Research article Special Issues

Browder spectra of closed upper triangular operator matrices

  • Let TB=[AB0D] be an unbounded upper operator matrix with diagonal domain, acting in HK, where H and K are Hilbert spaces. In this paper, some sufficient and necessary conditions are characterized under which TB is Browder (resp., invertible) for some closable operator B with D(B)D(D). Further, a sufficient and necessary condition is given under which the Browder spectrum (resp., spectrum) of TB coincides with the union of the Browder spectrum (resp., spectrum) of its diagonal entries.

    Citation: Qingmei Bai, Alatancang Chen, Jingying Gao. Browder spectra of closed upper triangular operator matrices[J]. AIMS Mathematics, 2024, 9(2): 5110-5121. doi: 10.3934/math.2024248

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  • Let TB=[AB0D] be an unbounded upper operator matrix with diagonal domain, acting in HK, where H and K are Hilbert spaces. In this paper, some sufficient and necessary conditions are characterized under which TB is Browder (resp., invertible) for some closable operator B with D(B)D(D). Further, a sufficient and necessary condition is given under which the Browder spectrum (resp., spectrum) of TB coincides with the union of the Browder spectrum (resp., spectrum) of its diagonal entries.



    A graph labeling is an assignment of integers to the vertices or edges, or both, subject to certain conditions. Gallian [3] has written a dynamic survey of graph labeling. MacDougall et al. [5] introduced the notion of a vertex magic total labeling of graphs. Let G be a graph of order n and size m. A vertex magic total labeling of G is defined as a one-to-one function

    f:V(G)E(G){1,2,,n+m}

    with the property that for each vertex u of G,

    f(u)+vN(u)f(uv)=k

    for some constant k where N(u) is the neighborhood of u. The constant k is called the magic constant for f. The vertex-magic total labelings of wheels and related graphs were studied in [6], and later in [11]. The properties of the general graphs such as cycles, paths, complete graphs, wheels, bipartite graphs and trees, which satisfy the vertex magic total labelings, were studied in [10]. MacDougall et al. [4] introduced the concept of a super vertex magic total labeling. They defined a vertex magic total labeling to be super if

    f[V(G)]={1,2,,n}.

    In 2017, Nagaraj et al. [7] introduced the concept of an even vertex magic total labeling. They called a vertex magic total labeling as even if

    f[V(G)]={2,4,,2n}.

    A graph G is called an even vertex magic if there exists an even vertex magic total labeling of G. We note that if G is an even vertex magic, then nm. The following results, which appeared in [7], are useful to us.

    Theorem 1.1. [7] Let G be a nontrivial graph of order n and size m. If G is an even vertex magic, then magic constant k is given by the following:

    k=m2+2mn+mn.

    A wheel Wn, n3, is a graph of order n+1 that contains a cycle Cn, for which every vertex in the cycle Cn is connected to one other vertex known as the hub. The edges of the wheel which are incident to the hub are called spokes. The vertices and edges of the cycle Cn in Wn are called rim vertices and rim edges, respectively. It was shown in [7] that a wheel Wn has no even vertex magic total labeling, as we state next.

    Theorem 1.2. [7] A wheel Wn is not even vertex magic.

    In this paper, the labeling problem is related to the work in [1]. In addition to the aforementioned vertex labeling by even numbers 2,4,,2n, they studied vertex labelings by using three consecutive numbers 0,1,2 with some specific properties. These labelings were referred to as a weak Roman dominating function and a perfect Roman dominating function.

    From the studies in [8,9,12], there exist graphs with the same order and size that are even vertex magics. Moreover, the wheel related graphs, namely fans, cycles and suns, having the even vertex magic total labelings were established in [7]. However, since these graphs have the same order and size, it is interesting and challenging to study wheel related graphs when the size is greater than the order, which have an even vertex magic total labeling.

    The t-fold wheel Wn,t, n3, t1, is a wheel related graph derived from a wheel Wn by duplicating the t hubs, each adjacent to all rim vertices, and not adjacent to each other. It is observed that the t-fold wheel Wn,t has a size nt+n that exceeds its order n+t. The goal of this paper is to study conditions for an even vertex magic Wn,t in terms of n and t. Furthermore, we also determine an even vertex magic total labeling of some t-fold wheel Wn,t.

    Since the 1-fold wheel Wn,1 is isomorphic to the wheel Wn and by Theorem 1.1, Wn is not an even vertex magic. In this section, we consider the t-fold wheel Wn,t, where n and t are integers with n3 and t2.

    In order to present the conditions for an even vertex magic Wn,t, we initially explore the magic constant of the t-fold wheel Wn,t of order n+t and size nt+n by employing Theorem 1.1.

    Proposition 2.1. Let n and t be integers with n3 and t2. If the t-fold wheel Wn,t is an even vertex magic, then the magic constant is defined as follows:

    k=2nt+3n+n2t2+2n2t+nn+t.

    We are able to show the bound of an integer t for the t-fold wheel having an even vertex magic total labeling as follows.

    Proposition 2.2. Let n and t be integers with n3 and t2. If the t-fold wheel Wn,t is an even vertex magic, then 2tn.

    Proof. Suppose that the t-fold wheel Wn,t is an even vertex magic with magic constant k. By Proposition 2.1, we obtain the following:

    k=2nt+3n+n2t2+2n2t+nn+t.

    On the contrary, assume that t>n. Let t=n+r, for some r1. Then,

    n2t2+2n2t+n=n4+2n3r+n2r2+2n3+2n2r+n

    and

    n+t=n+(n+r)=2n+r.

    Let

    P(n)=n4+2n3r+n2r2+2n3+2n2r+n.

    By using the remainder theorem, the remainder when P(n) is divided by 2n+r is as follows:

    P(r2)=r4+4r38r16.

    If

    P(r2)=0,

    then r=2, which is a contradiction. Thus,

    P(r2)0.

    Specifically, n2t2+2n2t+n is not divisible by n+t. Thus, k is not an integer, which is a contradiction. Therefore, 2tn.

    According to Proposition 2.2, the t-fold wheel W3,t is not an even vertex magic, where t4. Figure 1 shows the even vertex magics W3,2 and W3,3 with magic constants k=36 and k=50, respectively, where their vertices and edges are labeled by the even vertex magic total labelings. We present an even vertex magic total labeling of the t-fold wheel Wn,t by considering only the integer n as the following results.

    Figure 1.  Even vertex magic total labelings of W3,2 and W3,3.

    Proposition 2.3. For every integer n3, if the n-fold wheel Wn,n is an even vertex magic, then n is odd.

    Proof. Let n be an integer with n3. Suppose that the n-fold wheel Wn,n is an even vertex magic with a magic constant k. On the contrary, assume that n is even. There exists an integer q such that n=2q. By Proposition 2.1,

    k=2n2+3n+n3+2n2+12.

    Since

    n3+2n2+1=8q3+8q2+1

    is odd, n3+2n2+1 is not divisible by 2. Thus, k is not an integer, which is a contradiction. Therefore, n is odd.

    As we have seen in Figure 1, the 3-fold wheel W3,3 is an even vertex magic, as indicated by Proposition 2.3. By an argument similar to the one used in the proof of Proposition 2.3, we obtain the condition for an even vertex magic Wn,n2, as we now show.

    Proposition 2.4. For every integer n4, if the (n2)-fold wheel Wn,n2 is an even vertex magic, then n is even.

    The even vertex magic total labeling of the 2-fold wheel W4,2 with a magic constant k=50 is shown in Figure 2.

    Figure 2.  Even vertex magic total labeling of W4,2.

    In order to deduce an even vertex magic total labeling of the t-fold wheel for achieving the main result, we need some additional notation for the t-fold wheel Wn,t. For every pair of integers n3 and t2, let

    V(Wn,t)={u1,u2,,un,v1,v2,,vt}

    and

    E(Wn,t)={uiui+1|1in1}{unu1}{uivj|1in,1jt}.

    Suppose the t-fold wheel Wn,t is an even vertex magic. Then, for any even vertex magic total labeling f of Wn,t, let

    Srv=ni=1f(ui),   Sre=n1i=1f(uiui+1)+f(unu1)

    and

    Sh=tj=1f(vj),   Ss=tj=1ni=1f(uivj).

    Next, we present the following lemma to show the necessary condition for an even vertex magic Wn,t with the following magic constant:

    k=2nt+3n+n2t2+2n2t+nn+t.

    Note that

    Srv+2SreSh=(nt)k.

    Lemma 2.5. Let n and t be integers where n3 and t2. If the t-fold wheel Wn,t is an even vertex magic, then

    Srv+2SreSh=(t2+4t+3)n2+(2t36t23t+1)n+(t3+2t21)(2nt)n+t.

    With the aid of Lemma 2.5 and Proposition 2.2, the necessary condition for an even vertex magic total labeling of the t-fold wheel Wn,t can also be given in terms of n and t.

    Proposition 2.6. Let n and t be integers where n3 and t2. If the t-fold wheel Wn,t is an even vertex magic, then

    (t22t+1)n2+(2t3+6t2+7t+1)n(t2+t)(t3+2t21)(2nt)n+t0.

    Proof. Suppose that the t-fold wheel Wn,t is an even vertex magic. By Lemma 2.5,

    Srv+2SreSh=(t2+4t+3)n2+(2t36t23t+1)n+(t3+2t21)(2nt)n+t.

    Next, we consider the maximum of (Srv+2SreSh).

    By Proposition 2.2, 2tn, and then 2n+2t<nt+n+t+1. The maximum of

    (Srv+2SreSh)=2t+2i2n+2tiis eveni+2nt+2n+ti=nt+n+t+1i2i2tiis eveni=(2i2n+2tiis eveni2i2tiis eveni)+2(nt+2n+ti=1int+n+ti=1i)2i2tiis eveni=((2n+2t)(2n+2t+2)4(2t)(2t+2)4)+2((nt+2n+t)(nt+2n+t+1)2(nt+n+t)(nt+n+t+1)2)(2t)(2t+2)4=2n2t+4n2+4ntt2+2nt.

    Since Srv+2SreSh does not exceed the maximum of (Srv+2SreSh), the maximum of

    (Srv+2SreSh)(Srv+2SreSh)0.

    Therefore,

    (t22t+1)n2+(2t3+6t2+7t+1)n(t2+t)(t3+2t21)(2nt)n+t0.

    Now, we investigate the sufficient condition for a labeling f that can be an even vertex magic total labeling of Wn,n when n is odd.

    Theorem 2.7. Let n be an odd integer where n3. For every n-fold wheel Wn,n, let

    f:V(Wn,n)E(Wn,n){1,2,,n2+3n}

    be defined by the following:

    f(ui) = 2i,  if   1in,f(vj) = 2n+2j,  if   1jn,f(uiui+1) = 2n+12i,  if  1in1,f(unu1) = 1,  f(un+1jvj) = n2+3n+12j,  if   1jn,f[EH]f[{un+1jvj|1jn}] = {2n+1,2n+3,,n2+n1}    {4n+2,4n+4,,n2+3n},  if  EH={uivj|1i,jn}.

    If

    n1j=1f(u1vj)=n3+4n252,

    then f can be an even vertex magic total labeling of Wn,n.

    Proof. Assume that

    n1j=1f(u1vj)=n3+4n252.

    We have that

    Ssnj=1f(un+1jvj)=2n+1in2+n1iis oddi+4n+2in2+3niis eveni=(1in2+n1iis oddi1i2n1iis oddi)+(2in2+3niis eveni2i4niis eveni)=((n2+n)24(2n)24)+((n2+3n)(n2+3n+2)44n(4n+2)4)=n4+4n34n2n2,

    and then,

    (Ssnj=1f(un+1jvj))n1j=1f(u1vj)=n4+3n38n2n+52.

    Next, we consider the sum of the label of each vertex and the labels of all edges incident to this vertex. By the assumption, for 1jn,

    f(vj)+ni=1f(uivj)=f(vj)+f(un+1jvj)+1inin+1jf(uivj)=f(vj)+f(un+1jvj)+Ssnj=1f(un+1jvj)n=(2n+2j)+(n2+3n+12j)+n3+4n24n12=n3+6n2+6n+12,
    f(u1)+f(u1u2)+f(unu1)+nj=1f(u1vj)=f(u1)+f(u1u2)+f(unu1)+f(u1vn)+n1j=1f(u1vj)=2+(2n+12)+1+(n2+n1+2)+n3+4n252=n3+6n2+6n+12.

    For 2in,

    f(ui)+f(uiui+1)+f(ui1ui)+nj=1f(uivj)=f(ui)+f(uiui+1)+f(ui1ui)+f(uivn+1i)+1jnjn+1if(uivj)=f(ui)+f(uiui+1)+f(ui1ui)+f(uivn+1i)+(Ssnj=1f(un+1jvj))n1j=1f(u1vj)n1=2i+(2n+12i)+(2n+12i+2)+(n2+n1+2i)+n3+4n24n52=n3+6n2+6n+12.

    Therefore, f can be an even vertex magic total labeling of Wn,n with a magic constant

    k=n3+6n2+6n+12.

    Now, we investigate the sufficient condition for a labeling f that can be an even vertex magic total labeling of Wn,n2 when n is even.

    Theorem 2.8. Let n be an even integer with n4. For every (n2)-fold wheel Wn,n2, let

    f:V(Wn,n2)E(Wn,n2){1,2,,n2+n2}

    be defined by the following:

    f(ui)=2i,if   1in,f(vj)=2n+2j,if  1jn2,f[EC]={a1,a2,,an},if  EC={uiui+1,unu1|1in1},f[EH]={1,2,,n2+n2}{2,4,,2n+4,a1,a2,,an},if  EH={uivj|1in,1jn2}.

    If

    Ss=n4+n315n2+20n42,

    then f can be an even vertex magic total labeling of Wn,n2.

    Proof. Assume that

    Ss=n4+n315n2+20n42.

    It suffices to show that for each vertex u of Wn,n2,

    f(u)+vN(u)f(uv)=k,

    where

    k=n3+3n23n2.

    To do this, we consider the relevant sums, as follows.

    Since the sum of the labels of all rim edges is equal to the sum of the labels of all vertices and the labels of all edges subtracted by the sum of the labels of all vertices and the labels of all spokes, it follows that

    Sre=n2+n2i=1i2i4n4iis eveniSs=(n2+n2)(n2+n1)2(4n4)(4n2)4n4+n315n2+20n42=n3+5n211n+22.

    Since the sum of the labels of all hubs is equal to the sum of even integers from 2n+2 to 4n4,

    Sh+Ss=2n+2i4n4iis eveni+Ss=(2i4n4iis eveni2i2niis eveni)+Ss=(4n4)(4n2)4(2n)(2n+2)4+n4+n315n2+20n42=n4+n39n2+6n2.

    Since the sum of the labels of all rim vertices is equal to the sum of even integers from 2 to 2n,

    Srv+2Sre+Ss=2i2niis eveni+2Sre+Ss=(2n)(2n+2)4+2(n3+5n211n+22)+n4+n315n2+20n42=n4+3n33n22.

    Next, we consider the sum of the label of each vertex and the labels of all edges incident to this vertex. We have the sum of the label of each hub and the labels of all edges incident to this hub as follows.

    For 1jn2,

    f(vj)+ni=1f(uivj)=Sh+Ssn2=n3+3n23n2.

    We obtain the sum of the label of each rim vertex and the labels of all edges incident to this rim vertex as follows.

    For 2in1,

    f(ui)+f(uiui+1)+f(ui1ui)+n2j=1f(uivj)=Srv+2Sre+Ssn=n3+3n23n2.

    Similarly,

    f(un)+f(unu1)+f(un1un)+n2j=1f(unvj)=n3+3n23n2

    and

    f(u1)+f(u1u2)+f(unu1)+n2j=1f(u1vj)=n3+3n23n2.

    Therefore, f can be an even vertex magic total labeling of Wn,n2 with the following magic constant:

    k=n3+3n23n2.

    In this section, we establish a characterization of an even vertex magic Wn,t for an integer 3n9. First, we present an n-fold wheel Wn,n which has an even vertex magic total labeling for every odd integer 3n9 as follows.

    Theorem 3.1. For every odd integer 3n9, the n-fold wheel Wn,n is an even vertex magic.

    Proof. Let n be an odd integer where 3n9. We define

    f:V(Wn,n)E(Wn,n){1,2,,n2+3n},

    as the sufficient condition of Theorem 2.7, by

    f(ui)=2i,if   1in,f(vj)=2n+2j,if  1jn,f(uiui+1)=2n+12i,if  1in1,f(unu1)=1,

    and for 1i,jn, f(uivj) are shown in Tables 14,

    Table 1.  Labels of edges uivj of W3,3 by f, for 1i,j3.
    f(uivj) v1 v2 v3
    u1 11 18 13
    u2 14 15 9
    u3 17 7 16

     | Show Table
    DownLoad: CSV
    Table 2.  Labels of edges uivj of W5,5 by f, for 1i,j5.
    f(uivj) v1 v2 v3 v4 v5
    u1 15 21 36 38 31
    u2 13 30 29 33 28
    u3 40 24 35 19 17
    u4 34 37 11 23 32
    u5 39 27 26 22 25

     | Show Table
    DownLoad: CSV
    Table 3.  Labels of edges uivj of W7,7 by f, for 1i,j7.
    f(uivj) v1 v2 v3 v4 v5 v6 v7
    u1 15 51 64 40 29 68 57
    u2 66 43 17 38 19 59 70
    u3 25 45 56 48 61 32 47
    u4 52 31 33 63 54 53 30
    u5 42 35 65 34 60 36 46
    u6 55 67 23 58 49 27 41
    u7 69 50 62 37 44 39 21

     | Show Table
    DownLoad: CSV
    Table 4.  Labels of edges uivj of W9,9 by f, for 1i,j9.
    f(uivj) v1 v2 v3 v4 v5 v6 v7 v8 v9
    u1 73 33 102 69 19 79 63 86 91
    u2 43 56 35 62 49 89 85 93 87
    u3 75 61 48 53 104 92 95 23 50
    u4 80 58 72 25 108 97 77 47 39
    u5 51 96 59 81 99 37 55 57 70
    u6 60 100 44 101 67 40 41 90 64
    u7 38 83 103 82 74 78 29 76 46
    u8 88 105 94 65 42 66 52 31 68
    u9 107 21 54 71 45 27 106 98 84

     | Show Table
    DownLoad: CSV

    For every odd integer 3n9, the labeling f, as defined above, is an even vertex magic total labeling of the n-fold wheel Wn,n with magic constants k=50,153,340 and 635, respectively. Therefore, Wn,n is an even vertex magic.

    As a consequence of an even vertex magic W3,2, Proposition 2.2 and Theorem 3.1, in any t-fold wheel W3,t, we are able to show that both W3,t and W3,t are only even vertex magics.

    Theorem 3.2. For every integer t2, the t-fold wheel W3,t is an even vertex magic if and only if t=2,3.

    The following result gives the necessary and sufficient condition for the t-fold wheel Wn,t to be an even vertex magic for every odd integer 5n9.

    Theorem 3.3. For every odd integer 5n9 and an integer t2, the t-fold wheel Wn,t is an even vertex magic if and only if t=n.

    Proof. Let n be an odd integer where 5n9 and t is an integer where t2. Assume that the t-fold wheel Wn,t is an even vertex magic. By Proposition 2.2, 2tn.

    Case 1. n=5,7. If 2tn1, then n2t2+2n2t+n is not divisible by n+t, and hence k is not an integer, which is a contradiction. Therefore, t=n.

    Case 2. n=9. If either t=2 or 4tn1, then n2t2+2n2t+n is not divisible by n+t, and hence k is not an integer, which is a contradiction. If t=3, then,

    2nt3n2t22n2t+6nt2+7nt+n2t2+nt2nt4+4nt32ntn+t=174<0,

    which is a contradiction with Proposition 2.6. Therefore, t=n.

    Conversely, assume t=n. By Theorem 3.1, Wn,t is an even vertex magic.

    We show an even vertex magic total labeling of Wn,n2 for every even integer 4n8 as follows.

    Theorem 3.4. For every even integer 4n8, the (n2)-fold wheel Wn,n2 is an even vertex magic.

    Proof. Let n be an even integer with 4n8. We define

    f:V(Wn,n2)E(Wn,n2){1,2,,n2+n2}

    as the sufficient condition of Theorem 2.8, by

    f(ui)=2i,if  1in,f(vj)=2n+2j,if  1jn2,

    for 1in1, f(uiui+1) and f(unu1) are shown in Tables 57.

    Table 5.  Labels of edges uiui+1 and u4u1 of W4,2 by f, for 1i3.
    f(u1u2) f(u2u3) f(u3u4) f(u4u1)
    18 9 11 13

     | Show Table
    DownLoad: CSV
    Table 6.  Labels of edges uiui+1 and u6u1 of W6,4 by f, for 1i5.
    f(u1u2) f(u2u3) f(u3u4) f(u4u5) f(u5u6) f(u6u1)
    40 39 38 13 17 19

     | Show Table
    DownLoad: CSV
    Table 7.  Labels of edges uiui+1 and u8u1 of W8,6 by f, for 1i7.
    f(u1u2) f(u2u3) f(u3u4) f(u4u5) f(u5u6) f(u6u7) f(u7u8) f(u8u1)
    70 68 66 64 62 19 13 11

     | Show Table
    DownLoad: CSV

    And for 1in and 1jn2, f(uivj) are shown in Tables 810.

    Table 8.  Labels of edges uivj of W4,2 by f, for 1i4 and 1j2.
    f(uivj) v1 v2
    u1 16 1
    u2 14 5
    u3 7 17
    u4 3 15

     | Show Table
    DownLoad: CSV
    Table 9.  Labels of edges uivj of W6,4 by f, for 1i6 and 1j4.
    f(uivj) v1 v2 v3 v4
    u1 36 24 3 29
    u2 11 25 33 1
    u3 26 9 5 30
    u4 22 23 34 15
    u5 37 21 28 27
    u6 7 35 32 31

     | Show Table
    DownLoad: CSV
    Table 10.  Labels of edges uivj of W8,6 by f, for 1i8 and 1j6.
    f(uivj) v1 v2 v3 v4 v5 v6
    u1 69 3 40 56 39 50
    u2 1 67 5 59 31 35
    u3 21 17 65 7 44 46
    u4 42 38 23 33 9 57
    u5 61 27 37 34 30 15
    u6 41 53 49 32 47 25
    u7 29 60 54 52 63 36
    u8 58 55 45 43 51 48

     | Show Table
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    For every even integer 4n8, the labeling f, as defined above, is an even vertex magic total labeling of the (n2)-fold wheel Wn,n2 with magic constants k=50,153 and 340, respectively. Therefore, Wn,n2 is an even vertex magic.

    There is a similar methodology of the proof of Theorem 3.4, which is also used in the study of graph operations (see [2]). Next, we determine a characterization of the t-fold wheel Wn,t to be an even vertex magic for every even integer 4n8. In order to we need to present the following lemma involving a 3-fold wheel W8,3.

    Lemma 3.5. The 3-fold wheel W8,3 is not an even vertex magic.

    Proof. On the contrary, assume that the 3-fold wheel W8,3 is an even vertex magic with a magic constant k. Since W8,3 has an order 11 and a size 32 and by Proposition 2.1, k=160. We have that

    Srv=132,  2Sre=1,6282Ss

    and

    Srv+2Sre+Ss=8k=1,280.

    Thus, Ss=480. However, Sh+Ss=3k=480. This is a contradiction because Sh>0. Therefore, W8,3 is not an even vertex magic.

    We are able to show that the necessary and sufficient condition for the t-fold wheel Wn,t is an even vertex magic for every even integer 4n8.

    Theorem 3.6. For every even integer 4n8 and integer t2, the t-fold wheel Wn,t is an even vertex magic if and only if t=n2.

    Proof. Let n be an even integer where 4n8 and t is an integer where t2. Assume that the t-fold wheel Wn,t is an even vertex magic. By Proposition 2.2, 2tn.

    Case 1. n=4,6. If either 2tn3 or n1tn, then n2t2+2n2t+n is not divisible by n+t, and hence k is not an integer, which is a contradiction. Therefore, t=n2.

    Case 2. n=8. If either 4tn3 or n1tn, then n2t2+2n2t+n is not divisible by n+t, and hence k is not an integer, which is a contradiction. If t=3, then, by Lemma 3.5, Wn,t is not an even vertex magic, which is a contradiction. If t=2, then,

    2nt3n2t22n2t+6nt2+7nt+n2t2+nt2nt4+4nt32ntn+t=62<0,

    which is a contradiction with Proposition 2.6. Therefore, t=n2.

    Conversely, assume t=n2. By Theorem 3.4, Wn,t is an even vertex magic.

    In this paper, we have not only established the bound of an integer t for the even vertex magic total labeling of the t-fold wheel, but have also presented the necessary condition for such labeling in terms of n and t. Furthermore, we have conducted an investigation into the sufficient conditions for labelings that can serve as even vertex magic total labelings for Wn,n when n is odd, and Wn,n2 when n is even.

    Our research has led us to the following significant conclusions:

    ● For every integer t2, the t-fold wheel W3,t is an even vertex magic total labeling if and only if t=2,3.

    ● For every odd integer 5n9 and an integer t2, the t-fold wheel Wn,t is an even vertex magic total labeling if and only if t=n.

    ● For every even integer 4n8 and an integer t2, the t-fold wheel Wn,t is an even vertex magic total labeling if and only if t=n2.

    In essence, our work has discussed the characterizations of t-fold wheel Wn,t to possess an even vertex magic total labeling for an integer 3n9. It would be interesting to apply the results of this paper to further study under what conditions for Wn,t will be an even vertex magic, especially for a larger n.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors declare that they have no conflicts of interest.



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