
This is a systematic review and meta-analysis that evaluated the prevalence of Escherichia coli antibiotic-resistant genes (ARGs) in animals, humans, and the environment in South Africa. This study followed Preferred Reporting Items for Systematic Reviews and Meta-analyses (PRISMA) guidelines to search and use literature published between 1 January 2000 to 12 December 2021, on the prevalence of South African E. coli isolates' ARGs. Articles were downloaded from African Journals Online, PubMed, ScienceDirect, Scopus, and Google Scholar search engines. A random effects meta-analysis was used to estimate the antibiotic-resistant genes of E. coli in animals, humans, and the environment. Out of 10764 published articles, only 23 studies met the inclusion criteria. The obtained results indicated that the pooled prevalence estimates (PPE) of E. coli ARGs was 36.3%, 34.4%, 32.9%, and 28.8% for blaTEM-M-1, ampC, tetA, and blaTEM, respectively. Eight ARGs (blaCTX-M, blaCTX-M-1, blaTEM, tetA, tetB, sul1, sulII, and aadA) were detected in humans, animals and the environmental samples. Human E. coli isolate samples harboured 38% of the ARGs. Analyzed data from this study highlights the occurrence of ARGs in E. coli isolates from animals, humans, and environmental samples in South Africa. Therefore, there is a necessity to develop a comprehensive “One Health” strategy to assess antibiotics use in order to understand the causes and dynamics of antibiotic resistance development, as such information will enable the formulation of intervention strategies to stop the spread of ARGs in the future.
Citation: Tsepo Ramatla, Mpho Tawana, Kgaugelo E. Lekota, Oriel Thekisoe. Antimicrobial resistance genes of Escherichia coli, a bacterium of “One Health” importance in South Africa: Systematic review and meta-analysis[J]. AIMS Microbiology, 2023, 9(1): 75-89. doi: 10.3934/microbiol.2023005
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This is a systematic review and meta-analysis that evaluated the prevalence of Escherichia coli antibiotic-resistant genes (ARGs) in animals, humans, and the environment in South Africa. This study followed Preferred Reporting Items for Systematic Reviews and Meta-analyses (PRISMA) guidelines to search and use literature published between 1 January 2000 to 12 December 2021, on the prevalence of South African E. coli isolates' ARGs. Articles were downloaded from African Journals Online, PubMed, ScienceDirect, Scopus, and Google Scholar search engines. A random effects meta-analysis was used to estimate the antibiotic-resistant genes of E. coli in animals, humans, and the environment. Out of 10764 published articles, only 23 studies met the inclusion criteria. The obtained results indicated that the pooled prevalence estimates (PPE) of E. coli ARGs was 36.3%, 34.4%, 32.9%, and 28.8% for blaTEM-M-1, ampC, tetA, and blaTEM, respectively. Eight ARGs (blaCTX-M, blaCTX-M-1, blaTEM, tetA, tetB, sul1, sulII, and aadA) were detected in humans, animals and the environmental samples. Human E. coli isolate samples harboured 38% of the ARGs. Analyzed data from this study highlights the occurrence of ARGs in E. coli isolates from animals, humans, and environmental samples in South Africa. Therefore, there is a necessity to develop a comprehensive “One Health” strategy to assess antibiotics use in order to understand the causes and dynamics of antibiotic resistance development, as such information will enable the formulation of intervention strategies to stop the spread of ARGs in the future.
The graphs considered throughout this paper are finite and connected. The graph-theoretical terminology and notation that are used in this study without explaining here can be found in the books [1,2].
A graph invariant is a function f defined on the set of all graphs with the condition that f(G1)=f(G2) whenever G1 and G2 are isomorphic. The real-valued graph invariants are commonly known as topological indices in mathematical chemistry, particularly in chemical graph theory [20].
The Sombor index and the reduced Sombor index abbreviated as SO and SOred, respectively, are the topological indices introduced recently by mathematical chemist Ivan Gutman in his seminal paper [9]. The Sombor index SO has attained attention from many scientific groups all over the world in a very short time, which resulted in many publications; for example, see the review papers [8,15] and associated papers listed therein. The chemical applicability of the indices SO and SOred is also well documented. Redžepović [17] examined the discriminating and predictive ability of the indices SO and SOred on a large class of isomers and found that both of these indices have good discriminating and predictive potential. Deng et al. [7] compared the predictive ability of SO on octane isomers with that of similar kind of existing topological indices and showed that SO has a higher accuracy in predicting physico-chemical properties of the the aforementioned chemical compounds; Liu et al. [13] conducted a similar comparative study for SOred and concluded that SOred outperforms in several cases. Also, it was demonstrated in [14] that boiling points of benzenoid hydrocarbons are highly correlated with SO and SOred. By considering these chemical applications of SO and SOred, it make sense to study further these indices, particularly for molecular graphs (these are graphs of maximum degree at most 4).
In chemical graph theory, molecular trees play an important role because a certain class of chemical compounds can be viewed by using the concept of molecular trees. Thereby, in the present paper, we study the topological indices SO and SOred for molecular trees under certain constraints. Deng et al. [7] determined the trees possessing the maximum and minimum values of SO and SOred among all molecular trees of a given order; see also [5,18] where the same problem regarding SO was solved independently. Let us consider the following problem.
Problem 1. Characterize the trees possessing the maximum and minimum values of SO and SOred among all molecular trees of a given order and with a fixed number of (i) branching vertices (ii) vertices of degree 2.
The minimal part of Problem 1 concerning SO has already been solved in [3] where several other interesting extremal problems were also resolved. The main objective of this study is to give a solution to the maximal part of Problem 1. Detail about the mathematical study of the Sombor index for general trees can be found in [5,6,10,12,19,21,22].
Let V(G) and E(G) denote the set of vertices and edges, respectively, in the graph G. For the vertex v∈V(G), the degree of v is denoted by dG(v) (or simply by dv if only one graph is under consideration). A vertex u∈V(G) is said to be a pendent vertex or a branching vertex if du=1 or du≥3, respectively. The set NG(u) consists of the vertices of the graph G that are adjacent to the vertex u. Let ni(G) denotes the count of vertices having degree i in the graph G. Denote by xi,j(G) the cardinality of the set consisting of the edges connecting the vertex of degree i with the vertex of degree j in the graph G. A graph of order n is also known as an n-vertex graph.
For a graph G, the Sombor index and reduced Sombor index abbreviated as SO and SOred, respectively, are defined [9] as
SO(G)=∑uv∈E(G)√d2u+d2vandSOred(G)=∑uv∈E(G)√(du−1)2+(dv−1)2. |
We start this section with the following elementary result, noted in [16].
Lemma 1. The function f defined as
f(x,y)=√x2+y2−√(x−c)2+y2, |
with 1≤c<x and y>0, is strictly decreasing in y and strictly increasing in x.
For r≥3, each of the vertices u2,u3,⋯,ur−1 of the path P:u1u2⋯ur in a graph is called internal vertex of P. Denote by Pi,j(C) (or simply Pi,j) the path from a branching vertex of degree i to a branching vertex of degree j in a molecular tree C such that all the internal vertices (if exist) of Pi,j have degree 2.
Lemma 2. If the molecular tree C contains a path P4,4:v1v2⋯vs and an edge uw such that du=1 and dw=3 with the condition that v1 lies on the w−vs path, then for the molecular tree C′ obtained from C by deleting the edges v1v2, uw and adding new edges v2w, v1u, the inequalities SO(C)<SO(C′) and SOred(C)<SOred(C′) hold, where C and C′ have the same degree sequence.
Proof. Clearly, the trees C and C′ have the same degree sequence. Also, we note that
SO(C)=SO(C′)+√10−√17+√4+d2v2−√9+d2v2<SO(C′) |
and
SOred(C)=SOred(C′)+2−3+√9+(dv2−1)2−√4+(dv2−1)2<SO(C′). |
Lemma 3. For a molecular tree C, if x1,3(C)>0 such that C contains the paths P3,4:u1u2⋯ut and P′4,3:utut+1⋯us−1us, where t≥2, s≥3, and t<s, then a molecular tree C′ can be obtained with x1,3(C′)<x1,3(C) such that SO(C)<SO(C′) and SOred(C)<SOred(C′), where C and C′ have the same degree sequence.
Proof. Suppose that uv∈E(C) such that du=1 and dv=3. We prove the result by considering two possible cases.
Case I. s≤4.
Note that one of the paths P4,3 and P′4,3 contains exactly one internal vertex and the other contains no internal vertex. Thus, without loss of generality it can be assumed that t=2. If s=3, then we take C′=C−{uv,u1u2,u2u3}+{u1u3,uu2,u2v}, and for s=4, we take C′=C−{u1u2,u3u4,uv}+{u1u4,uu2,u3v}. In either case, we note that both C and C′ have the same degree sequence, x1,3(C′)<x1,3(C), and
SO(C)=SO(C′)+5+√10−3√2−√17<SO(C′) |
SOred(C)=SOred(C′)+√13−1−2√2<SOred(C′), |
as required.
Case II. s>4.
If t=2 or t=s−1, the result is proved in a fully analogous way as in Case I. In what follows, suppose that 2<t<s−2. When s=5 then (t=3 and) we take C′=C−{u2u3,u4u5,uv}+{u2u5,u3u,u4v}, and otherwise (that is, if s≥6 then) we take C′=C−{u2u3,ut−1ut,us−1us,uv}+{u2us,uut,us−1u3,ut−1v}. In either case, we observe that both C and C′ have the same degree sequence, x1,3(C′)<x1,3(C), and
SO(C)=SO(C′)+√10+2√5−√13−√17<SO(C′), |
SOred(C)=SOred(C′)+√10−1−√5<SOred(C′), |
as desired.
Lemma 4. Let C be a molecular tree with xy,uv,wz∈E(C), where the vertices x, y, u, v, w, z are chosen in such a way that du=1, dx=3=dy, dv=dw=dz=4, and w lies on one of the three paths v−z, x−z, y−z paths, and that x1,3(C)=0. If NC(z)={w,z1,z2,z3} and C′=C−{xy,z1z,z2z,z3z,uv}+{xu,uy,z1u,z2u,z3v} then SO(C)<SO(C′) and SOred(C)<SOred(C′), where C and C′ have the same degree sequence.
Proof. It can be easily observed that C and C′ have the same degree sequence. Also, after elementary calculations, we arrive at SO(C)=SO(C′)+7√2−10<SO(C′) and SOred(C)=SOred(C′)+5√2−2√13<0.
Lemma 5. Let C be a molecular tree with uv,z′w,wz∈E(C), where the vertices u,v,w,z,z′∈V(C) are chosen in such a way that dw=2 and min{du,dv,dz,dz′}≥3 provided that du+dv>dz+dz′. If C′=C−{uv,z′w,wz}+{z′z,uw,wv}, then SO(C)<SO(C′) and SOred(C)<SOred(C′), where C and C′ have the same degree sequence.
Proof. We observe that both the trees C and C′ have the same degree sequence and
SO(C)=SO(C′)+√d2u+d2v+√4+d2z′+√4+d2z−√4+d2v−√4+d2u−√d2z+d2z′=SO(C′)+I1 |
and
SOred(C)=SOred(C′)+√(du−1)2+(dv−1)2+√1+(dz′−1)2+√1+(dz−1)2−√1+(du−1)2−√1+(dv−1)2−√(dz−1)2+(dz′−1)2=SOred(C′)+I2, |
where
I1=√d2u+d2v+√4+d2z′+√4+d2z−√4+d2v−√4+d2u−√d2z+d2z′ |
and
I2=√(du−1)2+(dv−1)2+√1+(dz′−1)2+√1+(dz−1)2−√1+(du−1)2 −√1+(dv−1)2−√(dz−1)2+(dz′−1)2. |
To complete the proof, it is enough to show that I1<0 and I2<0. Recall that du+dv∈{7,8} and hence we consider two cases accordingly.
ⅰ) If du+dv=8 then du=4=dv and hence dz+dz′∈{6,7}, and therefore we get I1<0 and I2<0
ⅱ) If du+dv=7 then du=3 and dv=4 (or du=4 and dv=3) and hence dz=3=dz′ and thence we have I1<0 and I2<0.
Lemma 6. Let C be a molecular tree with uv,vw,xy∈E(C), where the vertices u,v,w,x,y are chosen in such a way that du=2=dv, dw≥2, and min{dx,dy}≥3. If C′=C−{uv,vw,xy}+{uw,xv,vy} then SO(C)<SO(C′) and SOred(C)<SOred(C′), where C and C′ have the same degree sequence.
Proof. Clearly, the degree sequences of C and C′ is the same. Since dx,dy∈{3,4}, we have
SO(C)=SO(C′)+2√2+√d2x+d2y−√4+d2x−√4+d2y<SO(C′) |
and
SOred(C)=SOred(C′)+√2+√(dx−1)2+(dy−1)2 −√1+(dx−1)2−√1+(dy−1)2<SOred(C′). |
For n≥4, denote by Cn,nb the collection of all n-vertex molecular trees with nb branching vertices, where nb≤12(n−2). As the path Pn is the unique graph in Cn,0, where SO(Pn)=2√2(n−3)+2√5 and SOred(Pn)=√2(n−3)+2, and C1 (depicted in Figure 1) is the unique graph in C4,1 (whose (reduced) Sombor index value is given in Table 1), in rest of the investigation we assume that n>4 and nb≥1. We also define the sub-classes Cp and C∗q of Cn,nb as follows:
Cp={C∈Cn,nb:n2(C)=0,n3(C)≥0}, | (4.1) |
C∗q={C∈Cn,nb:n3(C)=0,n2(C)≥0}. | (4.2) |
SO(Ci) | SOred(Ci) | |
C1 | 3√10 | 6 |
C2 | 4√17 | 12 |
C3 | 2√2(n−5)+2√10+√13+√5 | √2(n−5)+5+√5 |
C4 | 2√2(n−6)+3(√17+√5) | √2(n−6)+10+√10 |
C5 | 4√10+3√2 | 2√2+8 |
C6 | 2√10+3√17+5 | √13+13 |
Lemma 7. If Cb is a molecular tree with the maximum Sombor index (respectively, reduced Sombor index) over the class Cn,nb, then either Cb∈Cp or Cb∈C∗q.
Proof. Contrarily, assume that Cb∈Cn,nb∖(Cp∪C∗q). There must be vertices u and v in Cb with du=3 and dv=2. Let Nu(Cb)={u1,u2,u3} and Nv(Cb)={v1,v2}, where v2 and u3 lie on the u−v path. If uv∈E(C) then we take v2=u and u3=v, and also it possible that u3=v2. If C′ is the tree deduced from C by dropping the edges uu1, uu2 and adding two new edges vu1, vu2 in C, then C′∈Cn,nb and (by keeping in mind Lemma 1, we have)
SO(Cb)=SO(C′)+3∑i=1√9+d2ui+2∑j=1√4+d2vj−√1+d2u3−2∑i=1√16+d2ui−2∑j=1√16+d2vj<SO(C′)+√9+d2u3−√1+d2u3+2∑j=1(√4+d2vj−√16+d2vj)≤SO(C′)+2∑j=1(√4+d2vj−√16+d2vj)+√13−√5≤SO(C′)+3√5+√13−8√2<SO(C′), |
which is contradicting our assumption concerning the choice of Cb.
Similarly, we have
SOred(Cb)=SOred(C′)+3∑i=1√4+(dui−1)2+2∑j=1√1+(dvj−1)2−√(du3−1)2−2∑i=1√9+(dui−1)2−2∑j=1√9+(dvj−1)2<SOred(C′)+√4+(du3−1)2−√(du3−1)2+2∑j=1√1+(dvj−1)2−2∑j=1√9+(dvj−1)2≤SOred(C′)+2∑j=1√1+(dvj−1)2−2∑j=1√9+(dvj−1)2+√5−1≤SOred(C′)+√5+2√10−6√2−1<SOred(C′), |
a contradiction.
Lemma 8. For nb>1, if Cb is the tree with the maximum Sombor index (respectively reduced Sombor index) in the class Cn,nb, then x1,2(Cb)=0.
Proof. Contrariwise, assume that there is a path P:v1v2⋯vt in Cb, where t≥3 such that dv1=1, dvt>2 and dvj=2 for all 2≤j≤t−1. Since n2(Cb)≥1, by Lemma 7 we must have dvt=4. Also, at least one of the neighbors of vt different from vt−1 is non-pendent because nb>1. Take w∈NCb(vt)∖{vt−1} such that dw≥2. As n3(Cb)=0 by Lemma 7, we have dw∈{2,4}. If C′=Cb−{v1v2,vt−1v1,uw}+{v1vt,uv2,vt−1w}, then we have C′∈Cn,nb and
SO(Cb)=SO(C′)+√5−√17+√16+d2w−√4+d2w<SO(C′), |
and
SOred(Cb)=SOred(C′)+1−3+√9+(dw−1)2−√1+(dw−1)2<SOred(C′). |
a contradiction.
For a non-trivial molecular tree C of order n, the following identities hold:
n=n1(C)+n2(C)+n3(C)+n4(C), | (4.3) |
n1(C)+2n2(C)+3n3(C)+4n4(C)=2(n−1), | (4.4) |
nb=n3(C)+n4(C). | (4.5) |
The results in the following lemma directly follows from Eqs (4.3)–(4.5).
Lemma 9. For a molecular tree C∈Cn,nb, the following statements hold:
(i) If C∈Cp, then n1(C)=n−nb, n3(C)=3nb−n+2 and n4(C)=n−2nb−2.
(ii) If C∈C∗q, then n1(C)=2nb+2, n2(C)=n−3nb−2 and n4(C)=nb.
(iii) C∈Cp∩C∗q if and only if nb=n−23.
Lemma 10. For a molecular tree C with the maximum Sombor index (respectively reduced Sombor index) over the class Cn,nb, n2(C)≥1 if and only if 0≤nb<n−23 or n≥3nb+3.
Proof. If n2(C)≥1, then n3(C)=0 by Lemma 7 and hence C∈C∗q; now, by using the result n2(C)=n−3nb−2 of Lemma 9 it is deduced that nb<n−23 or n≥3nb+3.
Conversely, suppose that 0≤nb<n−23 or n≥3nb+3 with nb≥0. Contrarily, assume that n2(C)=0. From Eqs (4.3)–(4.5), it follows that 2nb+n4(C)=n−2, which together with the assumption n≥3nb+3 implies that n4(C)≥nb+1, which is a contradiction.
Note that C2 and C3 with n=5 (see Figure 1) are the only molecular trees in C5,1, and it can be easily observed that SO(C2)>SO(C3) (respectively SOred(C2)>SOred(C3)); see Table 1. For n≥5, we define the the following sub-classes of Cn,nb:
B0={C∈C∗q: nb=1 and x1,2(C)=1},
B1={C∈C∗q: 1<nb<n−14 and x1,2(C)=0=x4,4(C)},
B2={C∈C∗q: n−14≤nb<n−23 and x1,2(C)=0=x2,2(C)},
B3={C∈Cp: n−23<nb≤3n−78 and x1,3(C)=0=x3,3(C)},
B4={C∈Cp: 3n−78<nb≤2n−65 and x1,3(C)=0=x4,4(C) and x3,3≠0},
B5={C∈Cp: 2n−65<nb≤n−22 and x4,4=0 and x3,3=n3(C)−1},
where Cp and C∗q are defined in (4.1) and (4.2). For a molecular tree C, we have
∑1≤j≤4;j≠ixi,j(C)+2xi,i(C)=i⋅niwherei=1,2,3,4. | (4.6) |
Theorem 1. If n≥6 and C∈Cn,1, then
i) SO(C)≤2√2(n−6)+3√5+3√17,
ii) SOred(C)≤√2(n−6)+10+√10.
The equalities occur if and only if C≅B0.
Proof. Let n>5 and C1∈Cn,1 be a molecular tree with the maximum Sombor index (respectively reduced Sombor index). Let v be the unique branching vertex of C1.
Claim 1. dv=4.
Contrariwise, suppose that dv=3. The constraint n>5 ensures that there is a vertex u of degree 2 in C1 which is adjacent to a pendent vertex, say w. Let C′ be the new tree obtained from C1 by deleting the edge uw and adding the new edge vw. Certainly, we have C′∈Cn,1.
Note that
SO(C1)=SO(C′)+∑z∈NC1(v)√9+d2z+ξ−∑z∈NC1(v)√16+d2z−√17<SO(C′), |
and
SOred(C1)=SOred(C′)+∑z∈NC1(v)√4+(dz−1)2+η−∑z∈NC1(v)√9+(dz−1)2−3.<SOred(C′), |
which leads to the contradiction to our assumption concerning C1, where
ξ={√5 if uv∈V(C1),√8 if uv∉V(C1),andη={1 if uv∈V(C1),√2 if uv∉V(C1). |
Claim 2. The vertex v has exactly one non pendent neighbor.
Since n≥6, the vertex v has at least one non pendent neighbor. Contrarily assume that P1:v1v2⋯vrv and P2:w1w2⋯wsv are two paths in C1 with dv1=1=dw1 and dvi=2=dwj for 2≤i≤r and 2≤j≤s. If C′=C1−{v2v1,vrv}+{v1v,vrw1}, then we have C′∈Cn,1 and
SO(C1)=SO(C′)+√4+d2v−√1+d2v+√5−√8<SO(C′), |
and
SOred(C1)=SOred(C′)+√1+(dv−1)2−√(dv−1)2+1−√2<SOred(C′), |
a contradiction (where dv=4 by Claim 1).
Now, the desired result follows from Claims 1 and 2.
Theorem 2. If C∈Cn,nb and 1<nb<n−14, then
(i) SO(C)≤2√2(n−1)+2√17(nb+1)+4√5(nb−1)−8√2nb,
(ii) SOred(C)≤√2(n−4nb−1)+2√10(nb−1)+6(nb+1),
and the equalities occur if and only if C≅B1.
Proof. Denote by Cb the molecular tree with the maximum Sombor index (respectively reduced Sombor index) in the class Cn,nb, for 1<nb<n−14. Since nb<n−14<n−23, Lemma 10 ensures that n2(Cb)>0, which together with Lemma 7 implies that Cb∈C∗q, and hence by Lemma 9(b) one has n1(Cb)=2nb+2, n2(Cb)=n−3nb−2 and n4(Cb)=nb. Because of the constraint nb>1, Lemma 8 guaranties that
x1,2(Cb)=0, | (4.7) |
plugging it into (4.6) for i=1, we get
x1,4(Cb)=n1(Cb)=2nb+2. | (4.8) |
Since nb<n−14 or 4nb<n−1 which gives n4(Cb)−1=nb−1<n−3nb−2=n2(Cb) and therefore
n4(Cb)≤n2(Cb). | (4.9) |
We claim that x2,2(Cb)≠0. Contrarily, assume that x2,2(Cb)=0. Then, (4.6) with i=2 gives
x2,4(Cb)=2n2(Cb). | (4.10) |
Equations (4.3) and (4.4) implies that
n1(Cb)−2n4(Cb)=2 | (4.11) |
Also, (4.6) with i=4 yields
2x4,4(Cb)=4n4(Cb)−x1,4(Cb)−x2,4(Cb) | (4.12) |
Using (4.8)–(4.11) in (4.12), we have
2x4,4(Cb)=4n4(Cb)−n1(Cb)−2n2(Cb)≤4n4(Cb)−n1(Cb)−2n4(Cb)=−2, |
a contradiction. Hence, the claim x2,2(Cb)≠0 is true.
We also claim that
x4,4(Cb)=0. | (4.13) |
Suppose to the contrary that x4,4(Cb)≠0. Take xy∈E(Cb) such that dx=dy=4. Since x2,2(Cb)≠0, take uv,vw∈E(Cb) such that du=dv=2 and dw≥2. If C′ is the tree obtained by applying the transformation mentioned in the statement of Lemma 6, then Lemma 6 guaranties that SO(Cb)<SO(C′) and SOred(Cb)<SOred(C′), which is a contradiction to the definition of Cb. Therefore, x4,4(Cb)=0.
Now, by using Eqs (4.6)–(4.8) and (4.13), we get x2,4(Cb)=2nb−2 and x2,2(Cb)=n−4nb−1. Hence, SO(Cb)=2√2(n−1)+2√17(nb+1)+4√5(nb−1)−8√2nb and SOred(Cb)=√2(n−4nb−1)+2√10(nb−1)+6(nb+1), which completes the proof.
Theorem 3. If C∈Cn,nb such that n−14≤nb<n−23, then
i) SO(C)≤4√5(n−3nb−2)−4√2(n−4nb−1)+2√17(nb+1),
ii) SOred(C)≤2√10(n−3nb−2)−3√2(n−4nb−1)+6(nb+1),
and the equalities occur if and only if C≅B2.
Proof. Denote by Cb the molecular tree with the maximum Sombor index (respectively reduced Sombor index) from the class Cn,nb for n−14≤nb<n−23. By Lemma 10, the inequality n2(Cb)>0 holds for nb<n−23 and ultimately we have Cb∈C∗q or n3(Cb)=0 because of Lemma 7. The equations n1(Cb)=2nb+2, n2(Cb)=n−3nb−2 and n4(Cb)=nb follow from (b) part of Lemma 9, and (4.7) and (4.8) hold by Lemma 8. Also, the inequality n2(Cb)≤n4(Cb)−1 is obtained from n−14≤nb. By using the method used in the proof of Theorem 2, we get x2,2(Cb)=0, x2,4(Cb)=2n2(Cb)=2n−6nb−4 and x4,4(Cb)=nb−1−n2(Cb)=4nb−n+1. Hence, Cb≅B2 or SO(Cb)=4√5(n−3nb−2)−4√2(n−4nb−1)+2√17(nb+1) and SOred(Cb)=2√10(n−3nb−2)−3√2(n−4nb−1)+6(nb+1), which completes the proof.
Theorem 4. Let C∈Cn,nb be the molecular tree such that nb=n−23, then
i) SO(C)=2√173(n+1)+4√23(n−5),
ii) SOred(C)=2(n+1)+√2(n−5).
Proof. The part (c) of Lemma 9 ensures that C∈Cp∩C∗q, which implies that n3(C)=0=n2(C). Thus, x1,4(C)=n−nb=23(n+1) and x4,4(C)=nb−1=13(n−5).
Theorem 5. If C∈Cn,nb such that n−23<nb≤3n−78, then
i) SO(C)≤√17(n−nb)+5(9nb−3n+6)+4√2(3n−8nb−7),
ii) SOred(C)≤3(n−nb)+√13(9nb−3n+6)+3√2(3n−8nb−7),
and the equalities occur if and only if C≅B3.
Proof. Denote by Cb the molecular tree with the maximum Sombor index (respectively reduced Sombor index) in the class Cn,nb, for n−23<nb≤3n−78. By Lemma 10, we have n2(Cb)=0 (as n−23<nb), which implies that Cb∈Cp. Also, Lemma 9(a) guaranties that n1(Cb)=n−nb, n3(Cb)=3nb−n+2 and n4(Cb)=n−2nb−2. Now, the inequality nb≤3n−78 can be written as 6nb−2n+5<n−2nb−2, which leads us to the fact that 2n3(Cb)+1≤n4(Cb). Using Lemmas 2–4 and keeping in mind the fact n4(Cb)≥2n3(Cb)+1, we have
x1,3(Cb)=0 | (4.14) |
and
x1,4(Cb)=n−nb. | (4.15) |
Now, using Lemmas 2–4 and Eqs (4.6), (4.14) and (4.15), we have x3,3(Cb)=0, x3,4(Cb)=3n3(Cb)=9nb−3n+6, x4,4(Cb)=n4(Cb)−2n3(Cb)−1=3n−8nb−7. Hence, SO(Cb)=√17(n−nb)+5(9nb−3n+6)+4√2(3n−8nb−7), and SOred(Cb)=3(n−nb)+√13(9nb−3n+6)+3√2(3n−8nb−7). This completes the proof.
Theorem 6. If C∈Cn,nb such that 3n−78<nb≤2n−65, then
i) SO(C)≤√17(n−nb)+3√2(8nb−3n+7)+5(3n−7nb−8),
ii) SOred(C)≤3(n−nb)+2√2(8nb−3n+7)+√13(3n−7nb−8),
and the equalities occur if and only if C≅B4.
Proof. Denote by Cb the molecular tree with the maximum Sombor index (respectively reduced Sombor index) from the class Cn,nb for 3n−78<nb≤2n−65. By using Lemma 10 it is easy to check that n2(Cb)=0 as n−23<3n−78<nb, which implies that Cb∈Cp and further (a) part of Lemma 9 concludes that n1(Cb)=n−nb, n3(Cb)=3nb−n+2 and n4(Cb)=n−2nb−2. Note that nb≤2n−65 can be easily written as 3nb−n+4≤n−2nb−2, which leads us to the fact n3(Cb)+2≤n4(Cb).
From Lemmas 2–4 it is clear that we have to place the vertices as described in the proof of Theorem 5. Keeping in mind the fact 2n3(Cb)+1>n4(Cb)≥n3(Cb)+2 and Eqs (4.6), (4.14) and (4.15), we have x3,3(Cb)=2n3(Cb)+1−n4(Cb)=8nb−3n+7, x3,4(Cb)=n3(Cb)+2+2(n4(Cb)−(n3(Cb)+2))=3n−7nb−8 and x4,4(Cb)=0. Hence, SO(Cb)=√17(n−nb)+3√2(8nb−3n+7)+5(3n−7nb−8) and SOred(Cb)=3(n−nb)+2√2(8nb−3n+7)+√13(3n−7nb−8), which completes the proof.
Theorem 7. If C∈Cn,nb such that 2n−65<nb≤n−22, then
i) SO(C)≤(5+3√17)(n−2nb−2)−√10(2n−5nb−6)−3√2(n−3nb−1),
ii) SOred(C)≤(5n−8nb−6)+√13(n−2nb−2)−2√2(n−3nb−1),
and the equalities occur if and only if Cb≅B5.
Proof. Denote by Cb the molecular tree with the maximum Sombor index (respectively reduced Sombor index) from the class Cn,nb for 2n−65<nb≤n−22. By using Lemma 10 it is easy to check that n2(Cb)=0 as n−23<2n−65<nb, which implies that Cb∈Cp and further (a) part of Lemma 9 concludes that n1(Cb)=n−nb, n3(C)=3nb−n+2 and n4(Cb)=n−2nb−2. Note that nb>2n−65 can be easily written as 3nb−n+4>n−2nb−2, which leads us to the fact n3(Cb)+2>n4(Cb). From Lemmas 2–4 it is clear that we have to place the vertices of degree 4 between the pendent vertices and the vertices of degree 3. The fact n3(Cb)+2>n4(Cb) gives the result
x1,4(Cb)=3n4(Cb)=3n−6nb−6, | (4.16) |
and
x1,3(Cb)=5nb−2n+6. | (4.17) |
Now, using (4.6), (4.16) and (4.17), we have x3,3(Cb)=n3(Cb)−1=3nb−n+1, x3,4(Cb)=n4(Cb)=n−2nb−2 and x4,4(Cb)=0. Hence, SO(Cb)=(5+3√17)(n−2nb−2)−√10(2n−5nb−6)−3√2(n−3nb−1) and SOred(Cb)=(5n−8nb−6)+√13(n−2nb−2)−2√2(n−3nb−1), which completes the proof.
Denote by C∗n,q the class of all n-vertex molecular trees, where q is the number of vertices of degree 2. Next, we are going to obtain the upper bounds for the molecular trees with respect to Sombor index and reduced Sombor index from the collection of molecular trees C∗n,q for 0≤q≤n−2. It is obvious that the path graph Pn is the unique graph for q=n−2, and there does not exist any graph corresponding to the value q=n−3 in the collection C∗n,q. Hence, we proceed with the assumption that 0≤q≤n−4.
Lemma 11. If the molecular tree C∈C∗n,q is the tree with the maximum Sombor index (respectively reduced Somber index), then n3(C)≤2.
Proof. Suppose contrarily that n3(C)>2, and there are vertices u,w and z of degree 3 such that w is located on the unique u−z path. Consider Nz(C)={z1,z2,z3} with the assumption that the vertex z3 lies on the u−z path in C (z3 may coincide with w). Now, a tree C′ can be obtained from the collection C∗n,q such as C′=C−{zz1,zz2}+{uz1,wz2}, which gives the following result:
SO(C)=SO(C′)+∑x∈NC(u)√d2x+9+∑y∈NC(w)√d2y+9+3∑i=1√d2zi+9−∑x∈NC(u)√d2x+16−∑y∈NC(w)√d2y+16−2∑i=1√d2zi+16−√d2z3+1<SO(C′)+∑x∈NC(u)√d2x+9+3∑i=1√d2zi+9−∑x∈NC(u)√d2x+16−2∑i=1√d2zi+16−√d2z3+1≤SO(C′)+5(5)−20√2+√10−√2<SO(C′), |
a contradiction for the chosen C.
Similarly
SOred(C)=SOred(C′)+∑x∈NC(u)√(dx−1)2+4+∑y∈NC(w)√(dy−1)2+4+3∑i=1√(dzi−1)2+4−∑x∈NC(u)√(dx−1)2+9−∑y∈NC(w)√(dy−1)2+9−2∑i=1√(dzi−1)2+9−√(dz3−1)2<SOred(C′)+5√13−15√2+2<SOred(C′), |
which also leads to a contradiction.
Lemma 12. If C∈C∗n,q, then
i) n3(C)=0 if and only if n−q−2≡0 (mod 3), n4(C)=n−q−23 and n1(C)=23(n−q+1),
ii) n3(C)=1 if and only if n−q−1≡0 (mod 3), n4(C)=n−q−43 and n1(C)=23(n−q−1)+1,
iii) n3(C)=2 if and only if n−q≡0 (mod 3), n4(C)=n−q−63 and n1(C)=23(n−q).
Proof. The following equation can be drawn by using Eqs (4.3) and (4.4):
n1(C)=n3(C)+2n4(C)+2. | (4.18) |
Now, using Eqs (4.3) and (4.18), we get
n−q−2−2n3(C)=3n4(C) | (4.19) |
or
n−q−2−2n3(C)≡0(mod3). | (4.20) |
By solving the Eqs (4.4) and (4.18) for the values of n1(C) and n4(C), we get
n1(C)=2n−2q+2−n3(C)3, | (4.21) |
and
n4(C)=n−q−2−2n3(C)3. | (4.22) |
The required results are directly followed by Eqs (4.20)–(4.22).
Lemma 13. If C∈C∗n,q is the tree with the maximum Sombor index (respectively reduced Somber index) with n−4≤q≤n−5, then x1,2(C)≤1.
Proof. Note that there is a unique branching vertex (say) u in C with the fact that du=3 if q=n−4, and du=4 if q=n−5. Let us assume that, for l,m≥3, there are paths u1u2⋯ulu and u′1u′2⋯u′mu in C such that dui=2=du′j for all 2≤i≤l and 2≤j≤m, and du1=1=du′1. If a tree C′ in the class C∗n,q is chosen as C′=C−{ul−1ul}+{ul−1u′1}, then
SO(C)=SO(C′)+√4+d2u−√1+d2u+√5−2√2, |
and
SOred(C)=SOred(C′)+√1+(du−1)2−√(du−1)2+1−√2, |
by using the fact du=3 or du=4 in the above mentioned results it can easily be checked that SO(C)<SO(C′) and SOred(C)<SOred(C′), which is a contradiction.
Lemma 14. If C∈C∗n,q is the tree with the maximum Sombor index (respectively reduced Somber index) such that q≤n−6, then x1,2(C)=0.
Proof. The results can be proved by using the same process as done in Lemma 8.
Note that C∗4,0 and C∗5,0 contain unique trees C1 and C2, respectively, with the Sombor and reduced Sombor index values given in Table 1. Further more, by using Lemma 13 it can be observed that C5 and C6 are the molecular trees with maximum (reduced) Somber index values (given in Table 1) among the graphs in C∗6,0 and C∗7,0, respectively. Among all the molecular trees C∗n,q, where q≥1, If we consider C3={C∈C∗n,q:x1,3(C)=2,x1,2(C)=1,x2,3(C)=1,x2,2(C)=n−5} and C4={C∈C∗n,q:x1,4(C)=3,x1,2(C)=1,x2,4(C)=1,x2,2(C)=n−6} (given in Figure 1), the following result is observed:
Theorem 8. For the molecular tree C∈C∗n,q, where n≥5, the following results hold:
a) If C∈C∗n,n−4∖C3, then SO(C)<SO(C3).
b) If C∈C∗n,n−5∖C4, then SO(C)<SO(C4).
Proof. Using Lemma 13, it can be concluded that C3 among the class C∗n,n−4 and C4 among C∗n,n−5, respectively, contain the maximum (reduced) Sombor index value (given in Table 1), which completes the proof.
Consider the following subsets of C∗n,q:
Q0={C∈C∗n,q: q<n−5 and n3(C)=0 such that x1,4(C)=23(n−q+1) and x2,2(C)=0 whenever x4,4(C)≠0},
Q1={C∈C∗n,q: q<n−5 and n3(C)=1 such that x1,2(C)=0 and x1,3(C)≠0 implies that there does not exist P4,4 in C moreover if x2,j(C)≠0 for 2≤j≤3, then x4,4(C)=0},
Q2={C∈C∗n,q: n3(C)=2 such that x1,2(C)=0, x1,3(C)≠0 ⇒ P4,4=0 and P3,3≠0, and whenever x1,3(C)=0 along with P3,3≠0, then P4,4=0, furthermore x2,i(C)≠0 ⇒ xj,k=0, where 2≤i≤3 and 3≤j,k≤4}.
Theorem 9. Let C∈C∗n,q for q<n−5 such that n3(C)=0. Then
SO(C)≤{2√173(n−q+1)+4√23(n−4q−5)+4√5qifq<n−54,2√173(n−q+1)+4√53(n−q−5)−2√23(n−4q−5)ifq≥n−54. |
And
SOred(C)≤{√2(n−4q−5)+2(n−q+1)+2√10qifq<n−54,2(n−q+1)+2√103(n−q−5)−√23(n−4q−5)ifq≥n−54. |
Equalities hold if and only if C∈Q0.
Proof. Denote by Cq0 the molecular tree having maximum Sombor index (or reduced Sombor index) among the collection C∗n,q such that n3(Cq0)=0. By using Lemma 12 it follows that n−q−2≡0 (mod 3), n4(Cq0)=n−q−23 and n1(Cq0)=23(n−q+1). The vertices of degree 2 are to be placed according to the conditions proved in Lemmas 6 and 14, that is all the pendent vertices are to be attached to the vertices of degree 4 i.e., Lemma 14 concludes that
x1,2(Cq0)=0, | (4.23) |
which implies that
x1,4(Cq0)=23(n−q+1). | (4.24) |
Further more, the vertices of degree 2 are to be placed between the vertices of degree 4 in such a way that if there is an edge connecting the vertices of degree 4, then no two vertices of degree 2 are adjacent. The above discussion leads us to the fact that Cq0∈Q0. Now the following two cases arise here:
Case I. n2(Cq0)<n4(Cq0)−1 or q<n−54.
In this case, Lemma 6 implies that
x2,2(Cq0)=0. | (4.25) |
By using (4.6), (4.23)–(4.25), we have x2,4(Cq0)=2q and x4,4(Cq0)=n−4q−53. Hence, SO(Cq0)=2√173(n−q+1)+4√23(n−4q−5)+4√5q and SOred(Cq0)=√2(n−4q−5)+2(n−q+1)+2√10q.
Case II. n2(Cq0)≥n4(Cq0)−1 or q≥n−54.
In this case, by Lemma 6, we have
x4,4(Cq0)=0. | (4.26) |
Using (4.6), (4.23)–(4.26), we get x2,2(Cq0)=4q−n+53 and x2,4(Cq0)=2(n4(Cq0)−1)=23(n−q−5). Hence, SO(Cq0)=2√173(n−q+1)+4√53(n−q−5)−2√23(n−4q−5) and SOred(Cq0)=2(n−q+1)+2√103(n−q−5)−√23(n−4q−5), which completes the proof.
Theorem 10. Let C∈C∗n,q such that n3(C)=1. Then
SO(C)≤{2√10+3√17+5ifq=0andn=7,2√2(q−1)+2√5+2√10+3√17+√13ifq=n−7≥1,6√17+√10(√10+1)ifq=0andn=10,6√17+5+2√5+√13+√10ifq=1andn=11,2√2(q−2)+6√17+4√5+2√13+√10ifq≥2andn−q=10,(2√5−5)q+9√17+15ifq≤2andn−q=13,(2√2)(q−3)+9√17+3(2√5+√13)ifq>2andn−q=13,√173(2n−2q+1)+4√5q+4√23(n−4q−13)+15ifq<n−134,√173(2n−2q+1)+2√53(n+2q−13)+√133(4q−n+13)+53(n−4q−4)ifn−134≤q<n−43,√173(2n−2q+1)+2√23(4q−n+4)+2√53(2n−2q−15)+3√13ifq≥n−43. |
And
SOred(C)≤{√13(√13+1)ifq=0andn=7,√2(q−1)+√10+√5+13ifq=n−7≥1,20+2√13ifq=0andn=10,20+√13+√10+√5ifq=1andn=11,√2(q−2)+20+2√10+2√5ifq≥2andn−q=10,√13(3−q)+27+√10qifq≤2andn−q=13,√2(q−3)+27+3(√5+√10)ifq>2andn−q=13,(2n−2q+1)+2√10q+√2(n−4q−13)+3√13ifq<n−134,(2n−2q+1)+√103(n+2q−13)+√53(4q−n+13)+√133(n−4q−4)ifn−134≤q<n−43,(2n−2q+1)+√23(4q−n+4)+√103(2n−2q−15)+3√13ifq≥n−43. |
Equalities hold if and only if C∈Q1.
Proof. If Cq1 denotes the molecular tree having maximum Sombor index (or reduced Sombor index) among the collection C∗n,q, where n3(Cq1)=1, then by using Lemma 12 we have n−q−1≡0 (mod 3), n1(Cq1)=2n−2q+13, n4(Cq1)=n−q−43. Note that Lemmas 2, 14 and 5 show that Cq1∈Q1. This implies that
x1,2(Cq1)=0, | (4.27) |
and the vertices of degree 4 are to be placed in the three neighbors of the vertex of degree 3 in such a way that if a pendent vertex is present in Cq1 which is adjacent to the vertex of degree 3, then there must not exists a P4,4 path in Cq1. The following cases are possible:
Case 1. n4(Cq1)=1
Subcase 1.1. q=0.
In this case n=7, x2,j(Cq1)=0 for all 1≤j≤4, x3,4(Cq1)=1, x1,3(Cq1)=2 and x1,4(Cq1)=3. The graph here is C6 given in Figure 1 with (reduced) Sombor index value given in Table 1.
Subcase 1.2. q>0.
This holds for n≥8, by keeping in mind the Lemmas 5 and 6, and using the results in Eqs (4.6) and (4.27), we have x3,4(Cq1)=0, x2,3(Cq1)=1=x2,4(Cq1), x1,3(Cq1)=2, x1,4(Cq1)=3 and x2,2(Cq1)=q−1. Hence, SO(Cq1)=2√2(q−1)+2√5+2√10+3√17+√13 and SOred=√2(q−1)+√10+√5+13.
Case 2. n4(Cq1)=2.
Subcase 2.1. q=0 In this case we have n=10, where Lemma 2 shows that
x3,4(Cq1)=2. | (4.28) |
By using Eqs (4.6), (4.27) and (4.28), we have x1,3(Cq1)=1 and x1,4(Cq1)=6, which gives SO(Cq1)=6√17+√10(√10+1) and SOred(Cq1)=20+2√13.
Subcase 2.2. q=1
Here n=11, By using Eqs (4.6), (4.27), we have x1,3(Cq1)=1, x1,4(Cq1)=6 and x3,4(Cq1)=1=x2,3(Cq1)=x2,4(Cq1), which gives SO(Cq1)=6√17+5+2√5+√13+√10 and SOred(Cq1)=20+√13+√10+√5.
Subcase 2.3. q≥2
This holds for n≥12, and by keeping in mind the Lemmas 5, 6 and the results in Eqs (4.6) and (4.27), we have x1,3(Cq1)=1, x1,4(Cq1)=6, x2,4(Cq1)=2=x2,3(Cq1) and x2,2(Cq1)=q−2. Hence, SO(Cq1)=2√2(q−2)+6√17+4√5+2√13+√10 and SOred(Cq1)=√2(q−2)+20+2√10+2√5.
Case 3. n4(Cq1)=3
Lemma 2 gives
x1,3(Cq1)=0. | (4.29) |
Subcase 3.1. q≤2
Again Lemmas 5, 6 and the Eqs (4.6), (4.27) and (4.29) imply that, x1,4(Cq1)=9, x2,3(Cq1)=x2,4(Cq1)=q, x2,2(Cq1)=0=x4,4(Cq1) and x3,4(Cq1)=3−q. Hence, SO(Cq1)=(2√5−5)q+9√17+15 and SOred(Cq1)=√13(3−q)+27+√10q.
Subcase 3.2. q>2
Here x1,4(Cq1)=9, x2,3(Cq1)=x2,4(Cq1)=3, x2,2(Cq1)=q−3 and x3,4(Cq1)=0 are obtained by using Lemmas 5, 6 and the Eqs (4.6), (4.27) and (4.29). Hence, SO(Cq1)=(2√2)(q−3)+9√17+3(2√5+√13) and SOred(Cq1)=√2(q−3)+27+3(√5+√10).
Case 4. n4(Cq1)>3
Subcase 4.1. q<n−134
It is easy to check that q<n−134 implies that q<n4(Cq1)−3 or q<x4,4(Cq1). Lemmas 5, 6 and the Eqs (4.6), (4.27) and (4.29) follow the results x2,2(Cq1)=0=x2,3(Cq1), x2,4(Cq1)=q, x3,4(Cq1)=3, x1,4(Cq1)=2n−2q+13 and x4,4(Cq1)=n−4q−133. Hence, SO(Cq1)=√173(2n−2q+1)+4√5q+4√23(n−4q−13)+15 and SOred(Cq1)=(2n−2q+1)+2√10q+√2(n−4q−13)+3√13.
Subcase 4.2. n−134≤q<n−43
It gives n4(Cq1)−3≤q<n4, which implies that
x4,4(Cq1)=0, | (4.30) |
by keeping in mind the Lemmas 5, 6 and using the Eqs (4.27), (4.29) and (4.30) in (4.6), we have x2,2(Cq1)=0, x2,3(Cq1)=4q−n+133, x2,4(Cq1)=n+2q−133, x1,4(Cq1)=2n−2q+13 and x3,4(Cq1)=n−4q−43. Hence, SO(Cq1)=√173(2n−2q+1)+2√53(n+2q−13)+√133(4q−n+13)+53(n−4q−4) and SOred(Cq1)=(2n−2q+1)+√103(n+2q−13)+√53(4q−n+13)+√133(n−4q−4).
Subcase 4.3. q≥n−43
This imply that q≥n4(C), and Lemmas 5 and 6 conclude that
x3,4(Cq1)=0, | (4.31) |
and using the Eqs (4.27), (4.29)–(4.31) in (4.6), we have x1,4(Cq1)=2n−2q+13, x2,2(Cq1)=4q−n+43, x2,3(Cq1)=3 and x2,4(Cq1)=2n−2q−153. Hence, SO(Cq1)=√173(2n−2q+1)+2√23(4q−n+4)+2√53(2n−2q−15)+3√13 and SOred(Cq1)=(2n−2q+1)+√23(4q−n+4)+√103(2n−2q−15)+3√13.
Theorem 11. Let C∈C∗n,q, where n3(C)=2. Then
SO(C)≤{4√10+3√2ifn−q=6andq=0,2√2(q−1)+4√10+2√13ifn−q=6andq>0,√10(18−n+q)3+(3√17+5)(n−q−6)3+3√2if9≤n−q≤15andq=0,√103(18−n+q)+√17(n−q−6)+(√13+2√5−5)q+3√2+15if9≤n−q≤15and1≤q≤n−64,√103(18−n+q)+(3√17+2√5)3(n−q−6)+√133(n−q)+2√23(4q−n+3)if9≤n−q≤15andq>n−64,2√173(n−q)+(√13+2√5)q+5(4−q)+3√2ifn−q=18and1≤q≤4,2√173(n−q)+6√13+8√5+2√2(q−5)ifn−q=18andq>4,2√173(n−q)+(√13+2√5)q+5(6−q)ifn−q=21and0≤q≤6,2√173(n−q)+6(√13+2√5)+2√2(q−6)ifn−q=21andq>6,2√173(n−q)+4√5q+4√2(n−4q−21)3+30ifn−q>21andq≤n−214,2√173(n−q)+√13(4q−n+21)3+2√5(n+2q−21)3+5(n−4q−3)3ifn−q>21andn−214<q≤n−34,2√173(n−q)+2√2(4q−n+3)3+4√5(n−q−12)3+6√13ifn−q>21andq>n−34. |
And
SOred(C)≤{8+2√2ifn−q=6andq=0,√2(q−1)+8+2√5ifn−q=6andq>0,2(18−n+q)3+(9+√13)(n−q−6)3+2√2if9≤n−q≤15andq=0,2(18−n+q)3+3(n−q−6)3+(√5+√10−√13)q+2√2+3√13if9≤n−q≤15and1≤q≤n−64,2(18−n+q)3+(9+√10)(n−q−6)3+√53(n−q)+√23(4q−n+3)if9≤n−q≤15andq>n−64,2(n−q)+(√5+√10)q+√13(4−q)+2√2ifn−q=18and1≤q≤4,2(n−q)+6√5+4√10+√2(q−5)ifn−q=18andq>4,2(n−q)+(√5+√10)q+√13(6−q)ifn−q=21and0≤q≤6,2(n−q)+6(√5+√10)+√2(q−6)ifn−q=21andq>6,2(n−q)+2√10q+√2(n−4q−21)+6√13ifn−q>21andq≤n−214,2(n−q)+2√10q+√5(4q−n+21)3+√10(n+2q−21)3+√13(n−4q−3)3ifn−q>21andn−214<q≤n−34,2(n−q)+6√5+√2(4q−n+3)3+2√10(n−q−12)3ifn−q>21andq>n−34. |
Equalities hold if and only if C∈Q2.
Proof. Let Cq2 denotes the molecular tree with maximum (reduced) Sombor index value among the class C∗n,q, where n3(Cq2)=2. By Lemma 12, n−q≡0 (mod 3), n1(Cq2)=2n−2q3 and n4(Cq2)=n−q−63. By Lemma 14 it holds that
x1,2(Cq2)=0, | (4.32) |
and Lemmas 2–6 provide the consequence Cq2∈Q2. We consider the following cases:
Case 1. n4(Cq2)=0 or n−q=6
Subcase 1.1. q=0
Here, using Eq (4.6), we have x1,3(Cq2)=4 and x3,3(Cq2)=1, which give SO(Cq2)=4√10+3√2 and SOred(Cq2)=8+2√2.
Subcase 1.2. q>0
By using (4.32) in (4.6), it is easy to get x1,3(Cq2)=4, x2,2(Cq2)=q−1, x2,3(Cq2)=2 and x3,3(Cq2)=0, which give SO(Cq2)=2√2(q−1)+4√10+2√13 and SOred(Cq2)=√2(q−1)+8+2√5.
Case 2. 1≤n4(Cq2)≤3 or 9≤n−q≤15
By using Lemmas 2–4 and Eq (4.32), we have
x1,3(Cq2)=4−n4(Cq2)=18−n+q3, | (4.33) |
x1,4(Cq2)=n1(Cq2)−x1,3(Cq2)=n−q−6. | (4.34) |
and
x4,4(Cq2)=0. | (4.35) |
Subcase 2.1. q=0
Using Eqs (4.32)–(4.35) in (4.6), we have x3,3(Cq2)=1 and x3,4(Cq2)=n−q−63, which give SO(Cq2)=√103(18−n+q)+(√17+53)(n−q−6)+3√2 and SOred(Cq2)=2(18−n+q)3+(9+√13)(n−q−6)3+2√2.
Subcase 2.2. 1≤q≤n−64
This case holds for q≤n4(Cq2), so by using Lemmas 5, 6 and the Eqs (4.32)–(4.35) in (4.6), we have x3,3(Cq2)=1, x3,4(Cq2)=3−q, x2,3(Cq2)=q=x2,4(Cq2) and x2,2(Cq2)=0, which give SO(Cq2)=√103(18−n+q)+√17(n−q−6)+(√13+2√5−5)q+3√2+15 and SOred(Cq2)=2(18−n+q)3+3(n−q−6)3+(√5+√10−√13)q+2√2+3√13.
Subcase 2.3. q>n−64
This holds for q>n4(Cq2), and again by keeping in mind the results used in Subcase 2.2, we may easily get x3,3(Cq2)=0=x3,4(Cq2), x2,3(Cq2)=n−q3, x2,4(Cq2)=n−q−63 and x2,2(Cq2)=4q−n+33. Hence, SO(Cq2)=√103(18−n+q)+(3√17+2√5)3(n−q−6)+√133(n−q)+2√23(4q−n+3) and SOred(Cq2)=2(18−n+q)3+(9+√10)(n−q−6)3+√53(n−q)+√23(4q−n+3).
Case 3. n−q=18 or n4(Cq2)=4
In this case, Lemmas 2–4 and Eq. (4.32) show that (4.35) holds and
x1,3(Cq2)=0, | (4.36) |
also
x1,4(Cq2)=2n−2q3. | (4.37) |
Consider the following possibilities:
Subcase 3.1. 0≤q≤4
Keeping in mind Lemmas 5 and 6, also using the Eqs (4.6), (4.32) and (4.35)–(4.37), we have x2,2(Cq2)=0, x2,3(Cq2)=q=x2,4(Cq2), x3,3(Cq2)=1 and x3,4(Cq2)=4−q. Hence, SO(Cq2)=2√173(n−q)+(√13+2√5)q+5(4−q)+3√2 and SOred(Cq2)=2(n−q)+(√5+√10)q+√13(4−q)+2√2.
Subcase 3.2. q>4
Taking into account the facts used in Subcase 3.1, we may check that x2,2(Cq2)=q−5, x2,3(Cq2)=6, x2,4(Cq2)=4 and x3,3(Cq2)=0=x3,4(Cq2). Hence, SO(Cq2)=2√173(n−q)+6√13+8√5+2√2(q−5) and SOred(Cq2)=2(n−q)+6√5+4√10+√2(q−5).
Case 4. n−q>18 or n4(Cq2)>4
Lemmas 2–4 imply that (4.36) and (4.37) hold and
x3,3(Cq2)=0. | (4.38) |
Subcase 4.1. n−q=21 and 0≤q≤6
Note that (4.35) holds in this case, further more Lemmas 5, 6 and the Eqs (4.6), (4.32), (4.35) and (4.36)–(4.38), we have x2,2(Cq2)=0, x2,3(Cq2)=q=x2,4(Cq2) and x3,4(Cq2)=6−q. Hence, SO(Cq2)=2√173(n−q)+(√13+2√5)q+5(6−q) and SOred(Cq2)=2(n−q)+(√5+√10)q+√13(6−q).
Subcase 4.2. n−q=21 and q>6
Taking into account the facts used in Subcase 3.1, we have x2,2(Cq2)=q−6, x2,3(Cq2)=6=x2,4(Cq2) and x3,4(Cq2)=0. Hence, SO(Cq2)=2√173(n−q)+6(√13+2√5)+2√2(q−6) and SOred(Cq2)=2(n−q)+6(√5+√10)+√2(q−6).
Subcase 4.3. n−q>21 and q≤n−214
This case implies that q≤n4(Cq2)−5 for which x2,3(Cq2)=0 by Lemma 5. Now using Lemma 6 and Eqs (4.6), (4.32) and (4.36)–(4.38), we have x2,2(Cq2)=0, x2,4(Cq2)=2q, x4,4(Cq2)=n−4q−213 and x3,4(Cq2)=6. Hence, SO(Cq2)=2√173(n−q)+4√5q+4√2(n−4q−21)3+30 and SOred(Cq2)=2(n−q)+2√10q+√2(n−4q−21)+6√13.
Subcase 4.4. n−q>21 and n−214<q≤n−34
(4.35) holds in this case and also using (4.6), (4.32) and (4.36)–(4.38), we have x2,2(Cq2)=0, x2,3(Cq2)=4q−n+213, x2,4(Cq2)=n+2q−213 and x3,4(Cq2)=n−4q−33. Hence, SO(Cq2)=2√173(n−q)+√13(4q−n+21)3+2√5(n+2q−21)3+5(n−4q−3)3 and SOred(Cq2)=2(n−q)+2√10q+√5(4q−n+21)3+√10(n+2q−21)3+√13(n−4q−3)3.
Subcase 4.5. n−q>21 and q>n−34
Again using Lemmas 5, 6 and the Eqs (4.6), (4.32), (4.35) and (4.36)–(4.38), we have x2,2(Cq2)=4q−n+33, x2,3(Cq2)=6, x2,4(Cq2)=2(n−q−12)3 and x3,4(Cq2)=0. Hence, SO(Cq2)=2√173(n−q)+2√2(4q−n+3)3+4√5(n−q−12)3+6√13 and SOred(Cq2)=2(n−q)+6√5+√2(4q−n+3)3+2√10(n−q−12)3.
In this paper, an extremal chemical-graph-theoretical problem concerning the Sombor index and the reduced Sombor index is addressed. Particularly, the problem of characterizing trees possessing the maximum values of the aforementioned two indices from the class of all molecular trees of a given order and with a fixed number of (i) branching vertices (ii) vertices of degree 2, is solved in this paper. A solution to the minimal version of this problem regarding the Sombor index was reported in [3]. It is believed that the minimal version of the problem under consideration regarding the reduced Sombor index is not difficult and can be solved by using the technique used in [3]; nevertheless, it still seems to be interesting to find such a solution. Solving Problem 1 for the multiplicative and exponential versions of the Sombor and reduced Sombor indices (for example, see [11]) is another possible direction for a future work concerning the present study.
The authors have no conflict of interest.
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1. | Na Pang, Nonlinear neural networks adaptive control for a class of fractional-order tuberculosis model, 2023, 20, 1551-0018, 10464, 10.3934/mbe.2023461 |
SO(Ci) | SOred(Ci) | |
C1 | 3√10 | 6 |
C2 | 4√17 | 12 |
C3 | 2√2(n−5)+2√10+√13+√5 | √2(n−5)+5+√5 |
C4 | 2√2(n−6)+3(√17+√5) | √2(n−6)+10+√10 |
C5 | 4√10+3√2 | 2√2+8 |
C6 | 2√10+3√17+5 | √13+13 |