1 [6] and Theorem 1 [24], so that: So we got that the lower bound of rac(H 1 ) is rac(H 1 ) ≥ mn − m with m, n ≥ 2.
Next, we prove the correctness of the upper bound of rac(H 1 ) is rac(H 1 ) ≤ mn − m. We define the vertex function of H 1 graph with f : Obviously, we get the edge weights from a predefined vertex function where the edge weights will be used as edge coloring. The edge weights of H 1 graph are Based on the edge weight function that has been obtained, a rainbow path can be formed if there are as many colors as mn − m. The rainbow path of H 1 graph that is formed can be seen in the Table 1. So we got that the upper bound of rac(H 1 ) is rac(H 1 ) ≤ mn − m. Table 1. The rainbow path from x to y on H 1 graph.
As a result of the explanation above, we have established that the lower and upper bounds of rac(H 1 ) are mn − m ≤ rac(H 1 ) ≤ mn − m. So that the correctness of the formula rac(H 1 ) is mn − m with m, n ≥ 2 has been proven.
For illustration of rac(H 1 ) is provided in Figure 2 and the vertex labels on the graph are black numbers, while the edge weights on the graph are blue numbers.
We need to prove the correctness of the formula rac(H 2 ) is mn with m, n ≥ 2, it is necessary to prove it by using the lower and upper bounds, namely rac(H 2 ) ≥ mn and rac(H 2 ) ≤ mn.
First, we prove the correctness of the lower bound of rac(H 2 ) is rac(H 2 ) ≥ mn. We know that H 2 graph is classified as a tree graph. Based on Proposition 1 [6] and Theorem 1 [24], so that: So we got that the lower bound of rac(H 2 ) is rac(H 2 ) ≥ mn with m, n ≥ 2. Next, we prove the correctness of the upper bound of rac( Obviously, we get the edge weights from a predefined vertex function where the edge weights will be used as edge coloring. The edge weights of H 2 graph are Based on the edge weight function that has been obtained, a rainbow path can be formed if there are as many colors as mn. The rainbow path of H 2 graph that is formed can be seen in the  Table 2. The rainbow path from x to y on H 2 graph. As a result of the explanation above, we have established that the lower and upper bounds of rac(H 2 ) are mn ≤ rac(H 2 ) ≤ mn. So that the correctness of the formula rac(H 2 ) is mn with m, n ≥ 2 has been proven.
For illustration of rac(H 2 ) is provided in Figure 3 and the vertex labels on the graph are black numbers, while the edge weights on the graph are blue numbers.
We need to prove the correctness of the formula rac(H 3 ) is 2dm − 2m with d, m ≥ 2, it is necessary to prove it by using the lower and upper bounds, namely rac( First, we prove the correctness of the lower bound of rac(H 3 ) is rac(H 3 ) ≥ 2dm − 2m. We know that H 3 graph is classified as a tree graph. Based on Proposition 1 [6] and Theorem 1 [24], so that: Obviously, we get the edge weights from a predefined vertex function where the edge weights will be used as edge coloring. The edge weights of H 3 graph are Based on the edge weight function that has been obtained, a rainbow path can be formed if there are as many colors as 2dm − 2m. The rainbow path of H 3 graph that is formed can be seen in the Table 3. So we got that the upper bound of rac(H 3 ) is rac(H 3 ) ≤ 2dm − 2m. Table 3. The rainbow path from x to y on H 3 graph.
As a result of the explanation above, we have established that the lower and upper bounds of rac(H 3 ) are 2dm − 2m ≤ rac(H 3 ) ≤ 2dm − 2m. So that the correctness of the formula rac(H 3 ) is 2dm − 2m with d, m ≥ 2 has been proven. Figure 4 and the vertex labels on the graph are black numbers, while the edge weights on the graph are blue numbers.   (P (3,n) , v, m) graph with m, n ≥ 2, then rac(H 4 ) = mn + 2.

Proof. Let H 4 be a vertex amalgamation product of paw with vertex set
We need to prove the correctness of the formula rac(H 4 ) is mn + 2 with m, n ≥ 2, it is necessary to prove it by using the lower and upper bounds, namely rac(H 4 ) ≥ mn + 2 and rac(H 4 ) ≤ mn + 2.
First, we prove the correctness of the lower bound of rac(H 4 ) is rac(H 4 ) ≥ mn + 2. We know that H 4 graph is P (3,n) graph which is copied as many as m. Based on Theorem 3 [4], so that H 4 graph requires as many colors as mn + 2. So we got that the lower bound of rac(H 4 ) is rac(H 4 ) ≥ mn + 2 with m, n ≥ 2.
Next, we prove the correctness of the upper bound of rac(H 4 ) is rac(H 4 ) ≤ mn + 2. We define the vertex function of H 4 graph with f : Obviously, we get the edge weights from a predefined vertex function where the edge weights will be used as edge coloring. The edge weights of H 4 graph are Based on the edge weight function that has been obtained, a rainbow path can be formed if there are as many colors as mn + 2. The rainbow path of H 4 graph that is formed can be seen in the Table 4. So we got that the upper bound of rac(H 4 ) is rac(H 4 ) ≤ mn + 2. Table 4. The rainbow path from x to y on H 4 graph.
As a result of the explanation above, we have established that the lower and upper bounds of rac(H 4 ) are mn + 2 ≤ rac(H 4 ) ≤ mn + 2. So that the correctness of the formula rac(H 4 ) is mn + 2 with m, n ≥ 2 has been proven.
For illustration of rac(H 4 ) is provided in Figure 5 and the vertex labels on the graph are black numbers, while the edge weights on the graph are blue numbers. Theorem 9 If H 5 is Amal(F n , v, m) graph with m, n ≥ 2, then rac(H 5 ) = mn + 1.

Proof. Let H 5 be a vertex amalgamation product of star with vertex set
We need to prove the correctness of the formula rac(H 5 ) is mn + 1 with m, n ≥ 2, it is necessary to prove it by using the lower and upper bounds, namely rac(H 5 ) ≥ mn + 1 and rac(H 5 ) ≤ mn + 1.
Next, we prove the correctness of the upper bound of rac(H 5 ) is rac(H 5 ) ≤ mn + 1. We define the vertex function of H 5 graph with f : , if n = even and b = even Based on the edge weight function that has been obtained, a rainbow path can be formed if there are as many colors as mn + 1. The rainbow path of H 5 graph that is formed can be seen in the Table 5. So we got that the upper bound of rac(H 5 ) is rac(H 5 ) ≤ mn + 1. Table 5. The rainbow path from x to y on H 5 graph.
As a result of the explanation above, we have established that the lower and upper bounds of rac(H 5 ) are mn + 1 ≤ rac(H 5 ) ≤ mn + 1. So that the correctness of the formula rac(H 5 ) is mn + 1 with m, n ≥ 2 has been proven.
For illustration of rac(H 5 ) is provided in Figure 6 and the vertex labels on the graph are black numbers, while the edge weights on the graph are blue numbers.  Proof. Let H 6 be a vertex amalgamation product of star with vertex set V ( So, we get |V (H 6 )| = mn + m + 1 and |E(H 6 )| = 2mn + m.
We need to prove the correctness of the formula rac(H 6 ) is mn + 2m with m, n ≥ 2, it is necessary to prove it by using the lower and upper bounds, namely rac(H 6 ) ≥ mn + 2m and rac(H 6 ) ≤ mn + 2m.
First, we prove the correctness of the lower bound of rac(H 6 ) is rac(H 6 ) ≥ mn + 2m. We know that H 6 graph is T b n graph which is copied as many as m. Based on Theorem 4 [1], so that H 6 graph requires as many colors as mn + 2m. So we got that the lower bound of rac(H 6 ) is rac(H 6 ) ≥ mn + 2m with m, n ≥ 2.
Next, we prove the correctness of the upper bound of rac(H 6 ) is rac(H 6 ) ≤ mn + 2m. We define the vertex function of H 6 graph with f : Obviously, we get the edge weights from a predefined vertex function where the edge weights will be used as edge coloring. The edge weights of H 6 graph are w(zx a ) = a + 2 , 1 ≤ a ≤ m w(zy a,b ) = m + (n − 1)(a − 1) + b + 2 , 1 ≤ a ≤ m & 1 ≤ b ≤ n w(x i y a,b ) = m + (n − 1)(a − 1) + a + b + 2 , 1 ≤ a ≤ m & 1 ≤ b ≤ n Based on the edge weight function that has been obtained, a rainbow path can be formed if there are as many colors as mn + 2m. The rainbow path of H 6 graph that is formed can be seen in the Table 6. So we got that the upper bound of rac(H 6 ) is rac(H 6 ) ≤ mn + 2m. Table 6. The rainbow path from x to y on H 6 graph.
As a result of the explanation above, we have established that the lower and upper bounds of rac(H 6 ) are mn + 2m ≤ rac(H 6 ) ≤ mn + 2m. So that the correctness of the formula rac(H 6 ) is mn + 2m with m, n ≥ 2 has been proven.
For illustration of rac(H 6 ) is provided in Figure 7 and the vertex labels on the graph are black numbers, while the edge weights on the graph are blue numbers.

Conclusion
In this paper, we learned about the rainbow antimagic coloring of vertex amalgamation of graphs. We have concluded the exact value of rac(G) where G is vertex amalgamation of graphs namely path, star, broom, paw, fan, and triangular book graph. The open problems in this paper include. Open Problem (i) Determine rac(G) where G is graph with another operations in graph such as comb, corona, edge corona, etc. (ii) Determine rac(G) where G is another special graph or another graph family.