On a family of trees with minimal atom-bond connectivity index

Discrete Applied Mathematics (

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On a family of trees with minimal atom-bond connectivity index Zhibin Du a , Carlos M. da Fonseca b,∗ a

School of Mathematics and Statistics, Zhaoqing University, Zhaoqing 526061, Guangdong, China

b

Department of Mathematics, Kuwait University, Safat 13060, Kuwait

article

abstract

info

Let G = (V , E ) be a graph, di the degree of the vertex i, and ij the edge incident to the vertices i and  atom-bond connectivity index (or, simply, ABC index) is defined as  j. The (di + dj − 2)/(di dj ). While this vertex-degree-based graph invariant is ABC (G) = ij∈E relatively well-known in chemistry, only recently a significant number of results emerged among the mathematical community. Though, several important problems remained open. One of them is the characterization of the tree(s) with minimal ABC index. In this paper, we will present some structural properties of one family of trees containing a pendent path of length 3 which would minimize the ABC index, mainly including: it contains no the socalled Bk with k > 4, and contains at most two B2 ’s. © 2015 Elsevier B.V. All rights reserved.

Article history: Received 7 November 2014 Received in revised form 11 July 2015 Accepted 13 August 2015 Available online xxxx Keywords: ABC index Greedy trees

1. Introductions and preliminaries Let G = (V , E ) be a graph, di the degree of the vertex i, and ij the edge incident to the vertices i and j. The atom-bond connectivity index (or, simply, ABC index) is a vertex-degree-based graph topological index defined as

 ABC (G) =



di + dj − 2

ij∈E

di dj

,

and it can be seen as a modification of the Randić graph-theoretic invariant [17]. The importance of the ABC index was first revealed in 1998. In fact, it was showed in [6] that the ABC index can be a helpful analytical instrument for modeling thermodynamic properties of organic chemical compounds. A decade later, Estrada [5] based on a novel quantum-theory-like elucidation, exposed again the importance of this index on the stability of branched alkanes. Since then, a large number of mathematical directly related papers have emerged. According to Gutman [9], the ABC index happens to be the only topological index for which a theoretical, quantum-theorybased, foundation and justification have been found. As in many mathematical problems, characterizing the graph(s) and, in particular, the tree(s) with extremal values of an index is of crucial importance. While the study of the tree(s) with maximal ABC index has been completely determined – the star – in [7], for the tree(s) with minimal ABC index, some interesting questions remain open. Several conjectures (and sometimes surprising disproofs) are known, as well as some partial results have recently been found (see, e.g., [1,2,4,10,11,15,16]).

∗

Corresponding author. E-mail addresses: [email protected] (Z. Du), [email protected] (C.M. da Fonseca).

http://dx.doi.org/10.1016/j.dam.2015.08.017 0166-218X/© 2015 Elsevier B.V. All rights reserved.

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Fig. 1.1. An example of greedy tree.

For example, the catacondensed hexagonal systems with extremal ABC indices were characterized in [15] and it was shown that the ABC index of a graph decreases when any edge is deleted. In [14] the so-called breadth-first searching graphs were introduced in order to obtain the minimal value or lower bounds of ABC index of connected graphs. More recently, trees with minimal ABC index among the so-called Kragujevac trees were determined in [13]. A path v0 v1 . . . vr in a graph G is said to be a pendent path of length r, where dv0 > 3, dv1 = · · · = dvr −1 = 2, and dvr = 1. For the tree(s) with minimal ABC index, the lengths of its pendent paths are of crucial importance. In particular, the next lemma has become a key result in this area: Lemma 1.1 ([12,15]). If T is a tree with minimal ABC index, then every pendent path in T is of length 2 or 3, and there is at most one pendent path of length 3 in T . Recently, Dimitrov [3] presented a new approach for computing trees with minimal ABC index. By considering the degree sequences of trees and some known properties of the trees with minimal ABC index, and together with some existing counterexamples, Dimitrov set some new conjectures. One of them is the following: Conjecture 1 ([3]). If T is a tree of order n > 1178 with minimal ABC index, then there is no pendent path of length 3 in T . Before we proceed, let us recall that, according to Wang [18], for a given degree sequence of the vertices of degree more than 1, the greedy tree T is a rooted tree achieved by (i) Label the vertex with the largest degree as i; (ii) Label the neighbors of i as i1 , i2 , . . .; assign the largest degree available to them such that di1 > di2 > · · ·; (iii) Label the neighbors of i1 (except i) as i11 , i12 , . . . , such that they take all the largest degrees available and di11 > di12 > · · ·; (iv) Repeat (iii) for all the newly labeled vertices, always start with the neighbors of the labeled vertex with largest degree whose neighbors are not labeled yet. In particular, the vertex i is said to be the root of T , which is also the vertex lying on the first layer of T ; the vertices i1 , i2 , . . . are said to be the vertices lying on the second layer of T ; the vertices i11 , i12 , . . . are said to be the vertices lying on the third layer of T , and so on. For example, the tree as shown in Fig. 1.1 is a greedy tree with degree sequence

(4, 4, 3, 3, 2, 2, 2, 2, 2). In particular, the first vertex of degree 4 is the root of the tree. Recently, it was pointed out that the minimal ABC index among trees would be attained by some greedy tree. Lemma 1.2 ([8,19]). Given the degree sequence, the greedy tree minimizes the ABC index. By Lemma 1.2, we know that if we want to consider trees with minimal ABC index, we need only to focus on the greedy trees. In order to approach to a solution for Conjecture 1, in this paper we will reveal some structural properties of a greedy tree with minimal ABC index and containing a pendent path of length 3, mainly including: it contains no the so-called Bk with k > 4, and contains at most two B2 ’s. 2. Graph transformations In this section, we will introduce some graph transformations and consider the effect on ABC index under such transformations. The following lemma is clear and, therefore, we state it without a proof. Lemma 2.1 ([13]). Let T be a tree. Suppose that v1 v2 v3 is a pendent path of length 2 in T , and w u1 u2 u3 is a pendent path of length 3 in T , where v1 and w are the pendent vertices of T . Let T1 be the tree obtained from T by deleting the edge w u1 and adding the edge wv1 . Then ABC (T ) = ABC (T1 ). Let B1 , B∗1 , Bk with k > 2, and B∗k with k > 2 be, respectively, the branches of trees defined as shown in Fig. 2.1. For convenience, a branch Bk with k > 1 is said to be a Bk -type branch, and a branch B∗k with k > 1 is said to be a B∗k -type branch.

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Fig. 2.1. The structures of B1 , B∗1 , Bk with k > 2, and B∗k with k > 2.

Fig. 2.2. The trees T and T1 in Lemma 2.3.

Let

 f ( x, y ) =

x+y−2 xy

,

where x, y > 1. Clearly, ABC (G) =



f (di , dj ).

ij∈E

The next technical lemma is of particular interest for our aims. Lemma 2.2 ([4, Proposition A.3]). Let g (x, y) = f (x + ∆x, y − ∆y) − f (x, y), where x, y > 2, ∆x > 0, and 0 6 ∆y < y. Then g (x, y) is increasing in x and decreasing in y. Lemma 2.3. Suppose that T and T1 are the trees as shown in Fig. 2.2. Then ABC (T1 ) < ABC (T ). Proof. Note that du > 2. By Lemma 2.2, we have f (3, du ) − f (4, du ) < f (2, du ) − f (3, du ). Furthermore, we can get that ABC (T1 ) − ABC (T ) = 2f (3, du ) − f (4, du ) − f (2, du )

= (f (3, du ) − f (4, du )) − (f (2, du ) − f (3, du )) < 0, i.e., ABC (T1 ) < ABC (T ).



Lemma 2.4. Suppose that T and T1 are the trees as shown in Fig. 2.3, where dv > 2. (i) Suppose that du = r + 2. If r > 3, then ABC (T1 ) < ABC (T ). (ii) Suppose that du = r + 3. If r > 3, then ABC (T1 ) < ABC (T ). (iii) Suppose that du = r + 4. If r > 2, then ABC (T1 ) < ABC (T ). Proof. We only prove (i), and the proofs of (ii) and (iii) are similar.

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Fig. 2.3. The trees T and T1 in Lemma 2.4.

Note that r + 1, dv > 2. By Lemma 2.2, we have f (r + 1, dv ) − f (r + 2, dv ) is increasing in dv > 2. On the other hand, both √ 1 − √ 1 and f (r + 1, 3) are decreasing in r. r +1 r +2 So, we can get that ABC (T1 ) − ABC (T ) = (f (r + 1, dv ) − f (r + 2, dv )) + 6



1 f (r + 1, 3) − √ 2

lim (f (r + 1, dv ) − f (r + 2, dv )) +



dv →+∞



1 f (r + 1, 3) − √ 2  1



 + f (r + 1, 3) − √ r +1 r +2 2     1 1 1 −√ + f (3 + 1, 3) − √ 6 √ 3+1 3+2 2 < 0, =

i.e., ABC (T1 ) < ABC (T ).

√

1

−√

1







Lemma 2.5. Suppose that T and T1 are the trees as shown in Fig. 2.4. Then ABC (T1 ) < ABC (T ). Proof. First we show that 2f (x, 4) − 3f (x, 3) is decreasing in x > 3. Let g (x) = 2f (x, 4) − 3f (x, 3), where x > 3. Then 2x

2

 1+

1



x

1+

2 x



g (x) = ′

  3 1+

2 x





−2 1+

1 x

< 0,

i.e., g ′ (x) < 0, thus g (x) = 2f (x, 4) − 3f (x, 3) is decreasing in x > 3. Note that du > 3. So, we have 1 ABC (T1 ) − ABC (T ) = 2f (du , 4) − 3f (du , 3) + √ 2 1 6 2f (3, 4) − 3f (3, 3) + √ 2 < 0, i.e., ABC (T1 ) < ABC (T ).



Lemma 2.6. Suppose that T , T1 and T2 are the trees as shown in Fig. 2.5. (i) Suppose that du = 4. If dv > 3, then ABC (T1 ) < ABC (T ). (ii) Suppose that du = 5. If dv > 6, then ABC (T1 ) < ABC (T ). (iii) Suppose that du = 5. If dv = 5, then ABC (T2 ) < ABC (T ). Proof. First we prove (i). By Lemma 2.2, we know that f (5, dv ) − f (4, dv ) is decreasing in dv > 3. Then we have 1 ABC (T1 ) − ABC (T ) = f (5, dv ) − f (4, dv ) + f (5, 4) + √ − 2f (4, 3) 2 1 6 f (5, 3) − f (4, 3) + f (5, 4) + √ − 2f (4, 3) 2 < 0, i.e., ABC (T1 ) < ABC (T ).

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Fig. 2.4. The trees T and T1 in Lemma 2.5.

Fig. 2.5. The trees T , T1 and T2 in Lemma 2.6.

Clearly, we can prove (ii) by the similar arguments of (i). Now we prove (iii). It is easily verified that 1 ABC (T2 ) − ABC (T ) = f (5, 4) + 3f (4, 3) − f (5, 5) − 2f (5, 3) − √ 2 < 0, i.e., ABC (T2 ) < ABC (T ).



Lemma 2.7. Suppose that T and T1 are the trees as shown in Fig. 2.6. If dv > 4, then ABC (T1 ) < ABC (T ). Proof. By Lemma 2.2, we know that f (5, dv ) − f (3, dv ) is decreasing in dv > 4. Then we have 1 ABC (T1 ) − ABC (T ) = f (5, dv ) − f (3, dv ) + √ − f (3, 3) 2 1 6 f (5, 4) − f (3, 4) + √ − f (3, 3) 2 < 0, i.e., ABC (T1 ) < ABC (T ).



Lemma 2.8. For s > 1, 1

(s − 1) · f (s + 2, 4) − s · f (s + 3, 4) < − . 2

Proof. Let g (x) = x · f (x + 3, 4), where x > 0. Then

(s − 1) · f (s + 2, 4) − s · f (s + 3, 4) = g (s − 1) − g (s). Furthermore, by Lagrange mean value theorem, we have

(s − 1) · f (s + 2, 4) − s · f (s + 3, 4) = g (s − 1) − g (s) = −g ′ (ϵ), where s − 1 < ϵ < s.

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Fig. 2.6. The trees T and T1 in Lemma 2.7.

So,

(s − 1) · f (s + 2, 4) − s · f (s + 3, 4) < − is equivalent to g ′ (ϵ) > First we get that g ′ (ϵ) =

1 . 2

1 2

We are left to show that g ′ (ϵ) >

1 . 2

ϵ 2 + 7ϵ + 15 . 2(ϵ + 3)3/2 (ϵ + 5)1/2

Next, we can verify that

(ϵ 2 + 7ϵ + 15)2 − ((ϵ + 3)3/2 (ϵ + 5)1/2 )2 = 7ϵ 2 + 48ϵ + 90 > 0, ϵ+15 which implies that (ϵ+ϵ3)3+/27(ϵ+ > 1. So, we have g ′ (ϵ) > 5)1/2 Then the result follows easily.  2

1 . 2

Lemma 2.9. Suppose that T , T1 and T2 are the trees as shown in Fig. 2.7. (i) (ii) (iii) (iv)

Suppose that Suppose that Suppose that Suppose that

du du du du

= s + 3. If = s + 2. If = s + 3. If = s + 2. If

dv > 2 and s > 15, then ABC (T1 ) < ABC (T ). s > 15, then ABC (T1 ) < ABC (T ). dv > s + 3 and 1 6 s 6 14, then ABC (T2 ) < ABC (T ). 1 6 s 6 14, then ABC (T2 ) < ABC (T ).

Proof. First we prove (i). By Lemma 2.2, we know that f (s + 2, dv ) − f (s + 3, dv ) is increasing in dv > 2. On the other hand, by Lemma 2.8, 1

(s − 1) · f (s + 2, 4) − s · f (s + 3, 4) < − . 2

Then we have ABC (T1 ) − ABC (T ) = (f (s + 2, dv ) − f (s + 3, dv )) + (s − 1) · f (s + 2, 4) 1 − s · f (s + 3, 4) + 2f (s + 2, 5) − 2f (s + 3, 3) + √ 2 6 lim (f (s + 2, dv ) − f (s + 3, dv )) + (s − 1) · f (s + 2, 4) dv →+∞

1

− s · f (s + 3, 4) + 2f (s + 2, 5) − 2f (s + 3, 3) + √ 2   1 1 = √ −√ + (s − 1) · f (s + 2, 4) s+2 s+3 1

− s · f (s + 3, 4) + 2f (s + 2, 5) − 2f (s + 3, 3) + √ 2   1 1 1 −√ − < √ 2 s+2 s+3 1

+ 2f (s + 2, 5) − 2f (s + 3, 3) + √ . 2

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Fig. 2.7. The trees T , T1 and T2 in Lemma 2.9.

Let g ( s) =

 √

1

−√

s+2

for s > 15. Clearly, √ 1

s+2

−

√1 s+3

1 s+3

 −

1 2

1

+ 2f (s + 2, 5) − 2f (s + 3, 3) + √

2

is decreasing in s > 15. On the other hand, by Lemma 2.2, we have f (s + 2, 5) − f (s + 3, 3) is

decreasing in s > 15. So, we can deduce that g (s) is decreasing in s > 15. Then it follows that ABC (T1 ) − ABC (T ) < g (s) 6 g (15) < 0, i.e., ABC (T1 ) < ABC (T ), for s > 15. By the similar arguments of (i), we can also get (ii). Now we prove (iii). By Lemma 2.2, we have f (s + 4, dv ) − f (s + 3, dv ) is decreasing in dv > s + 3. So, we can get that ABC (T2 ) − ABC (T ) = (f (s + 4, dv ) − f (s + 3, dv )) + (s − 2) · f (s + 4, 4) 1 − s · f (s + 3, 4) + 5f (s + 4, 3) − 2f (s + 3, 3) − √ 2 6 (f (s + 4, s + 3) − f (s + 3, s + 3)) + (s − 2) · f (s + 4, 4) 1 − s · f (s + 3, 4) + 5f (s + 4, 3) − 2f (s + 3, 3) − √ . 2 Let 1 g (s) = f (s + 4, s + 3) − f (s + 3, s + 3) + (s − 2) · f (s + 4, 4) − s · f (s + 3, 4) + 5f (s + 4, 3) − 2f (s + 3, 3) − √ 2

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Fig. 2.8. The trees T and T1 in Lemma 2.10.

Fig. 2.9. The trees T and T1 in Lemma 2.11.

for 1 6 s 6 14. It can be verified that g (s) < 0 for 1 6 s 6 14 directly. So, we can get ABC (T2 ) − ABC (T ) 6 g (s) < 0, i.e., ABC (T2 ) < ABC (T ) for 1 6 s 6 14. By the similar arguments of (iii), we can also get (iv).



Lemma 2.10. Suppose that T and T1 are the trees as shown in Fig. 2.8. If dv > 2, then ABC (T1 ) < ABC (T ). Proof. By Lemma 2.2, we know that f (4, dv ) − f (3, dv ) is decreasing in dv > 2. Then we have 1 ABC (T1 ) − ABC (T ) = f (4, dv ) − f (3, dv ) + f (4, 4) − 2f (3, 3) + √ 2 1 6 f (4, 2) − f (3, 2) + f (4, 4) − 2f (3, 3) + √ 2 < 0, i.e., ABC (T1 ) < ABC (T ).



Lemma 2.11. Suppose that T and T1 are the trees as shown in Fig. 2.9. (i) Suppose that dw = 4. If s > 0, then ABC (T1 ) < ABC (T ). (ii) Suppose that dw = 5. If s > 1, then ABC (T1 ) < ABC (T ). Proof. We only prove (i), and the proof of (ii) is similar.

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Fig. 3.1. The tree T in Proposition 3.1 and Theorem 3.6.

By Lemma 2.2, we know that f (s + 4, 4) − f (s + 4, 3) is decreasing in s > 0. Then we have 1 ABC (T1 ) − ABC (T ) = 2(f (s + 4, 4) − f (s + 4, 3)) + √ − f (4, 3) 2 1 6 2(f (0 + 4, 4) − f (0 + 4, 3)) + √ − f (4, 3) 2 < 0, i.e., ABC (T1 ) < ABC (T ).



3. The structural properties In this final section, we will present our main results: the structural properties of a greedy tree with minimal ABC index containing a pendent path of length 3. First, we present an easy but important observation for the trees with a pendent path of length 3, which will be used in our proofs frequently. Suppose that there is a pendent path of length 2 and a pendent path of length 3 in a tree T . From Lemma 2.1, we can graft the pendent edge of a pendent path of length 3 to a pendent path of length 2, in this process, the lengths of the two pendent paths are interchanged, but without changing the ABC index. Proposition 3.1. Suppose that T is the tree as shown in Fig. 3.1, where dv > du and r > 0. (i) If du = r + 4, then T is not a tree with minimal ABC index. (ii) If du = r + 3, and there is a pendent path of length 3 in T , then T is not a tree with minimal ABC index. Proof. First we prove (i). If r = 0, i.e., dv > 4, then by Lemma 2.6(i), we can construct another tree with a smaller ABC index than ABC (T ). If r = 1, i.e., dv > 5, then by Lemmas 2.6(iii) and (ii), we can construct another tree with a smaller ABC index than ABC (T ). If r > 2, then by Lemma 2.4(iii), we can construct another tree with a smaller ABC index than ABC (T ). Then (i) follows easily. Next we prove (ii). Suppose that r = 0, i.e., dv > 3. Note that there is a pendent path of length 3 in T . Then by Lemma 2.10, we can construct another tree with a smaller ABC index than ABC (T ). If r > 1, then note that there is a pendent path of length 3 in T , by Lemma 2.1 and the proof of (i), we can get (ii) easily.  The following proposition can be obtained from Lemma 2.5. Proposition 3.2. If there is a B∗2 and two B2 ’s attached to the same vertex in a tree T , then T is not a tree with minimal ABC index. Proposition 3.3. Suppose that there is a pendent path of length 3 in a tree T . If there is three B2 ’s attached to the same vertex in T , then T is not a tree with minimal ABC index. Proof. Note that there is a pendent path of length 3 in T . Then by Lemmas 2.1 and 2.5, we can get the result.



By Lemma 2.3, we get the following proposition. Proposition 3.4. If there is a B∗1 and a B3 attached to the same vertex in a tree T , then T is not a tree with minimal ABC index.

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Fig. 3.2. The structure of T in the proof of Theorem 3.6 (Subcase 1.1).

Theorem 3.5. Suppose that T is a tree with minimal ABC index. If there is a pendent path of length 3 in T , then there is no Bk or B∗k in T , where k > 4. Proof. Suppose to the contrary that there is some Bk or B∗k in T , where k > 4. Since there is a pendent path of length 3 in T , by Lemma 2.1, we may assume that there is some B∗k in T , where k > 4. Note that there is no pendent path of length 1 in T from Lemma 1.1. Now by Lemma 2.4(i), we can obtain another tree with a smaller ABC index than ABC (T ), which is a contradiction.  Each component obtained from the deletion of a given vertex v of a tree T is commonly called a branch of T at v . Theorem 3.6. Suppose that T is a tree of order n > 18 with minimal ABC index. If T is a greedy tree and there is a pendent path of length 3 in T , then two B2 ’s cannot be attached to the same vertex in T . Proof. Suppose to the contrary that there are two B2 ’s attached to the same vertex, say u, in T . Since T is a greedy tree, du > 3. Note that there is no pendent path of length 1 in T from Lemma 1.1. If du = 3, then recall that there is a pendent path of length 3 in T , and by Lemmas 2.1 and 2.10, we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. Suppose in the following that du > 4. Assume that u lies on the ith layer of T . Case 1. u is not the root of T . Denote by v the neighbor of u in T lying on the (i − 1)th layer. Let B1 (u) be the set of the branches of T at u, except the three branches containing v , a B2 and a B2 , respectively. Clearly, B1 (u) is not an empty set since du > 4. If B1 (u) contains B∗2 , i.e., there is at least one B∗2 and two B2 ’s attached to u, then by Proposition 3.2, we can get a contradiction. If B1 (u) contains B2 , i.e., there are at least three B2 ’s attached to u, then by Proposition 3.3, we can get again a contradiction. Thus B1 (u) contains no B2 or B∗2 . Subcase 1.1. Every branch of B1 (u) is a Bk type or B∗k type branch. Obviously, every branch of B1 (u) is possibly B1 , B∗1 , B3 or B∗3 from Theorem 3.5. First suppose that B1 (u) contains B∗1 . More precisely, B1 (u) contains exactly one B∗1 and no B∗3 from Lemma 1.1, and contains no B3 from Proposition 3.4. So, the remaining branches in B1 (u) are all B1 , i.e., T is of the form as shown in Fig. 3.1 with du = r + 4. Note that dv > du = r + 4 since T is a greedy tree. Then by Proposition 3.1(i), we can deduce a contradiction. Next suppose that B1 (u) contains no B∗1 . If B1 (u) contains B1 , then recall that there is a pendent path of length 3 in T , we can deduce that T is not a greedy tree, which is a contradiction. Thus B1 (u) contains no B1 . So, we know that every branch of B1 (u) is B3 or B∗3 from Theorem 3.5. Recall that there is a pendent path of length 3 in T , by Lemma 2.1, we may assume that T is of the form as shown in Fig. 3.2. Now by Lemmas 2.1, 2.9(i) and (iii) with s = du − 3 > 1, we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. Subcase 1.2. There is some branch of B1 (u) which is not a Bk type or B∗k type branch. Denote by w a neighbor of u in T such that the branch of T at u containing w is not a Bk type or B∗k type branch. Note that T is a greedy tree, and there is no pendent path of length 1 in T , and at most one pendent path of length 3 in T from Lemma 1.1. So, dw > 3, and T is of the form as shown in Fig. 3.3, where the B∗1 and B1 ’s attached to w may not exist. Let B(w) be the set of the branches of T at w , except the one containing u. Clearly, the branches of B(w) are possibly B2 , B∗1 or B1 . In particular, B(w) contains at least one B2 from the selection of w , and it may contain no B∗1 or B1 . Since dw > 3, B(w) contains at least two branches. Without loss of generality, we may assume that the vertex w is chosen such that the B2 ’s attached to w as many as possible. If B(w) contains at least three B2 ’s, then by Proposition 3.3, we can get a contradiction. Suppose that B(w) contains exactly two B2 ’s. Note that du > dw since T is a greedy tree. Then by Proposition 3.1, we can get a contradiction.

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Fig. 3.3. The structure of T in the proof of Theorem 3.6 (Subcase 1.2).

Fig. 3.4. The structure of T in the proof of Theorem 3.6 (Subcase 1.2).

Suppose in the following that B(w) contains exactly one B2 . First suppose that B(w) contains B∗1 , more precisely, B(w) contains exactly one B∗1 from Lemma 1.1. Now B(w) contains exactly one B2 and one B∗1 , and the remaining branches in B(w) are all B1 , i.e., T is of the form as shown in Fig. 3.4, where the B1 ’s attached to w may not exist. ¯ 1 (u) be the set (of branches) obtained from B1 (u) by deleting the branch containing w. Recall that T is a greedy tree, Let B ¯ 1 (u) is a Bk type branch. More precisely, every branch of B¯ 1 (u) is and by the chosen of w , we can get that every branch of B possibly B1 or B3 from Theorem 3.5. ¯ 1 (u) contains no B1 since T is a greedy tree. So, every branch of B¯ 1 (u) is B3 . Note that B Suppose that B(w) contains no B1 . Recall that du > 4. Then by Lemma 2.7, we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. ¯ 1 (u) is B3 , i.e., T is of the form as shown in Fig. 2.9 with If B(w) contains exactly one B1 , then noting that every branch of B s = du − 4 > 0, then by Lemma 2.11(i), we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. ¯ 1 (u) is B3 , i.e., T is of the form Suppose that B(w) contains exactly two B1 ’s and take into account that every branch of B as shown in Fig. 2.9, with s = du − 4. Observe that du > dw = 5 since T is a greedy tree, thus s = du − 4 > 1. Now by Lemma 2.11(ii), we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. If B(w) contains at least three B1 ’s, then by Lemma 2.4(ii), we can construct another tree with a smaller ABC index than ABC (T ), which is once again a contradiction. Next suppose that B(w) contains no B∗1 . Now B(w) contains exactly one B2 and no B∗1 , and the remaining branches in B(w) are all B1 , i.e., T is of the form as shown in Fig. 3.5. Note that B(w) contains at least one B1 since dw > 3 (i.e., B(w) contains at least two branches). Recall that there is a pendent path of length 3 in T . Then, by Lemma 2.1 and the above arguments on the case when B(w) contains B∗1 , we can also deduce a contradiction. So, we can also always deduce a contradiction in this subcase. Case 2. u is the root of T . Let B2 (u) be the set of the branches of T at u, except the two branches containing a B2 and a B2 , respectively. Similar to the arguments in Case 1, by Propositions 3.2 and 3.3, we know that B2 (u) contains no B2 or B∗2 . Subcase 2.1. B2 (u) contains B∗1 . Actually, in the subcase, since T is a greedy tree, every branch of B2 (u) is a Bk type or B∗k type branch. Moreover, every branch of B2 (u) is possibly B1 , B∗1 , B3 or B∗3 from Theorem 3.5. Note that B2 (u) contains exactly one B∗1 and no B∗3 from Lemma 1.1, and contains no B3 from Proposition 3.4. So, the remaining branches in B2 (u) are all B1 , i.e., T is of the form as shown in Fig. 3.6. If B2 (u) contains at most two B1 ’s, then n 6 18, which is a contradiction.

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Fig. 3.5. The structure of T in the proof of Theorem 3.6 (Subcase 1.2).

Fig. 3.6. The structure of T in the proof of Theorem 3.6 (Subcase 2.1).

Fig. 3.7. The structure of T in the proof of Theorem 3.6 (Subcase 2.2).

If B2 (u) contains at least three B1 ’s, then by Lemma 2.4(iii), we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. Subcase 2.2. B2 (u) contains no B∗1 .

If B2 (u) contains B1 , then recall that there is a pendent path of length 3 in T , we can deduce that T is not a greedy tree, which is a contradiction. Thus B2 (u) contains no B1 . First suppose that every branch of B2 (u) is a Bk type or B∗k type branch. Then every branch of B2 (u) is B3 or B∗3 from Theorem 3.5, i.e., T is of the form as shown in Fig. 3.7. Now by Lemmas 2.1, 2.9(ii) and (iv) with s = du − 2 > 2, we can construct another tree with a smaller ABC index than ABC (T ), which is a contradiction. Next suppose that there is some branch of B2 (u) which is not a Bk type or B∗k type branch. Since T is a greedy tree, we get that T is of the form as shown in Fig. 3.3. By the arguments in Subcase 1.2, we can deduce a contradiction again. Combining Cases 1 and 2, the result now follows.  Similarly to the proof of Theorem 3.6, we have Theorem 3.7. Suppose that T is a tree with minimal ABC index. If T is a greedy tree and there is a pendent path of length 3 in T , then B2 and B∗2 cannot be attached to the same vertex in T . Finally, also by Theorem 3.6, we have the following corollary. Corollary 3.8. Suppose that T is a tree of order n > 18 with minimal ABC index. If T is a greedy tree and there is a pendent path of length 3 in T , then there are at most two B2 ’s in T .

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Acknowledgments This work was supported by the National Natural Science Foundation of China (Grant No. 11426199), Guangdong Provincial Natural Science Foundation of China (Grant No. 2014A030310277), and the Department of Education of Guangdong Province Natural Science Foundation of China (Grant No. 2014KQNCX224). References [1] M.B. Ahmadi, S.A. Hosseini, M. Zarrinderakht, On large trees with minimal atom-bond connectivity index, MATCH Commun. Math. Comput. Chem. 69 (2013) 565–569. [2] K.Ch. Das, I. Gutman, B. Furtula, On atom-bond connectivity index, Filomat 26 (4) (2012) 733–738. [3] D. Dimitrov, Efficient computation of trees with minimal atom-bond connectivity index, Appl. Math. Comput. 224 (2013) 663–670. [4] D. Dimitrov, On structural properties of trees with minimal atom-bond connectivity index, Discrete Appl. Math. 172 (2014) 28–44. [5] E. Estrada, Atom-bond connectivity and the energetic of branched alkanes, Chem. Phys. Lett. 463 (2008) 422–425. [6] E. Estrada, L. Torres, L. Rodríguez, I. Gutman, An atom-bond connectivity index: Modelling the enthalpy of formation of alkanes, Indian J. Chem. 37A (1998) 849–855. [7] B. Furtula, A. Graovac, D. Vukičević, Atom-bond connectivity index of trees, Discrete Appl. Math. 157 (2009) 2828–2835. [8] L. Gan, B. Liu, Z. You, The ABC index of trees with given degree sequence, MATCH Commun. Math. Comput. Chem. 68 (2012) 137–145. [9] I. Gutman, Degree-based topological indices, Croat. Chem. Acta 86 (4) (2013) 351–361. [10] I. Gutman, B. Furtula, Trees with smallest atom-bond connectivity index, MATCH Commun. Math. Comput. Chem. 68 (2012) 131–136. [11] I. Gutman, B. Furtula, M.B. Ahmadib, S.A. Hosseini, P.S. Nowbandegani, M. Zarrinderakht, The ABC index conundrum, Filomat 27 (2013) 1075–1083. [12] I. Gutman, B. Furtula, M. Ivanović, Notes on trees with minimal atom-bond connectivity index, MATCH Commun. Math. Comput. Chem. 67 (2012) 467–482. [13] S.A. Hosseini, M.B. Ahmadi, I. Gutman, Kragujevac trees with minimal atom-bond connectivity index, MATCH Commun. Math. Comput. Chem. 71 (2014) 5–20. [14] W. Lin, T. Gao, Q. Chen, X. Lin, On the minimal ABC index of connected graphs with given degree sequence, MATCH Commun. Math. Comput. Chem. 69 (2013) 571–578. [15] W. Lin, X. Lin, T. Gao, X. Wu, Proving a conjecture of Gutman concerning trees with minimal ABC index, MATCH Commun. Math. Comput. Chem. 69 (2013) 549–557. [16] J. Liu, J. Chen, Further properties of trees with minimal atom-bond connectivity index, Abstr. Appl. Anal. 2014 (2014) 609208. [17] M. Randić, Characterization of molecular branching, J. Am. Chem. Soc. 97 (1975) 6609–6615. [18] H. Wang, The extremal values of the Wiener index of a tree with given degree sequence, Discrete Appl. Math. 156 (2008) 2647–2654. [19] R. Xing, B. Zhou, Extremal trees with fixed degree sequence for atom-bond connectivity index, Filomat 26 (2012) 683–688.