# On the Extremal Trees for Some Bond Incident Degree Indices with a Fixed Number of Segments

Authors: Muzamil Hanif; Akhlaq Ahmad Bhatti; Muhammad Javaid; Ebenezer Bonyah
Journal: Journal of Chemistry
 (2022)

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## Abstract

The Gourava indices and hyper-Gourava indices are introduced by Kulli in 2017. These graph invariants are related to the degree of vertices of a graphG. Let Tn,r be the class of all n−vertex chemical trees with r segments. In this paper, we characterize the graphs with the maximum value of the above indices in the class of chemical trees. In addition to different degree sequences, sharp upper bounds on those indices of trees with a fixed number of segments are determined, and the corresponding extremal trees are characterized.


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## Body

## 1. Introduction

LetG be a simple graph of order n with vertex set VG=vi,i=1,2,3,…,n and edge set EG=ej,j=1,2,3,…,m. An edge of G, connecting the vertices u and v, is denoted by u∼v. Let NuG be the neighbor set of the vertex u in graph G. A vertex u adjacent to its neighbors in G is called its degree, and it is denoted by duG. A vertex which is adjacent to only one neighbor in a graph is said to be pendent, and a vertex which is adjacent to its more than two neighbors in a graph is said to be a branching vertex. If u,v∈G, a pendent vertex u and branching vertex v are adjacent, then vertex u is called a starlike pendent vertex. A segment of a tree T (see [1]) is a path-subtree S whose terminal vertices are pendent or branching vertices of T, i.e., an internal vertex of a segment S has degree two. Let Pn and Sn be the path and star graph of order n, respectively. A starlike tree T of order n is a tree with exactly one branching vertex or central vertex of degree greater than or equal to three. Topological indices are widely used in mathematical chemistry to predict properties of chemical compounds. They have been studied intensively in recent years and among the oldest and the most studied being the first and second Zagreb indices M1G and M2G respectively, which were used in the study of π−electron energy of molecular structures. In 1972, Gutman and Trinajstić [2, 3] defined the first and second Zagreb indices as(1)M1G=∑v∈VGdv2G,M2G=∑u∼vduGdvG.The first Zagreb index is also defined as [4](2)M1G=∑u∼vduG+dvG.The extremal Zagreb indices have been studied for certain classes of trees in last some decades. For more details, see [1, 5–15].Goubko and Gutman [16] characterized the trees with the minimum first Zagreb index among the trees with a fixed number of pendent vertices. Lin [17] characterized the tree that can maximize and minimize the first Zagreb index with fixed number of segments. After that, Borovićanin, Das, Furtula, and Gutman [18, 19] characterized the tree with maximum and minimum Zagreb indices for certain classes of trees with fixed number of segments or branching vertices. In 2011, Azari and Iranmanesh [20] defined the generalized Zagreb index of graphs as(3)Mα,βG=∑u∼vduGαdvGβ+duGβdvGα,∀α,β∈N.By the motivation of the first and second Zagreb indices and its wide applications in science, Kulli [21] defined the first Gourava index of a graph G as(4)GO1G=∑u∼vduG+dvG+duGdvG.Then, by motivation of the generalized Zagreb index and the first Gourava index, Kulli [21] defined the second Gourava index as(5)GO2G=∑u∼vduG+dvGduGdvG,which is also written in the form of generalized Zagreb index as(6)GO2G=∑u∼vduG2dvG+duGdvG2.After that, Kulli defined first and second hyper Gourava indices as [22](7)HGO1G=∑u∼vduG+dvG+duGdvG2,HGO2G=∑u∼vduG+dvGduGdvG2.In this paper, we characterize the maximum Gourava indices and hyper-Gourava indices among all chemical trees of prescribed degree sequences with fixed number of segments. A vertex degree-based topological index of a functionTI is defined as TI:Tn,r⟶R induced by numbers ψi,ji,j∈Γ, where(8)Γ=i,j∈N×N:1≤i≤j≤4defined for every tree T∈Tn,r as(9)TIT=∑i,j∈Γθi,jTψi,j.Here,θi,jT denotes the number of edges of adjacent vertices of degree i and j in graph T. For GO1,GO2,HGO1, and HGO2, the values of ψi,j are i+j+ij,i+j×ij,i+j+ij2, and i+j×ij2, respectively.LetTn,r be the set of all classes of chemical trees with order n and exactly r segments. Let Tn,r1,Tn,r2,Tn,r3, and Tn,r4 be the subclasses of Tn,r. It is noted that Pn and Sn are the unique elements of Tn,1 and Tn,n−1, respectively. It is also observed that the class Tn,2 is empty because there is no such graph which has exactly two segments. We denote sT be the number of segments in a tree T and the number of vertices by Vi of degree i,1≤i≤4, also Vi=0 for i>4. For chemical trees, the following relations are well known, where 1≤j≤4 and Δ=4.(10)2θj,j+∑i=1i≠jΔθj,i=jVj,(11)∑1≤i≤j≤Δθi,j=n−1.From (10), we have the following system of equations:(12)2θ1,1+θ1,2+θ1,3+θ1,4=V1,(13)θ2,1+2θ2,2+θ2,3+θ2,4=2V2,(14)θ3,1+θ3,2+2θ3,3+θ3,4=3V3,(15)θ4,1+θ4,2+θ4,3+2θ4,4=4V4.Furthermore, we will use this system of equations to find the number of edges of different types in upcoming theorems.Lemma 1.
For any chemical treeT∈Tn,rwith3≤r≤n−1, the following result holds.(16)DST=4,4,…,4︸r−1/3,2,2,…,2︸n−r−1,1,1,…,1︸2r+2/3,ifr≡1mod3,4,4,…,4︸r/3−1,3,2,2,…,2︸n−r−1,1,1,…,1︸2r/3+1,ifr≡0mod3,4,4,…,4︸r−5/3,3,3,2,2,…,2︸n−r−1,1,1,…,1︸2r+1/3,ifr≡2mod3.Proof.
A sequenced1,d2,…,dn of n positive integers, where n≥2, is a degree sequence of a tree of order n if and only if ∑i=1ndi=2n−1 and ∑i=1ndi=V1+2V2+3V3+⋯+nVn. In addition to this, if we have a chemical tree T, the order of the tree n is equal to the sum of the number of vertices of degrees i,i∈1,2,3,4, i.e.,(17)n=∑i=14Vi,(18)2n−1=∑i=14iVi.
From (17) and (18), we get(19)V1=2+∑i=34i−2Vi,where r is a number of segments in Tn,r. Using (19) in the equation r=V1+V3+V4−1, we get r≡2V3+1mod3. Using V2=n−r−1 in (18), we get(20)V1=2r−∑i=34iVi.
By solving (19) and (20), we get the values of unknown V1 and V4 as(21)V1=2r−V3+43,V4=r−2V3−13.
ForV3=0, V1=2r+2/3,V2=n−r−1, and V4=r−1/3 with r≡1mod3.
ForV3=1, V1=2r/3+1,V2=n−r−1, and V4=r/3−1 with r≡0mod3.
ForV3=2, V1=2r+1/3,V2=n−r−1, and V4=r−5/3 with r≡2mod3. ■


## 2. Main Result

RecallTn,r1 be a class of all n−vertex chemical trees of degree sequence r,2,2,…,2︸n−r−1,1,1,…,1︸r. Let Tmax1∈Tn,r1 be a tree for 3≤r≤n−2, which maximizes the Gourava indices and hyper-Gourava indices. For proving Theorem 1, we determine the structure of Tmax1 from the following lemmas for the above-mentioned indices.Lemma 2.
LetTmax1∈Tn,r1 be a maximal tree. If a pendent vertex u is adjacent to a vertex v of dv∈VTmax1≥3, then Tmax1 does not have a vertex w of dw∈VTmax1=2 with adjacent to both non-pendent neighbors.Proof.
Suppose, on the contrary, thatu∼v such that u and v are pendent and branching vertices, respectively. There exists a vertex w∈Tmax1 of degree two, such that w∼w1 and w∼w2, where w1,w2 are non-pendent vertices, i.e., 2≤k=dw1≤dw2. Let T∗=Tmax1−uv,ww1,ww2+uw,vw,w1,w2, then T∗∈Tn,r1. We have(22)GO1Tmax1−GO1T∗=du+dv+dudv+∑i=1dv−2dv+dvi+dvdvi+dv+dw1+dvdw1+∑i=1dw1−2dw1i+dw1+dw1idw1+dw+dw1+dwdw1+dw+dw2+dwdw2+∑i=1dw2−1dw2i+dw2+dw2idw2−du+dw+dudw−∑i=1dv−2dv+dvi+dvdvi−dv+dw+dvdw−∑i=1dw1−2dw1+dw1i+dw1dw1i−dv+dw1+dvdw1−dw1+dw2+dw1dw2−∑i=1dw2−1dw2+dw2i+dw2dw2i=dudv−dudw−dvdw+dwdw1+dwdw2−dw1dw2=4k−k2−dv−2<0ifk=dw1=dw2,dv≥3−k2−dv+3k<0ifk=dw1<dw2,dv≥3GO1Tmax1−GO1T∗<0,GO2Tmax1−GO2T∗=du+dv∗dudv+∑i=1dv−2dv+dvi∗dvdvi+dv+dw1∗dvdw1+∑i=1dw1−2dw1i+dw1∗dw1idw1+dw+dw1∗dwdw1+dw+dw2∗dwdw2+∑i=1dw2−1dw2i+dw2∗dw2idw2−du+dw∗dudw−∑i=1dv−2dv+dvi∗dvdvi−dv+dw∗dvdw−∑i=1dw1−2dw1+dw1i∗dw1dw1i−dv+dw1∗dvdw1−dw1+dw2∗dw1dw2−∑i=1dw2−1dw2+dw2i∗dw2dw2i=dv1+dv−dw1+dw−dvdwdv+dw+dwdw1dw+dw1+dwdw2dw+dw2−dw1dw2dw1+dw2=8k+4k2−2k3−dv2−3dv−60ifk=dw1=dw2,dv≥311k−2k3+k2−dv2−3dv<0ifk=dw1<dw2,dv≥3GO2Tmax1−GO2T∗<0,HGO1Tmax1−HGO1T∗=du+dv+dudv2+∑i=1dv−2dv+dvi+dvdvi2+dv+dw1+dvdw12+∑i=1dw1−2dw1i+dw1+dw1idw12+dw+dw1+dwdw12+dw+dw2+dwdw22+∑i=1dw2−1dw2i+dw2+dw2idw22−du+dw+dudw2−∑i=1dv−2dv+dvi+dvdvi2−dv+dw+dvdw2−∑i=1dw1−2dw1+dw1i+dw1dw1i2−dv+dw1+dvdw12−dw1+dw2+dw1dw22−∑i=1dw2−1dw2+dw2i+dw2dw2i2,=du+dv+dudv2+dw+dw1+dwdw12+dw+dw2+dwdw22−du+dw+dudw2−dv+dw+dvdw2−dw1+dw2+dw1dw22=24k+14k2−4k3−k4−8dv−5dv2−20<0ifk=dw1=dw2,dv≥336k+7k2−6k3−k4−8dv−5dv2<0ifk=dw1<dw2,dv≥3HGO1Tmax1−HGO1T∗<0,HGO2Tmax1−HGO2T∗=du+dv∗dudv2+∑i=1dv−2dv+dvi∗dvdvi2+dv+dw1∗dvdw12+∑i=1dw1−2dw1i+dw1∗dw1idw12+dw+dw1∗dwdw12+dw+dw2∗dwdw22+∑i=1dw2−1dw2i+dw2∗dw2idw22−du+dw∗dudw2−∑i=1dv−2dv+dvi∗dvdvi2−dv+dw∗dvdw2−∑i=1dw1−2dw1+dw1i∗dw1dw1i2−dv+dw1∗dvdw12−dw1+dw2∗dw1dw22−∑i=1dw2−1dw2+dw2i∗dw2dw2i2=dv21+dv2−dw21+dw2−dv2dw2dv+dw2+dw2dw12dw+dw12+dw2dw22dw+dw22−dw12dw22dw1+dw22=8k4+32k3+32k2−4k6−15dv2−14dv3−3dv4−36<0ifk=dw1=dw2,dv≥396k+103k2+42k3−5k4−12k5−4k6−15dv2−14dv3−3dv4<0ifk=dw1<dw2,dv≥3HGO2Tmax1−HGO2T∗<0,
For each caseGO1Tmax1<GO1T∗,GO2Tmax1<GO2T∗,HGO1Tmax1<HGO1T∗, and HGO2Tmax1<HGO2T∗, which is a contradiction to the maximality of Tmax1.Lemma 3.
If a vertexx∈VTmax1 of maximum degree Δ,3≤Δ≤n−2, is the only branching vertex in Tmax1. Then, Tmax1 is a starlike tree.Proof.
InTmax1, we notice that a vertex has maximum degree greater than two. This implies Tmax1≠Pn. Suppose, on the contrary, that a branching vertex y∈VTmax1 different from x with 3≤k=dy∈VTmax1≤dx∈VTmax1. Let P:xu1u2…uly be a path in Tmax1 also x∼xi,1≤i≤dx−1,y∼ul and y∼yj,1≤j≤dy−1. Here, all xi and yj are pendent vertices. Now, we obtained another tree T∗∈Tn,r1 such that T∗=Tmax1−y∼yj|1≤j≤dy−1+x∼yj|1≤j≤dy−1. Then, we have two cases:Case 1: ifx∼y, we have(23)GO1Tmax1−GO1T∗=∑i=1dx−1dx+dxi+dxdxi+dx+dy+dxdy+∑j=1dy−1dy+dyj+dydyj−∑i=1dx−1dx+dy−1+dxi+dx+dy−1dxi−dx+dy−1+dy+dx+dy−1dy−∑j=1dy−1dx+dy−1+dyj+dx+dy−1dyj=1−dy2+dx−12dx+1−dx+dy−22dx+2dy−1+dy−12dy+1,sincedxi=dyj=1=−dy−14dx+dy−3=−k−15k−3<0ifdx=dy=k≥3−k−15k+1<0ifdx>dy=k≥3GO1Tmax1−GO1T∗<0,GO2Tmax1−GO2T∗=∑i=1dx−1dx+dxi∗dxdxi+dx+dy∗dxdy+∑j=1dy−1dy+dyj∗dydyj−∑i=1dx−1dx+dy−1+dxi∗dx+dy−1dxi−dx+dy−1+dy∗dx+dy−1dy−∑j=1dy−1dx+dy−1+dyj∗dx+dy−1dyj=dxdydx+dy−dydx+dy−1dx+2dy−1+∑i=1dx−1dxdx+1−∑i=1dx−1dx+dydx+dy−1+∑j=1dy−1dydy+1−∑j=1dy−1dx+dydx+dy−1,sincedxi=dyj=1=−dy−13dx2+2dydy−2+dx5dy−3=−kk−110k−7<0ifdx=dy=k≥3−2kk−15k+2<0ifdx>dy=k≥3GO2Tmax1−GO2T∗<0,HGO11Tmax−HGO1T∗=∑i=1dx−1dx+dxi+dxdxi2+dx+dy+dxdy2+∑j=1dy−1dy+dyj+dydyj2−∑i=1dx−1dx+dy−1+dxi+dx+dy−1dxi2−dx+dy−1+dy+dx+dy−1dy2−∑j=1dy−1dx+dy−1+dyj+dx+dy−1dyj2=dx−12dx+12+dy−12dy+12−dx−12dx+2dy−12−dy−12dx+2dy−12+dx+dy+dxdy2−dx+2dy−1+dydx+dy−12,sincedxi=dyj=1=−dy−1dy3+3dy2−11dy+12dx2+2dxdy2+8dy−5−1=−k−13k3+31k2−21k−1<0,ifdx=dy=k≥3,−k−13k3+33k2+19k+1<0,ifdx>dy=k≥3.HGO1Tmax1−HGO1T∗=∑i=1dx−1dx+dxi+dxdxi2+dx+dy+dxdy2+∑j=1dy−1dy+dyj+dydyj2−∑i=1dx−1dx+dy−1+dxi+dx+dy−1dxi2−dx+dy−1+dy+dx+dy−1dy2−∑j=1dy−1dx+dy−1+dyj+dx+dy−1dyj2=dx−12dx+12+dy−12dy+12−dx−12dx+2dy−12−dy−12dx+2dy−12+dx+dy+dxdy2−dx+2dy−1+dydx+dy−12,sincedxi=dyj=1=−dy−1dy3+3dy2−11dy+12dx2+2dxdy2+8dy−5−1=−k−13k3+31k2−21k−1<0ifdx=dy=k≥3−k−13k3+33k2+19k+1<0ifdx>dy=k≥3HGO1Tmax1−HGO1T∗<0,HGO2Tmax1−HGO2T∗=∑i=1dx−1dx+dxi∗dxdxi2+dx+dy∗dxdy2+∑j=1dy−1dy+dyj∗dydyj2−∑i=1dx−1dx+dy−1+dxi∗dx+dy−1dxi2−dx+dy−1+dy∗dx+dy−1dy2−∑j=1dy−1dx+dy−1+dyj∗dx+dy−1dyj2=−dy−15dx4+4dy4dy−2+2dx32dy2+5dy−3+dx212dy3+4dy2−14dy+1+dxdy12dy3−9dy2−7dy+4,sincedxi=dyj=1=−k2k−132k3+2k2−27k+5<0ifdx=dy=k≥3−2kk−116k4+25k3+23k2+10k+2<0ifdx>dy=k≥3HGO2Tmax1−HGO2T∗<0,contradiction to the choice ofTmax1.Case 2: ifx≁y, we have(24)GO1Tmax1−GO1T∗=∑i=1dx−1dx+dxi+dxdxi+dx+du1+dxdu1+∑j=1dy−1dy+dyj+dydyj+∑i=1l−1dui+dui+1+duidui+1+dy+dul+dydul−∑i=1dx−1dx+dy−1+dxi+dx+dy−1dxi−∑j=1dy−1dx+dy−1+dxi+dx+dy−1dxi−dx+dy−1+du1+dx+dy−1du1−dy−dy+1+dul+dy−dy+1dul−∑i=1l−1dui+dui+1+duidui+1=dx−12dx+1+dy−12dy+1−dx+dy−22dx+2dy−1,sincedxi=dyj=1withdu1=dul=2=−4k−12<0ifdx=dy=k≥3−4kk−1<0ifdx>dy=k≥3GO1Tmax1−GO1T∗<0,GO2Tmax1−GO2T∗=∑i=1dx−1dx+dxi∗dxdxi+dx+du1∗dxdu1+∑j=1dy−1dy+dyj∗dydyj+∑i=1l−1dui+dui+1∗duidui+1+dy+dul∗dydul−∑i=1dx−1dx+dy−1+dxi∗dx+dy−1dxi−∑j=1dy−1dx+dy−1+dyj∗dx+dy−1dyj−dx+dy−1+du1∗dx+dy−1du1−dy−dy+1+dul∗dy−dy+1dul−∑i=1l−1dui+dui+1∗duidui+1=−dx−1dy−13dx+3dy+4,sincedxi=dyj=1withdu1=dul=2=−2k−123k+2<0ifdx=dy=k≥3−kk−16k+7<0ifdx>dy=k≥3GO2Tmax1−GO2T∗<0,HGO1Tmax1−HGO1T∗=∑i=1dx−1dx+dxi+dxdxi2+dx+du1+dxdu12+∑j=1dy−1dy+dyj+dydyj2+∑i=1l−1dui+dui+1+duidui+12+dy+dul+dydul2−∑i=1dx−1dx+dy−1+dxi+dx+dy−1dxi2−∑j=1dy−1dx+dy−1+dyj+dx+dy−1dyj2−dx+dy−1+du1+dx+dy−1du12−dy−dy+1+dul+dy−dy+1dul2−∑i=1l−1dui+dui+1+duidui+12,=−6dx−1dy−12dx+2dy+3,sincedxi=dyj=1withdu1=dul=2=−6k−124k+3<0ifdx=dy=k≥3−6kk−14k+5<0ifdx>dy=k≥3HGO1Tmax1−HGO1T∗<0,HGO2Tmax1−HGO2T∗=∑i=1dx−1dx+dxi∗dxdxi2+dx+du1∗dxdu12+∑j=1dy−1dy+dyj∗dydyj2+∑i=1l−1dui+dui+1∗duidui+12+dy+dul∗dydul2−∑i=1l−1dui+dui+1∗duidui+12−∑i=1dx−1dx+dy−1+dxi∗dx+dy−1dxi2−∑j=1dy−1dx+dy−1+dyj∗dx+dy−1dyj2−dx+dy−1+du1+dx+dy−1du1−dy−dy+1+dul∗dy−dy+1dul2,−5dx−1dy−1dx3+dx22dy+3+2dxdy2+2dy+4+dy3+3dy2+8dy+8,sincedxi=dyj=1withdu1=dul=2=−10k−123k3+5k2+8k+4<0ifdx=dy=k≥3−5kk−16k3+19k2+31k+20<0ifdx>dy=k≥3HGO2Tmax1−HGO2T∗<0.
Again contradiction to the choice ofTmax1. For each case GO1Tmax1<GO1T∗,GO2Tmax1<GO2T∗,HGO1Tmax1<HGO1T∗, and HGO2Tmax1<HGO2T∗. So, Tmax1 contains only one branching vertex. Hence, Tmax1 is a starlike tree.Theorem 1.
LetTmax1∈Tn,r1, where 3≤r≤n−2, then for a degree sequence r,2,2,…,2︸n−r−1,1,1,…,1︸r, we have(25)GO1Tmax1≤r2+rn−6+6n−1,ifn<2r+1,3r2−9r+8n−1,ifn≥2r+1,GO2Tmax1≤r2n−3+3rn−3+6n−1,ifn<2r+1,2r3+4r2−26r+16n−1,ifn≥2r+1,HGO1Tmax1≤−r3+r25n−9+r8n−35+28n−1,ifn<2r+1,9r3+12r2−99r+64n−1,ifn≥2r+1,HGO2Tmax1≤−2r5+3nr4+14r3n−2+15r2n−1−36r+36n−1,ifn<2r+1,4r5+16r4+16r3−476r+256n−1,ifn≥2r+1.
The equality holds if and only ifTmax1 has degree sequence r,2,2,…,2︸n−r−1,1,1,…,1︸r.Proof.
Case 1: ifV2<r and n<2r+1, then by Lemmas 2 and 3, we have θ1,2=θ2,r=n−r−1,θ2,2=0, and θ1,r=−n+2r+1. Now, solving (12)–(15), we get(26)GO1Tmax1=n−r−12+1+2∗1+2r−n+11+r+1∗r+n−r−12+r+2∗r=5n−r−1+2r+12r−n+1+n−r−13r+2=r2+rn−6+6n−1,GO2Tmax1=n−r−12+1∗2∗1+2r−n+11+r∗1∗r+n−r−12+r∗2∗r=6n−r−1+2rr+2n−r−1+rr+12r−n+1=r2n−3+3rn−3+6n−1,HGO1Tmax1=n−r−12+1+2∗12+2r−n+11+r+1∗r2+n−r−12+r+2∗r2=25n−r−1+2r+122r−n+1+3r+22n−r−1=−r3+r25n−9+r8n−35+28n−1,HGO2Tmax1=n−r−12+1∗2∗12+2r−n+11+r∗1∗r2+n−r−12+r∗2∗r2=36n−r−1+4r2r+22n−r−1+r2r+122r−n+1=−2r5+3nr4+14r3n−2+15r2n−1−36r+36n−1.Case 2: ifV2≥r and n≥2r+1, then by Lemmas 2 and 3, we have θ1,r=0,θ2,2=n−2r−1, and θ1,2=θ2,r=r. Now, solving (12)–(15), we get(27)GO1Tmax1=r1+2+1∗2+n−2r−12+2+2∗2+r2+r+2∗r=8n−2r−1+5r+r3r+2=3r2−9r+8n−1,GO2Tmax1=r1+2∗1∗2+n−2r−12+2∗2∗2+r2+r∗2∗r=16n−2r−1+6r+2r2r+2=2r3+4r2−26r+16n−1,HGO1Tmax1=r1+2+1∗22+n−2r−12+2+2∗22+r2+r+2∗r2=64n−2r−1+25r+r3r+22=9r3+12r2−99r+64n−1,HGO2Tmax1=r1+2∗1∗22+n−2r−12+2∗2∗22+r2+r∗2∗r2=256n−2r−1+36r+4r3r+22=4r5+16r4+16r3−476r+256n−1,which completes the proof.
For proving Theorem2, Theorem 3, and Theorem 4, we determine the structures of Tn,r2,  Tn,r3, and Tn,r4 from the following lemmas.Lemma 4.
For3≤r≤n−1, let Tmax∈Tn,r be a tree with maximal Gourava indices and hyper-Gourava indices which contain maximum two vertices of degree three.Proof.
Suppose, on the contrary, thatTmax contains three vertices of degree three. Let P:u1u2u3…ul be the longest path in Tmax. Assume that vertices ui,uj, and ul−1 are branching vertices of degree three. Suppose two pendent vertices v and w=ul are adjacent to ul−1. Let T∗=Tmax−ul−1v,ul−1w+uiv,ujw. For ui,uj,ul−1∈Tmax, we have four cases.Case 1: ifui≁uj and uj≁ul−1.(28)GO1Tmax−GO1T∗=−2dui+duj−2dul−1+7<0,sincedi=dj=dl−1=3,GO2Tmax−GO2T∗=−dui2−13dui−duj2−13duj+2dul−12+5dul−1−16<0,sincedi=dj=dl−1=3,HGO1Tmax−HGO1T∗=−4dui2+duj2+14dui+duj−2dul−12−11dul−1+32<0,sincedi=dj=dl−1=3,HGO2Tmax−HGO2T∗=−dui4+duj4−42dui3+duj3−169dui2+duj2−216dui+duj+dul−14+34dul−13+2dul−12+dul−1−160<0,sincedi=dj=dl−1=3,a contradiction toTmax.Case 2: ifui≁uj and uj∼ul−1.(29)GO1Tmax−GO1T∗=−2dui+3dul−1−14<0,sincedi=dj=dl−1=3,GO2Tmax−GO2T∗=−dui2−13dui+duj2+duj2dul−1−9+dul−12+5dul−1−28<0,sincedi=dj=dl−1=3,HGO1Tmax−HGO1T∗=−4dui2−56dui+4duj2dul−1−1−2dujdul−12−6dul−1+22+5dul−12+20dul−1−104<0,sincedi=dj=dl−1=3,HGO2Tmax−HGO2T∗=−dui4−42dui3−169dui2−216dui−5duj4−26duj3−89duj2−120duj+dul−14−2duj+1+dul−132duj2+10duj+10+dul−128duj3+6duj2−16duj−11+dul−14duj4−8duj3−16duj2+8duj+12−128<0,sincedi=dj=dl−1=3,a contradiction withTmax.Case 3: ifui∼uj and uj≁ul−1.(30)GO1Tmax−GO1T∗=−9−2dui−3duj+4dul−1<0,sincedi=dj=dl−1=3,GO2Tmax−GO2T∗=−dui2−dui2duj+9−2duj2+5duj−dul−12−5dul−1+10<0,sincedi=dj=dl−1=3,HGO1Tmax−HGO1T∗=−4dui2−2duiduj2+2duj+20−7duj2−42duj+8dul−12+44dul−1−87<0,sincedi=dj=dl−1=3,HGO2Tmax−HGO2T∗=−dui4−2dui32duj2+13−dui26duj3+6duj2+97−2duiduj4+3duj3+2duj2+60−2duj4+14duj3+49duj2+60duj−dul−14−18dul−13−dul−12+44<0,sincedi=dj=dl−1=3,a contradiction withTmax.Case 4: ifui∼uj∼ul−1.(31)GO1Tmax−GO1T∗=−3duj−dul−1+3<0,sincedi=dj=dl−1=3,GO2Tmax−GO2T∗=dui2−dui2duj+1−2duj2−2dujdul−1+3+dul−12+dul−1−8<0,sincedi=dj=dl−1=3,HGO1Tmax−HGO1T∗=4dui2−2duiduj2+2duj+6−7duj2−2dujdul−12+2dul−1+13+5dul−12+4dul−1−31<0,sincedi=dj=dl−1=3,HGO2Tmax−HGO2T∗=3dui4+dui3−4duj2+6−3dui22duj3+2duj2+3−2duiduj4+3duj3+2duj2+12−2duj4−4duj3dul−12+3−2duj23dul−13+3dul−12+13−2dujdul−14+3dul−13+2dul−12+12+dul−14+2dul−13+dul−12−16<0,sincedi=dj=dl−1=3,a contradiction with Tmax.
For each caseGO1Tmax<GO1T∗,GO2Tmax<GO2T∗,HGO1Tmax<HGO1T∗, and HGO2Tmax<HGO2T∗, which is a contradiction with the maximality of Tmax. So, Tmax contains at most two vertices of degree three.Lemma 5.
LetTmax∈Tn,r be a tree which contains interval path of length one only where 3≤r≤n−1.Proof.
Suppose, on the contrary, thatTmax has an internal path of length greater than or equal to two. Let P:u1u2u3…uk be an internal path of length greater than or equal to two in Tmax, where u1 and uk be the branching vertices and ∀duj=2,1<j<k. Let w be a pendent vertex adjacent to some ui∈VTmax other then uj,1<j<k. Let T∗=Tmax−uiw,u1u2,uk−1uk+u1uk,u2w,uk−1ui, then T∗∈Tn,r and(32)GO1Tmax−GO1T∗=∑m=1du1−1du1+dum+du1dum+du1+du2+du1du2+∑j=2k−2duj+duj+1+dujduj+1+duk−1+duk+duk−1duk+∑l=1duk−2duk+dul+dukdul+duk+dui+dukdui+dw+dui+dwdui−∑m=1du1−2du1+1+dum+du1+1dum−du1+1+duk−1+du1+1duk−1−∑l=1duk−1duk−1+dul+duk−1dul−du1+1+du2+1+du1+1du2+1−dw+du2+1+dwdu2+1−du2+1+du3+du2+1du3−∑j=3k−2duj+duj+1+dujduj+1−duk−1+dui−1+duk−1dui−1=−3−2du2−du3+dui+dwdui−du2−1+duk−1−duiduk−1+duiduk+duk−1duk−du1duk+1+∑l=1duk−2duk+dul+dukdul−∑l=1duk−1−1+duk+dul+dulduk−1+∑m=1du1−1du1+dum+du1dum−∑m=1du1−21+dum+du1+dumdu1+1=4duk−du1−du1duk−8<0,since3≤duk<du1,GO2Tmax−GO2T∗=∑m=1du1−1du1+dum∗du1dum+du1+du2∗du1du2+∑j=2k−2duj+duj+1∗dujduj+1+duk−1+duk∗duk−1duk+∑l=1duk−2duk+dul∗dukdul+duk+dui∗dukdui+dw+dui∗dwdui−∑m=1du1−2du1+1+dum∗du1+1dum−du1+1+duk−1∗du1+1duk−1−∑l=1duk−1duk−1+dul∗duk−1dul−du1+1+du2+1∗du1+1du2+1−dw+du2+1∗dwdu2+1−du2+1+du3∗du2+1du3−∑j=3k−2duj+duj+1∗dujduj+1−duk−1+dui−1∗duk−1dui−1=−dwdu2+1du2+dw+1+du1du2du1+du2−du1+1du2+1du1+du2+2−du3du2+1du2+du3+1+dwduidw+dui−duk−1dui−1dui+duk−1−1−du1−1duk−1du1+duk+duidukdui+duk+duk−1dukduk−1+duk+du2du3du2+du3+∑l=1duk−2dukdulduk+dul−∑l=1duk−1dulduk−1duk+dul−1+∑m=1du1−1du1dumdu1+dum−∑m=1du1−2dumdu1+1duk+dul+1=−du12duk+1−du1duk2+7+4duk2+6duk−34<0,since3≤duk<du1,HGO1Tmax−HGO1T∗=∑m=1du1−1du1+dum+du1dum2+du1+du2+du1du22+∑j=2k−2duj+duj+1+dujduj+12+duk−1+duk+duk−1duk2+∑l=1duk−2duk+dul+dukdul2+duk+dui+dukdui2+dw+dui+dwdui2−∑m=1du1−2du1+1+dum+du1+1dum2−du1+1+duk−1+du1+1duk−12−∑l=1duk−1duk−1+dul+duk−1dul2−du1+1+du2+1+du1+1du2+12−dw+du2+1+dwdu2+12−du2+1+du3+du2+1du32−∑j=3k−2duj+duj+1+dujduj+12−duk−1+dui−1+duk−1dui−12=du2+du1du2+12−1+du2+dwdu2+22−3+2du2+du1du2+22−1+2du3+du2du3+12+dui+dwdui+12−duiduk−1+1−12−dukdu1+2−12+duk+duiduk+12+duk+duk−1duk+12+du2+du3+du2du32−∑l=1duk−1−1+duk+dulduk2+∑l=1duk−2duk+dul+dukdul2+∑m=1du1−1du1+dum+du1dum2−∑m=1du1−21+du1+du1dum2=−du12duk2+11−du14duk2−2duk+32+29duk2+8duk−64<0,since3≤duk<du1,HGO2Tmax−HGO2T∗=∑m=1du1−1du1+dum∗du1dum2+du1+du2∗du1du22+∑j=2k−2duj+duj+1∗dujduj+12+duk−1+duk∗duk−1duk2+∑l=1duk−2duk+dul∗dukdul2+duk+dui∗dukdui2+dw+dui∗dwdui2−∑m=1du1−2du1+1+dum∗du1+1dum2−du1+1+duk−1∗du1+1duk−12−∑l=1duk−1duk−1+dul∗duk−1dul2−du1+1+du2+1∗du1+1du2+12−dw+du2+1∗dwdu2+12−du2+1+du3∗du2+1du32−∑j=3k−2duj+duj+1∗dujduj+12−duk−1+dui−1∗duk−1dui−12=−dw2du2+12du2+dw+12du12du22du1+du22−du1+12du2+12du1+du2+22−du32du2+12du2+du3+12+dw2dui2dui2+dw2−duk−12dui−12dui+duk−1−12−du1+12duk−1−12du1+duk2+dui2duk2dui+duk2+duk−12duk2duk−1+duk2+du22du32du2+du32+∑l=1duk−2duk2dul2duk+dul2+∑m=1du1−1du12dum2du1+dum2−∑l=1duk−1dul2duk−12duk+dul−12−∑m=1du1−2dum2du1+12du1+dum+12=du14duk2−2duk+9−2du13duk3−duk2−duk+39−du12duk4+2duk3−6duk2+2duk+269−2du1duk4−duk3−duk2+duk+170+10duk4+28duk3+30duk2−924<0,since3≤duk<du1,a contradiction to Tmax, due to the fact du1≥4 and duk≥3. Hence, Tmax contains internal path of length one only.Lemma 6.
For3≤r≤n−1, let Tmax∈Tn,r be a maximal tree containing a pendent vertex w∈VTmax adjacent to some vertex u∈VTmax of du∈VTmax≥3, then Tmax contains a pendent path of length at most two.Proof.
Suppose, on the contrary, thatTmax has a pendent path of length greater than or equal to three. Let P:u1u2u3…uk be a pendent path of length greater than or equal to three in Tmax and a pendent vertex w∈VTmax adjacent to some u∈VTmax of du∈VTmax≥3. Let T∗=Tmax−uw,u1u2,u2u3+u2w,u2u,u1u3, then T∗∈Tn,r and(33)GO1Tmax−GO1T∗=2−du<0,sincedu≥3,GO2Tmax−GO2T∗=10−3du−du2<0,sincedu≥3,HGO1Tmax−HGO1T∗=36−8du−5du2<0,sincedu≥3,HGO2Tmax−HGO2T∗=220−15du2−14du3−3du4<0,sincedu≥3,a contradiction to Tmax. Hence, Tmax contains a pendent path of length at most two.Lemma 7.
For3≤r≤n−1, if Tmax∈Tn,r be a maximal tree contains a pendent vertex u∈VTmax adjacent to a vertex v∈VTmax of dv∈VTmax=4, then Tmax does not contain any vertex of degree two adjacent with any vertex of degree three.Proof.
Suppose, on the contrary, that a vertexs∈VTmax of degree two is adjacent with a branching vertex t∈VTmax of degree three and a pendent vertex u∈VTmax is adjacent to a branching vertex v∈VTmax of degree four. Let NsTmax=q,t. Let T∗=Tmax−qs,st,vu+qt,vs,su, then T∗∈Tn,r and(34)GO1Tmax−GO1T∗=dt−dv<0,since3=dt<dv=4,GO2Tmax−GO2T∗=dt2−dv2+3dt−dv<0,since3=dt<dv=4,HGO1Tmax−HGO1T∗=5dt2−dv2+8dt−dv<0,since3=dt<dv=4,HGO2Tmax−HGO2T∗=3dt4−dv4+14dt3−dv3+15dt2−dv2<0,since3=dt<dv=4,a contradiction to Tmax. Hence, Tmax does not contain any vertex of degree two adjacent with any vertex of degree three if Tmax a pendent vertex is adjacent with a vertex of degree four.Lemma 8.
For3≤r≤n−1, if Tmax∈Tn,r be a maximal tree containing a vertex u∈VTmax of du=3, then vertex u is adjacent to at most one branching vertex in Tmax.Proof.
Suppose, on the contrary, thatx,y, and z are three branching vertices in VTmax. Let P:u1u2u3…uj−1ujuj+1…ul be the longest path in Tmax containing x,y, and z. By Lemma 4, there are maximum two vertices of degree three in P. Let uj−1=x,uj=y, and uj+1=z. If there are at most two vertices of degree three including y in P, then suppose a vertex ui∈VTmax of degree three for some i,1≤i≤j−1, and a vertex uk∈VTmax of degree four for some k,j+1≤k≤l, is adjacent only one branching vertex. Now, bearing in mind duk+1=2,dz=4, and dx=4. Let T∗=Tmax−xy,yz,ukuk+1+xz,uky,yuk+1, then T∗∈Tn,r and(35)GO1Tmax−GO1T∗=−20−dx−duk+1,≤−24−duk+1<0,GO2Tmax−GO2T∗=54−dx−7dx2−5duk+1−duk+12,≤−62−5duk+1−duk+12<0,HGO1Tmax−HGO1T∗=228−16dx−9dx2−12dk+1−7dk+12,≤20−12dk+1−7dk+12<0,HGO2Tmax−HGO2T∗=6156−175dx2−74dx3−7dx4+175dk+12−38dk+13−5dk+14,≤−3172+175dk+12−38dk+13−5dk+14<0,a contradiction to Tmax. Hence, a vertex of degree three in Tmax is adjacent at most one branching vertex in Tmax.Corollary 1 (see [23]).
IfΔ=4 of Tmax∈Tn,r, then the graph induced by the vertices of degree four of Tmax is a tree where 3≤r≤n−1.Theorem 2.
LetTmax2∈Tn,r2, where 3≤r≤n−1, then for a degree sequence 4,4,…,4︸r−1/3,2,2,…,2︸n−r−1,1,1,…,1︸2r+2/3, we have(36)GO1Tmax2≤10n+4r−30,ifn<5r+73,2312n+11r−38,ifn≥5r+73,GO2Tmax2≤34n+22r−178,ifn<5r+73,16n+52r−136,ifn≥5r+73,HGO1Tmax2≤140n+106r−800,ifn<5r+73,2396n+349r−934,ifn≥5r+73,HGO2Tmax2≤1940n+3788r−23252,ifn<5r+73,8396n+2473r−7246,ifn≥5r+73.The equality holds if and only ifTmax2 has a degree sequence 4,4,…,4︸r−1/3,2,2,…,2︸n−r−1,1,1,…,1︸2r+4/3.Proof.
ForDS=4,4,…,4︸r−1/3,2,2,…,2︸n−r−1,1,1,…,1︸2r+2/3 with r≥5, we have V4≥1. By Corollary 1, we get θ4,4=V4−1=r−1/3−1=r−4/3.⇒θ4,4=r−4/3.Case 1: whenV1>V2, we have n<5r+7/3. By Lemmas 5 and 6, we have θ2,2=0. Also, from (12)–(15), we get(37)θ1,2+θ1,4=2r+43,(38)θ2,1+θ2,4=2n−r−1,(39)θ4,1+θ4,2+2r−83=4r−13.From (37), (38), and (39), we get θ1,4=5r−3n+7/3,θ1,2=θ2,4=n−r−1. Hence,(40)GO1Tmax2=n−r−11+2+1∗2+5r−3n+731+4+1∗4+n−r−12+4+2∗4+r−434+4+4∗4=19n−r−1+8r−4+35r−3n+7=10n+4r−30,GO2Tmax2=n−r−11+2∗1∗2+n−r−12+4∗2∗4+5r−3n+731+4∗1∗4+r−434+4∗4∗4=54n−r−1+1283r−4+205r−3n+73=34n+22r−178,HGO1Tmax2=n−r−11+2+1∗22+n−r−12+4+2∗42+5r−3n+731+4+1∗42+r−434+4+4∗42=221n−r−1+192r−4+275r−3n+7=140n+106r−800,HGO2Tmax2=n−r−11+2∗1∗22+n−r−12+4∗2∗42+5r−3n+731+4∗1∗42+r−434+4∗4∗42=2340n−r−1+163843r−4+4005r−3n+73=1940n+3788r−23252.Case 2: whenV1≤V2, we have n≥5r+7/3. By Lemmas 5 and 6, we have θ1,4=0. Also, from (12)–(15), we get(41)θ1,2=2r+43,(42)θ2,1+2θ2,2+θ2,4=2n−r−1,(43)θ4,2+2r−43=4r−13.
From (41), (42), and (43), we get θ4,2=2r+4/3,θ2,2=3n−5r−7/3. Hence,(44)GO1Tmax2=2r+431+2+1∗2+3n−5r−732+2+2∗2+2r+432+4+2∗4+r−434+4+4∗4=83n−5r−73+8r−4+38r+23=2312n+11r−38,GO2Tmax2=2r+431+2∗1∗2+3n−5r−732+2∗2∗2+2r+432+4∗2∗4+r−434+4∗4∗4=163n−5r−73+1283r−4+36r+2=16n+52r−136,HGO1Tmax2=2r+431+2+1∗22+3n−5r−732+2+2∗22+2r+432+4+2∗42+r−434+4+4∗42=643n−5r−73+192r−4+442r+23=2396n+349r−934,HGO2Tmax2=2r+431+2∗1∗22+3n−5r−732+2∗2∗22+2r+432+4∗2∗42+r−434+4∗4∗42=2563n−5r−73+163843r−4+1560r+2=8396n+2473r−7246,which completes the proof.Theorem 3.
LetTmax3∈Tn,r3, where 3≤r≤n−1, then for a degree sequence 4,4,…,4︸r/3−1,3,2,2,…,2︸n−r−1,1,1,…,1︸2r/3+1, we have(45)GO1Tmax3≤10n+4r−34,ifn<5r+33,1327n+17r−102,ifn=5r+33,2312n+11r−48,ifn>5r+33,GO2Tmax3≤34n+22r−202,ifn<5r+33,2336n+58r−303,ifn=5r+33,16n+52r−186,ifn>5r+33,HGO1Tmax3≤140n+106r−914,ifn<5r+33,13291n+533r−2742,ifn=5r+33,2396n+349r−1272,ifn>5r+33,HGO2Tmax3≤1940n+3788r−27764,ifn<5r+33,43594n+4276r−20823,ifn=5r+33,43192n+4946r−20019,ifn>5r+33.The equality holds if and only ifTmax3 has a degree sequence 4,4,…,4︸r/3−1,3,2,2,…,2︸n−r−1,1,1,…,1︸2r/3+1.Proof.
ForDS=4,4,…,4︸r/3−1,3,2,2,…,2︸n−r−1,1,1,…,1︸2r/3+1 with V4≥1 and r≥6. By Corollary 1, we have θ4,4=V4−1=r/3−2. ⇒θ4,4=r/3−2, θ3,3=0 and by Lemmas 5 and 8, we have θ3,4=1.Case 1: whenV2<2V4+2, we have n<5r+3/3⇒θ2,2=0 and θ1,4≠0, then by Lemma 7, θ2,3=0. From (12)–(15), we get(46)θ1,2+θ1,3+θ1,4=2r3+1,(47)θ2,1+θ2,4=2n−2r−2,(48)θ3,1=2,(49)θ4,1+θ4,2=2r3−1.From (46)–(48) and (49), we get θ1,4=5r/3−n,θ1,2=θ2,4=n−r−1. Hence,(50)GO1Tmax3=n−r−11+2+1∗2+21+3+1∗3+5r3−n1+4+1∗4+n−r−12+4+2∗4+3+4+3∗4+r3−24+4+4∗4=33+19n−r−1+8r−6+35r−3n=10n+4r−34,GO2Tmax3=n−r−11+2∗1∗2+21+3∗1∗3+5r3−n1+4∗1∗4+n−r−12+4∗2∗4+3+4∗3∗4+r3−24+4∗4∗4=108+54n−r−1+1283r−6+205r−3n3=34n+22r−202,HGO1Tmax3=n−r−11+2+1∗22+21+3+1∗32+5r3−n1+4+1∗42+n−r−12+4+2∗42+3+4+3∗42+r3−24+4+4∗42=459+221n−r−1+192r−6+275r−3n=140n+106r−914,HGO2Tmax3=n−r−11+2∗1∗22+21+3∗1∗32+5r3−n1+4∗1∗42+n−r−12+4∗2∗42+3+4∗3∗42+r3−24+4∗4∗42=7344+2340n−r−1+163843r−6+4005r−3n3=1940n+3788r−27764.Case 2: whenV2=2V4+2, we have n=5r+3/3. It follows that θ2,2=0=θ1,4. From (12)–(15), we get(51)θ1,2+θ1,3=2r3+1,(52)θ2,1+θ2,3+θ2,4=2n−2r−2,(53)θ3,1+θ3,2=2,(54)θ4,2+2r3−2=4r3−5.From (51)–(53) and (54), we get θ1,2=n−r−1,θ1,3=5r/3−n+2,θ2,3=3n−5r/3, and θ2,4=2r/3−1. Hence,(55)GO1Tmax3=n−r−11+2+1∗2+5r3−n+21+3+1∗3+3n−5r32+3+2∗3+2r3−12+4+2∗4+3+4+3∗4+r3−24+4+4∗4,=19+5n−r−1+8r−6+115r−3n3+75r−3n+63+1432r−3=13−39n+127r−102,GO2Tmax3=n−r−11+2∗1∗2+5r3−n+21+3∗1∗3+3n−5r32+3∗2∗3+2r3−12+4∗2∗4+3+4∗3∗4+r3−24+4∗4∗4=84+6n−r−1+1283r−6+105r−3n+45r−3n+6+162r−3=2336n+58r−303,HGO1Tmax=n−r−11+2+1∗22+5r3−n+21+3+1∗32+3n−5r32+3+2∗32+2r3−12+4+2∗42+3+4+3∗42+r3−24+4+4∗42=361+25n−r−1+192r−6+1215r−3n3+495r−3n+63+19632r−3=13291n+533r−2742,HGO2Tmax3=n−r−11+2∗1∗22+5r3−n+21+3∗1∗32+3n−5r32+3∗2∗32+2r3−12+4∗2∗42+3+4∗3∗42+r3−24+4∗4∗42=7056+36n−r−1+163843r−6+3005r−3n+485r−3n+6+7682r−3=43594n+4276r−20823.Case 3: whenV2>2V4+2, we have n>5r+3/3. It further implies θ1,4=0 and θ2,2≠0; then, by Lemmas 5 and 6, we have θ1,3=0. From (12)–(15), we get(56)θ1,2=2r3+1,(57)θ2,1+2θ2,2+θ2,3+θ2,4=2n−2r−2,(58)θ3,2=2,(59)θ4,2+2r3−2=4r3−5.
From (56)–(58) and (59), we get θ2,2=−5r/3+n−2 and θ2,4=2r/3−1. Hence,(60)GO1Tmax3=2r3+11+2+1∗2+−5r3+n−22+2+2∗2+23+2+3∗2+2r3−12+4+2∗4+3+4+3∗4+r3−24+4+4∗4=41+83n−5r−63+8r−6+1432r−3+532r+3=2312n+11r−48,GO2Tmax3=2r3+11+2∗1∗2+−5r3+n−22+2∗2∗2+23+2∗3∗2+2r3−12+4∗2∗4+3+4∗3∗4+r3−24+4∗4∗4=144+163n−5r−63+1283r−6+162r−3+22r+3=16n+52r−186,HGO1Tmax3=2r3+11+2+1∗22+−5r3+n−22+2+2∗22+23+2+3∗22+2r3−12+4+2∗42+3+4+3∗42+r3−24+4+4∗42=603+643n−5r−63+192r−6+19632r−3+2532r+3=2396n+349r−1272,HGO2Tmax3=2r3+11+2∗1∗22+−5r3+n−22+2∗2∗22+23+2∗3∗22+2r3−12+4∗2∗42+3+4∗3∗42+r3−24+4∗4∗42=8856+2563n−5r−63+163843r−6+7682r−3+122r+3=43192n+4946r−20019,which completes the proof.Theorem 4.
LetTmax4∈Tn,r4, where 3≤r≤n−1, then for a degree sequence 4,4,…,4︸r−5/3,3,3,2,2,…,2︸n−r−1,1,1,…,1︸2r+2/3, we have(61)GO1Tmax4≤10n+4r−38,ifn≤5r−73,1327n+17r−121,if5r−43≤n≤5r+23,2312n+11r−58,ifn>5r+23,GO2Tmax4≤34n+22r−226,ifn≤5r−73,4318n+29r−187,if5r−43≤n≤5r+23,16n+52r−236,ifn>5r+23,HGO1Tmax4≤140n+106r−1028,ifn≤5r−73,13291n+533r−3385,if5r−43≤n≤5r+23,2396n+349r−1610,ifn>5r+23,HGO2Tmax4≤1940n+3788r−32276,ifn≤5r−73,83297n+2138r−13108,if5r−43≤n≤5r+23,8396n+2473r−12773,ifn>5r+23.The equality holds if and only ifTmax4 has a degree sequence 4,4,…,4︸r−5/3,3,3,2,2,…,2︸n−r−1,1,1,…,1︸2r+2/3.Proof.
ForDS=4,4,…,4︸r−5/3,3,3,2,2,…,2︸n−r−1,1,1,…,1︸2r+2/3 with V4≥1 and r≥8. By Corollary 1, it follows θ4,4=V4−1=r−8/3. By Lemmas 5 and 8, we get θ3,3=0 and θ3,4=2.Case 1: whenV2≤2V4, we have n≤5r−7/3. Hence, θ2,2=0,⇒θ1,4≠0, and by Lemma 7, θ2,3=0. From (12)–(15), we get(62)θ1,2+θ1,3+θ1,4=2r+23,(63)θ2,1+θ2,4=2n−2r−2,(64)θ3,1=4,(65)θ4,1+θ4,2+2r−83=4r−53−2.From (62)–(64) and (65), we get θ1,2=θ2,4=n−r−1,θ1,4=5r−7/3−n. Hence,(66)GO1Tmax4=n−r−11+2+1∗2+41+3+1∗3+5r−73−n1+4+1∗4+n−r−12+4+2∗4+23+4+3∗4+r−834+4+4∗4=66+19n−r−1+8r−8+35r−3n−7=10n+4r−38,GO2Tmax4=n−r−11+2∗1∗2+41+3∗1∗3+5r−73−n1+4∗1∗4+n−r−12+4∗2∗4+23+4∗3∗4+r−834+4∗4∗4=216+54n−r−1+1283r−8+205r−3n−73=34n+22r−226,HGO1Tmax4=n−r−11+2+1∗22+41+3+1∗32+5r−73−n1+4+1∗42+n−r−12+4+2∗42+23+4+3∗42+r−834+4+4∗42=918+221n−r−1+192r−8+275r−3n−7=140n+106r−1028,HGO2Tmax4=n−r−11+2∗1∗22+41+3∗1∗32+5r−73−n1+4∗1∗42+n−r−12+4∗2∗42+23+4∗3∗42+r−834+4∗4∗42=14688+2340n−r−1+163843r−8+4005r−3n−73=1940n+3788r−32276.Case 2: when2V4+1≤V2≤2V4+3, we have 5r−4/3≤n≤5r+2/3 and θ1,4=0=θ2,2. From (12)–(15), we get(67)θ1,2+θ1,3=2r+23,(68)θ2,1+θ2,3+θ2,4=2n−2r−2,(69)θ3,1+θ3,2=4,(70)θ4,2+2r−83=4r−53−2.From (67)–(69) and (70), we get θ1,2=n−r−1,θ1,3=5r+5/3−n,θ2,3=−5r+7/3+n,θ2,4=2r−10/3. Hence,(71)GO1Tmax4=n−r−11+2+1∗2+5r+531+3+1∗3+−5r+73+n2+3+2∗3+2r−1032+4+2∗4+23+4+3∗4+r−834+4+4∗4=38+113n−5r+73+5n−r−1+8r−8+283r−5+75r−3n+53=1327n+17r−121,GO2Tmax4=n−r−11+2∗1∗2+5r+531+3∗1∗3+−5r+73+n2+3∗2∗3+2r−1032+4∗2∗4+23+4∗3∗4+r−834+4∗4∗4=168+103n−5r+7+6n−r−1+1283r−8+32r−5+125r−3n+53=4318n+29r−187.HGO1Tmax4=n−r−11+2+1∗22+5r+531+3+1∗32+−5r+73+n2+3+2∗32+2r−1032+4+2∗42+23+4+3∗42+r−834+4+4∗42=722+1213n−5r+73+25n−r−1+192r−8+3923r−5+495r−3n+53=13291n+533r−3385,HGO2Tmax4=n−r−11+2∗1∗22+5r+531+3∗1∗32+−5r+73+n2+3∗2∗32+2r−1032+4∗2∗42+23+4∗3∗42+r−834+4∗4∗42=14112+3003n−5r+7+36n−r−1+163843r−8+1536r−5+1445r−3n+53=83297n+2138r−13108.Case 3: whenV2>2V4+3, we have n>5r+2/3 which implies θ1,4=0 and θ2,2≠0. Hence, by Lemmas 5 and 6, θ1,3=0. From (12)–(15), we get(72)θ1,2=2r+23,(73)2θ2,2+θ2,3+θ2,4=6n−8r−83,(74)θ3,2=4,(75)θ4,2+2r−83=4r−53−2.
From (72)–(74) and (75), we get θ2,2=3n−5r−5/3,θ2,4=2r−10/3. Hence,(76)GO1Tmax3=2r3+11+2+1∗2+−5r3+n−22+2+2∗2+23+2+3∗2+2r3−12+4+2∗4+3+4+3∗4+r3−24+4+4∗4=41+83n−5r−63+8r−6+1432r−3+532r+3=2312n+11r−48,GO2Tmax3=2r3+11+2∗1∗2+−5r3+n−22+2∗2∗2+23+2∗3∗2+2r3−12+4∗2∗4+3+4∗3∗4+r3−24+4∗4∗4=144+163n−5r−63+1283r−6+162r−3+22r+3=16n+52r−186,HGO1Tmax3=2r3+11+2+1∗22+−5r3+n−22+2+2∗22+23+2+3∗22+2r3−12+4+2∗42+3+4+3∗42+r3−24+4+4∗42=603+643n−5r−63+192r−6+19632r−3+2532r+3=2396n+349r−1272,HGO2Tmax3=2r3+11+2∗1∗22+−5r3+n−22+2∗2∗22+23+2∗3∗22+2r3−12+4∗2∗42+3+4∗3∗42+r3−24+4∗4∗42=8856+2563n−5r−63+163843r−6+7682r−3+122r+3=43192n+4946r−20019,which completes the proof.


## 3. Conclusions

Nowadays, many researchers are introduced to investigating properties of different molecular descriptors through topological indices. In this way, we have investigated upper bonds on Gourava indices and hyper-Gourava indices for some classes ofn−vertex chemical trees with a fixed number of segments and vertices of degree two.At this point, we left the lower bonds on Gourava and hyper-Gourava indices for the above classes ofn−vertex chemical trees with a fixed number of segments as an open problem.



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*Source: 4032709-2022-03-02.xml*