# On Edge Magic Total Labeling of (7, 3)-Cycle Books

Authors: Baki Swita; Ulfasari Rafflesia; Nova Henni Ms; Dolly Stio Adji; Mudin Simanihuruk
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

A graphGis called (a, b)-cycle books B[(Ca, m),(Cb, n), Pt] ifGconsists ofmcyclesCa andncyclesCb with a common pathPt. In this article we show that the (7,3)-cycle booksB[(C7,1),(C3, n), P2] admits edge-magic total labeling. In addition we prove that (a,3)-cycle booksB[(Ca,1),(C3, n), P2] admits super edge-magic total labeling fora = 7 and extends the values ofato 4x− 1 for any positive integerx. Moreover we prove that the (7, 3)-cycle booksB[(C7,2),(C3, n), P2] admits super edge-magic total labeling.


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## Body

## 1. Introduction

Let G be a graph such that|V (G)| = p and |E(G)| = q. Anedge-magic total labelingof G is a bijective function f: V (G)∪E(G) → {1, 2,..., p+q} such that f (w) + f (wz) + f (z) = k for any edge wz∈ E(G). Moreover G is said to be anedge-magic total. If f (V (G)) = {1, 2,..., p}, then f isthe super edge-magic total labelingof G and G is said to be asuper edge-magic total.In this paper we generalize the definition of cycle books [1] and we investigate their edge-magic total labeling. Let m, n, a, b, and t be any positive integers and let Ca and Cb be the cycles of order a and b, respectively. A graph G is called (a, b)-cycle booksB[(Ca, m), (Cb, n), Pt] if G consists of m cycles Ca and n cycles Cb with a common path Pt.From now on the graphs (a, b)-cycle books B[(Ca, m), (Cb, n), Pt] is denoted by B(a, m, b, n, t). If a = b = 4, m + n = r, and t = 2, then the graphs B(4, m, 4, n, t) are the graphs cycle books Br [2]. If a = b, m + n = r, and t = 2, then the graphs B(a, m, b, n, t) are the graphs cycle books Ba,r [1] and it is the graphs cycle books Θ(Ca)r [3]. If m = n = 1 and t = 2, then the graphs B(a,m,b,n,t) are a cycle Ca+b-2x with a chord x=min(a,b)-1 [4]. If a = b = 2r + 1, m + n ≥ 2, and t = 2r, then the graphs B(a, m, b, n, t) are called a graph with many odd cycles P2r(+)Nm+n [5].It is an open problem to determine the edge-magic total labeling of B(a, m, b, n, t) (see Research Problem 2.7, p.39 [1] for the case B(a, m, a, n, 2)). However some authors provided a partial solution to this problem. Figueroa-Centeno, Ichishima, and Muntaner-Batle [2] proved that B(4, m, 4, n, 2) is an edge-magic total. Moreover they proved that B(4, m, 4, n, 2) is not super edge-magic total for m + n ≡ 1, 3, 7 mod 8 and m + n = 4, B(4, m, 4, n, 2) is super edge-magic total for m + n = 2, 5, 6, 8, 10, 11 and conjecture that B(4, m, 4, n, 2) is super edge-magic total if and only if m + n is even or m + n ≡ 5(mod 8). Gallian [6] pointed out that Yuansheng et al. [7] proved this conjecture when m + n is even. Later on Yegnanarayanan and Vaidhyanathan [8] proved that B(4, m, 4, n, 2) is an edge-magic total.Note that C(2x+1)(2y+1) [+] Nn ([9], p.7) is isomorphic to B((2x + 1)(2y + 1), 1, 4, n, 3) and P2x(+)Nn [5] is isomorphic to B(2x+1, μ, 2x+1, ν, 2x), μ+ν = n. Singgih provided a new method to construct super edge-magic total labeling of graph C(2x+1)(2y+1) [+] Nn ([9], Theorem 4.13.1,p.93) from the super edge-magic total labeling of P2x(+)Nn. Hence super edge-magic total labeling of B((2x+1)(2y+1), 1, 4, n, 3) is deduced from the super edge-magic total labeling of B(2x + 1, μ, 2x + 1, ν, 2x), μ + ν = n.This paper is organized as follows. In Section3.1 we prove that the graphs B(7, m, 3, n, 2) are an edge-magic total for any integer n ≥ 1 and m = 1. In Section 3.2 we prove that B[(Ca, 1), (C3, n), P2] admits super edge-magic total for a = 7 and extends the values of a to 4x − 1 for any positive integers x. Moreover we prove that the graphs B[(C7, 2), (C3, n), P2] admit super edge- magic total. The results of this paper are developed from [10, 11].


## 2. Preliminary Notes

In this section we provide some previous results on super edge-magic total labeling of a graph. Figueroa-Centeno, Ichishima, and Muntaner-Batle [2] proved some necessary conditions for super edge-magic total labeling of a graph. We need them to prove the main results of this paper.Theorem 1 (see [2]).
A(p,q)-graph G is super edge-magic if and only if there exists a bijective function (1)f:VG→1,2,...,psuch that the set S = {f(w) + f(z): wz ∈ E(G)} exists of q consecutive integers. In such a case, f extends to a super edge-magic labeling of G with magic constant k = p + q + s, where s = min⁡(S) and S = {k – (p+1), k – (p + 2),...., k – (p + q)}.Theorem 2 (see [2]).
Let G be super edge-magic labeling(p,q)-graph and f be a super edge-magic labeling of G. Then (2)∑v∈VGfvdegv=qs+12q-1q.In particular 2∑v∈V(G)f(v)deg(v)=0(modq).


## 3. Main Results

We sometimes interchange the role of vw and (v, w) as an element of E(G). From now on we assume that B(a, m, b, n, 2) is defined as follows. Let G be the graphs B(a, m, b, n, 2). Let A ={v10, va1}, B = {v21, v31,..., va-11, v22, v32,..., va-12,..., v2m, v3m,..., va-1m}, and D = {w21, w31,..., wb-11, w22, w32,..., wb-12,..., w2n, w3n,..., wb-1n}. We define the vertex set V (G) = A∪ B∪ D and the edge set E(G) = {(v10,va1)}∪{(vi-1j, vij ), 3 ≤ i ≤ a-1, 1 ≤ j ≤ m}∪{(wi-1j, wij ), 3 ≤ i ≤ b-1, 1 ≤ j ≤ n}∪{(v10, v2i), (v10, w2j ), 1 ≤ i ≤ m, 1 ≤ i ≤ n}∪{(va0, va-1i), (va0, wb-1j), 1 ≤ i ≤ m, 1 ≤ j ≤ n}. A graph B(a, m, b, n, 2) is shown in Figure 1.Figure 1
Graphs B(a,m,b,n,2).


### 3.1. Edge Magic Total Labeling of (7,3)-Cycle Books

In this section we provide an edge-magic total labeling of (7, 3)-cycle books B(7, 1, 3, n, 2).Theorem 3.
Let G be a(7, 3)-cycle books B(7, 1, 3, n, 2). Then G is an edge-magic with magic constant k = 4n + 20.Proof.
Let G be a (7,3)-cycle books B(7, 1, 3, n, 2) in Figure1 with |V(G)| = p and |E(G)| = q. We first notice that p = n + 7 and q = 2n + 7. We define the bijective function f: V (G)∪ E(G) → {1, 2,..., p + q} or f: V (G)∪ E(G) → {1, 2,..., 3n + 14} as in (3):(3)fx=1,ifx=v10,3n+14,ifx=v10v21,n+5,ifx=v21,2n+11,ifx=v21v31,n+4,ifx=v31,2n+9,ifx=v31v41,n+7,ifx=v41,n+3,ifx=v41v51,2n+10,ifx=v51,2n+8,ifx=v51v61,2,ifx=v61,3n+12,ifx=v61v70,n+6,ifx=v70,3n+13,ifx=v70v10,n+i+7,ifx=v10w2i,1≤i≤n,3n-i+12,ifx=w2i,1≤i≤n,i+2,ifx=v70w2i,1≤i≤n.It is easy to verify thatf(w)+ f(wz)+ f(z)= 4n + 20 for any edge wz∈E(G), hence the theorem.


### 3.2. Super Edge Magic Total Labeling of (7,3)-Cycle Books

In this section we prove that the graphs B(a, 1, 3, n, 2) are super edge-magic total for a = 4x−1, x ≥ 2, x is integer, and the graphs B(7, 2, 3, n, 2) are super edge-magic total for any positive integer n. First we provide some lemmas.Lemma 4.
Let G be the graphs B(a, m, b, n, 2) in Figure 1, |V(G)| = p, |E(G)| = q, and let f be a super edge-magic total labeling of G.
Let S = {f (w) + f (z): f or any edge wz ∈ E(G)}beqconsecutive integers and s=min⁡(S). Then(4)fv10+fva0=qs+1/2qq-1-pp-1m+n-1.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1. Let |V(G)| = p and |E(G)| = q. We first notice that deg(v10) = deg(va0) = m + n + 1, deg(v21) = deg(v31) =... = deg(va-11) = 2,..., deg(v2m) = deg(v3m) =... = deg(va-1m) = 2, deg(w21) = deg(w31) =... = deg(wb-11) = 2,..., deg(w2n) = deg(w3n) =... = deg(wb-1n) = 2.
By Theorem2 we have (m+n+1)[f(v10)+f(va0)] + ∑v∈V(G)-{v10,va0}f(v)2 = qs+ q(q-1)/2. Hence (m+n-1)[f(v10)+f(va0)] + ∑v∈V(G)f(v)2 = qs+ q(q-1)/2. Moreover (m+n-1)[f(v10)+f(va0)] + 2(1 + 2 +...+ p) = qs+ q(q-1)/2. From the last equation we conclude that (5)fv10+fva0=qs+1/2qq-1-pp+1m+n-1hence the lemma.Lemma 5.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, and b = 3 and let f be a super edge-magic total labeling of G. Let S = {f(w)+f (z): f or any edge wz ∈ E(G)}beqconsecutive integers andsands = min⁡(S). Then(6)3m+n-1+pp+1q-12qq-1≤s≤m+n-12p-1+pp+1q-12qq-1Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, and b = 3. Let |V(G)| = p and |E(G)| = q. We first prove the upper bound of s. Notice that f (v10) + f (v70) ≤ p + p – 1. By Lemma 4 and the last inequality we have (7)s≤m+n-12p-1+pp+1q-12qq-1hence the upper bound of s. Next we prove the lower bound of s. Notice that f (v10) + f (v70) ≥ 3. By Lemma 4 and the last inequality we have s ≥ (3(m+n-1)+p(p+1))/q - 1/2q(q-1), hence the lemma.Lemma 6.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, b = 3, m = 1, and n = 1 and let f be an edge-magic total labeling of G. Let S = {f(w) + f (z): f or any edge wz ∈ E(G)} be q consecutive integers and s = min⁡(S). Then G is a super edge-magic total if and only if(i)
f(v10) + f(v70) =9s – 36,(ii)
3 ≤ s ≤ 5,(iii)
k = p + q + s.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, b = 3, m = 1, and n = 1 and let G be a super edge-magic total.
Note that p = (a−2)m+2+(b−2)n and q = (a−3)+2m+1+(b−3)n+2n. Let f be an edge-magic total labeling of G. If f is a super edge-magic total labeling of G, the conditions (i) and (ii) follow from (4) and (6) of Lemmas 4 and 5, respectively, and the condition (iii) follows from Theorem 1.
Let f satisfy the conditions (i), (ii), and (iii). We will prove that G is super edge-magic total for a = 7, b = 3, m = 1, and n = 1. We first notice that p = 8 and q = 9.
Claim 1 (s = 5). If s = 4, then f (v10) + f (v70) = 0, a contradiction to f (v10) + f (v70) ≥ 3, hence s = 5 and the claim.
By (iii) and Claim 1, we have k = 22 and by (i) we have f (v10) + f (v70) = 9. Hence either f (v10) = 1 and f (v70) = 8, f (v10) = 2 and f (v70) = 7, f (v10) = 3 and f (v70) = 6, or f (v10) = 4 and f (v70) = 5.
Case 1 (f (v10) = 1 and f (v70) = 8). Using k = 22, we define the bijection f1(x) in (8) such that f1(v10) = f (v10) = 1 and f1(v70) = f (v70) = 8.(8)f1x=1,ifx=v10,6,ifx=v21,2,ifx=v31,3,ifx=v41,7,ifx=v51,4,ifx=v61,8,ifx=v70,5,ifx=w21,The function f1(x) is depicted in Figure 2.
Using the definition of f1(x) we get the following: f1 (v31) +f (v41) = 5, f1 (v10) + f1 (w21) = 6, f1 (v10) + f1 (v21) = 7, f1 (v21) + f1 (v31) = 8, f1 (v10) + f1 (v70) = 9, f1 (v41) + f1 (v51) = 10, f1 (v51) + f1 (v61) = 11, f1 (v61) + f1 (v70) = 12, and f1 (v70) + f1 (w21) = 13. Hence S = {f1(u) + f1(w): uw∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f1(uv) =k − (f1(u) + f1(v)) for all uv∈ E(G).
Case 2 (f (v10) = 2 and f (v70) = 7). Using k = 22, we define the bijection f2(x) in (9) such that f2(v10) = f (v10) = 2 and f2(v70) = f (v70) = 7.(9)f2x=2,ifx=v10,6,ifx=v21,1,ifx=v31,5,ifx=v41,8,ifx=v51,4,ifx=v61,7,ifx=v70,3,ifx=w21.Using the definition of f2(x) we get the following: f2(v10) + f2(w21) = 5, f2 (v31) + f2(v41) = 6, f2 (v21) + f2 (v31) = 7, f2 (v10) + f2 (v21) = 8, f2 (v10) + f2 (v70) = 9, f2 (v70)+f2 (w21) = 10, f2 (v61)+ f2 (v70) = 11, f2(v51)+f2 (v61) = 12, and f2 (v41)+f2 (v51) = 13. Hence S = {f2(u) + f2(w): uw ∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f2(uv) = k − (f2(u) + f2(v)) for all uv ∈ E(G) and it is shown in Figure 3.
Case 3 (f (v10) = 3 and f (v70) = 6). Using k = 22, we define the bijection f3(x) in (10) such that f3(v10) = f (v10) = 3 and f3(v70) = f (v70) = 6.(10)f3x=3,ifx=v10,7,ifx=v21,4,ifx=v31,8,ifx=v41,5,ifx=v51,1,ifx=v61,6,ifx=v70,2,ifx=w21,The function f3(x) is depicted in Figure 4. Using the definition of f3(x) we get the following: f3(v10) + f3(w21) = 5, f3 (v51) + f3 (v61) = 6, f3 (v61) + f3 (v70) = 7, f3 (v70) + f3 (w21) = 8, f3 (v10) + f3 (v70) = 9, f3 (v10) + f3 (v21) = 10, f3 (v21) + f3 (v31) = 11, f3 (v31) + f3 (v41) = 12, and f3 (v41) + f3 (v51) = 13. Hence S = {f3(u) + f3(w): uw ∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f3(uv) = k − (f3(u) + f3(v)) for all uv ∈ E(G).
Case 4 (f(v10) = 4 and f(v70) = 5). Using k = 22, we define the bijection f4(x) in (11) such that f4(v10) = f (v10) = 4 and f4(v70) = f (v70) = 5.(11)f4x=4,ifx=v10,7,ifx=v21,3,ifx=v31,2,ifx=v41,6,ifx=v51,1,ifx=v61,5,ifx=v70,8,ifx=w21,The function f4(x) is depicted in Figure 5. Using the definition of f4(x) we get the following: f4 (v31) + f4 (v41) = 5, f4 (v61) + f4 (v70) = 6, f4 (v51) + f4 (v61) = 7, f4 (v41) + f4 (v51) = 8, f4 (v10) + f4 (v70) = 9, f4 (v21) + f4 (v31) = 10, f4 (v10) + f4 (v21) = 11, f4 (v10) + f4 (w21) = 12, and f4 (v70) + f4 (w21) = 13. Hence S = {f4(u) + f4(w): uw ∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f4(uv) = k − (f4(u) + f4(v)) for all uv ∈ E(G), hence the lemma.Figure 2
Bijective functionf1(x).Figure 3
Bijective functionf2(x).Figure 4
Bijective functionf3(x).Figure 5
Bijective functionf4(x).Now we are ready to state our main result.Theorem 7.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, m = 1,andb = 3. ThenGis a super edge-magic total. Moreover there are two bijective functionsg1(V (G))andg2(V (G))with the magic constant k = 3n + 19.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, m = 1, and b = 3. We first notice that p = (a − 2)m + (b − 2)n + 2 and q = (a − 3)m + 2m + 1 + (b − 3)n + 2n. For a = 7, m = 1, and b = 3 and n = 1 we have p = 8 and q = 9. Using the definition of the bijective function f1(x) in Case 1 of Lemma 6 together with s = 5 in Claim 1 of Lemma 6, we define the following bijective function g1(x) in (12):(12)g1x=1,ifx=v10,n+5,ifx=v21,2,ifx=v31,3,ifx=v41,n+6,ifx=v51,4,ifx=v61,n+7,ifx=v70,i+4,ifx=w2i,1≤i≤nThe function g1(x) is depicted in Figure 6. Using the definition of g1(x) we get the following: g1(v31) + g1(v41) = 5, g1(v10) + g1(w21) = 6, g1(v10) + g1(w2i) = i + 5, i∈{2, 3,..., n}, g1(v10) + g1(v21) = n + 6, g1(v21)+g(v31) = n+7, g1(v10) + g1(v70) = n+8, g1(v41)+g(v51) = n+9, g1(v51)+g1(v61) = n + 10, g1(v61) + g1(v70) = n + 11, g1(v70) + g1 (w2i) = n + 11 + i, i∈{1, 2, 3,..., n}. Hence S = {g1(u) + g1(w): uw∈ E(G} = {5, 6,..., i + 5, n + 7, n +8, n+9, n+10, n+11, n+i+11, i∈{1,..., n}} consists of q = 2n+7 consecutive integers. Thus by Theorem 1 and Claim 1 of Lemma 6, we conclude that G is super edge-magic total with magic constant k = p+q+s = n+7+2n+7+5 = 3n + 19. Moreover, by Theorem 1, we conclude g1(uv) = k − (g1(u) + g1(v)) for all uv∈ E(G).
Using the definition of the bijective function f2(x) in Case 2 of Lemma 6 together with s = 5 in Claim 1 of Lemma 6, we define the following bijective function g2(x) in (13):(13)g2x=2,ifx=v10,n+5,ifx=v21,1,ifx=v31,n+4,ifx=v41,n+7,ifx=v51,n+3,ifx=v61,n+6,ifx=v70,i+2,ifx=w2i,1≤i≤nThe function g2(x) is depicted in Figure 7. Using the definitions of g2(x) we get the following: g2(v10) + g2(w2i) = i +4, i∈{1, 2,..., n}, g2(v31) + g2(v41) = n + 5, g2(v21) + g2(v31) = n + 6, g2(v10) + g2(v21) = n+7, g2(v10)+g2(v70) = n+8, g2(v70)+g2(w2i) = n+8+i, i∈{1, 2,..., n}, g2(v70) + g2(v61) = 2n + 9, g2(v61) + g2(v51) = 2n + 10, g2(v51) + g2(v41) = 2n + 11. Hence S = {g2(u) + g2(w): uw∈ E(G} = {5, 6,..., i + 4, n + 5, n + 6, n + 7, n + 8, n + 8 + i, 2n + 9, 2n + 10, 2n + 11, i∈{1, 2,..., n}} consists of q = 2n + 7 consecutive integers. Thus by Theorem 1 and Claim 1 of Lemma 6, we conclude that G is super edge-magic total with magic constant k = p + q + s = n + 7 + 2n + 7 + 5 = 3n + 19.
Moreover, by Theorem1, we conclude g2(uv) = k − (g2(u) + g2(v)) for all uv∈ E(G), hence the theorem.Figure 6
Bijective functiong1(x).Figure 7
Bijective functiong2(x).Theorem 8.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 such that |V(G)| = p, |E(G)| = q, a = 4x − 1, m = 1, and b = 3.
Letα2: V (G) → {1, 2,..., p} be a bijective function such that(14)α2vij=n-1+2x+12i+2,ifiiseven,2≤i≤2x-2,j=1,α2vij=12i+2,ifiiseven,2x≤i≤4x-2,j=1,α2vij=12i+2,i=1,j=0,α2vij=12i+1,ifiisodd,1≤i≤2x-1,j=1,α2vij=n-1+2x+12i+1,ifiisodd,2x+1≤i≤4x-3,j=1,α2vij=n-1+2x+12i+1,ifi=4x-1,j=0,α2wij=2x+j,2≤i≤b-2,1≤j≤n.Then α2(v) is a super edge-magic total labeling of G with the magic constant k = 10x + 3n − 1, v ∈ V (G).Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p and |E(G)| = q. Let a = 4x − 1, m = 1, and b = 3. Notice that p = 4x − 1 + n and q = 2n + 4x − 1. Let S = {α2(u) + α2(w): uw∈ E(G}. We show that S consists of q consecutive integers.
Using the definition ofα2(v) in (14), v∈ V (G), we get the following:(15)α2v2x-11+α2v2x1=2x+1,α2v10+α2w2i=2x+1+i,1≤i≤n,α2v10+α2v21=n+2x+2,α2v21+α2v31=n+2x+3,⋯α2vx-21+α2vx-11=n+3x-1,α2vx-11+α2vx1=n+3x,α2vx1+α2vx+11=n+3x+1,⋯α2v2x-21+α2v2x-11=n+4x-1,α2v10+α2v4x-10=n+4x,α2v2x1+α2v2x+11=n+4x+1,⋯α2v3x-21+α2v3x-11=n+5x-1,α2v3x-11+α2v3x1=n+5x,α2v3x1+α2v3x+11=n+5x+1,⋯α2v4x-21+α2v4x-11=n+6x-1,α2v4x-10+α2w2i=n+6x-1+i,1≤i≤n.Hence by (15) we have S = {2x + 1, 2x + 2, 2x + 3,..., 2x + n + 1, 2x + n + 2,..., 3x + n −1, 3x + n, 3x + n + 1,..., 4x + n − 1, 4x + n, 4x + n + 1,..., 5x + n − 1, 5x + n, 5x +n + 1,..., 6x + n − 1, 6x + n, 6x + n + 1,..., 6x + 2n − 2, 6x + 2n − 1} consists of q consecutive integers. Thus by Theorem 1 we conclude that G is super edge-magic total. Notice that s = 2x + 1. Hence k = p + q + s = 10x + 3n − 1. Moreover, by Theorem 1, we conclude α2(uv) = k − (α2(u) + α2(v)) for all uv∈ E(G), hence the theorem.Lemma 9.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, m = 2, b = 3, and n = 1 and let f be an edge-magic total labeling of G. Let S = {f(w) + f (z): f or any edge wz ∈ E(G)} be q consecutive integers and s = min⁡(S). Then G is a super edge-magic total if and only if(i)
f(v10)+f(v70)=(15s-77)/2,(ii)
6 ≤ s ≤ 7,(iii)
k = p + q + s.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, m = 2, b = 3, and n = 1 and let G be a super edge-magic total. Let f be an edge-magic total labeling of G. If f is a super edge-magic total labeling of G, the conditions (i) and (ii) follow from Lemmas 4 and 5, respectively, and the condition (iii) follows from Theorem 1.
Let f satisfy the conditions (i), (ii), and (iii). We first notice that p = (a − 2)m + (b − 2)n + 2 and q = (a − 3)m + 2m + 1 + (b − 3)n + 2n. We will prove that G is super edge-magic total for a = 7, m = 2, b = 3, and n = 1. Notice that p = 13 and q = 15.
Claim 1 (s = 7). If s = 6, then f(v10)+f(v70)= 6,5 (by (i)); a contradiction to f(v10)+f(v70) is an integer, hence s = 7 and the claim. By (iii) and Claim 1, we have k =35 and by (i) we have f(v10)+f(v70) = 14.
Next we will show that there is a bijection h1(x) such that h1(v10)= f(v10), h1(v70)=f(v70), h1(v10)+h1(v70) = 14, and h1(x) is a super edge-magic total labeling of B(7, 2, 3, 1). Let h1(v10)= f(v10) = 2 and h1(v70)=f(v70)=12. Using k = 35, we define the bijection h1(x) in (16):(16)h1x=2,ifx=v10,6,ifx=v21,13,ifx=v31,3,ifx=v41,10,ifx=v51,8,ifx=v61,12,ifx=v70,7,ifx=v22,4,ifx=v32,11,ifx=v42,1,ifx=v52,9,ifx=v62,5,ifx=w21.
Using the definition of h1(x) we get the following: h1(v10) + h1(w21) = 7, h1(v10)+h1(v21) = 8, h1(v10)+h1(v22) = 9, h1(v52)+h1(v62) = 10, h1(v22)+h1(v32) = 11, h1(v42) + h1(v52) = 12, h1(v41) + h1(v51) = 13, h1(v10) + h1(v70) = 14, h1(v32) + h1(v42) = 15, h1(v31) + h1(v41) = 16, h1(v70) + h1(w21) = 17, h1(v51) + h1(v61) = 18, h1(v21) + h1(v31) = 19, h1(v61) + h1(v70) = 20, h1(v62) + h1(v70) = 21. Hence S = {h1(u) + h1(w): uw∈ E(G} = {7, 8,..., 21} consists of q = 15 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude h1(uv) = k − (h1(u) + h1(v)) for all uv∈ E(G) and it is shown in Figure 8, hence the theorem.Figure 8
Bijective functionh1(x).Theorem 10.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p and |E(G)| = q, a = 7, m = 2, and b = 3. Then G is a super edge-magic total with the magic constant k = 3n + 32.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, m = 2, and b = 3. We first notice that p = (a − 2)m + (b − 2)n + 2 and q = (a − 3)m + 2m + 1 + (b − 3)n + 2n. Using the definition of the bijective function h1(x) and s = 7 in Lemma 9, we define the following bijective function h2(x) in (17):(17)h2x=2,ifx=v10,n+5,ifx=v21,n+12,ifx=v31,3,ifx=v41n+9,ifx=v51n+7,ifx=v61,n+11,ifx=v70,n+6,ifx=v22,4,ifx=v32,n+10,ifx=v42,1,ifx=v52,n+8,⋯ifx=v62i+4,1≤i≤n.The function h2(x) is depicted in Figure 9. Using the definition of h2(x) we get the following: h2(v10)+h2(w2i) = i + 6, i∈{1,..., n}, h2(v10) + h2(v21) = n + 7, h2(v10)+h2(v22) = n+8, h2(v52)+h2(v62) = n+9, h2(v22) + h2(v32) = n + 10, h2(v42) + h2(v52) = n + 11, h2(v41) + h2(v51) = n + 12, h2(v10) + h2(v70) = n + 13, h2(v32) + h2(v42) = n + 14, h2(v31) + h2(v41) = n + 15, h2(v70) + h2(w2i) = n + 15 + i, i∈{1,..., n}, h2(v51) + h2(v61) = 2n + 16, h2(v21) +h2(v31) = 2n + 17, h2(v70) + h2(v61) = 2n + 18, h2(v70) + h2(v62) = 2n + 19. Hence S = {h2(u) + h2(w): uw∈ E(G} = {7, 8,..., n + 6, n + 7, n + 8, n + 9, n +10, n + 11, n + 12, n + 13, n + 14, n + 15, n + 16,..., 2n + 15, 2n + 16, 2n + 17, 2n +18, 2n + 19} consists of q = 2n + 13 consecutive integers. Thus by Theorem 1 and Claim 1 of Lemma 9, we conclude that G is super edge-magic total with magic constant k = p + q + s = n + 12 + 2n + 13 + 7 = 3n + 32. Moreover, by Theorem 1, we conclude h2(uv) = k − (h2(u) + h2(v)) for all uv∈ E(G), hence the theorem.Figure 9
Bijective functionh2(x).


## 3.1. Edge Magic Total Labeling of (7,3)-Cycle Books

In this section we provide an edge-magic total labeling of (7, 3)-cycle books B(7, 1, 3, n, 2).Theorem 3.
Let G be a(7, 3)-cycle books B(7, 1, 3, n, 2). Then G is an edge-magic with magic constant k = 4n + 20.Proof.
Let G be a (7,3)-cycle books B(7, 1, 3, n, 2) in Figure1 with |V(G)| = p and |E(G)| = q. We first notice that p = n + 7 and q = 2n + 7. We define the bijective function f: V (G)∪ E(G) → {1, 2,..., p + q} or f: V (G)∪ E(G) → {1, 2,..., 3n + 14} as in (3):(3)fx=1,ifx=v10,3n+14,ifx=v10v21,n+5,ifx=v21,2n+11,ifx=v21v31,n+4,ifx=v31,2n+9,ifx=v31v41,n+7,ifx=v41,n+3,ifx=v41v51,2n+10,ifx=v51,2n+8,ifx=v51v61,2,ifx=v61,3n+12,ifx=v61v70,n+6,ifx=v70,3n+13,ifx=v70v10,n+i+7,ifx=v10w2i,1≤i≤n,3n-i+12,ifx=w2i,1≤i≤n,i+2,ifx=v70w2i,1≤i≤n.It is easy to verify thatf(w)+ f(wz)+ f(z)= 4n + 20 for any edge wz∈E(G), hence the theorem.


## 3.2. Super Edge Magic Total Labeling of (7,3)-Cycle Books

In this section we prove that the graphs B(a, 1, 3, n, 2) are super edge-magic total for a = 4x−1, x ≥ 2, x is integer, and the graphs B(7, 2, 3, n, 2) are super edge-magic total for any positive integer n. First we provide some lemmas.Lemma 4.
Let G be the graphs B(a, m, b, n, 2) in Figure 1, |V(G)| = p, |E(G)| = q, and let f be a super edge-magic total labeling of G.
Let S = {f (w) + f (z): f or any edge wz ∈ E(G)}beqconsecutive integers and s=min⁡(S). Then(4)fv10+fva0=qs+1/2qq-1-pp-1m+n-1.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1. Let |V(G)| = p and |E(G)| = q. We first notice that deg(v10) = deg(va0) = m + n + 1, deg(v21) = deg(v31) =... = deg(va-11) = 2,..., deg(v2m) = deg(v3m) =... = deg(va-1m) = 2, deg(w21) = deg(w31) =... = deg(wb-11) = 2,..., deg(w2n) = deg(w3n) =... = deg(wb-1n) = 2.
By Theorem2 we have (m+n+1)[f(v10)+f(va0)] + ∑v∈V(G)-{v10,va0}f(v)2 = qs+ q(q-1)/2. Hence (m+n-1)[f(v10)+f(va0)] + ∑v∈V(G)f(v)2 = qs+ q(q-1)/2. Moreover (m+n-1)[f(v10)+f(va0)] + 2(1 + 2 +...+ p) = qs+ q(q-1)/2. From the last equation we conclude that (5)fv10+fva0=qs+1/2qq-1-pp+1m+n-1hence the lemma.Lemma 5.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, and b = 3 and let f be a super edge-magic total labeling of G. Let S = {f(w)+f (z): f or any edge wz ∈ E(G)}beqconsecutive integers andsands = min⁡(S). Then(6)3m+n-1+pp+1q-12qq-1≤s≤m+n-12p-1+pp+1q-12qq-1Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, and b = 3. Let |V(G)| = p and |E(G)| = q. We first prove the upper bound of s. Notice that f (v10) + f (v70) ≤ p + p – 1. By Lemma 4 and the last inequality we have (7)s≤m+n-12p-1+pp+1q-12qq-1hence the upper bound of s. Next we prove the lower bound of s. Notice that f (v10) + f (v70) ≥ 3. By Lemma 4 and the last inequality we have s ≥ (3(m+n-1)+p(p+1))/q - 1/2q(q-1), hence the lemma.Lemma 6.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, b = 3, m = 1, and n = 1 and let f be an edge-magic total labeling of G. Let S = {f(w) + f (z): f or any edge wz ∈ E(G)} be q consecutive integers and s = min⁡(S). Then G is a super edge-magic total if and only if(i)
f(v10) + f(v70) =9s – 36,(ii)
3 ≤ s ≤ 5,(iii)
k = p + q + s.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, b = 3, m = 1, and n = 1 and let G be a super edge-magic total.
Note that p = (a−2)m+2+(b−2)n and q = (a−3)+2m+1+(b−3)n+2n. Let f be an edge-magic total labeling of G. If f is a super edge-magic total labeling of G, the conditions (i) and (ii) follow from (4) and (6) of Lemmas 4 and 5, respectively, and the condition (iii) follows from Theorem 1.
Let f satisfy the conditions (i), (ii), and (iii). We will prove that G is super edge-magic total for a = 7, b = 3, m = 1, and n = 1. We first notice that p = 8 and q = 9.
Claim 1 (s = 5). If s = 4, then f (v10) + f (v70) = 0, a contradiction to f (v10) + f (v70) ≥ 3, hence s = 5 and the claim.
By (iii) and Claim 1, we have k = 22 and by (i) we have f (v10) + f (v70) = 9. Hence either f (v10) = 1 and f (v70) = 8, f (v10) = 2 and f (v70) = 7, f (v10) = 3 and f (v70) = 6, or f (v10) = 4 and f (v70) = 5.
Case 1 (f (v10) = 1 and f (v70) = 8). Using k = 22, we define the bijection f1(x) in (8) such that f1(v10) = f (v10) = 1 and f1(v70) = f (v70) = 8.(8)f1x=1,ifx=v10,6,ifx=v21,2,ifx=v31,3,ifx=v41,7,ifx=v51,4,ifx=v61,8,ifx=v70,5,ifx=w21,The function f1(x) is depicted in Figure 2.
Using the definition of f1(x) we get the following: f1 (v31) +f (v41) = 5, f1 (v10) + f1 (w21) = 6, f1 (v10) + f1 (v21) = 7, f1 (v21) + f1 (v31) = 8, f1 (v10) + f1 (v70) = 9, f1 (v41) + f1 (v51) = 10, f1 (v51) + f1 (v61) = 11, f1 (v61) + f1 (v70) = 12, and f1 (v70) + f1 (w21) = 13. Hence S = {f1(u) + f1(w): uw∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f1(uv) =k − (f1(u) + f1(v)) for all uv∈ E(G).
Case 2 (f (v10) = 2 and f (v70) = 7). Using k = 22, we define the bijection f2(x) in (9) such that f2(v10) = f (v10) = 2 and f2(v70) = f (v70) = 7.(9)f2x=2,ifx=v10,6,ifx=v21,1,ifx=v31,5,ifx=v41,8,ifx=v51,4,ifx=v61,7,ifx=v70,3,ifx=w21.Using the definition of f2(x) we get the following: f2(v10) + f2(w21) = 5, f2 (v31) + f2(v41) = 6, f2 (v21) + f2 (v31) = 7, f2 (v10) + f2 (v21) = 8, f2 (v10) + f2 (v70) = 9, f2 (v70)+f2 (w21) = 10, f2 (v61)+ f2 (v70) = 11, f2(v51)+f2 (v61) = 12, and f2 (v41)+f2 (v51) = 13. Hence S = {f2(u) + f2(w): uw ∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f2(uv) = k − (f2(u) + f2(v)) for all uv ∈ E(G) and it is shown in Figure 3.
Case 3 (f (v10) = 3 and f (v70) = 6). Using k = 22, we define the bijection f3(x) in (10) such that f3(v10) = f (v10) = 3 and f3(v70) = f (v70) = 6.(10)f3x=3,ifx=v10,7,ifx=v21,4,ifx=v31,8,ifx=v41,5,ifx=v51,1,ifx=v61,6,ifx=v70,2,ifx=w21,The function f3(x) is depicted in Figure 4. Using the definition of f3(x) we get the following: f3(v10) + f3(w21) = 5, f3 (v51) + f3 (v61) = 6, f3 (v61) + f3 (v70) = 7, f3 (v70) + f3 (w21) = 8, f3 (v10) + f3 (v70) = 9, f3 (v10) + f3 (v21) = 10, f3 (v21) + f3 (v31) = 11, f3 (v31) + f3 (v41) = 12, and f3 (v41) + f3 (v51) = 13. Hence S = {f3(u) + f3(w): uw ∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f3(uv) = k − (f3(u) + f3(v)) for all uv ∈ E(G).
Case 4 (f(v10) = 4 and f(v70) = 5). Using k = 22, we define the bijection f4(x) in (11) such that f4(v10) = f (v10) = 4 and f4(v70) = f (v70) = 5.(11)f4x=4,ifx=v10,7,ifx=v21,3,ifx=v31,2,ifx=v41,6,ifx=v51,1,ifx=v61,5,ifx=v70,8,ifx=w21,The function f4(x) is depicted in Figure 5. Using the definition of f4(x) we get the following: f4 (v31) + f4 (v41) = 5, f4 (v61) + f4 (v70) = 6, f4 (v51) + f4 (v61) = 7, f4 (v41) + f4 (v51) = 8, f4 (v10) + f4 (v70) = 9, f4 (v21) + f4 (v31) = 10, f4 (v10) + f4 (v21) = 11, f4 (v10) + f4 (w21) = 12, and f4 (v70) + f4 (w21) = 13. Hence S = {f4(u) + f4(w): uw ∈ E(G} = {5, 6,..., 13} consists of q = 9 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude f4(uv) = k − (f4(u) + f4(v)) for all uv ∈ E(G), hence the lemma.Figure 2
Bijective functionf1(x).Figure 3
Bijective functionf2(x).Figure 4
Bijective functionf3(x).Figure 5
Bijective functionf4(x).Now we are ready to state our main result.Theorem 7.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, m = 1,andb = 3. ThenGis a super edge-magic total. Moreover there are two bijective functionsg1(V (G))andg2(V (G))with the magic constant k = 3n + 19.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, m = 1, and b = 3. We first notice that p = (a − 2)m + (b − 2)n + 2 and q = (a − 3)m + 2m + 1 + (b − 3)n + 2n. For a = 7, m = 1, and b = 3 and n = 1 we have p = 8 and q = 9. Using the definition of the bijective function f1(x) in Case 1 of Lemma 6 together with s = 5 in Claim 1 of Lemma 6, we define the following bijective function g1(x) in (12):(12)g1x=1,ifx=v10,n+5,ifx=v21,2,ifx=v31,3,ifx=v41,n+6,ifx=v51,4,ifx=v61,n+7,ifx=v70,i+4,ifx=w2i,1≤i≤nThe function g1(x) is depicted in Figure 6. Using the definition of g1(x) we get the following: g1(v31) + g1(v41) = 5, g1(v10) + g1(w21) = 6, g1(v10) + g1(w2i) = i + 5, i∈{2, 3,..., n}, g1(v10) + g1(v21) = n + 6, g1(v21)+g(v31) = n+7, g1(v10) + g1(v70) = n+8, g1(v41)+g(v51) = n+9, g1(v51)+g1(v61) = n + 10, g1(v61) + g1(v70) = n + 11, g1(v70) + g1 (w2i) = n + 11 + i, i∈{1, 2, 3,..., n}. Hence S = {g1(u) + g1(w): uw∈ E(G} = {5, 6,..., i + 5, n + 7, n +8, n+9, n+10, n+11, n+i+11, i∈{1,..., n}} consists of q = 2n+7 consecutive integers. Thus by Theorem 1 and Claim 1 of Lemma 6, we conclude that G is super edge-magic total with magic constant k = p+q+s = n+7+2n+7+5 = 3n + 19. Moreover, by Theorem 1, we conclude g1(uv) = k − (g1(u) + g1(v)) for all uv∈ E(G).
Using the definition of the bijective function f2(x) in Case 2 of Lemma 6 together with s = 5 in Claim 1 of Lemma 6, we define the following bijective function g2(x) in (13):(13)g2x=2,ifx=v10,n+5,ifx=v21,1,ifx=v31,n+4,ifx=v41,n+7,ifx=v51,n+3,ifx=v61,n+6,ifx=v70,i+2,ifx=w2i,1≤i≤nThe function g2(x) is depicted in Figure 7. Using the definitions of g2(x) we get the following: g2(v10) + g2(w2i) = i +4, i∈{1, 2,..., n}, g2(v31) + g2(v41) = n + 5, g2(v21) + g2(v31) = n + 6, g2(v10) + g2(v21) = n+7, g2(v10)+g2(v70) = n+8, g2(v70)+g2(w2i) = n+8+i, i∈{1, 2,..., n}, g2(v70) + g2(v61) = 2n + 9, g2(v61) + g2(v51) = 2n + 10, g2(v51) + g2(v41) = 2n + 11. Hence S = {g2(u) + g2(w): uw∈ E(G} = {5, 6,..., i + 4, n + 5, n + 6, n + 7, n + 8, n + 8 + i, 2n + 9, 2n + 10, 2n + 11, i∈{1, 2,..., n}} consists of q = 2n + 7 consecutive integers. Thus by Theorem 1 and Claim 1 of Lemma 6, we conclude that G is super edge-magic total with magic constant k = p + q + s = n + 7 + 2n + 7 + 5 = 3n + 19.
Moreover, by Theorem1, we conclude g2(uv) = k − (g2(u) + g2(v)) for all uv∈ E(G), hence the theorem.Figure 6
Bijective functiong1(x).Figure 7
Bijective functiong2(x).Theorem 8.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 such that |V(G)| = p, |E(G)| = q, a = 4x − 1, m = 1, and b = 3.
Letα2: V (G) → {1, 2,..., p} be a bijective function such that(14)α2vij=n-1+2x+12i+2,ifiiseven,2≤i≤2x-2,j=1,α2vij=12i+2,ifiiseven,2x≤i≤4x-2,j=1,α2vij=12i+2,i=1,j=0,α2vij=12i+1,ifiisodd,1≤i≤2x-1,j=1,α2vij=n-1+2x+12i+1,ifiisodd,2x+1≤i≤4x-3,j=1,α2vij=n-1+2x+12i+1,ifi=4x-1,j=0,α2wij=2x+j,2≤i≤b-2,1≤j≤n.Then α2(v) is a super edge-magic total labeling of G with the magic constant k = 10x + 3n − 1, v ∈ V (G).Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p and |E(G)| = q. Let a = 4x − 1, m = 1, and b = 3. Notice that p = 4x − 1 + n and q = 2n + 4x − 1. Let S = {α2(u) + α2(w): uw∈ E(G}. We show that S consists of q consecutive integers.
Using the definition ofα2(v) in (14), v∈ V (G), we get the following:(15)α2v2x-11+α2v2x1=2x+1,α2v10+α2w2i=2x+1+i,1≤i≤n,α2v10+α2v21=n+2x+2,α2v21+α2v31=n+2x+3,⋯α2vx-21+α2vx-11=n+3x-1,α2vx-11+α2vx1=n+3x,α2vx1+α2vx+11=n+3x+1,⋯α2v2x-21+α2v2x-11=n+4x-1,α2v10+α2v4x-10=n+4x,α2v2x1+α2v2x+11=n+4x+1,⋯α2v3x-21+α2v3x-11=n+5x-1,α2v3x-11+α2v3x1=n+5x,α2v3x1+α2v3x+11=n+5x+1,⋯α2v4x-21+α2v4x-11=n+6x-1,α2v4x-10+α2w2i=n+6x-1+i,1≤i≤n.Hence by (15) we have S = {2x + 1, 2x + 2, 2x + 3,..., 2x + n + 1, 2x + n + 2,..., 3x + n −1, 3x + n, 3x + n + 1,..., 4x + n − 1, 4x + n, 4x + n + 1,..., 5x + n − 1, 5x + n, 5x +n + 1,..., 6x + n − 1, 6x + n, 6x + n + 1,..., 6x + 2n − 2, 6x + 2n − 1} consists of q consecutive integers. Thus by Theorem 1 we conclude that G is super edge-magic total. Notice that s = 2x + 1. Hence k = p + q + s = 10x + 3n − 1. Moreover, by Theorem 1, we conclude α2(uv) = k − (α2(u) + α2(v)) for all uv∈ E(G), hence the theorem.Lemma 9.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p, |E(G)| = q, a = 7, m = 2, b = 3, and n = 1 and let f be an edge-magic total labeling of G. Let S = {f(w) + f (z): f or any edge wz ∈ E(G)} be q consecutive integers and s = min⁡(S). Then G is a super edge-magic total if and only if(i)
f(v10)+f(v70)=(15s-77)/2,(ii)
6 ≤ s ≤ 7,(iii)
k = p + q + s.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, m = 2, b = 3, and n = 1 and let G be a super edge-magic total. Let f be an edge-magic total labeling of G. If f is a super edge-magic total labeling of G, the conditions (i) and (ii) follow from Lemmas 4 and 5, respectively, and the condition (iii) follows from Theorem 1.
Let f satisfy the conditions (i), (ii), and (iii). We first notice that p = (a − 2)m + (b − 2)n + 2 and q = (a − 3)m + 2m + 1 + (b − 3)n + 2n. We will prove that G is super edge-magic total for a = 7, m = 2, b = 3, and n = 1. Notice that p = 13 and q = 15.
Claim 1 (s = 7). If s = 6, then f(v10)+f(v70)= 6,5 (by (i)); a contradiction to f(v10)+f(v70) is an integer, hence s = 7 and the claim. By (iii) and Claim 1, we have k =35 and by (i) we have f(v10)+f(v70) = 14.
Next we will show that there is a bijection h1(x) such that h1(v10)= f(v10), h1(v70)=f(v70), h1(v10)+h1(v70) = 14, and h1(x) is a super edge-magic total labeling of B(7, 2, 3, 1). Let h1(v10)= f(v10) = 2 and h1(v70)=f(v70)=12. Using k = 35, we define the bijection h1(x) in (16):(16)h1x=2,ifx=v10,6,ifx=v21,13,ifx=v31,3,ifx=v41,10,ifx=v51,8,ifx=v61,12,ifx=v70,7,ifx=v22,4,ifx=v32,11,ifx=v42,1,ifx=v52,9,ifx=v62,5,ifx=w21.
Using the definition of h1(x) we get the following: h1(v10) + h1(w21) = 7, h1(v10)+h1(v21) = 8, h1(v10)+h1(v22) = 9, h1(v52)+h1(v62) = 10, h1(v22)+h1(v32) = 11, h1(v42) + h1(v52) = 12, h1(v41) + h1(v51) = 13, h1(v10) + h1(v70) = 14, h1(v32) + h1(v42) = 15, h1(v31) + h1(v41) = 16, h1(v70) + h1(w21) = 17, h1(v51) + h1(v61) = 18, h1(v21) + h1(v31) = 19, h1(v61) + h1(v70) = 20, h1(v62) + h1(v70) = 21. Hence S = {h1(u) + h1(w): uw∈ E(G} = {7, 8,..., 21} consists of q = 15 consecutive integers. Thus by Theorem 1 and Claim 1, we conclude that G is super edge-magic total. Moreover we conclude h1(uv) = k − (h1(u) + h1(v)) for all uv∈ E(G) and it is shown in Figure 8, hence the theorem.Figure 8
Bijective functionh1(x).Theorem 10.
Let G be the graphs B(a, m, b, n, 2) in Figure 1 with |V(G)| = p and |E(G)| = q, a = 7, m = 2, and b = 3. Then G is a super edge-magic total with the magic constant k = 3n + 32.Proof.
Let G be the graphs B(a, m, b, n, 2) in Figure1 with |V(G)| = p, |E(G)| = q, a = 7, m = 2, and b = 3. We first notice that p = (a − 2)m + (b − 2)n + 2 and q = (a − 3)m + 2m + 1 + (b − 3)n + 2n. Using the definition of the bijective function h1(x) and s = 7 in Lemma 9, we define the following bijective function h2(x) in (17):(17)h2x=2,ifx=v10,n+5,ifx=v21,n+12,ifx=v31,3,ifx=v41n+9,ifx=v51n+7,ifx=v61,n+11,ifx=v70,n+6,ifx=v22,4,ifx=v32,n+10,ifx=v42,1,ifx=v52,n+8,⋯ifx=v62i+4,1≤i≤n.The function h2(x) is depicted in Figure 9. Using the definition of h2(x) we get the following: h2(v10)+h2(w2i) = i + 6, i∈{1,..., n}, h2(v10) + h2(v21) = n + 7, h2(v10)+h2(v22) = n+8, h2(v52)+h2(v62) = n+9, h2(v22) + h2(v32) = n + 10, h2(v42) + h2(v52) = n + 11, h2(v41) + h2(v51) = n + 12, h2(v10) + h2(v70) = n + 13, h2(v32) + h2(v42) = n + 14, h2(v31) + h2(v41) = n + 15, h2(v70) + h2(w2i) = n + 15 + i, i∈{1,..., n}, h2(v51) + h2(v61) = 2n + 16, h2(v21) +h2(v31) = 2n + 17, h2(v70) + h2(v61) = 2n + 18, h2(v70) + h2(v62) = 2n + 19. Hence S = {h2(u) + h2(w): uw∈ E(G} = {7, 8,..., n + 6, n + 7, n + 8, n + 9, n +10, n + 11, n + 12, n + 13, n + 14, n + 15, n + 16,..., 2n + 15, 2n + 16, 2n + 17, 2n +18, 2n + 19} consists of q = 2n + 13 consecutive integers. Thus by Theorem 1 and Claim 1 of Lemma 9, we conclude that G is super edge-magic total with magic constant k = p + q + s = n + 12 + 2n + 13 + 7 = 3n + 32. Moreover, by Theorem 1, we conclude h2(uv) = k − (h2(u) + h2(v)) for all uv∈ E(G), hence the theorem.Figure 9
Bijective functionh2(x).



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