On atom–bond connectivity index ofconnected graphs

Rundan Xinga, Bo Zhoua∗, Fengming Dongb†

aDepartment of Mathematics, South China Normal University,Guangzhou 510631, P. R. China

email: zhoubo@scnu.edu.cnbMathematics and Mathematics Education, National Institute of Education,

Nanyang Technological University,Singapore 637616, Singapore

email: fengming.dong@nie.edu.sg

Abstract

The atom–bond connectivity (ABC) index of a graph G is defined as

ABC(G) =∑

uv∈E(G)

√du + dv − 2

dudv,

where E(G) is the edge set and du is the degree of vertex u of G. We give a up-per bound for ABC index of connected graphs with fixed number of vertices andmaximum degree, and characterize the extremal graphs. We determine the n-vertex unicyclic graphs with the maximum, the second, the third and the fourthmaximum ABC indices, and the n-vertex bicyclic graphs with the maximum andthe second maximum ABC indices respectively for n ≥ 5.

Keywords Atom–bond connectivity index; Maximum degree; Unicyclic graphs;Bicyclic graphs; Pendent vertices

1 Introduction

Let G be a simple graph with vertex set V (G) and edge set E(G). For u ∈ V (G), Γ(u)

denotes the set of its neighbors in G and the degree of u is du = |Γ(u)|. The atom–bond

∗Corresponding author.†Supported by NIE AcRf funding (RI 5/06 DFM) of Singapore.

1

connectivity (ABC) index of G, proposed by Estrada et al. in [1], is defined as

ABC(G) =∑

uv∈E(G)

√du + dv − 2

dudv

.

It found applications in chemical research [1, 2]. Upper bounds for the ABC index

of general graphs using some other graph parameters have been given in [4]. The

properties of ABC index for trees have been studied in [3, 4, 5]. More properties for

the ABC index may be found in [6].

In this paper, we give a upper bound for ABC indices of connected graphs with

fixed number of vertices and maximum degree and characterize the extremal graphs,

and determine the n-vertex unicyclic graphs with the maximum, the second, the third

and the fourth maximum ABC indices, and the n-vertex bicyclic graphs with the

maximum and the second maximum ABC indices respectively for n ≥ 5.

2 Preliminaries

First we give some lemmas.

Let f(x, y) =√

x+y−2xy

=√

1x

+ 1y− 2

xy, where x, y are positive integers. Let ga(x) =

f(x, a) − f(x, a − 1), where a and x are positive integers. Let l1(x) = xf(1, x + 2) −(x− 1)f(1, x+ 1) = x

√x+1x+2− (x− 1)

√x

x+1with x > 0. Let l2(x) = xf(1, x+ 3)− (x−

1)f(1, x+ 2) with x > 0.

Lemma 2.1 [5] f(x, 1) is increasing for x, f(x, 2) =√

12, and f(x, y) is decreasing

for x if y ≥ 3.

Lemma 2.2 [5] If a ≥ 2, then ga(x) ≤ ga(2) < ga(1) for x ≥ 2, and for fixed x, ga(x)

is decreasing for a if x = 1, and increasing for a if x ≥ 2.

Lemma 2.3 l1(x) is increasing for x > 0.

Proof. Let m1(x) = x√

x+1x+2

. Then l1(x) = m1(x)−m1(x− 1). By direct calculation,

we have

m′1(x) =

√x+ 1

x+ 2+ x · 1

2

√x+ 2

x+ 1· 1

(x+ 2)2

2

=

√x+ 1

x+ 2

(1− 1

2(x+ 1)+

1

x+ 2

),

and

m′′1(x) =1

2

√x+ 2

x+ 1· 1

(x+ 2)2

(1− 1

2(x+ 1)+

1

x+ 2

)+

√x+ 1

x+ 2

(1

2· 1

(x+ 1)2− 1

(x+ 2)2

)=

√x+ 1

x+ 2

(1− 1

2(x+1)+ 1

x+2

2(x+ 1)(x+ 2)+

1

2(x+ 1)2− 1

(x+ 2)2

)=

7x+ 8

4(x+ 1)32 (x+ 2)

52

> 0.

By Lagrange’s mean value theorem, l′1(x) = m′1(x)−m′1(x− 1) = m′′1(ξ) > 0 for some

ξ with x− 1 ≤ ξ ≤ x. Thus l1(x) is increasing for x. �

Lemma 2.4 The function f(3, x) is convex decreasing for x.

Proof. Since f(3, x) =√

x+13x

, we have

f ′(3, x) =1√3· 1

2

√x

x+ 1·(− 1

x2

)= − 1

2√

3x32 (x+ 1)

12

< 0,

and

f ′′(3, x) = − 1

2√

3

[(−3

2x−

52

)(x+ 1)−

12 + x−

32 ·(−1

2

)(x+ 1)−

32

]=

4x+ 3

4√

3x52 (x+ 1)

32

> 0.

The result follows. �

Lemma 2.5 l2(x) is increasing for x > 0.

Proof. Let m2(x) = xf(1, x+ 3) = x√

x+2x+3

. Then l2(x) = m2(x)−m2(x− 1). Since

m′2(x) =

√x+ 2

x+ 3+ x · 1

2

√x+ 3

x+ 2· 1

(x+ 3)2

=

√x+ 2

x+ 3

(1 +

3

2(x+ 3)− 1

x+ 2

),

3

and then for x > 0, we have

m′′2(x) =1

2

√x+ 3

x+ 2· 1

(x+ 3)2

(1 +

3

2(x+ 3)− 1

x+ 2

)+

√x+ 2

x+ 3

(− 3

2(x+ 3)2+

1

(x+ 2)2

)=

√x+ 2

x+ 3

(1 + 3

2(x+3)− 1

x+2

2(x+ 2)(x+ 3)− 3

2(x+ 3)2+

1

(x+ 2)2

)=

11x+ 24

4(x+ 2)32 (x+ 3)

52

> 0.

By Lagrange’s mean value theorem, l′2(x) = m′2(x) − m′2(x − 1) = m′′2(ξ) > 0 with

x− 1 ≤ ξ ≤ x. Then the result follows. �

Let Cn and Sn be the cycle and the star on n vertices, respectively.

For a graph G with u, v ∈ V (G), G−u denotes the graph formed from G by deleting

vertex u (and its incident edges), G+ uv denotes the graph formed from G by adding

the edge uv if uv 6∈ E(G), and G − uv denotes the graph formed from G by deleting

the edge uv if uv ∈ E(G).

3 Upper bound for ABC index of connected graphs

with fixed maximum degree

In this section, we obtain the upper bound of ABC index among the connected graphs

with n vertices, m edges and maximum degree ∆, and characterize the extremal graphs.

We use the technique from [5].

For positive integers i, j, ∆ with 1 ≤ i ≤ j ≤ ∆ and ∆ ≥ 3, let

h(i, j,∆) = 2

(1√2−√

∆− 1

∆

)(1

i+

1

j− 1

2− 1

∆

)+

√i+ j − 2

ij− 1√

2.

Lemma 3.1 [5] Let i, j,∆ be positive integers with 1 ≤ i ≤ j ≤ ∆ and ∆ ≥ 3. Then

h(i, j,∆) < 0 for (i, j) 6= (1,∆), (2,∆).

For a connected graph G with n vertices, m edges and maximum degree ∆, denote

by ni the number of vertices with degree i in G for i = 1, 2, . . . ,∆, and xi,j the number

4

of edges of G connecting vertices of degree i and j, where 1 ≤ i ≤ j ≤ ∆ and n ≥ 3.

Then

n1 + n2 + · · ·+ n∆ = n

n1 + 2n2 + · · ·+ ∆n∆ = 2m

x1,2 + x1,3 + · · ·+ x1,∆ = n1

x1,2 + 2x2,2 + · · ·+ x2,∆ = 2n2

· · ·

x1,∆ + x2,∆ + · · ·+ 2x∆,∆ = ∆n∆.

(1)

For positive integers n,m and ∆ with 3 ≤ ∆ ≤ n − 1 ≤ m, let Gn,m,∆ be the set

of connected simple graphs G of order n, size m and maximum degree ∆ such that

xi,j = 0 for all i, j with i ≥ 3 or j 6= ∆.

Lemma 3.2 Let n,m and ∆ be positive integers with 3 ≤ ∆ ≤ n− 1 ≤ m.

(i) If G ∈ Gn,m,∆, then x1,∆ = 2n−m− 2m∆

, x2,∆ = 2(m− n) + 2m∆

, n∆ = m∆

and

ABC(G) =

√∆− 1

∆

(2n−m− 2m

∆

)+√

2(m− n+

m

∆

). (2)

(ii) Gn,m,∆ 6= ∅ if and only if m ≤ 2n− 2m∆

and m ≡ 0 (mod ∆).

Proof. (i) Assume that G ∈ Gn,m,∆. By definition of Gn,m,∆, xi,j = 0 if either i ≥ 3

or j < ∆, which implies that

n1 + n2 + n∆ = n

n1 + 2n2 + ∆n∆ = 2m

x1,∆ = n1

x2,∆ = 2n2

x1,∆ + x2,∆ = ∆n∆.

(3)

Solving the above equations gives that x1,∆ = n1 = 2n − m − 2m∆

, x2,∆ = 2n2 =

2m− 2n+ 2m∆

and n∆ = m∆

. Expression (2) follows from the definition of ABC(G).

(ii) By (i), if there exists G in Gn,m,∆, then x1,∆ = 2n −m − 2m∆≥ 0 and n∆ = m

∆

is an integer, implying that the necessity of (ii) holds.

5

Assume that m ≤ 2n − 2m∆

and m ≡ 0 (mod ∆). Let m = s∆. So s is a positive

integer. Let H be a connected graph (may have loops and parallel edges) of order

s, size m − n + s and maximum degree at most ∆. Since m ≤ 2n − 2s, we have

s∆ = m ≥ 2(m − n + s), implying that such graph H exists. Let {vi : 1 ≤ i ≤ s}and {ei : 1 ≤ i ≤ m − n + s} be the vertex set and edge set of H respectively. Let

G be the graph obtained from H by subdividing each edge ei in H and then adding

pendent edges to each vertex vi so that the degree of vi in G is ∆. It suffices to show

that G ∈ Gn,m,∆.

It is clear that G is a connected simple graph of maximum degree ∆. Note that

every vertex in G is of degree in {1, 2,∆} and xi,j = 0 for all i, j if (i, j) 6= (1,∆) or

(2,∆). So G is a bipartite graph with bipartition (A,B), where A = {vi : 1 ≤ i ≤ s}.As each vi has degree ∆ in G, the size of G is ∆s = m. The order of G is

n1 + n2 + n∆ = (m− 2(m− n+ s)) + (m− n+ s) + s = n.

Therefore G ∈ Gn,m,∆. �

Now we can establish an upper bound for ABC(G) in terms of n,m and ∆, where

G is a connected graph of order n, size m and maximum degree ∆. Especially when

m ≤ 2n − 2m∆

and m ≡ 0 (mod ∆), this upper bound can be obtained if and only if

G ∈ Gn,m,∆.

Theorem 3.1 Let G be a connected graph with n vertices, m edges and maximum

degree ∆ ≥ 3. Then

ABC(G) ≤√

∆− 1

∆

(2n−m− 2m

∆

)+√

2(m− n+

m

∆

)with equality if and only if G ∈ Gn,m,∆.

Proof. Since G is a connected graph with n vertices, m edges and maximum degree

∆, we have (1). Using the abbreviations

f1 = x1,2 + x1,3 + · · ·+ x1,∆−1

f2 = x1,2 + 2x2,2 + · · ·+ x2,∆−1

f3 = x1,3 + x2,3 + · · ·+ x3,∆

6

· · ·

f∆−1 = x1,∆−1 + x2,∆−1 + · · ·+ x∆−1,∆

f∆ = x3,∆ + x4,∆ + · · ·+ 2x∆,∆,

i.e.,

f1 = n1 − x1,∆

f2 = 2n2 − x2,∆

f3 = 3n3

· · ·

f∆−1 = (∆− 1)n∆−1

f∆ = ∆n∆ − x1,∆ − x2,∆,

we have

∆∑i=1

fi = 2m− 2(x1,∆ + x2,∆)

∆∑i=1

1

ifi = n−

(1 +

1

∆

)x1,∆ −

(1

2+

1

∆

)x2,∆,

implying that

x1,∆ = 2n−m− 2m

∆−

∆∑i=1

(2

i− 1

2− 1

∆

)fi

= 2n−m− 2m

∆−

∑1≤i≤j≤∆

(i,j)6=(1,∆),(2,∆)

(2

i+

2

j− 1− 2

∆

)xi,j

x2,∆ = 2(m− n) +2m

∆+

∆∑i=1

(2

i− 1− 1

∆

)fi

= 2(m− n) +2m

∆+

∑1≤i≤j≤∆

(i,j)6=(1,∆),(2,∆)

(2

i+

2

j− 2− 2

∆

)xi,j.

Substituting them back into the formula for ABC(G), we get

ABC(G) =

√∆− 1

∆x1,∆ +

1√2x2,∆ +

∑1≤i≤j≤∆

(i,j)6=(1,∆),(2,∆)

√i+ j − 2

ijxi,j

=

√∆− 1

∆

(2n−m− 2m

∆

)+√

2(m− n+

m

∆

)7

+∑

1≤i≤j≤∆(i,j)6=(1,∆),(2,∆)

h(i, j,∆)xi,j,

and by Lemma 3.1, we have h(i, j,∆) < 0 for all 1 ≤ i ≤ j ≤ ∆ with (i, j) 6=(1,∆), (2,∆), and thus

ABC(G) ≤√

∆− 1

∆

(2n−m− 2m

∆

)+√

2(m− n+

m

∆

),

where the equality holds if and only if xi,j = 0 for all i, j with 1 ≤ i ≤ j ≤ ∆ with

(i, j) 6= (1,∆), (2,∆), i.e., G ∈ Gn,m,∆ by definition. �

By Lemma 3.2, Gn,m,∆ = ∅ if m > 2n − 2m∆

or m 6≡ 0 (mod ∆). Thus the upper

bound in Theorem 3.1 for ABC(G) cannot be obtained when m > 2n− 2m∆

or m 6≡ 0

(mod ∆). So the problem of finding a sharp upper bound for ABC(G) when m >

2n− 2m∆

or m 6≡ 0 (mod ∆) remains open.

4 Large ABC indices of unicyclic graphs

In this section, we determine the n-vertex unicyclic graphs with the maximum, the

second, the third and the fourth maximum ABC indices for n ≥ 5.

Let Φ∆ be the class of connected graphs whose pendent vertices are adjacent to the

maximum degree vertices and all other edges have at least one end-vertex of degree

two. And let Ψ2 be the class of connected graphs whose vertices are of degree at least

two and all the edges have at least one end-vertex of degree two.

Lemma 4.1 [6] Let G be an n-vertex connected graph with m edges, p pendent vertices,

maximum vertex degree ∆ and minimum non-pendent vertex degree δ1. Then

ABC(G) ≤ p

√1− 1

∆+m− pδ1

√2(δ1 − 1)

with equality if and only if G is a (∆, 1)-semiregular graph or G is a regular graph or

G ∈ Φ∆ or G ∈ Ψ2.

Let Un be the set of n-vertex unicyclic graphs. Let Un,p be the set of unicyclic

graphs with n vertices and p pendent vertices, and Sn,p the unicyclic graph formed by

attaching p pendent vertices to a vertex of the cycle Cn−p, where 0 ≤ p ≤ n− 3.

8

Corollary 4.1 Let G ∈ Un,p, where 0 ≤ p ≤ n− 3. Then

ABC(G) ≤ p

√p+ 1

p+ 2+

√2

2(n− p)

with equality if and only if G = Sn,p.

Proof. By Lemma 4.1, we have

ABC(G) ≤ p

√1− 1

∆+n− pδ1

√2(δ1 − 1)

= p

√1− 1

∆+ (n− p)

√−2

(1

δ1

− 1

2

)2

+1

2

≤ p

√1− 1

∆+

√2

2(n− p)

≤ p

√1− 1

p+ 2+

√2

2(n− p)

with equality if and only if G is a (∆, 1)-semiregular graph or G is a regular graph

or G ∈ Φ∆ or G ∈ Ψ2, where δ1 = 2 is the minimum non-pendent vertex degree and

∆ = p+ 2 is the maximum vertex degree. Then we easily get the result. �

Let h(p) = p√

p+1p+2

+√

22

(n− p), where 0 ≤ p ≤ n− 3. Note that

h′(p) =

√p+ 1

p+ 2+

p

2(p+ 2)2

√p+ 2

p+ 1−√

2

2≥ p

2(p+ 2)2

√p+ 2

p+ 1> 0

for p ≥ 1, and h(1) =√

23

+√

22

(n− 1) > h(0) =√

22n. Then we have

ABC(Sn,n−3) > ABC(Sn,n−2) > · · · > ABC(Sn,1) > ABC(Sn,0). (4)

Theorem 4.1 Among all graphs in Un with n ≥ 3, Sn,n−3 is the unique graph with the

maximum ABC index, which is equal to (n− 3)√

n−2n−1

+ 3√

22

.

Proof. By Corollary 4.1, among all graphs in Un,p with 0 ≤ p ≤ n − 3, Sn,p is the

unique graph with the maximum ABC index h(p) = p√

p+1p+2

+√

22

(n − p). Then the

result follows easily from Ineq. (4). �

Label by v1, v2, . . . , vr the vertices of Cr consecutively. Let Sn(n1, n2, . . . , nr) be the

unicyclic graph formed by attaching ni − 1 pendent vertices to vi, where ni ≥ 1 for

i = 1, 2, . . . , r, n1 = max{n1, n2, . . . , nr}, and∑r

i=1 ni = n.

9

Lemma 4.2 Let G = Sn(n1, n2, n3) with ni ≥ nj ≥ 2, where 1 ≤ i 6= j ≤ 3. Let b be a

pendent neighbor of vj. Then ABC(G− vjb+ vib) > ABC(G).

Proof. Note that

ABC(Sn(n1, n2, n3)) =3∑

i=1

(ni − 1)f(1, ni + 1) +∑

1≤i<j≤3

f(ni + 1, nj + 1).

Suppose without loss of generality that i = 1 and j = 3. Then

ABC(Sn(n1 + 1, n2, n3 − 1))− ABC(Sn(n1, n2, n3))

= n1f(1, n1 + 2) + (n3 − 2)f(1, n3)

+f(n1 + 2, n2 + 1) + f(n1 + 2, n3) + f(n2 + 1, n3)

−[(n1 − 1)f(1, n1 + 1) + (n3 − 1)f(1, n3 + 1)

+f(n1 + 1, n2 + 1) + f(n1 + 1, n3 + 1) + f(n2 + 1, n3 + 1)]

= n1f(1, n1 + 2)− (n1 − 1)f(1, n1 + 1)

−[(n3 − 1)f(1, n3 + 1)− (n3 − 2)f(1, n3)]

+f(n1 + 2, n2 + 1)− f(n1 + 1, n2 + 1)

−[f(n2 + 1, n3 + 1)− f(n2 + 1, n3)]

+f(n1 + 2, n3)− f(n1 + 1, n3 + 1)

= l1(n1)− l1(n3 − 1) + gn1+2(n2 + 1)− gn3+1(n2 + 1)

+f(n1 + 2, n3)− f(n1 + 1, n3 + 1).

By Lemmas 2.2 and 2.3, since n1 > n3−1, we have l1(n1) > l1(n3−1), and n2 + 1 ≥ 2,

we have gn1+2(n2 + 1) > gn3+1(n2 + 1). Note that (n1 + 2)n3 < (n1 + 1)(n3 + 1), then

f(n1 + 2, n3)− f(n1 + 1, n3 + 1) =√

n1+n3

(n1+2)n3−√

n1+n3

(n1+1)(n3+1)> 0. Thus ABC(Sn(n1 +

1, n2, n3 − 1))− ABC(Sn(n1, n2, n3)) > 0. The result follows. �

Theorem 4.2 Among all graphs in Un with n ≥ 4, Sn(n − 3, 2, 1) for 5 ≤ n ≤ 15

and Sn,n−4 for n = 4 or n ≥ 16 are the unique graphs with the second maximum ABC

index, which is equal to (n − 4)√

n−3n−2

+√

n−13(n−2)

+√

2 +√

23

for 5 ≤ n ≤ 15, and

(n− 4)√

n−3n−2

+ 2√

2 for n = 4 or n ≥ 16.

Proof. By Theorem 4.1 and the monotonicity (4) of ABC(Sn,p) with 0 ≤ p ≤ n − 3,

the second maximum ABC index of graphs in Un with n ≥ 4 is achieved exactly by the

10

graphs in Un \ {Sn,n−3} with the maximum ABC index, and by Corollary 4.1, is only

achieved possibly by graphs in Un,n−3 \ {Sn,n−3} with the maximum ABC index and

Sn,n−4.

The case n = 4 is trivial. Suppose that n ≥ 5. Note that the unicyclic graphs

in Un,n−3 are of the form Sn(n1, n2, n3) with n1 + n2 + n3 = n. If n2 ≥ 2 or n3 ≥ 2,

then by Lemma 4.2, we can get another graph in Un,n−3 with larger ABC index. Then

Sn(n− 3, 2, 1) is the unique graph with the maximum ABC index among all graphs in

Un,n−3 \ {Sn,n−3}, which is equal to (n − 4)√

n−3n−2

+√

n−13(n−2)

+√

2 +√

23. Note that

ABC(Sn,n−4) = h(4) = (n− 4)√

n−3n−2

+ 2√

2. It is easily seen that

ABC(Sn,n−4)− ABC(Sn(n− 3, 2, 1))

= (n− 4)

√n− 3

n− 2+ 2√

2

−

[(n− 4)

√n− 3

n− 2+

√n− 1

3(n− 2)+√

2 +

√2

3

]

=√

2−√

2

3−

√n− 1

3(n− 2)

=√

2−√

2

3− f(3, n− 2).

By Lemma 2.1, f(3, n−2) is decreasing for n. By simple calculation, we haveABC(Sn,n−4) <

ABC(Sn(n − 3, 2, 1)) for 5 ≤ n ≤ 15 and ABC(Sn,n−4) > ABC(Sn(n − 3, 2, 1)) for

n ≥ 16. The result follows easily. �

Lemma 4.3 Let G = Sn(n1, n2, n3, n4). If n3 ≥ 2, then ABC(Sn(n1 + 1, n2, n3 −1, n4)) > ABC(G). If n1 > n2 ≥ n4 ≥ 2, then ABC(Sn(n1, n2 + 1, n3, n4 − 1)) >

ABC(G).

Proof. Note that

ABC(Sn(n1, n2, n3, n4))

=4∑

i=1

(ni − 1)f(1, ni + 1) + f(n1 + 1, n2 + 1) + f(n2 + 1, n3 + 1)

+f(n3 + 1, n4 + 1) + f(n1 + 1, n4 + 1).

If n3 ≥ 2, then by Lemmas 2.2 and 2.3, we have

ABC(Sn(n1 + 1, n2, n3 − 1, n4))− ABC(Sn(n1, n2, n3, n4))

11

= n1f(1, n1 + 2) + (n3 − 2)f(1, n3) + f(n1 + 2, n2 + 1)

+f(n2 + 1, n3) + f(n3, n4 + 1) + f(n1 + 2, n4 + 1)

−[(n1 − 1)f(1, n1 + 1) + (n3 − 1)f(1, n3 + 1) + f(n1 + 1, n2 + 1)

+f(n2 + 1, n3 + 1) + f(n3 + 1, n4 + 1) + f(n1 + 1, n4 + 1)]

= l1(n1)− l1(n3 − 1) + gn1+2(n2 + 1)− gn3+1(n2 + 1)

+gn1+2(n4 + 1)− gn3+1(n4 + 1) > 0,

for n1 > n3 − 1 and n2 + 1, n4 + 1 ≥ 2.

The proof of the second part is similar. �

Lemma 4.4 Let G = Sn(n1, n2, 1, 1) with n1 ≥ n2 ≥ 2. Then ABC(Sn(n1 + 1, n2 −1, 1, 1)) > ABC(G).

Proof. By the expression of ABC(Sn(n1, n2, n3, n4)) as in the proof of Lemma 4.3 and

Lemma 2.3, we have

ABC(Sn(n1 + 1, n2 − 1, 1, 1))− ABC(Sn(n1, n2, 1, 1))

= n1f(1, n1 + 2) + (n2 − 2)f(1, n2) + f(n1 + 2, n2) +3√2

−[(n1 − 1)f(1, n1 + 1) + (n2 − 1)f(1, n2 + 1) + f(n1 + 1, n2 + 1) +

3√2

]= l1(n1)− l1(n2 − 1) +

√n1 + n2

(n1 + 2)n2

−√

n1 + n2

(n1 + 1)(n2 + 1)> 0.

Thus the result follows. �

Denote by v1, v2, v3, v4 the vertices of S4,1 with dv1 = 1, dv2 = 3, dv3 = dv4 = 2. Let

Bn(n1, n2, n3, n4) be the unicyclic graph formed by attaching ni − 1 pendent vertices

to the vertex vi in S4,1, where n1 ≥ 2, n2, n3, n4 ≥ 1 and∑4

i=1 ni = n.

Lemma 4.5 Let G = Bn(n1, n2, n3, n4) for n ≥ 5 be a graph with maximum ABC

index for n1 ≥ 2, n2, n3, n4 ≥ 1 and∑4

i=1 ni = n. Then G = Bn(n − 3, 1, 1, 1) with

ABC index (n− 4)√

n−4n−3

+√

n−23(n−3)

+ 3√

22

.

Proof. The case n = 5 is trivial. Suppose that n ≥ 6. Suppose without loss of

generality that n3 ≥ n4. By direct calculation, we have

ABC(G) = (n1 − 1)f(1, n1) + (n2 − 1)f(1, n2 + 2) + (n3 − 1)f(1, n3 + 1)

12

+(n4 − 1)f(1, n4 + 1) + f(n1, n2 + 2) + f(n2 + 2, n3 + 1)

+f(n2 + 2, n4 + 1) + f(n3 + 1, n4 + 1).

Claim 1. n4 = 1.

Suppose that n4 ≥ 2. Then let G′ = Bn(n1, n2, n3 + 1, n4 − 1). Similarly to the

proof of Lemma 4.2, by noting that n3 > n4 − 1 and n2 + 2 ≥ 3, and Lemma 2.3, we

have

ABC(G′)− ABC(G)

= n3f(1, n3 + 2) + (n4 − 2)f(1, n4) + f(n2 + 2, n3 + 2) + f(n2 + 2, n4)

+f(n3 + 2, n4)− [(n3 − 1)f(1, n3 + 1) + (n4 − 1)f(1, n4 + 1)

+f(n2 + 2, n3 + 1) + f(n2 + 2, n4 + 1) + f(n3 + 1, n4 + 1)]

= l1(n3)− l1(n4 − 1) + gn3+2(n2 + 2)− gn4+1(n2 + 2)

+

√n3 + n4

(n3 + 2)n4

−√

n3 + n4

(n3 + 1)(n4 + 1)> 0,

which is a contradiction. This proves Claim 1.

Claim 2. n2 = 1.

Suppose that n2 ≥ 2. First suppose that n3 ≥ 2. Let G′ = Bn(n1, 1, n2 +n3− 2, 1).

Note that n1 ≥ 2. By Lemma 2.1, we have

ABC(G′)− ABC(G)

= (n2 + n3 − 2)f(1, n2 + n3) + f(3, n1) + f(3, n2 + n3)

−[(n2 − 1)f(1, n2 + 2) + (n3 − 1)f(1, n3 + 1) + f(n1, n2 + 2)

+f(n2 + 2, n3 + 1)]

= (n2 − 1)(f(1, n2 + n3)− f(1, n2 + 2))

+(n3 − 1)(f(1, n2 + n3)− f(1, n3 + 1))

+f(n1, 3)− f(n1, n2 + 2) +

√n2 + n3 + 1

3(n2 + n3)−

√n2 + n3 + 1

(n2 + 2)(n3 + 1)> 0,

which is a contradiction. Thus, if n3 ≥ 2, then n2 = 1.

Now suppose that n3 = 1, i.e., G = Bn(n1, n2, 1, 1). If n1 = 2, then G = Bn(2, n−4, 1, 1). If n1 ≥ 3, then let G′′ = Bn(n1 + n2 − 1, 1, 1, 1). By Lemma 2.1, we have

ABC(G′′)− ABC(G)

13

= (n1 + n2 − 2)f(1, n1 + n2 − 1) + f(3, n1 + n2 − 1)

−[(n1 − 1)f(1, n1) + (n2 − 1)f(1, n2 + 2) + f(n1, n2 + 2)

= (n1 − 1)(f(1, n1 + n2 − 1)− f(1, n1))

+(n2 − 1)(f(1, n1 + n2 − 1)− f(1, n2 + 2))

+

√n1 + n2

3(n1 + n2 − 1)−√

n1 + n2

n1(n2 + 2)> 0,

which is also a contradiction. Thus, if n1 ≥ 3, then n2 = 1, i.e., G = Bn(n− 3, 1, 1, 1).

By direct calculation, we have

ABC(Bn(2, n− 4, 1, 1))− ABC(Bn(n− 3, 1, 1, 1))

= (n− 5)

√n− 3

n− 2+

5√2−

[(n− 4)

√n− 4

n− 3+

√n− 2

3(n− 3)+

3√2

]

= (n− 5)

√n− 3

n− 2− (n− 4)

√n− 4

n− 3−

√n− 2

3(n− 3)+√

2.

Let q(n) be the right–most expression in the above equation. Note that

2x2 − 9x+ 7

2(x− 2)32

− 2x2 − 15x+ 29

2(x− 4)32

=−8(x− 6)5 − 115(x− 6)4 − 648(x− 6)3 − 1803(x− 6)2 − 2490(x− 6)− 1372

(x− 2)32 (x− 4)

32 [(2x2 − 9x+ 7)(x− 4)

32 + (2x2 − 15x+ 29)(x− 2)

32 ]

.

We have

q′(x) =

√x− 3

x− 2+ (x− 5) · 1

2

√x− 2

x− 3· 1

(x− 2)2−√x− 4

x− 3

−(x− 4) · 1

2

√x− 3

x− 4· 1

(x− 4)2− 1

2√

3

√x− 3

x− 2· 1

(x− 3)2

=1

(x− 3)12

·

[2x2 − 9x+ 7

2(x− 2)32

− 2x2 − 15x+ 29

2(x− 4)32

]− 1

2√

3(x− 3)32 (x− 2)

12

< 0

for x ≥ 6, implying that q(n) is decreasing for n ≥ 6. Then q(n) ≤ q(6) =√

32− 2√

23−

23+√

2 < 0. ThusG = Bn(n−3, 1, 1, 1) with the ABC index (n−4)√

n−4n−3

+√

n−23(n−3)

+ 3√2.

Claim 3. n3 = 1.

Suppose that n3 ≥ 2. Let G′ = Bn(n1 + n3 − 1, 1, 1, 1). By Lemma 2.4, f(3, x)

is convex decreasing for x, which implies that f(3, a) − f(3, b) > f(3, c) − f(3, d) for

14

b − a = d − c > 0. Note that n1 − 2 = (n1 + n3 − 1) − (n3 + 1) and (n3 + 1) − 2 =

(n1 + n3 − 1)− n1. And by Lemma 2.1, we have

ABC(G′)− ABC(G)

= (n1 + n3 − 2)f(1, n1 + n3 − 1) + f(3, n1 + n3 − 1) +1√2

−[(n1 − 1)f(1, n1) + (n3 − 1)f(1, n3 + 1) + f(3, n1) + f(3, n3 + 1)]

= (n1 − 1)(f(1, n1 + n3 − 1)− f(1, n1))

+(n3 − 1)(f(1, n1 + n3 − 1)− f(1, n3 + 1))

+f(3, n1 + n3 − 1) + f(3, 2)− f(3, n1)− f(3, n3 + 1)

> f(3, n1 + n3 − 1) + f(3, 2)− f(3, n1)− f(3, n3 + 1) > 0,

a contradiction again.

Now the result follows from Claims 1–3. �

Theorem 4.3 Among all graphs in Un with n ≥ 5,

(i) Sn,n−4 for 5 ≤ n ≤ 15, and Sn(n− 3, 2, 1) for n ≥ 16 are the unique graphs with

the third maximum ABC index, which is equal to (n− 4)√

n−3n−2

+ 2√

2 for 5 ≤ n ≤ 15,

and (n− 4)√

n−3n−2

+√

n−13(n−2)

+√

2 +√

23

for n ≥ 16;

(ii) B5(2, 1, 1, 1) and C5 for n = 5, S6(2, 1, 2, 1) for n = 6, Sn(n − 4, 3, 1) for

7 ≤ n ≤ 16, and Bn(n − 3, 1, 1, 1) for n ≥ 17 are the unique graphs with the fourth

maximum ABC index, which is equal to 5√2

for n = 5, 2(√

23

+√

2)

for n = 6,

(n− 5)√

n−4n−3

+ 12

√n−1n−3

+√

2 +√

3 for 7 ≤ n ≤ 16, and (n− 4)√

n−4n−3

+√

n−23(n−3)

+ 3√2

for n ≥ 17.

Proof. Let G ∈ Un with n ≥ 5.

If G ∈ Un,p with 0 ≤ p ≤ n− 5, then by Theorem 4.1 and the monotonicity (4) of

ABC(Sn,p) with 0 ≤ p ≤ n − 5, we have ABC(G) ≤ h(n − 5) = (n − 5)√

n−4n−3

+ 5√2

with equality if and only if G = Sn,n−5.

A graph in Un,n−3 is of the form Sn(n1, n2, n3). Among all graphs in Un,n−3 with

n ≥ 6, by Lemma 4.2, Sn(n − 3, 2, 1) is the unique graphs with the second maximum

15

ABC index, and S6(2, 2, 2) for n = 2 and Sn(n−4, 3, 1) for n ≥ 7 are the unique graphs

with the second maximum ABC index.

A graph in Un,n−4 is of the form Sn(n1, n2, n3, n4) or Bn(n1, n2, n3, n4). Among

all graphs of the form Sn(n1, n2, n3, n4) different from Sn,n−4, if n2 = n4 = 1, then

by Lemma 4.3, Sn(n − 4, 1, 2, 1) is the unique graph with the maximum ABC index,

which is equal to (n − 5)√

n−4n−3

+√

23

+ 2√

2, and if n2 ≥ 2, then by Lemmas 4.3

and 4.4, Sn(n − 4, 2, 1, 1) is the unique graph with the maximum ABC index, which

is equal to (n− 5)√

n−4n−3

+√

n−23(n−3)

+√

23

+ 3√2. Obviously, ABC(Sn(n− 4, 1, 2, 1))−

ABC(Sn(n−4, 2, 1, 1)) =√

22−√

n−23(n−3)

= f(2, n−3)−f(3, n−3) > 0 for n ≥ 6. Thus

Sn(n− 4, 1, 2, 1) is the unique graph with the maximum ABC index among all graphs

of the form Sn(n1, n2, n3, n4) different from Sn,n−4 with n ≥ 6. Among all graphs of the

form Bn(n1, n2, n3, n4) with n ≥ 5, by Lemma 4.5, Bn(n−3, 1, 1, 1) is the unique graph

with the maximum ABC index, which is equal to (n− 4)√

n−4n−3

+√

n−23(n−3)

+ 3√2. Note

that ABC(Bn(n− 3, 1, 1, 1))−ABC(Sn(n− 4, 1, 2, 1)) =√

n−4n−3

+√

n−23(n−3)

− 1√2−√

23.

Let r(n) be the right expression in the equation. Note that

r′(x) =1

2

√x− 3

x− 4· 1

(x− 3)2+

1

2√

3

√x− 3

x− 2·(− 1

(x− 3)2

)=

x− 1√

3(x− 3)32 (x− 2)

12 (x− 4)

12 [√

3(x− 2)12 + (x− 4)

12 ]> 0.

Then r(n) < 0 for n = 6, 7, and r(n) > 0 for n ≥ 8. Thus, among all graphs in Un,n−4

with n ≥ 5, Bn(n − 3, 1, 1, 1) for n = 5 and n ≥ 8, and Sn(n − 4, 1, 2, 1) for n = 6, 7

are the unique graphs with the second maximum ABC index.

By Theorems 4.1 and 4.2, the third maximum ABC index of graphs in Un with

n ≥ 5 is achieved exactly by the graphs in Un \ {Sn,n−3, Sn(n− 3, 2, 1)} for 5 ≤ n ≤ 15,

and Un \{Sn,n−3, Sn,n−4} for n ≥ 16. Let G0 be a graph in Un with the third maximum

ABC index. For 5 ≤ n ≤ 15, we are to determine the maximum ABC index of graphs

in Un\{Sn,n−3, Sn(n−3, 2, 1)}. If G ∈ Un,n−3, then G0 = S6(2, 2, 2) for n = 6 with ABC

index 3√

23+2, andG0 = Sn(n−4, 3, 1) with ABC index (n−5)

√n−4n−3

+ 12

√n−1n−3

+√

2+√

3

for n ≥ 7, and if G ∈ Un,p with p ≤ n − 4, then G0 = Sn,n−4 with ABC index

(n−4)√

n−3n−2

+2√

2 for n ≥ 5. The case n = 6 is easy to check. Similarly, ABC(Sn,n−4)−

ABC(Sn(n−4, 3, 1)) = (n−4)√

n−3n−2−(n−5)

√n−4n−3− 1

2

√n−1n−3

+√

2−√

3 > 0 for n ≥ 7.

Thus G0 = Sn,n−4 for 5 ≤ n ≤ 15. For n ≥ 16, we are to determine the maximum ABC

index of graphs in Un \ {Sn,n−3, Sn,n−4}. If G ∈ Un,n−3, then G0 = Sn(n− 3, 2, 1) with

16

ABC index (n−4)√

n−3n−2

+√

n−13(n−2)

+√

2+√

23, if G ∈ Un,n−4, then G0 = Bn(n−3, 1, 1, 1)

with ABC index (n − 4)√

n−4n−3

+√

n−23(n−3)

+ 3√2, and if G ∈ Un,p with p ≤ n − 5, then

G0 = Sn,n−5 with ABC index (n − 5)√

n−4n−3

+ 5√2. By similar argument as above, we

have G0 = Sn(n− 3, 2, 1) for n ≥ 16. Thus we get (i).

By Theorems 4.1, 4.2 and (i), the fourth maximum ABC index of graphs in Un

with n ≥ 5 is achieved exactly by the graphs in Un \ {Sn,n−3, Sn(n − 3, 2, 1), Sn,n−4},and then is achieved exactly by S6(2, 2, 2) for n = 6, and Sn(n − 4, 3, 1) for n ≥ 7,

Bn(n − 3, 1, 1, 1) for n = 5 and n ≥ 8, Sn(n − 4, 1, 2, 1) for n = 6, 7, and Sn,n−5 for

n ≥ 5 with the maximum ABC index respectively. Note that ABC(Sn(n− 4, 3, 1)) =

(n−5)√

n−4n−3

+ 12

√n−1n−3

+√

2+√

3, ABC(Bn(n−3, 1, 1, 1)) = (n−4)√

n−4n−3

+√

n−23(n−3)

+ 3√2,

ABC(Sn(n−4, 1, 2, 1)) = (n−5)√

n−4n−3

+√

23+2√

2, ABC(Sn,n−5) = (n−5)√

n−4n−3

+ 5√2.

The cases n = 5, 6, 7 may be checked directly by calculation. Suppose that n ≥ 8. Note

that

ABC(Sn(n− 4, 3, 1))− ABC(Sn,n−5) =1

2

√n− 1

n− 3+√

3− 3√2>

1

2+√

3− 3√2> 0.

By Lemma 2.4, f(3, x) is convex decreasing for x ≥ 2, we have

ABC(Bn(n− 3, 1, 1, 1))− ABC(Sn,n−5)

=

√n− 4

n− 3+

√n− 2

3(n− 3)−√

2

= f(1, n− 3) + f(3, n− 3)− 2f(2, n− 3) > 0.

It is easily seen that

ABC(Bn(n− 3, 1, 1, 1))− ABC(Sn(n− 4, 3, 1))

=

√n− 4

n− 3+

√n− 2

3(n− 3)− 1

2

√n− 1

n− 3+

1√2−√

3.

Let s(n) be the right equation above. Similarly as above we have s′(x) > 0. And then

s(n) < 0 for 8 ≤ n ≤ 16, and s(n) > 0 for n ≥ 17. Thus we get (ii). �

5 Large ABC indices of bicyclic graphs

In this section, we determine the n-vertex bicyclic graphs with the maximum and the

second maximum ABC indices for n ≥ 5.

17

Let Bn,p be the set of bicyclic graphs with n vertices and p pendent vertices for

0 ≤ p ≤ n− 4.

Let Sr,tn be the n-vertex bicyclic graphs by identifying one vertex of two cycles Cr

and Ct and attaching n + 1 − r − t pendent vertices to the common vertex, where

t ≥ s ≥ 3 and r + s ≤ n+ 1. For 0 ≤ p ≤ n− 5, let Sn,p be the set of graphs Sr,tn with

3 ≤ r ≤ t ≤ n−2−p and r+t = n+1−p. It is easily seen that the graphs in Sn,p have the

same ABC index, which is equal to p·√

1+(p+4)−21·(p+4)

+ 1√2(n+1−p) = n+1√

2+p(√

p+3p+4− 1√

2

).

Similarly to the proof of Corollary 4.1 by using Lemma 4.1, we have

Corollary 5.1 Let G ∈ Bn,p with 0 ≤ p ≤ n− 5. Then

ABC(G) ≤ n+ 1√2

+ p

(√p+ 3

p+ 4− 1√

2

)with equality if and only if G ∈ Sn,p.

Let Q4 be the bicyclic graph obtained by adding an edge to the cycle C4. Label the

vertices of Q4 by v1, v2, v3, v4 with dv1 = dv2 = 3, dv3 = dv4 = 2. Let Qn(n1, n2, n3, n4)

be the graph formed from Q4 by attaching ni − 1 pendent vertices to vi, where ni ≥ 1

for i = 1, 2, 3, 4, n1 ≥ n2, n3 ≥ n4 and∑4

i=1 ni = n.

Lemma 5.1 Let G = Qn(n1, n2, n3, n4). If n2 ≥ 2, then ABC(Qn(n1 + 1, n2 −1, n3, n4)) > ABC(G). If n4 ≥ 2, then ABC(Qn(n1, n2, n3 + 1, n4 − 1)) > ABC(G).

Proof. Note that

ABC(Qn(n1, n2, n3, n4))

=2∑

i=1

(ni − 1)f(1, ni + 2) +4∑

i=3

(ni − 1)f(1, ni + 1)

+f(n1 + 2, n2 + 2) + f(n1 + 2, n3 + 1) + f(n1 + 2, n4 + 1)

+f(n2 + 2, n3 + 1) + f(n2 + 2, n4 + 1).

Since (n1 +3)(n2 +1) < (n1 +2)(n2 +2), we have f(n1 +3, n2 +1)−f(n1 +2, n2 +2) =√n1+n2+2

(n1+3)(n2+1)−√

n1+n2+2(n1+2)(n2+2)

> 0. Note that n1 ≥ n2. And since n3 + 1 ≥ n4 + 1 ≥ 2,

18

by Lemma 2.2, ga(n3 + 1) and ga(n4 + 1) are both increasing for a. Together with

Lemma 2.5, if n2 ≥ 2, then

ABC(Qn(n1 + 1, n2 − 1, n3, n4))− ABC(Qn(n1, n2, n3, n4))

= n1f(1, n1 + 3) + (n2 − 2)f(1, n2 + 1) + f(n1 + 3, n2 + 1)

+f(n1 + 3, n3 + 1) + f(n1 + 3, n4 + 1) + f(n2 + 1, n3 + 1)

+f(n2 + 1, n4 + 1)− [(n1 − 1)f(1, n1 + 2) + (n2 − 1)f(1, n2 + 2)

+f(n1 + 2, n2 + 2) + f(n1 + 2, n3 + 1) + f(n1 + 2, n4 + 1)

+f(n2 + 2, n3 + 1) + f(n2 + 2, n4 + 1)]

= [n1f(1, n1 + 3)− (n1 − 1)f(1, n1 + 2)]

−[(n2 − 1)f(1, n2 + 2)− (n2 − 2)f(1, n2 + 1)]

+f(n1 + 3, n3 + 1)− f(n1 + 2, n3 + 1)

−[f(n2 + 2, n3 + 1)− f(n2 + 1, n3 + 1)]

+f(n1 + 3, n4 + 1)− f(n1 + 2, n4 + 1)

−[f(n2 + 2, n4 + 1)− f(n2 + 1, n4 + 1)]

+

√n1 + n2 + 2

(n1 + 3)(n2 + 1)−

√n1 + n2 + 2

(n1 + 2)(n2 + 2)

> l2(n1)− l2(n2 − 1) + gn1+3(n3 + 1)− gn2+2(n3 + 1)

+gn1+3(n4 + 1)− gn2+2(n4 + 1) > 0.

Similarly note that n3 ≥ n4. Since n1 + 2 ≥ n2 + 2 ≥ 3, we have by Lemma 2.2 that

ga(n1 +2) and ga(n2 +2) are both increasing for a. Together with Lemma 2.3, if n4 ≥ 2,

then

ABC(Qn(n1, n2, n3 + 1, n4 − 1))− ABC(Qn(n1, n2, n3, n4))

= n3f(1, n3 + 2) + (n4 − 2)f(1, n4) + f(n1 + 2, n3 + 2)

+f(n1 + 2, n4) + f(n2 + 2, n3 + 2) + f(n2 + 2, n4)

−[(n3 − 1)f(1, n3 + 1) + (n4 − 1)f(1, n4 + 1) + f(n1 + 2, n3 + 1)

+f(n1 + 2, n4 + 1) + f(n2 + 2, n3 + 1) + f(n2 + 2, n4 + 1)]

= l1(n3)− l1(n4 − 1) + gn3+2(n1 + 2)− gn4+1(n1 + 2)

+gn3+2(n2 + 2)− gn4+1(n2 + 2) > 0.

Then the result follows. �

19

Lemma 5.2 Let G = Qn(n1, 1, n3, 1) with n1, n3 ≥ 2. Then ABC(G) ≤ min{ABC(Qn(n−3, 1, 1, 1)), ABC(Qn(1, 1, n− 3, 1))}.

Proof. By the expression of ABC(Qn(n1, n2, n3, n4)) as in the proof of Lemma 5.1,

we have

ABC(Qn(n1 + n3 − 1, 1, 1, 1))− ABC(Qn(n1, 1, n3, 1))

= (n1 + n3 − 2)f(1, n1 + n3 + 1) +√

2 + f(3, n1 + n3 + 1)

−[(n1 − 1)f(1, n1 + 2) + (n3 − 1)f(1, n3 + 1)

+f(3, n1 + 2) + f(3, n3 + 1) + f(n1 + 2, n3 + 1)]

= (n1 − 1)(f(1, n1 + n3 + 1)− f(1, n1 + 2))

+(n3 − 1)(f(1, n1 + n3 + 1)− f(1, n3 + 1))

+f(3, 2) + f(3, n1 + n3 + 1)− f(3, n1 + 2)− f(3, n3 + 1)

+1√2− f(n1 + 2, n3 + 1).

Since n3 + 1 ≥ 3 and n1 ≥ 2, and by Lemma 2.1, f(1, x) is increasing for x, we have

f(1, n1 +n3 + 1)− f(1, n1 + 2) > 0, f(1, n1 +n3 + 1)− f(1, n3 + 1) > 0. By Lemma 2.4

and the fact that (n1 + n3 + 1)− (n1 + 2) = (n3 + 1)− 2, we have f(3, 2) + f(3, n1 +

n3 +1)−f(3, n1 +2)−f(3, n3 +1) > 0, which together with Lemma 2.1, implying that

1√2− f(n1 + 2, n3 + 1) > 0. Thus ABC(Qn(n− 3, 1, 1, 1)) > ABC(Qn(n1, 1, n3, 1)).

Since 3(n1 + n3) ≤ (n1 + 2)(n3 + 1), we have f(3, n1 + n3) − f(n1 + 2, n3 + 1) =√n1+n3+13(n1+n3)

−√

n1+n3+1(n1+2)(n3+1)

≥ 0. Similarly as above, we have

ABC(Qn(1, 1, n1 + n3 − 1, 1))− ABC(Qn(n1, 1, n3, 1))

= (n1 + n3 − 2)f(1, n1 + n3) +2

3+ 2f(3, n1 + n3)

−[(n1 − 1)f(1, n1 + 2) + (n3 − 1)f(1, n3 + 1)

+f(3, n1 + 2) + f(3, n3 + 1) + f(n1 + 2, n3 + 1)]

= (n1 − 1)(f(1, n1 + n3)− f(1, n1 + 2))

+(n3 − 1)(f(1, n1 + n3)− f(1, n3 + 1))

+f(3, 3) + f(3, n1 + n3)− f(3, n1 + 2)− f(3, n3 + 1)

+f(3, n1 + n3)− f(n1 + 2, n3 + 1) > 0.

Thus the result follows. �

20

Theorem 5.1 Let G ∈ Bn,n−4 with n ≥ 5. Then

ABC(G) ≤ (n− 4)

√n− 2

n− 1+

√n

3(n− 1)+ 2√

2

with equality if and only if G = Qn(n− 3, 1, 1, 1).

Proof. Let G ∈ Bn,n−4 with n ≥ 5. Then G is of the form Qn(n1, n2, n3, n4) with

n1 ≥ n2, n3 ≥ n4 and∑4

i=1 ni = n.

Suppose that G0 = Qn(n1, n2, n3, n4) is a graph in Bn,n−4 with the maximum ABC

index. If n2 ≥ 2 or n4 ≥ 2, then from Lemma 5.1, we can get another graph in Bn,n−4

with larger ABC index. Thus G0 = Qn(n1, 1, n3, 1). Now by Lemma 5.2, we have

G0 = Qn(n1 +n3− 1, 1, 1, 1) or G0 = Qn(1, 1, n1 +n3− 1, 1). By direct calculation and

Lemmas 2.1, 2.4, we have

ABC(Qn(n1 + n3 − 1, 1, 1, 1))− ABC(Qn(1, 1, n1 + n3 − 1, 1))

= ABC(Qn(n− 3, 1, 1, 1))− ABC(Qn(1, 1, n− 3, 1))

= (n− 4)

√n− 2

n− 1+

√n

3(n− 1)+ 2√

2

−

[(n− 4)

√n− 3

n− 2+ 2

√n− 1

3(n− 2)+√

2 +2

3

]= (n− 4)(f(1, n− 1)− f(1, n− 2))

+f(3, n− 1) + f(3, 2)− f(3, n− 2)− f(3, 3) +1√2− f(3, n− 2) > 0.

Thus G0 = Qn(n− 3, 1, 1, 1). �

Corollary 5.2 Among the graphs in Bn,n−4 with n ≥ 5. Qn(1, 1, n−3, 1) is the unique

graph with the second maximum ABC indices, which are equal to (n − 4)√

n−3n−2

+

2√

n−13(n−2)

+√

2 + 23.

Proof. Suppose that G1 = Qn(n1, n2, n3, n4) is a graph in Bn,n−4 with the second

maximum ABC index, which, by Theorem 5.1, is achieved by the graphs in Bn,n−4 \{Qn(n − 3, 1, 1, 1)} with the maximum ABC index. By Lemma 5.1, we have G1 =

(n1, 1, n3, 1). Since G1 6= Qn(n − 3, 1, 1, 1), n3 ≥ 2, then by Lemma 5.2, we have

G1 = Qn(1, 1, n− 3, 1). �

21

Let Bn be the set of n-vertex unicyclic graphs.

Let k(p) = n+1√2

+ p(√

p+3p+4− 1√

2

), where 0 ≤ p ≤ n− 5. Note that

k′(p) =

√p+ 3

p+ 4+

p

2(p+ 4)2

√p+ 4

p+ 3−√

2

2>

p

2(p+ 4)2

√p+ 4

p+ 3≥ 0.

Then we have

ABC(Gn,n−5) > ABC(Gn,n−6) > · · · > ABC(Gn,1) > ABC(Gn,0), (5)

where Gn,i denotes any graph in the set Sn,i.

Theorem 5.2 Among all graphs in Bn with n ≥ 4, Qn(n − 3, 1, 1, 1) is the unique

graph with the maximum ABC index, which is equal to (n− 4)√

n−2n−1

+√

n3(n−1)

+ 2√

2.

Proof. The case n = 4 is trivial. Suppose in the following that n ≥ 5.

By Corollary 5.1, among all graphs in Bn,p with 0 ≤ p ≤ n − 5, the graphs in Sn,p

are the unique graphs with the maximum ABC index k(p) = n+1√2

+ p(√

p+3p+4− 1√

2

).

Note that Sn,n−5 = {S3,3n }. By Ineq. (5), S3,3

n is the unique graph with the maximum

ABC index in⋃n−5

p=0 Bn,p, which is equal to k(n− 5) = n+1√2

+ (n− 5)(√

n−2n−1− 1√

2

)=

(n − 5)√

n−2n−1

+ 3√

2. By Theorem 5.1, among all graphs in Bn,n−4, Qn(n − 3, 1, 1, 1)

is the unique graph with the maximum ABC index, which is equal to (n− 4)√

n−2n−1

+√n

3(n−1)+ 2√

2. Note that

ABC(Qn(n− 3, 1, 1, 1))− ABC(S3,3n )

= (n− 4)

√n− 2

n− 1+

√n

3(n− 1)+ 2√

2−

[(n− 5)

√n− 2

n− 1+ 3√

2

]

=

√n− 2

n− 1+

√n

3(n− 1)−√

2.

Let t(n) be the right-most equation of above expression. Since

t′(x) =1

2

√x− 1

x− 2· 1

(x− 1)2+

1

2√

3

√x− 1

x·(− 1

(x− 1)2

)=

x+ 1√

3x(x− 1)32 (x− 2)

12 (√

3x+√x− 2)

> 0

for x ≥ 3. Then t(n) ≥ t(5) =√

34

+√

512−√

2 > 0. The result follows. �

22

Theorem 5.3 Among all graphs in Bn with n ≥ 5, S3,3n is the unique graph with the

second maximum ABC index, which is equal to (n− 5)√

n−2n−1

+ 3√

2.

Proof. By Theorems 5.1, 5.2 and Corollaries 5.1, 5.2, the second maximum ABC

index of graphs in Bn is achieved exactly by the graphs in Bn \{Qn(n− 3, 1, 1, 1)}, and

together with the monotonicity (5), is thus only achieved by graphs in Sn,n−5, which

only contains S3,3n , and Qn(1, 1, n− 3, 1). For n ≥ 5, by Lemma 2.1, we have

ABC(Qn(1, 1, n− 3, 1))− ABC(S3,3n )

= (n− 4)

√n− 3

n− 2+ 2

√n− 1

3(n− 2)+√

2 +2

3−

[(n− 5)

√n− 2

n− 1+ 3√

2

]

= (n− 5) (f(1, n− 2)− f(1, n− 1)) +

√n− 3

n− 2+ 2

√n− 1

3(n− 2)+

2

3− 2√

2

≤√n− 3

n− 2+ 2

√n− 1

3(n− 2)+

2

3− 2√

2.

Let d(n) be the right-most equation of the above expression. Since

d′(x) =1

2

√x− 2

x− 3· 1

(x− 2)2+

2√3· 1

2

√x− 2

x− 1·(− 1

(x− 2)2

)=

−(x− 9)

2√

3(x− 2)32 (x− 1)

12 (x− 3)

12 (2√x− 3 +

√3(x− 1))

,

d(x) is increasing for 5 ≤ x ≤ 9, and decreasing for x ≥ 9. Then d(n) ≤ d(9) =√67

+ 2√

821

+ 23− 2√

2 < 0 for n ≥ 5. Thus the result follows. �

Acknowledgement. This work was supported by the Guangdong Provincial Natural

Science Foundation of China (no. 8151063101000026).

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