The minimum exponential atom-bond connectivity energy of trees

Wei Gao

doi:10.1515/spma-2023-0108

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The minimum exponential atom-bond connectivity energy of trees

Wei Gao

Published/Copyright: January 9, 2024

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From the journal Special Matrices Volume 12 Issue 1

Abstract

Let G = ( V ( G ) , E ( G ) ) be a graph of order n . The exponential atom-bond connectivity matrix A e ABC ( G ) of G is an n × n matrix whose ( i , j ) -entry is equal to e d ( v i ) + d ( v j ) − 2 d ( v i ) d ( v j ) if v i v j ∈ E ( G ) , and 0 otherwise. The exponential atom-bond connectivity energy of G is the sum of the absolute values of all eigenvalues of the matrix A e ABC ( G ) . It is proved that among all trees of order n , the star S n is the unique tree with the minimum exponential atom-bond connectivity energy.

Keywords: tree; exponential atom-bond connectivity index; exponential atom-bond connectivity energy

MSC 2010: 05C50; 05C09; 05C92

1 Introduction

Let G = ( V ( G ) , E ( G ) ) be a simple graph. By d G ( v ) (or d ( v ) for short), we denote the degree of v ∈ V ( G ) . N G ( v ) is the set of neighbors of vertex v in G . As usual, P n and S n denote the path and the star of order n , respectively. For an edge subset E 0 ⊂ E ( G ) , we use G − E 0 to represent the graph obtained from G by removing all edges in E 0 .

The adjacency matrix of G is A ( G ) = ( a i j ) n × n , where a i j = 1 if v i v j ∈ E ( G ) , and a i j = 0 otherwise. The energy ℰ ( G ) of G is

ℰ ( G ) = ∑ i = 1 n ∣ λ i ( G ) ∣ ,

where λ i ( G ) , i = 1 , 2 , … , n , are the eigenvalues of A ( G ) . For the research studies on ℰ ( G ) , one may refer to [7,9,10,12,14,16].

The atom-bond connectivity (ABC) index of G is [3]

ABC ( G ) = ∑ u v ∈ E ( G ) d ( u ) + d ( v ) − 2 d ( u ) d ( v ) .

The ABC index and the heat of formation of alkanes are well correlated [3]. Recently, the ABC matrix, ABC spectral radius, and ABC energy of a graph have been introduced and studied [1,4–6,15].

The exponential ABC index of G is [11]

e ABC ( G ) = ∑ u v ∈ E ( G ) e d ( u ) + d ( v ) − 2 d ( u ) d ( v ) .

As far as we know, the tree with maximum exponential ABC index is S n [2], and the problem of finding the trees with minimum exponential ABC index is open.

The exponential ABC matrix of G , denoted by A e ABC ( G ) , is the matrix whose ( i , j ) -entry is equal to e d ( v i ) + d ( v j ) − 2 d ( v i ) d ( v j ) if v i v j ∈ E ( G ) , and 0 otherwise. The characteristic polynomial of A e ABC ( G ) , denoted by ϕ e ABC ( G , λ ) = ∣ λ I − A e ABC ( G ) ∣ , is called the exponential ABC characteristic polynomial of G . The exponential ABC energy of G is defined as:

ℰ e ABC ( G ) = ∑ i = 1 n ∣ μ i ( G ) ∣ ,

where μ i ( G ) , i = 1 , 2 , … , n , are the eigenvalues of A e ABC ( G ) .

In this work, we find that among all trees of order n , the star S n is the unique tree having the minimum exponential ABC energy. This result is a little bit unexpected.

The rest of this article is arranged as follows. In Section 2, some useful lemmas are provided. In Section 3, we show that ℰ e ABC ( T ) > ℰ e ABC ( S n ) when T is a starlike tree. Then, we prove that this is true for all trees in Section 4. Finally in Section 5, an open question is given.

2 Lemmas

For a real square matrix B , recall that ℰ ( B ) is the sum of the absolute values of eigenvalues of B . It is clear that ℰ ( G ) = ℰ ( A ( G ) ) and ℰ e ABC ( G ) = ℰ ( A e ABC ( G ) ) .

Let T be a tree of order n and ℳ k ( T ) be the set of all k -matchings of T . For e = v i v j ∈ E ( T ) and α = { e 1 , e 2 , … , e k } ∈ ℳ k ( T ) , we denote ℱ T ( e ) = e d ( u ) + d ( v ) − 2 d ( u ) d ( v ) 2 , and ℱ T ( α ) = ∏ i = 1 k ℱ T ( e i ) . Then, the exponential ABC characteristic polynomial of T is of the form:

ϕ e ABC ( T , λ ) = ∑ k = 0 ⌊ n 2 ⌋ ( − 1 ) k b 2 k λ n − 2 k ,

where b 0 = 1 , b 2 = ∑ e ∈ E ( T ) ℱ T ( e ) , and b 2 k = ∑ α ∈ ℳ k ( T ) ℱ T ( α ) for i = 2 , … , ⌊ n 2 ⌋ . By equation (2.3) in [13], we have the Coulson-type integral formula for the exponential ABC energy of T as follows:

ℰ e ABC ( T ) = 1 π ∫ − ∞ + ∞ 1 x 2 ln 1 + ∑ k = 1 ⌊ n 2 ⌋ b 2 k x 2 k d x .

So the exponential ABC energy of a tree T is a monotonically increasing function of the parameters b 2 k and k = 1 , 2 , … , ⌊ n 2 ⌋ . Thus, the following lemma is clear.

Let A = ( a i j ) and B = ( b i j ) be two nonnegative real symmetric matrices. We write A < B or B > A if a i j ≤ b i j for any 1 ≤ i , j ≤ n , and A ≠ B .

Lemma 2.1

Let T be a tree, and B be a nonnegative real symmetric matrix. If A e ABC ( T ) > B , then

ℰ e ABC ( T ) = ℰ ( A e ABC ( T ) ) > ℰ ( B ) .

Lemma 2.2

Let T be a tree with V ( T ) = { v 1 , v 2 , … , v n } and v s v t ∈ E ( T ) with d T ( v s ) ≥ 2 , d T ( v t ) ≥ 2 , and d T ( v ) ≤ 2 for all v ∈ ( N T ( v s ) ∪ N T ( v t ) ) \ { v s , v t } . Let T ′ = T − v s v t . Then,

ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) .

Proof

Denote A e ABC ( T ) = ( a i j ) and A e ABC ( T ′ ) = ( a i j ′ ) . Then, both A e ABC ( T ) and A e ABC ( T ′ ) are nonnegative real symmetric matrices of order n , where a s t > 0 = a s t ′ , a i i = a i i ′ = 0 for i = 1 , 2 , … , n , and a i j = a i j ′ for i ∉ { s , t } and j ∉ { s , t } .

Note that d T ( v s ) ≥ 2 , d T ( v t ) ≥ 2 , and d T ( v ) ≤ 2 for all v ∈ ( N T ( v s ) ∪ N T ( v t ) ) \ { v s , v t } . Then, for j = 1 , 2 , … , n and j ≠ t ,

a s j = a j s = 0 , if v s v j ∉ V ( T ) , e 1 2 , if v s v j ∈ V ( T ) and d T ( v j ) = 2 , e d ( v s ) − 1 d ( v s ) , if v s v j ∈ V ( T ) and d T ( v j ) = 1 ,

a s j ′ = a j s ′ = 0 , if v s v j ∉ V ( T ) , e 1 2 , if v s v j ∈ V ( T ) and d T ( v j ) = 2 , e d ( v s ) − 2 d ( v s ) − 1 , if v s v j ∈ V ( T ) and d T ( v j ) = 1 ,

and so a s j ≥ a s j ′ . Similarly, a t j = a j t ≥ a t j ′ = a j t ′ for j = 1 , 2 , … , n and j ≠ s .

Thus, A e ABC ( T ) > A e ABC ( T ′ ) . By Lemma 2.1, ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) . □

By definition, the following lemma is a direct result.

Lemma 2.3

Let k ≥ 2 and G = G 1 ∪ G 2 ∪ ⋯ ∪ G k . Then,

ℰ e ABC ( G ) = ℰ e ABC ( G 1 ) + ℰ e ABC ( G 2 ) + ⋯ + ℰ e ABC ( G k ) .

Lemma 2.4

[1,8]

For n ≥ 3 , ℰ ( S n ) = 2 n − 1 .
For n ≥ 2 , ℰ ( P n ) = 2 csc π 2 ( n + 1 ) − 2 , i f n i s e v e n , 2 cot π 2 ( n + 1 ) − 2 , i f n i s o d d .

Lemma 2.5

For n ≥ 3 , ℰ e ABC ( S n ) = 2 n − 1 e n − 2 n − 1 .
ℰ e ABC ( P 2 ) = 2 , and ℰ e ABC ( P n ) = e 1 2 ℰ ( P n ) for n ≥ 3 . In particular, ℰ e ABC ( P 3 ) = 2 2 e 1 2 , and ℰ e ABC ( P 5 ) = 2 ( 3 + 1 ) e 1 2 .

Proof

(1) Note that e d ( u ) + d ( v ) − 2 d ( u ) d ( v ) = e n − 2 n − 1 for each edge u v ∈ E ( S n ) , and so A e ABC ( S n ) = e n − 2 n − 1 A ( S n ) . Thus, ℰ e ABC ( S n ) = e n − 2 n − 1 ℰ ( S n ) . By Lemma 2.4, (1) holds.

(2) For each u v ∈ E ( P n ) ,

e d ( u ) + d ( v ) − 2 d ( u ) d ( v ) = 1 , if n = 2 , e 1 2 , if n ≥ 3 .

A e ABC ( P n ) = A ( P n ) , if n = 2 , e 1 2 A ( P n ) , if n ≥ 3 .

Then, ℰ e ABC ( P 2 ) = ℰ ( P 2 ) and ℰ e ABC ( P n ) = e 1 2 ℰ ( P n ) for n ≥ 3 . By (2) of Lemma 2.4, ℰ e ABC ( P 2 ) = 2 , ℰ e ABC ( P 3 ) = 2 2 e 1 2 , and ℰ e ABC ( P 5 ) = 2 ( 3 + 1 ) e 1 2 .□

Lemma 2.6

For x ≥ 2 , g 1 ( x ) = x − 1 e x − 2 x − 1 that is monotonically increasing.
For x ≥ 2 , g 2 ( x ) = e x − 1 x x is monotonically decreasing.
For x ≥ 2 , g 3 ( x ) = e x − 1 x x x − 1 is monotonically decreasing.

Proof

It is clear that x − 2 x − 1 is monotonically increasing on x ≥ 2 . Noting that both x − 1 and e x − 2 x − 1 are monotonically increasing on x ≥ 2 , we have g 1 ( x ) that is monotonically increasing on x ≥ 2 .
Note that g 2 ′ ( x ) = e x − 1 x ( 1 − x ( x − 1 ) ) 2 x 2 x − 1 . For x ≥ 2 , g 2 ( x ) is monotonically decreasing.
For x ≥ 2 , noting that both e x − 1 x x and 1 x ( x − 1 ) are monotonically decreasing, we have that g 3 ( x ) is monotonically decreasing.□

Lemma 2.7

For x ≥ 6 , f 1 ( x ) = e 1 2 2 π x + 2 π − 1 − x − 1 e > 0 .
For x ≥ 5 , f 2 ( x ) = e 1 2 ( 8 ( x + 1 ) 2 − π 2 − 4 ( x + 1 ) π ) − 4 π e ( x + 1 ) x − 1 > 0 .
For x ≥ 10 , f 3 ( x ) = 2 ( x − 1 ) e 1 2 3 − x − 1 e > 0 .
For x ≥ 8 , f 4 ( x ) = ( 3 + 1 ) e 1 2 + x − 5 3 ⋅ 2 e 1 2 − x e > 0 .
For x ≥ 7 , f 5 ( x ) = 2 e 1 2 x − 3 x − 1 e > 0 .
For x ≥ 6 , f 6 ( x ) = e 1 2 x − 2 x − 1 e > 0 .
For 0 ≤ x ≤ n − 5 , f 7 ( x ) = 2 2 3 e 1 2 x + 2 n − x − 1 e n − x − 2 n − x − 1 ≥ 2 n − 1 e n − 2 n − 1 .

Proof

For x ≥ 6 , f ′ ( x ) = 4 x − 1 e 1 2 − π e 2 π x − 1 > 0 , and so f 1 ( x ) ≥ f 1 ( 6 ) ≐ 0.93 > 0 .
For x ≥ 5 ,
f 2 ′ ( x ) = 2 e π + 16 e 1 2 x − 1 − 4 e 1 2 π x − 1 − 6 e π x + 16 e 1 2 x − 1 x x − 1 > 16 e 1 2 x − 1 x − 6 e π x x − 1 = 2 x ( 8 e 1 2 x − 1 − 3 e π ) x − 1 > 0 .
Then, f 2 ( x ) ≥ f 2 ( 5 ) ≐ 1.26 > 0 .
For x ≥ 10 , f 3 ′ ( x ) = 2 e 1 2 3 − e 2 x − 1 ≥ 2 e 1 2 3 − e 6 > 0 . Note that f 3 ( 10 ) = 3 2 e 1 2 − 3 e ≐ 0.45 > 0 . Then, f 3 ( x ) ≥ f 3 ( 10 ) > 0 .
For x ≥ 8 , f 4 ′ ( x ) = 2 2 x e 1 2 − 3 e 6 x > 0 . Therefore, f 4 ( x ) ≥ f 4 ( 8 ) > 0 .
For x ≥ 7 , f 5 ′ ( x ) = − 3 e + 2 2 e 1 2 x − 1 2 x − 1 > 0 . Therefore, f 5 ( x ) ≥ f 5 ( 7 ) > 0 .
For x ≥ 6 , f 6 ′ ( x ) = − e + e 1 2 x − 1 x − 1 > 0 . Therefore, f 6 ( x ) ≥ f 6 ( 6 ) > 0 .
For 0 ≤ x ≤ n − 5 ,
f 7 ′ ( x ) = 2 2 3 e 1 2 − e n − x − 2 n − x − 1 n − x − 2 n − x − 1 + n − x − 2 n − x − 1 ( n − x − 2 ) ≥ 2 2 3 e 1 2 − e ( n − x − 1 ) n − x − 1 ( n − x − 2 ) = 2 2 3 e 1 2 − e n − x − 1 n − x − 2 .

Note that − t + 1 t is monotonically increasing for t > 0 . Then,

f 7 ′ ( x ) = 2 2 3 e 1 2 − e n − x − 1 n − x − 2 ≥ 2 2 3 e 1 2 − e n − x − 1 n − x − 2 x = n − 5 = 2 2 3 e 1 2 − 2 e 3 ≐ 0.0999373 > 0 .

It implies that for 0 ≤ x ≤ n − 5 , f 7 ( x ) ≥ f 7 ( 0 ) = 2 n − 1 e n − 2 n − 1 .□

Lemma 2.8

For n = 2 , 3 , ℰ e ABC ( P n ) = ℰ e ABC ( S n ) , and for n ≥ 4 , ℰ e ABC ( P n ) > ℰ e ABC ( S n ) .
For n ≥ 2 , ℰ e ABC ( S n + 1 ) > ℰ e ABC ( S n ) .

Proof

(1) For n = 2 , 3 , it is clear that ℰ e ABC ( P n ) = ℰ e ABC ( S n ) .

For n ≥ 4 and n is even, noting that sin x < x for 0 < x < π 2 , we obtain

csc π 2 ( n + 1 ) − 1 > 1 π 2 ( n + 1 ) − 1 = 2 π ( n + 1 ) − 1 .

If n = 4 , it is easy to see that ℰ e ABC ( P 4 ) > ℰ e ABC ( S 4 ) . If n ≥ 6 and n is even, then by (1) of Lemma 2.7,

ℰ e ABC ( P n ) = e 1 2 ℰ ( P n ) = e 1 2 2 csc π 2 ( n + 1 ) − 2 > 2 e 1 2 2 π ( n + 1 ) − 1 > 2 n − 1 e > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

For n ≥ 5 and n is odd, noting that cos x > 1 − 1 2 x 2 and sin x < x for 0 < x < π 2 , we have

cot x > 1 − 1 2 x 2 x = 2 − x 2 2 x .

By (2) of Lemma 2.7,

ℰ e ABC ( P n ) = e 1 2 ℰ ( P n ) = e 1 2 2 cot π 2 ( n + 1 ) − 2 > 2 e 1 2 2 − ( π 2 ( n + 1 ) ) 2 π n + 1 − 1 = 2 e 1 2 ( 8 ( n + 1 ) 2 − π 2 − 4 π ( n + 1 ) ) 4 π ( n + 1 ) > 8 π e ( n + 1 ) n − 1 4 π ( n + 1 ) = 2 e n − 1 > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

(2) By (1) of Lemma 2.6, ℰ e ABC ( S n + 1 ) > ℰ e ABC ( S n ) .□

Lemma 2.9

Let ℓ 1 ≥ 3 and ℓ 2 ≥ 3 . Then,

ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) > ℰ e ABC ( S ℓ 1 + ℓ 2 ) .

In particular, if ℓ 1 ≥ 4 and ℓ 2 ≥ 3 , then

ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) > ℰ e ABC ( S ℓ 1 + ℓ 2 + 1 ) .

Proof

Let x ≥ 3 , y ≥ 3 , and

f ( x , y ) = x − 1 e x − 2 x − 1 + y − 1 e y − 2 y − 1 − x + y − 1 e x + y − 2 x + y − 1 .

Then, by (2) and (3) of Lemma 2.6,

∂ f ( x , y ) ∂ x = e x − 2 x − 1 2 x − 1 − e x + y − 2 x + y − 1 2 x + y − 1 + e x − 2 x − 1 2 ( x − 1 ) x − 2 − e x + y − 2 x + y − 1 2 ( x + y − 1 ) x + y − 2 > 0 .

Similarly, ∂ f ( x , y ) ∂ y > 0 . Thus,

f ( x , y ) ≥ f ( 3 , 3 ) = 2 2 e 1 2 − 5 e 4 5 ≐ 0.2671 > 0 ,

i.e.,

x − 1 e x − 2 x − 1 + y − 1 e y − 2 y − 1 > x + y − 1 e x + y − 2 x + y − 1 .

By Lemma 2.5, we have that for ℓ 1 ≥ 3 and ℓ 2 ≥ 3 ,

ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) > ℰ e ABC ( S ℓ 1 + ℓ 2 ) .

Similarly, we can prove that if x ≥ 4 and y ≥ 3 , then

x − 1 e x − 2 x − 1 + y − 1 e y − 2 y − 1 > x + y e x + y − 1 x + y .

It implies that if ℓ 1 ≥ 4 and ℓ 2 ≥ 3 , then

□ ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) > ℰ e ABC ( S ℓ 1 + ℓ 2 + 1 ) .

3 Exponential ABC energies of starlike trees

Recall that a tree T is a starlike tree if there exists a unique vertex v ∈ V ( T ) with d T ( v ) ≥ 3 ( v is called the center vertex of T ).

Let T be a starlike tree of order n . If the graph obtained from T by deleting its center vertex consists of ℓ 1 isolated vertices, ℓ 2 paths of order 2, and ℓ 3 paths of order 3, where 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 = n , then T is denoted as S n ( ℓ 1 , ℓ 2 , ℓ 3 ) . So S n ( n − 1 , 0 , 0 ) = S n , S n ( 0 , n − 1 2 , 0 ) = X n (where n is odd) as depicted in Figure 1, and S n ( ℓ , n − ℓ − 1 2 , 0 ) = Y n ( ℓ ) (where 1 ≤ ℓ ≤ n − 3 , and n − ℓ − 1 is even) as depicted in Figure 2.

$Figure 1 Tree X n {X}_{n} .$

Figure 1

Tree X n .

$Figure 2 Tree Y n ( ℓ ) {Y}_{n}\left(\ell ) .$

Figure 2

Tree Y n ( ℓ ) .

Lemma 3.1

Let n ≥ 7 be odd and X n be a starlike tree as depicted in Figure 1. Then,

ℰ e ABC ( X n ) > ℰ e ABC ( S n ) .

Proof

For u v ∈ E ( X n ) , e d ( u ) + d ( v ) − 2 d ( u ) d ( v ) = e 1 2 . Then, A e ABC ( X n ) = e 1 2 A ( X n ) , and so

ℰ e ABC ( X n ) = e 1 2 ℰ ( X n ) .

Let X n ′ = X n − { v 1 v i ∣ i = 4 , 6 , … , n − 1 } . Then, X n ′ = P 3 ∪ P 2 ∪ ⋯ ∪ P 2 ︸ n − 3 2 (for simplicity, let us write X n ′ = P 3 ∪ n − 3 2 P 2 ). So by Lemmas 2.2 and 2.4,

ℰ ( X n ) > ℰ ( X n ′ ) = ℰ ( P 3 ) + n − 3 2 ℰ ( P 2 ) = n + 2 2 − 3 , ℰ e ABC ( X n ) = e 1 2 ℰ ( X n ) > ( n + 2 2 − 3 ) e 1 2 .

By (5) of Lemma 2.7, n e 1 2 > 3 2 n − 1 e . So

□ ℰ e ABC ( X n ) > ( 2 2 − 3 ) e 1 2 + 3 2 n − 1 e = ( 3 − 2 2 ) e 2 n − 1 − e 1 2 + 2 n − 1 e > 2 e n − 1 > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Lemma 3.2

Let n ≥ 5 , and Y n ( ℓ ) be a starlike tree as depicted in Figure 2, where 1 ≤ ℓ ≤ n − 3 and n − ℓ − 1 is even. Then,

ℰ e ABC ( Y n ( ℓ ) ) > ℰ e ABC ( S n ) .

Proof

If n = 5 , then ℓ = 2 , and

A e ABC ( Y 5 ( 2 ) ) = 0 0 e 2 3 0 0 0 0 e 2 3 0 0 e 2 3 e 2 3 0 e 1 2 0 0 0 e 1 2 0 e 1 2 0 0 0 e 1 2 0 .

By a direct calculation,

ℰ e ABC ( Y 5 ( 2 ) ) = 2 e 2 2 3 + e 2 − e 4 2 3 + e 2 2 + 2 e 2 2 3 + e 2 + e 4 2 3 + e 2 2 ≐ 11.2149 > 4 e 3 2 = ℰ e ABC ( S 5 ) .

Let n ≥ 6 and A e ABC ( Y n ( ℓ ) ) = ( ω i , j ) n × n . Let B be the matrix obtained from A e ABC ( Y n ( ℓ ) ) by replacing ω ℓ + 1 , j and ω j , ℓ + 1 with 0’s, where j = ℓ + 2 , ℓ + 4 , … , n − 1 . It is clear that A e ABC ( Y n ( ℓ ) ) > B , and B = B ℓ + 1 ⊕ B 2 ⊕ ⋯ ⊕ B 2 ︸ n − ℓ − 1 2 , where

B ℓ + 1 = 0 ⋯ 0 e n + ℓ − 3 n + ℓ − 1 ⋮ ⋱ ⋮ ⋮ 0 ⋯ 0 e n + ℓ − 3 n + ℓ − 1 e n + ℓ − 3 n + ℓ − 1 ⋯ e n + ℓ − 3 n + ℓ − 1 0 , B 2 = 0 e 1 2 e 1 2 0 .

Obviously, ℰ ( B ℓ + 1 ) = 2 ℓ e n + ℓ − 3 n + ℓ − 1 and ℰ ( B 2 ) = 2 e 1 2 . By Lemma 2.1,

ℰ e ABC ( Y n ( ℓ ) ) = ℰ ( A e ABC ( Y n ( ℓ ) ) ) > ℰ ( B ) = ℰ ( B ℓ + 1 ) + n − ℓ − 1 2 ℰ ( B 2 ) = 2 ℓ e n + ℓ − 3 n + ℓ − 1 + ( n − ℓ − 1 ) e 1 2 .

For ℓ = 1 , noting that e n − 2 n > e 1 2 , by (6) of Lemma 2.7,

ℰ e ABC ( Y n ( ℓ ) ) = 2 e n − 2 n + ( n − 2 ) e 1 2 ≥ 2 e 1 2 + ( n − 2 ) e 1 2 = n e 1 2 > 2 n − 1 e ≥ 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

For ℓ = 2 , noting that 2 2 e n − 1 n + 1 ≥ 2 2 e 5 7 > 3 e 1 2 ,

ℰ e ABC ( Y n ( ℓ ) ) = 2 2 e n − 1 n + 1 + ( n − 3 ) e 1 2 > 3 e 1 2 + ( n − 3 ) e 1 2 = n e 1 2 > ℰ e ABC ( S n ) .

For ℓ = 3 , noting that 2 3 e n n + 2 ≥ 2 3 e 6 8 > 4 e 1 2 ,

ℰ e ABC ( Y n ( ℓ ) ) = 2 3 e n n + 2 + ( n − 4 ) e 1 2 > 4 e 1 2 + ( n − 4 ) e 1 2 = n e 1 2 > ℰ e ABC ( S n ) .

For 4 ≤ ℓ ≤ n − 3 , take x = n − ℓ − 1 . Then, 2 ≤ x ≤ n − 5 and ℰ e ABC ( Y n ( ℓ ) ) > x e 1 2 + 2 n − x − 1 e 2 n − x − 4 2 n − x − 2 . Noting that 1 > 2 2 3 and 2 n − x − 4 2 n − x − 2 ≥ n − x − 2 n − x − 1 , by (7) of Lemma 2.7:

□ ℰ e ABC ( Y n ( ℓ ) ) = x e 1 2 + 2 n − x − 1 e 2 n − x − 4 2 n − x − 2 > 2 2 3 e 1 2 x + 2 n − x − 1 e n − x − 2 n − x − 1 ≥ 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Lemma 3.3

Let 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 = n and ℓ 1 + ℓ 2 + ℓ 3 ≥ 3 . Then,

ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) ≥ ℰ e ABC ( S n ) ,

and the equality holds if and only if S n ( ℓ 1 , ℓ 2 , ℓ 3 ) = S n .

Proof

Lemmas 3.1 and 3.2 tell us that the result is obvious when ℓ 3 = 0 . So let us assume that ℓ 3 ≥ 1 , and say that the center vertex of S n ( ℓ 1 , ℓ 2 , ℓ 3 ) is v .

Case 1. ℓ 1 = ℓ 2 = 0 .

Then, ℓ 3 ≥ 3 and n ≥ 10 . Let T ′ = S n ( 0 , 0 , ℓ 3 ) − { v v j ∣ v j ∈ N S n ( 0 , 0 , ℓ 3 ) ( v ) } . Then, T ′ = { v } ∪ ℓ 3 P 3 . By Lemmas 2.2, 2.3, 2.5, and (3) of Lemma 2.7,

ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > ℰ e ABC ( T ′ ) = ℓ 3 ℰ e ABC ( P 3 ) = 2 2 ( n − 1 ) e 1 2 3 > 2 n − 1 e > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Case 2. ℓ 1 = 1 and ℓ 2 = 0 .

Then, ℓ 3 ≥ 2 and n ≥ 8 . Let T ′ = P 5 ∪ ( ℓ 3 − 1 ) P 3 . Then, T ′ can be derived by removing some edges from S n ( 1 , 0 , ℓ 3 ) . By Lemmas 2.2, 2.3, 2.5 and (4) of Lemma 2.7,

ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > ℰ e ABC ( T ′ ) = ℰ e ABC ( P 5 ) + ( ℓ 3 − 1 ) ℰ e ABC ( P 3 ) = 2 ( 3 + 1 ) e 1 2 + n − 2 3 − 1 ⋅ 2 2 e 1 2 > 2 n e > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Case 3. ℓ 1 ≥ 2 or ℓ 2 ≥ 1 .

Let T ′ = S 1 + ℓ 1 + 2 ℓ 2 ( ℓ 1 , ℓ 2 , 0 ) ∪ ℓ 3 P 3 . Then, T ′ can be derived by removing some edges from S n ( ℓ 1 , ℓ 2 , ℓ 3 ) . Note that ℓ 3 ≥ 1 , 1 + ℓ 1 + 2 ℓ 2 ≥ 3 , and n = 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 ≥ 3 ℓ 3 + 3 . Then, 1 ≤ ℓ 3 ≤ n − 3 3 . By Lemmas 2.2, 2.3, 2.5, and 3.2,

ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > ℰ e ABC ( T ′ ) = ℰ e ABC ( S 1 + ℓ 1 + 2 ℓ 2 ( ℓ 1 , ℓ 2 , 0 ) ) + ℓ 3 ℰ e ABC ( P 3 ) = ℰ e ABC ( Y 1 + ℓ 1 + 2 ℓ 2 ( ℓ 1 ) ) + ℓ 3 ℰ e ABC ( P 3 ) > 2 ℓ 1 + 2 ℓ 2 e ℓ 1 + 2 ℓ 2 − 1 ℓ 1 + 2 ℓ 2 + 2 2 e 1 2 ℓ 3 = 2 n − 3 ℓ 3 − 1 e n − 3 ℓ 3 − 2 n − 3 ℓ 3 − 1 + 2 2 e 1 2 ℓ 3 .

Note that n ≥ 6 . Consider the following cases.

Subcase 3.1 ℓ 3 = n − 3 3 .

If n = 6 , then ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > 4 2 e 1 2 > 2 5 e 4 5 = ℰ e ABC ( S 6 ) . For n ≥ 7 , by (5) of Lemma 2.7,

ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > 2 2 e 1 2 + 2 2 e 1 2 ⋅ n − 3 3 = 2 2 e 1 2 n 3 > 2 n − 1 e > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Subcase 3.2 ℓ 3 = n − 4 3 .

Since 2 3 e 2 3 − 8 2 3 e 1 2 > 0 , we have

ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > 2 3 e 2 3 + 2 2 e 1 2 ⋅ n − 4 3 > 2 2 e 1 2 n 3 > 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Subcase 3.3 1 ≤ ℓ 3 ≤ n − 5 3 .

Then, 3 ≤ 3 ℓ 3 ≤ n − 5 . By (7) of Lemma 2.7,

□ ℰ e ABC ( S n ( ℓ 1 , ℓ 2 , ℓ 3 ) ) > 2 n − 3 ℓ 3 − 1 e n − 3 ℓ 3 − 2 n − 3 ℓ 3 − 1 + 2 2 3 e 1 2 3 ℓ 3 ≥ 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Theorem 3.4

Let T be a starlike tree of order n ≥ 4 . If T ≠ S n , then ℰ e ABC ( T ) > ℰ e ABC ( S n ) .

Proof

Let v be the center vertex of T . Then, T is derived by merging v and ℓ i endpoints of ℓ i ( ℓ i ≥ 0 ) pendent paths of length i ( i = 1 , 2 , … , m ), where ∑ i = 1 m ℓ i ≥ 3 and 1 + ∑ i = 1 m i ℓ i = n .

If m ≤ 3 , by Lemma 3.3, the result holds. We now let m ≥ 4 .

Case 1. 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 = 1 , i.e., ℓ 1 = ℓ 2 = ℓ 3 = 0 .

Let T ′ = T − { v v j ∣ j ∈ N T ( v ) } . Then, T ′ = { v } ∪ ℓ 4 P 4 ∪ ⋯ ∪ ℓ m P m . So by Lemmas 2.2, 2.3, and (1) of Lemma 2.8,

ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) = ∑ i = 4 m ℓ i ℰ e ABC ( P i ) > ∑ i = 4 m ℓ i ℰ e ABC ( S i ) .

By Lemma 2.9, for 4 ≤ i ≤ m , ℓ i ℰ e ABC ( S i ) ≥ ℰ e ABC ( S i ℓ i + ℓ i − 1 ) . Note that ∑ i = 4 m ℓ i ≥ 3 and 1 + ∑ i = 4 m i ℓ i = n . Then,

ℰ e ABC ( T ) > ∑ i = 4 m ℓ i ℰ e ABC ( S i ) > ℰ e ABC ( S ∑ i = 4 m i ℓ i + ∑ i = 4 m ℓ i − 1 ) > ℰ e ABC ( S n + 1 ) > ℰ e ABC ( S n ) .

The last two inequalities are due to (2) of Lemma 2.8.

Case 2. 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 = 2 .

In this case, ℓ 1 = 1 , ℓ 2 = ℓ 3 = 0 , ∑ i = 4 m ℓ i ≥ 2 , and n = 2 + ∑ i = 4 m i ℓ i ≥ 10 . Let T ′ = P 2 ∪ ℓ 4 P 4 ∪ ⋯ ∪ ℓ m P m . Then, T ′ can be derived by removing some edges from T . By Lemmas 2.2, 2.3, 2.5, 2.8, and 2.9,

ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) = ℰ e ABC ( P 2 ) + ∑ i = 4 m ℓ i ℰ e ABC ( P i ) > 2 + ∑ i = 4 m ℓ i ℰ e ABC ( S i ) > 2 + ℰ e ABC ( S ∑ i = 4 m i ℓ i + ∑ i = 4 m ℓ i − 1 ) ≥ 2 + ℰ e ABC ( S n − 1 ) = 2 + 2 n − 2 e n − 3 n − 2 .

Since 2 > 2 2 3 e 1 2 , by (7) of Lemma 2.7,

ℰ e ABC ( T ) > 2 + 2 n − 2 e n − 3 n − 2 > 2 2 3 e 1 2 + 2 n − 2 e n − 3 n − 2 ≥ 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Case 3. 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 ≥ 3 .

Let T ′ = S 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 ( ℓ 1 , ℓ 2 , ℓ 3 ) ∪ ℓ 4 P 4 ∪ ⋯ ∪ ℓ m P m . Then, T ′ can be derived by removing some edges from T . Note that m ≥ 4 and 1 + ∑ i = 1 m i ℓ i = n . By Lemmas 2.2, 2.3, 2.8, 2.9, and 3.3,

ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) = ℰ e ABC ( S 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 ( ℓ 1 , ℓ 2 , ℓ 3 ) ) + ∑ i = 4 m ℓ i ℰ e ABC ( P i ) ≥ ℰ e ABC ( S 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 ) + ∑ i = 4 m ℓ i ℰ e ABC ( S i ) > ℰ e ABC ( S 1 + ℓ 1 + 2 ℓ 2 + 3 ℓ 3 ) + ℰ e ABC ( S ∑ i = 4 m i ℓ i ) > ℰ e ABC ( S 1 + ∑ i = 1 m i ℓ i ) = ℰ e ABC ( S n ) .

The theorem follows.□

4 Main result

Theorem 4.1

Let T be a tree of order n ≥ 3 . Then,

ℰ e ABC ( T ) ≥ ℰ e ABC ( S n ) ,

and the equality holds if and only if T = S n .

Proof

Denote b ( T ) = { v ∈ V ( T ) ∣ d T ( v ) ≥ 3 } . The theorem is proved by induction on ∣ b ( T ) ∣ .

When ∣ b ( T ) ∣ = 0 or 1, it is easy to see that the theorem holds by Lemma 2.8 or Theorem 3.4, respectively. Assume that the theorem holds for ∣ b ( T ) ∣ < k , where k ≥ 2 . We will check if it is true for ∣ b ( T ) ∣ = k .

Case 1. There is e = u v ∈ E ( T ) with u , v ∈ b ( T ) .

Set E 0 = { u v ∈ E ( T ) ∣ u , v ∈ b ( T ) } , and T ′ = T − E 0 . Without loss of generality, let T ′ be the graph as shown in Figure 3, where

v 1 , … , v m are m isolated vertices, d T ( v i ) ≥ 3 for i = 1 , 2 , … , m ;
T 1 ′ , … , T s ′ are paths of order 2, d T ( v m + i ) ≥ 3 and d T ( v m + i + 1 ) = 1 for i = 1 , 3 , … , 2 s − 1 ;
for j = 1 , … , t , T s + j ′ is a tree of order ℓ j ≥ 3 , and ∑ v ∈ V ( T s + j ′ ) d T ( v ) ≥ ∑ v ∈ V ( T s + j ′ ) d T ′ ( v ) + 1 ;
n = m + 2 s + ∑ j = 1 t ℓ j .

First, we will show t ≥ m + 2 . Note that for j = 1 , … , t ,

1 2 ∑ v ∈ V ( T s + j ′ ) d T ′ ( v ) = ∣ E ( T s + j ′ ) ∣ = ∣ V ( T s + j ′ ) ∣ − 1 = ℓ j − 1 .

Then,

n − 1 = ∣ E ( T ) ∣ = 1 2 ∑ v ∈ V ( T ) d T ( v ) = 1 2 ∑ i = 1 m + 2 s d T ( v i ) + 1 2 ∑ j = 1 t ∑ v ∈ V ( T s + j ′ ) d T ( v ) ≥ 1 2 ( 3 m + 4 s ) + 1 2 ∑ j = 1 t ∑ v ∈ V ( T s + j ′ ) d T s + j ′ ( v ) + 1 = 1 2 ( 3 m + 4 s ) + 1 2 ∑ j = 1 t ( 2 ℓ j − 1 ) = 1 2 ( 3 m + 4 s ) + ∑ j = 1 t ℓ j − t 2 .

In other words,

2 n − 1 − ∑ j = 1 t ℓ j ≥ 3 m + 4 s − t .

Since n = m + 2 s + ∑ j = 1 t ℓ j , we obtain 2 ( m + 2 s − 1 ) ≥ 3 m + 4 s − t , and thus, t ≥ m + 2 .

Denote A e ABC ( T ) = ( ω i j ) n × n . Replace ω i j and ω j i with 0 if v i v j ∈ E ( T ) with v i , v j ∈ b ( T ) . Let B be the resulting matrix. Then, A e ABC ( T ) > B > A e ABC ( T ′ ) . Note that B = B 0 ⊕ B 1 ⊕ ⋯ ⊕ B s ⊕ B ′ , where B 1 is a zero matrix of order m , B i ( i = 1 , … , s ) is a matrix of order 2 with B i ≥ 0 e 2 3 e 2 3 0 , and B ′ is a matrix of n − m − 2 s with B ′ ≥ A e ABC ( T s + 1 ′ ∪ ⋯ ∪ T s + t ′ ) . Then, ℰ ( B i ) ≥ 2 e 2 3 for i = 1 , … , s , and by the inductive hypothesis, ℰ e ABC ( T s + j ′ ) ≥ ℰ e ABC ( S ℓ j ) for j = 1 , … , t . By Lemmas 2.1 and 2.3,

ℰ e ABC ( T ) = ℰ ( A e ABC ( T ) ) > ℰ ( B ) = ∑ i = 1 s ℰ ( B i ) + ℰ ( B ′ ) ≥ ∑ i = 1 s ℰ ( B i ) + ∑ j = 1 t ℰ e ABC ( T s + j ′ ) ≥ 2 s e 2 3 + ∑ j = 1 t ℰ e ABC ( S ℓ j ) .

Since ℓ j ≥ 3 for j = 1 , … , t , by Lemma 2.9, ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) ≥ ℰ e ABC ( S ℓ 1 + ℓ 2 ) . Noting that ℓ 1 + ℓ 2 ≥ 6 , we have ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) + ℰ e ABC ( S ℓ 3 ) ≥ ℰ e ABC ( S ℓ 1 + ℓ 2 + ℓ 3 + 1 ) . Continuing with Lemma 2.9, we obtain ∑ j = 1 t ℰ e ABC ( S ℓ j ) ≥ ℰ e ABC ( S ∑ j = 1 t ℓ j + t − 2 ) . Note that n = m + 2 s + ∑ j = 1 t ℓ j , and t ≥ m + 2 . By Lemma 2.8,

ℰ e ABC ( T ) > 2 s e 2 3 + ℰ e ABC ( S ∑ j = 1 t ℓ j + t − 2 ) = 2 s e 2 3 + ℰ e ABC ( S n − m − 2 s + t − 2 ) ≥ 2 s e 2 3 + ℰ e ABC ( S n − 2 s ) .

If s = 0 , it is clear that ℰ e ABC ( T ) > ℰ e ABC ( S n ) . If s ≥ 1 , since ℓ j ≥ 3 for j = 1 , … , t , and t ≥ m + 2 ≥ 2 , we obtain 2 ≤ 2 s ≤ n − 6 . Note that e 2 3 > 2 2 3 e 1 2 . By (7) of Lemma 2.7,

ℰ e ABC ( T ) > 2 s e 2 3 + ℰ e ABC ( S n − 2 s ) > 2 2 3 e 1 2 2 s + 2 n − 2 s − 1 e n − 2 s − 2 n − 2 s − 1 ≥ 2 n − 1 e n − 2 n − 1 = ℰ e ABC ( S n ) .

Case 2. There is no u v ∈ E ( T ) such that u , v ∈ b ( T ) .

Let P r = v 1 v 2 … v r be an internal path of T with r ≥ 3 , v 1 , v r ∈ b ( T ) . Note that n ≥ 7 since ∣ b ( T ) ∣ = k ≥ 2 .

If r = 3 and n = 7 , then T is as depicted in Figure 4. By calculating directly,

ℰ e ABC ( T ) = 2 2 e 2 3 + 2 2 e 2 2 3 + e 2 > 2 6 e 5 6 = ℰ e ABC ( S 7 ) .

If r = 3 and n ≥ 8 , letting T ′ = T − { v 1 v 2 , v 2 v 3 } , then T ′ = { v 2 } ∪ T 1 ′ ∪ T 2 ′ , where T 1 ′ and T 2 ′ are two trees of orders ℓ 1 and ℓ 2 , respectively, ℓ 1 ≥ 3 , ℓ 2 ≥ 3 , and ℓ 1 + ℓ 2 = n − 1 ≥ 7 . By Lemmas 2.2 and 2.9, and the inductive hypothesis,

ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) = ℰ e ABC ( T 1 ′ ) + ℰ e ABC ( T 2 ′ ) ≥ ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) > ℰ e ABC ( S ℓ 1 + ℓ 2 + 1 ) = ℰ e ABC ( S n ) .

If r ≥ 4 , letting T ′ = T − v 1 v 2 , then T ′ = T 1 ′ ∪ T 2 ′ , where T 1 ′ and T 2 ′ are two trees of orders ℓ 1 and ℓ 2 , respectively, ℓ 1 ≥ 3 , ℓ 2 ≥ 3 , and ℓ 1 + ℓ 2 = n ≥ 8 . By Lemmas 2.2 and 2.9, and the inductive hypothesis,

ℰ e ABC ( T ) > ℰ e ABC ( T ′ ) = ℰ e ABC ( T 1 ′ ) + ℰ e ABC ( T 2 ′ ) ≥ ℰ e ABC ( S ℓ 1 ) + ℰ e ABC ( S ℓ 2 ) > ℰ e ABC ( S n ) .

The proof is now completed.□

$Figure 3 Tree T ′ T^{\prime} .$

Figure 3

Tree T ′ .

Figure 4

Tree T .

5 Conclusion

In this study, it is shown that the star S n is the unique tree with the minimum exponential atom-bond connectivity energy among all trees of order n . For further exploration, a natural question arises: What is the maximum value of the exponential atom-bond connectivity energy of trees of order n ?

Acknowledgement

The author would like to thank the anonymous reviewers for valuable suggestions that greatly improved the exposition of this manuscript.

Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2023-05-31

Revised: 2023-11-03

Accepted: 2023-11-03

Published Online: 2024-01-09

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https://doi.org/10.1515/spma-2023-0108

Keywords for this article

tree; exponential atom-bond connectivity index; exponential atom-bond connectivity energy

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