# Bicyclic graphs with maximum sum of the two largest Laplacian eigenvalues

## Abstract

Let G be a simple connected graph and $$S_{2}(G)$$ be the sum of the two largest Laplacian eigenvalues of G. In this paper, we determine the bicyclic graph with maximum $$S_{2}(G)$$ among all bicyclic graphs of order n, which confirms the conjecture of Guan et al. (J.Â Inequal. Appl. 2014:242, 2014) for the case of bicyclic graphs.

## 1 Introduction

Let $$G=(V(G), E(G))$$ be a simple connected graph with vertex set $$V(G)=\{v_{1}, v_{2}, \ldots, v_{n}\}$$. The numbers of its vertices and edges are denoted by $$n(G)$$ and $$m(G)$$ (or n and m for short). For a vertex $$v \in V(G)$$, let $$N(v)$$ be the set of all neighbors of v in G. The degree of v, denoted by $$d(v)$$, is the cardinality of $$\vert N(v) \vert$$, that is, $$d(v)=\vert N(v) \vert$$. A vertex with degree one is called pendant vertex. Particularly, denote by $$\Delta(G)$$ (or Î” for short) the maximum degree of G. The matrix $$L(G)=D(G)-A(G)$$ is called Laplacian matrix of G, where $$A(G)$$ is the adjacency matrix of G and $$D(G)=\operatorname{diag}(d(v_{1}), d(v_{2}), \ldots, d(v_{n}))$$ is the diagonal matrix of vertex degrees of G. We use the notation $$I_{n}$$ for the identity matrix of order n and denote by $$\phi(G, x) = \det(xI_{n}-L(G))$$ the Laplacian characteristic polynomial of G. It is well known that $$L(G)$$ is positive semidefinite and so its eigenvalue are nonnegative real number. The eigenvalues of $$L(G)$$ are called the Laplacian eigenvalues of G and are denoted by $$\mu_{1}(G) \geq\mu_{2}(G)\geq\cdots\geq\mu_{n}(G)$$ (or $$\mu _{1} \geq\mu_{2} \geq\cdots\geq\mu_{n}$$ for short), which are always enumerated in non-increasing order and repeated according to their multiplicity. It is well known that $$\sum_{i=1}^{i=n}\mu _{i}=2m(G)$$. Note that each row sum of $$L(G)$$ is 0 and, therefore, $$\mu_{n}(G)=0$$. Fiedler [2] showed that the second smallest eigenvalue $$\mu_{n-1}(G)$$ of $$L(G)$$ is 0 if and only if G is disconnected. Thus the second smallest eigenvalue of $$L(G)$$ is popularly known as the algebraic connectivity of G. The largest eigenvalue $$\mu_{1}(G)$$ of $$L(G)$$ is usually called the Laplacian spectral radius of the graph G. The investigation of Laplacian spectra of graphs is an interesting topic on which much literature focused in the last two decades (see [2â€“9]).

Let $$S_{k}(G)=\sum_{i=1}^{i=k}\mu_{i}(G)$$ be the sum of the k largest Laplacian eigenvalues of G. Hamers in [10] mentions that Brouwer conjectured that $$S_{k}(G) \leq m(G)+ \binom {k+1}{2}$$ for $$k=1,2,\ldots, n$$. This conjecture is interesting and still open. Up to now, little progress on it has been made (see [1, 10â€“13]). When $$k = 2$$, Haemers et al. [10] proved the conjecture by showing $$S_{2}(G) \leq m(G)+3$$ for any graph G. Especially when G is a tree, Fritscher et al. [14] improve this bound by showing $$S_{2}(T) \leq m(T)+3-\frac{2}{n(T)}$$, which indicates that Haemersâ€™ bound is always not attainable for trees. Therefore, it is interesting to determine which tree has maximum value of $$S_{2}(T)$$ among all trees of order n. Let $$S^{k}_{a,b}$$ be the tree of order n obtained from two stars $$S_{a+1}$$, $$S_{b+1}$$ by joining a path of length k between their central vertices. Guan et al. [1] proved that $$S_{2}(T)\leq S_{2}(T^{1}_{\lceil\frac{n-2}{2}\rceil, \lfloor\frac{n-2}{2} \rfloor})$$ for any tree of order $$n \geq4$$ and the equality holds if and only if $$T \cong T^{1}_{\lceil\frac {n-2}{2}\rceil, \lfloor\frac{n-2}{2} \rfloor}$$. Note that $$\mu_{1}(G)\leq n(G)$$ for any graph G of order n. When $$2n(G)< m(G)+3$$, it follows that $$S_{2}(G) \leq2n(G) < m(G)+3$$, which means that Haemersâ€™ bound is not attainable. When $$2n(G) \geq m(G)+3$$, Guan et al. [1] showed that $$S_{2}(G_{m,n})=m(G_{m,n})+3$$, where $$G_{m,n}$$ is a graph with n vertices and m edges which has $$m-n+1$$ triangles with a common edge and $$2n-m-3$$ pendant edges incident with one end vertex of the common edge (illustrated in FigureÂ 1 are $$G_{n,n}$$ and $$G_{n+1,n}$$). This indicates that Haemersâ€™ bound is always sharp for connected graphs $$(n \leq m \leq2n-3)$$. The following conjecture on the uniqueness of the extremal graph is also presented inÂ [1].

### Conjecture 1.1

[1]

Among all connected graphs with n vertices and m edges $$(n\leq m\leq2n -3)$$, $$G_{m,n}$$ is the unique graph with maximal value of $$S_{2}(G)$$, that is, $$S_{2}(G_{m,n})= m(G_{m,n})+3$$.

Zheng et al. [15] determined the uicyclic graph with maximum $$S_{2}(G)$$ among all uicyclic graphs of order n, which confirms ConjectureÂ 1.1 for $$m=n$$.

### Theorem 1.2

[15]

For any unicyclic graph G of order n, $$S_{2}(G) \leq m(G)+3$$ with equality if and only if $$G \cong G_{n, n}$$.

In this paper, we prove that ConjectureÂ 1.1 holds for $$m=n+1$$. The main result is as follows.

### Theorem 1.3

For any bicyclic graph G of order n, $$S_{2}(G) \leq m(G)+3$$ with equality if and only if $$G \cong G_{n+1, n}$$.

In SectionÂ 2 of this paper, we give some well known lemmas which are useful in the proof of TheoremÂ 1.3. In SectionÂ 3, we will give the proof of TheoremÂ 1.3.

## 2 Preliminaries

We first present some well-known results on $$\mu_{1}(G)$$.

### Lemma 2.1

[3]

Let G be a connected graph of order n, then $$\mu_{1}(G) \leq n(G)$$ with equality if and only if the complement of G is disconnected.

### Lemma 2.2

[8]

Let G be a connected graph of order n, $$d_{i}=d(v_{i})$$ and $$m_{i}=\sum_{v_{j} \in N(v_{i})}d_{j}/d_{i}$$. Then

$$\mu_{1}(G) \leq \max\bigl\{ d_{i}+m_{i}\mid v_{i} \in V(G)\bigr\} .$$

### Lemma 2.3

[7]

Let G be a connected graph with $$n \geq4$$ vertices and m edges. Then

\begin{aligned} \mu_{1}(G) < \max \biggl\{ \triangle(G), m-\frac {n-1}{2} \biggr\} +2. \end{aligned}

Let M be a real symmetric matrix of order n. Then all eigenvalues of M are real and can be denoted by $$\lambda_{1}(M)\geq\lambda _{2}(M)\geq\cdots \geq\lambda_{n}(M)$$ in non-increasing order. The following result in matrix theory plays a key role in our proofs.

### Lemma 2.4

[16]

Let A and B be two real symmetric matrices of order n. Then, for any $$1\leq k \leq n$$,

\begin{aligned} \sum_{i=1}^{k}\lambda_{i}(A+B) \leq\sum_{i=1}^{k}\lambda_{i}(A)+ \sum_{i=1}^{k}\lambda_{i}(B). \end{aligned}

The next lemma follows from LemmaÂ 2.4 immediately.

### Lemma 2.5

Suppose $$G_{1}, \ldots, G_{r}$$ are edge disjoint graphs. Then, for any k,

\begin{aligned} S_{k}(G_{1}\cup\cdots\cup G_{r})\leq\sum _{i=1}^{r}S_{k}(G_{i}). \end{aligned}

The following lemma can be found in [17] and is known as the interlacing theorem of Laplacian eigenvalues.

### Lemma 2.6

[17]

Let G be a graph of order n and $$G^{\prime}$$ be the graph obtained from G by deleting an edge of G. Then the Laplacian eigenvalues of G and $$G^{\prime}$$ interlace, that is,

\begin{aligned} \mu_{1}(G) \geq\mu_{1}\bigl(G^{\prime}\bigr) \geq\mu _{2}(G) \geq\cdots\geq\mu_{n-1}\bigl(G^{\prime}\bigr) \geq\mu_{n}(G) \geq \mu_{n}\bigl(G^{\prime}\bigr)=0. \end{aligned}

### Lemma 2.7

[17]

Let A be a real symmetric matrix of order n with eigenvalues $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{n}$$ and B be a principal submatrix of A of order m with eigenvalues $$\lambda^{\prime}_{1} \geq\lambda^{\prime}_{2} \geq\cdots\geq \lambda^{\prime}_{m}$$. Then the eigenvalues of B interlace the eigenvalues of A, that is, $$\lambda_{i} \geq\lambda^{\prime}_{i} \geq\lambda_{n-m+i}$$, for $$i=1, \ldots,m$$. Specially, for $$v \in V(G)$$, let $$L_{v}(G)$$ be the principal submatrix of $$L(G)$$ formed by deleting the row and column corresponding to vertex v. Then the eigenvalues of $$L_{v}(G)$$ interlace the eigenvalues of $$L(G)$$.

The multiplicities of an eigenvalue Î» for $$L(G)$$ is denoted by $$m_{G}(\lambda)$$. For a graph G of order n, it is well known that $$m_{G}(1)=n-r(I_{n}-L(G))$$, where $$r(I_{n}-L(G))$$ is the rank of $$I_{n}-L(G)$$.

### Lemma 2.8

Let v be a pendant vertex of graph G with n vertices and $$G^{\prime}$$ be the graph obtained from G by adding a new vertex $$v^{\prime}$$ which has a unique neighbor u, where u is the neighbor of v. (See FigureÂ  2.) Then

\begin{aligned} m_{G^{\prime}}(1)= m_{G}(1)+1. \end{aligned}

### Proof

Let $$L(G)$$ and $$L(G^{\prime})$$ be the Laplacian matrix of G and $$G^{\prime}$$, respectively. It is not difficult to check that $$r(I_{n}-L(G))= r(I_{n+1}-L(G^{\prime }))$$. Thus the result follows from the facts that $$m_{G}(1)=n-r(I_{n}-L(G))$$ and $$m_{G^{\prime }}(1)=(n+1)-r(I_{n+1}-L(G^{\prime}))$$.â€ƒâ–¡

Let $$P_{n}$$ and $$C_{n}$$ be the path and cycle of order n, respectively. A connected graph with n vertices and $$n+1$$ edges is called a bicyclic graph. Let $$\mathcal{B}_{n}$$ be the set of bicyclic graphs of order n. There are two basic bicyclic graphs: âˆž-graph and Î¸-graph. More concisely, an âˆž-graph, denoted by $$\infty(p,q,l)$$-graph, is obtained from two vertex-disjoint cycles $$C_{p}$$ and $$C_{q}$$ by connecting one vertex of $$C_{p}$$ and one vertex of $$C_{q}$$ with a path $$P_{l+1}$$ of length l, where $$q \geq p \geq3$$ and $$l\geq0$$ (in the case of $$l=0$$, we identify the above two vertices). A Î¸-graph, denoted by $$\theta(p,q,l)$$, is a union of three internally disjoint paths $$P_{p}$$, $$P_{q}$$, $$P_{l}$$ of length $$p-1$$, $$q-1$$, $$l-1$$, respectively with common end vertices, where $$l \geq q \geq p \geq2$$ and at most one of them is 2. Observe that any bicyclic graph G is obtained from a basic bicyclic graph $$\infty(p,q,l)$$ or $$\theta(p,q,l)$$ by attaching trees to some of its vertices. For any bicyclic graph G, we call its basic bicyclic graph $$\infty(p,q,l)$$ or $$\theta(p,q,l)$$ the kernel of G. For a vertex v of the kernel of G, if there is a tree $$T_{v}$$ attached to it, we denote by $$e(v)$$ the maximum distance between v and any vertex of $$T_{v}$$ (that is, $$e(v)=\max\{d(u,v)\mid u \in V(T_{v})\}$$); if there is no tree attached to it, we define $$e(v)=0$$. Let $$\mathcal{B}^{\infty}_{n}(p,q,l)$$ and $$\mathcal{B}^{\theta }_{n}(p,q,l)$$ be the sets of bicyclic graphs of order n with $$\infty(p,q,l)$$ and $$\theta(p,q,l)$$ as their kernel, respectively. Clearly, $$\mathcal{B}_{n}=\mathcal{B}^{\infty}_{n}(p,q,l)\cup\mathcal{B}^{\theta}_{n}(p,q,l)$$. Let S be a multiset of some nonnegative integers, denote by $$\Vert S \Vert$$ the number of nonzero elements in S, that is, $$\Vert S \Vert = \vert \{a \in S \mid a \geq1 \} \vert$$, where elements are counted according to their multiplicity.

### Lemma 2.9

Let G be the union of some disjoint graphs $$G_{1}, G_{2}, \ldots, G_{r}$$, where $$G_{i}$$ $$(i \in\{1,\ldots,r \})$$ is a tree or an unicyclic graph of order $$n_{i}$$ which is not isomorphic to $$G_{n_{i},n_{i}}$$ ($$G_{n, n}$$ is the unicyclic graph shown in FigureÂ  1). Then $$S_{2}(G)< m(G)+3$$.

### Proof

By TheoremÂ 1.2 and the fact $$S_{2}(T) < m(T)+3$$ for any tree, we have $$S_{2}(G_{i}) < m(G_{i})+3$$ for $$i \in\{1, \ldots, r \}$$. If $$\mu_{1}$$ and $$\mu_{2}$$ of G attain the same component, say $$G_{i_{0}}$$ $$(1 \leq i_{0} \leq r)$$, then $$S_{2}(G) = S_{2}(G_{i_{0}})< m(G_{i_{0}})+3 \leq m(G)+3$$. Otherwise, $$\mu_{1}$$ and $$\mu_{2}$$ of G attain two different components, without loss of generality, we assume that $$\mu_{1}$$ and $$\mu_{2}$$ attain $$G_{1}$$ and $$G_{2}$$, respectively, that is, $$\mu_{1}=\mu _{1}(G_{1})$$ and $$\mu_{2}=\mu_{1}(G_{2})$$. Then $$S_{2}(G)= \mu _{1}(G_{1})+ \mu_{1}(G_{2}) \leq n(G_{1})+ n(G_{2}) \leq(m(G_{1})+1)+ (m(G_{2})+1) \leq m(G)+2 < m(G)+3$$.â€ƒâ–¡

For any subgraph H of G, let $$G-H$$ be the graph obtained from G by deleting all edges from H.

### Lemma 2.10

Let G be a simple graph with at least 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that $$G-4P_{2}= \bigcup G_{i}$$ (or $$G-3P_{3}= \bigcup G_{i}$$), where $$G_{i}$$ is a tree or an unicyclic graph of order $$n_{i}$$ which is not isomorphic to $$G_{n_{i},n_{i}}$$ ($$G_{n, n}$$ is the unicyclic graph shown in FigureÂ  1). Then $$S_{2}(G)< m(G)+3$$.

### Proof

By direct calculation, we have $$S_{2}(4P_{2})=m(4P_{2})=4$$ and $$S_{2}(3P_{3})=m(3P_{3})=6$$. Using LemmasÂ 2.5 and 2.9, we have $$S_{2}(G) \leq S_{2}(G-4P_{2})+S_{2}(4P_{2}) < (m(G-4P_{2})+3)+m(4P_{2}) = m(G)+3$$ (or $$S_{2}(G) \leq S_{2}(G-3P_{3})+S_{2}(3P_{3}) < (m(G-3P_{3})+3)+m(3P_{3}) = m(G)+3$$), this completes the proof.â€ƒâ–¡

## 3 Proof of TheoremÂ 1.3

First of all, using the facts that $$S_{2}(G) \leq m(G)+3$$ for any graph G and that $$\mathcal{B}_{n}=\mathcal{B}^{\infty}_{n}(p,q,l)\cup \mathcal{B}^{\theta}_{n}(p,q,l)$$, we give the main idea of the proof.

1. 1.

For each class, we show that $$S_{2}(G) < m(G)+3$$ for the majority of graphs of $$\mathcal{B}^{\infty}_{n}(p,q,l)$$ or $$\mathcal {B}^{\theta}_{n}(p,q,l)$$ by LemmaÂ 2.10.

2. 2.

For the remaining graphs in $$\mathcal{B}^{\infty}_{n}(p,q,l)$$ or $$\mathcal{B}^{\theta}_{n}(p,q,l)$$ (except for $$G_{n+1, n}$$), we prove that $$S_{2}(G) < m(G)+3$$ by discussing case by case.

3. 3.

We show that $$S_{2}(G_{n+1, n}) = m(G_{n+1, n})+3$$, that is, the condition that equality holds.

Next, we discuss according to the following two subsections.

### 3.1 Bicyclic graphs in $$\mathcal{B}^{\infty}_{n}(p,q,l)$$

In this subsection, we prove that $$S_{2}(G) < m(G)+3$$ for $$G\in \mathcal{B}^{\infty}_{n}(p,q,l)$$.

### Lemma 3.1

Let $$G \in\mathcal{B}^{\infty}_{n}(p,q,l)$$. If $$(p,q,l)$$ satisfies one of the following conditions:

1. (1)

$$l\geq1$$,

2. (2)

$$(p,q,l)=(p,q,0)$$ and $$q \geq p \geq4$$,

3. (3)

$$(p,q,l)=(3,q,0)$$ and $$q \geq5$$,

then $$S_{2}(G)< m(G)+3$$.

### Proof

Direct calculation shows that $$S_{2}(C_{p} \cup C_{q}) \leq m(C_{p} \cup C_{q})$$, where $$q \geq p \geq3$$. For (1) $$l \geq1$$, using LemmasÂ 2.5 andÂ 2.9, we have $$S_{2}(G) \leq S_{2}(G-C_{p}-C_{q})+ S_{2}(C_{p} \cup C_{q}) < (m(G-C_{p}-C_{q})+3)+m(C_{p} \cup C_{q})=m(G)+3$$, since each component of $$G-C_{p}-C_{q}$$ is tree. For both (2) and (3), it is not difficult to check that G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree. Thus the result follows from LemmaÂ 2.10.â€ƒâ–¡

By LemmaÂ 3.1, it suffices to consider the following two cases: $$G\in\mathcal{B}^{\infty}_{n}(3,4,0)$$ and $$G\in\mathcal{B}^{\infty }_{n}(3,3,0)$$.

### Lemma 3.2

For $$G\in\mathcal{B}^{\infty}_{n}(3,4,0)$$, we have $$S_{2}(G)< m(G)+3$$.

### Proof

For $$G\in\mathcal{B}^{\infty}_{n}(3,4,0)$$, its kernel is $$\infty(3,4,0)$$, denoted by $$G^{0}_{1}$$ for short shown in FigureÂ 3. Let $$V(G^{0}_{1})= \{v_{1},\ldots,v_{6}\}$$. If there exists a vertex $$v_{i} \in V(G^{0}_{1})$$ such that $$e(v_{i}) \geq2$$, then G has 4 vertex-disjoint $$P_{2}$$ such that each component of $$G-4P_{2}$$ is a tree and the result follows from LemmaÂ 2.10. Thus it suffices to consider the case that G is isomorphic to $$G_{1}$$ shown in FigureÂ 3.

When $$\Vert \{n_{1},n_{2},\ldots, n_{6}\} \Vert \geq3$$, if G is not isomorphic to $$G^{1}_{1}$$ shown in FigureÂ 3, then G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree or a unicyclic graph of order $$n_{i}$$ which is not isomorphic to $$G_{n_{i},n_{i}}$$. Thus the result follows from LemmaÂ 2.10. If G is isomorphic to $$G^{1}_{1}$$, then we have $$m_{G}(1) \geq n-9$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{8} > \lambda_{9}=0$$ be the other night eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{9} =n+11$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda_{4}+\lambda _{5} \geq\mu_{3}(G^{2}_{1})+\mu_{4}(G^{2}_{1})+\mu _{5}(G^{2}_{1})=3.22+2.22+2.00>7$$, since $$G^{1}_{1}$$ contains $$G^{2}_{1}$$ shown in FigureÂ 3 as a subgraph. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert =2$$, then we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$.

When $$n_{3}=0$$, G contains a subgraph isomorphic to $$G^{i}_{1}$$ $$(i \in\{3,4,5\})$$ shown in FigureÂ 3. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(G^{i}_{1})+\mu_{4}(G^{i}_{1})+\mu_{5}(G^{i}_{1}) > 6$$ for $$i \in\{3,4,5\}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

When $$n_{3} \geq1$$, it can be checked that $$m_{G}(1) \geq n-7$$ by direct calculation and LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7}=n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda _{4} \geq\mu_{3}(G^{0}_{1})+\mu_{4}(G^{0}_{1})=5$$, since $$G_{1}$$ contains $$G^{0}_{1}$$ as a subgraph. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert =1$$, then we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda _{2} + \cdots+\lambda_{7}=n+9$$, since $$\sum_{i=1}^{i=n}\mu _{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculation, we have $$\lambda_{3}+\lambda_{4} \geq\mu_{3}(G^{0}_{1})+\mu _{4}(G^{0}_{1})=5$$, since G contains $$G^{0}_{1}$$ as a subgraph. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert =0$$, then G is isomorphic to $$G^{0}_{1}$$. Direct calculation shows that $$S_{2}(G)< m(G)+3$$.

The proof is completed.â€ƒâ–¡

### Lemma 3.3

For $$G\in\mathcal{B}^{\infty}_{n}(3, 3, 0)$$, we have $$S_{2}(G)< m(G)+3$$.

### Proof

For $$G\in\mathcal{B}^{\infty}_{n}(3,3,0)$$, its kernel is $$\infty(3,3,0)$$, denoted by $$G^{0}_{2}$$ for short shown in FigureÂ 4. Let $$V(G^{0}_{2})= \{v_{1},\ldots,v_{5}\}$$. If there exists a vertex $$v_{i} \in V(G^{0}_{2})$$ such that $$e(v_{i}) \geq2$$, then it suffices to consider the case that G is isomorphic to $$G^{1}_{2}$$ or $$G^{2}_{2}$$ shown in FigureÂ 4, otherwise G has 4 vertex-disjoint $$P_{2}$$ such that each component of $$G-4P_{2}$$ is a tree and the result follows from LemmaÂ 2.10.

When G is isomorphic to $$G^{i}_{2}$$ $$(i \in\{1,2\})$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{i}_{2})+\mu _{4}(H^{i}_{2})+\mu_{5}(H^{i}_{2})>6$$, where $$H^{i}_{2}$$ $$(i \in\{ 1,2\})$$ is a subgraph of $$G^{i}_{2}$$ shown in FigureÂ 4. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

Now, it suffices to consider the case that G is isomorphic to $$G_{2}$$ shown in FigureÂ 4.

When $$\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert \geq3$$, if G is isomorphic to $$G^{3}_{2}$$ shown in FigureÂ 4, then G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10. If G is isomorphic to $$G^{3}_{2}$$, we have $$m_{G}(1) \geq n-7$$ by direct calculation and LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations we have $$\lambda_{3}+\lambda_{4} \geq\mu _{3}(H^{3}_{2})+\mu_{4}(H^{3}_{2})=3.00+2.33 > 5$$, where $$H^{3}_{2}$$ is a subgraph of G shown in FigureÂ 4. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

When $$\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert = 2$$, it suffices to consider the case that G is isomorphic to $$G^{i}_{2}$$ $$(i \in\{4,5,6\})$$ shown in FigureÂ 4. When G is isomorphic to $$G^{4}_{2}$$, we have $$m_{G}(1) \geq n-6$$ by direct calculation and LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$$ be the other six eigenvalues of G. We have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations we have $$\lambda_{3}+\lambda_{4} \geq\mu_{3}(H^{4}_{2})+\mu _{4}(H^{4}_{2})=3.00+1.51 > 4$$, where $$H^{4}_{2}$$ shown in FigureÂ 4 is a subgraph of G. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$. When G is isomorphic to $$G^{i}_{2}$$ $$(i \in\{5,6\})$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq \cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations we have $$\lambda _{3}+\lambda_{4} \geq\mu_{3}(H^{i}_{2})+\mu_{4}(H^{i}_{2})> 5$$, where $$H^{i}_{2}$$ shown in FigureÂ 4 is a subgraph of $$G^{i}_{2}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert =1$$, then it suffices to consider the case that G is isomorphic to $$G^{7}_{2}$$ or $$G^{8}_{2}$$ shown in FigureÂ 4. When G is isomorphic to $$G^{7}_{2}$$, we have $$m_{G}(1) \geq n-6$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{5} > \lambda_{6}=0$$ be the other six eigenvalues of G. We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations we have $$\lambda_{3}+\lambda_{4} \geq\mu_{3}(H^{7}_{2})+\mu _{4}(H^{7}_{2})=3.00+1.4 > 4$$, where $$H^{7}_{2}$$ shown in FigureÂ 4 is a subgraph of $$G^{7}_{2}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$. When G is isomorphic to $$G^{8}_{2}$$, by an elementary calculation, we have $$\phi (G^{8}_{2},\lambda)= \lambda(\lambda-1)^{n-4}(\lambda -3)^{2}(\lambda-n)$$. It follows that $$S_{2}(G^{8}_{2})=n+3 < m+3$$.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert =0$$, then G is isomorphic to $$G^{0}_{2}$$. Direct calculation shows that $$S_{2}(G)< m(G)+3$$.â€ƒâ–¡

We safely come to the following result by the above discussion.

### Theorem 3.4

For $$G\in\mathcal{B}^{\infty}_{n}(p,q,l)$$, where $$q \geq p \geq3$$ and $$l \geq0$$, we have $$S_{2}(G) < m(G)+3$$.

### 3.2 Bicyclic graphs in $$\mathcal{B}^{\theta}_{n}(p,q,l)$$

In this subsection, we prove that $$S_{2}(G) \leq m(G)+3$$ for $$G\in \mathcal{B}^{\theta}_{n}(p,q,l)$$ and equality holds if and only if $$G\cong G_{n+1,n}$$. We begin with the following lemma, which follows from LemmaÂ 2.10 immediately.

### Lemma 3.5

For $$G\in\mathcal{B}^{\theta}_{n}(p,q,l)$$, if $$(p,q,l)$$ satisfies one of the following conditions:

1. (1)

$$4 \leq p \leq q \leq l$$,

2. (2)

$$(p,q,l)=(3, q, l)$$ and $$4 \leq q \leq l$$,

3. (3)

$$(p,q,l)=(3, 3, l)$$ and $$5 \leq l$$,

then $$S_{2}(G)< m(G)+3$$.

By LemmaÂ 3.5, it suffices to consider the following three cases: (1) $$G\in\mathcal{B}^{\theta}_{n}(3,3,4)$$, (2) $$G\in\mathcal {B}^{\theta}_{n}(3,3,3)$$, and (3) $$G\in\mathcal{B}^{\theta }_{n}(2,q,l)$$, where $$l \geq q \geq3$$.

First, we consider the case $$G\in\mathcal{B}^{\theta}_{n}(3,3,4)$$.

### Lemma 3.6

For $$G\in\mathcal{B}^{\theta}_{n}(3,3,4)$$, we have $$S_{2}(G)< m(G)+3$$.

### Proof

For $$G\in\mathcal{B}^{\theta}_{n}(3,3,4)$$, its kernel is $$\theta(3,3,4)$$, denoted by $$G^{0}_{3}$$ for short shown in FigureÂ 5. Let $$V(G^{0}_{3})= \{v_{1},\ldots,v_{6}\}$$. If there exists a vertex $$v_{i}$$ of $$G^{0}_{3}$$ such that $$e(v_{i}) \geq2$$, then G has 4 vertex-disjoint $$P_{2}$$ such that each component of $$G-4P_{2}$$ is a tree and the result follows from LemmaÂ 2.10. Thus it suffices to consider the case that G is isomorphic to $$G_{3}$$ shown in FigureÂ 5. For $$n(G) \leq8$$, it is easy to check that $$S_{2}(G) < m(G)+3$$ by a direct calculation. Thus we assume that $$n(G) \geq9$$ in the following.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \geq3$$, then G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \leq2$$, then we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. When G is not isomorphic to $$G_{3}^{1}$$ shown in FigureÂ 5, it contains a subgraph which is isomorphic to $$G^{i}_{3}$$ $$(i \in\{2,3,4\})$$ shown in FigureÂ 5. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda_{4}+\lambda_{5} \geq \lambda_{3}(G^{i}_{3})+\lambda_{4}(G^{i}_{3})+\lambda_{5}(G^{i}_{3}) > 6$$ for $$i \in\{2,3,4\}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$. When G is isomorphic to $$G_{3}^{1}$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} = n+9$$, since $$\sum^{i=n}_{i=1}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(G^{0}_{3})+\mu _{4}(G^{0}_{3})+\mu_{5}(G^{0}_{3})=2.00+2.00+1.26 > 5$$, since G contains $$G^{0}_{3}$$ as a subgraph. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

The proof is completed.â€ƒâ–¡

Second, we consider the case $$G\in\mathcal{B}^{\theta}_{n}(3,3,3)$$.

### Lemma 3.7

For $$G\in\mathcal{B}^{\theta}_{n}(3,3,3)$$, we have $$S_{2}(G)< m(G)+3$$.

### Proof

For $$G\in\mathcal{B}^{\theta}_{n}(3,3,3)$$, its kernel is $$\theta(3,3,3)$$, denoted by $$G^{0}_{4}$$ for short shown in FigureÂ 6. Let $$V(G^{0}_{4})= \{v_{1},\ldots,v_{5}\}$$. If there exists a vertex $$v_{i} \in V(G^{0}_{4})$$ such that $$e(v_{i}) \geq2$$, then by LemmaÂ 2.10, it suffices to consider the case that G is isomorphic to $$G^{1}_{4}$$ or $$G^{2}_{4}$$ shown in FigureÂ 6.

When G is isomorphic to $$G^{1}_{4}$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} = n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculation, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{1}_{4})+\mu _{4}(H^{1}_{4})+\mu_{5}(H^{1}_{4})=3.00+2.00+2.00 > 5$$, where $$H^{1}_{4}$$ shown in FigureÂ 6 is a subgraph of $$G^{1}_{4}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

When G is isomorphic to $$G^{2}_{4}$$, we have $$m_{G}(1) \geq n-9$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{8} > \lambda_{9}=0$$ be the other nine eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{9} = n+11$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. If $$n_{4} \geq1$$ and $$n_{5} \geq1$$, then $$G^{2}_{4}$$ contains $$H^{2}_{4}$$ shown in FigureÂ 6 as a subgraph. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5}+\lambda_{6} \geq\mu _{3}(H^{2}_{4})+\mu_{4}(H^{2}_{4})+\mu_{5}(H^{2}_{4})+\mu _{6}(H^{2}_{4})=2.47+2.00+2.00+1.22 > 7$$. Therefore $$S_{2}(G)=\lambda _{1} + \lambda_{2} < n+4 = m(G)+3$$. If $$n_{4} = 0$$ or $$n_{5} = 0$$, then we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} = n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculation, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{3}_{4})+\mu _{4}(H^{3}_{4})+\mu_{5}(H^{3}_{4})=2.32+2.00+2.00 > 6$$, where $$H^{3}_{4}$$ shown in FigureÂ 6 is a subgraph of $$G^{2}_{4}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

Now, we consider the case that G is isomorphic to $$G_{4}$$ shown in FigureÂ 6.

First, we consider the case that $$\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert \geq3$$. If G is not isomorphic to $$G^{4}_{4}$$ shown in FigureÂ 6, then G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10. When G is isomorphic to $$G^{4}_{4}$$, we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq \lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculation, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{4}_{4})+\mu _{4}(H^{4}_{4})+\mu_{5}(H^{4}_{4}) > 3.08+2.00+1.29 > 6$$, where $$H_{4}^{4}$$ shown in FigureÂ 6 is a subgraph of $$G_{4}^{4}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

When $$\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert = 2$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. Then by LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4} \geq\mu_{3}( H^{i}_{4})+\mu_{4}(H^{i}_{4}) \geq5$$ $$(i \in\{5, 6\})$$, where $$H_{4}^{i}$$ $$(i \in\{5, 6\})$$ shown in FigureÂ 6 is a subgraph of G. Therefore $$S_{2}(G)=\lambda _{1} + \lambda_{2}< n+4 = m(G)+3$$.

When $$\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert = 1$$, we have $$m_{G}(1) \geq n-6$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$$ be the other six eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculation, we have $$\lambda _{3}+\lambda_{4} \geq\mu_{3}(G^{0}_{4})+\mu _{4}(G^{0}_{4})=2.00+2.00=4$$, since G contains $$G^{0}_{4}$$ as a subgraph. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$, as required.

When $$\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert = 0$$, G is isomorphic to $$G^{0}_{4}$$. Direct calculation shows that $$S_{2}(G)< m(G)+3$$.

From the above discussion, we complete the proof.â€ƒâ–¡

Finally, we consider the case $$G\in\mathcal{B}^{\theta}_{n}(2,q,l)$$, where $$l \geq q \geq3$$.

### Lemma 3.8

If $$G\in\mathcal{B}^{\theta}_{n}(2,q,l)$$ with $$l \geq q \geq4$$, then $$S_{2}(G)< m(G)+3$$.

### Proof

If $$q \geq5$$, it is obviously that G has 4 vertex-disjoint $$P_{2}$$ such that $$G-4P_{2}$$ is a forest and the result follows immediately from LemmaÂ 2.10. When $$q = 4$$ and $$l \geq 5$$, the result follows immediately from LemmaÂ 2.10, since G has 4 vertex-disjoint $$P_{2}$$ such that $$G-4P_{2}$$ is a forest. Thus it suffices to consider the case that $$G\in\mathcal{B}^{\theta}_{n}(2,4,4)$$.

For $$G\in\mathcal{B}^{\theta}_{n}(2,4,4)$$, its kernel is $$\theta (2,4,4)$$, denoted by $$G^{0}_{5}$$ for short shown in FigureÂ 7. Let $$V(G^{0}_{5})=\{v_{1},\ldots,v_{6} \}$$. If there exists a vertex $$v_{i} \in V(G^{0}_{5})$$ such that $$e(v_{i}) \geq2$$, then G has 4 vertex-disjoint $$P_{2}$$ such that each component of $$G-4P_{2}$$ is a tree and the result follows from LemmaÂ 2.10. Now, it suffices to consider the case that G is isomorphic to $$G_{5}$$ shown in FigureÂ 7.

When $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \geq3$$, G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10.

When $$1 \leq \Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \leq2$$, we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. If G is not isomorphic to $$G^{1}_{5}$$ shown in FigureÂ 7, then G contains a subgraph which is isomorphic to $$G^{2}_{5}$$ shown in FigureÂ 7. By LemmaÂ 2.6 and direct calculation, we have $$\lambda_{3}+\lambda_{4}+\lambda_{5} \geq\mu _{3}(G^{2}_{5})+\mu_{4}(G^{2}_{5})+\mu_{5}(G^{2}_{5})=3.00+2.22+1.38 > 6$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$. When G is isomorphic to $$G^{1}_{5}$$, we have $$m_{G}(1)=n-6$$ by direct calculation and LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$$ be the other six eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculation, we have $$\lambda _{3}+\lambda_{4} \geq\mu_{3}(G^{0}_{5})+\mu _{4}(G^{0}_{5})=3.00+2.00> 4$$, since $$G^{1}_{5}$$ contains $$G^{0}_{5}$$ as a subgraph. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

When $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert = 0$$, G is isomorphic to $$G^{0}_{5}$$. Direct calculation shows that $$S_{2}(G) < m(G)+3$$.

The proof is completed.â€ƒâ–¡

### Lemma 3.9

If $$G\in\mathcal{B}^{\theta}_{n}(2,3,l)$$ with $$l \geq5$$, then $$S_{2}(G)< m(G)+3$$.

### Proof

When $$l \geq6$$, it is obviously that G has 4 vertex-disjoint $$P_{2}$$ such that $$G-4P_{2}$$ is a forest and the result follows immediately from LemmaÂ 2.10. Thus it suffices to consider the case that $$G\in\mathcal{B}^{\theta }_{n}(2,3,5)$$. For $$n(G) \leq7$$, it is easy to check that $$S_{2}(G) < m(G)+3$$ by a direct calculation. Thus we assume that $$n(G) \geq8$$ in the following.

For $$G\in\mathcal{B}^{\theta}_{n}(2,3,5)$$, its kernel is $$\theta (2,3,5)$$, denoted by $$G^{0}_{6}$$ for short shown in FigureÂ 7. Let $$V(G^{0}_{6})=\{ v_{1},\ldots,v_{6}\}$$. If there exists a vertex $$v_{i}$$ of $$G^{0}_{6}$$ such that $$e(v_{i}) \geq2$$, then G has 4 vertex-disjoint $$P_{2}$$ such that $$G-4P_{2}$$ is a forest and the result follows from LemmaÂ 2.10. Now, we can assume that G is isomorphic to $$G_{6}$$ shown in FigureÂ 7.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{6}\} \Vert \geq3$$, then G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10.

If $$\Vert \{n_{1},n_{2},\ldots, n_{6}\} \Vert \leq2$$, then we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. Note that G contains a subgraph which is isomorphic to $$G^{i}_{6}$$ $$(i \in\{1,2,3,4,5 \})$$ shown in FigureÂ 8. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(G^{i}_{6})+\mu _{4}(G^{i}_{6})+\mu_{5}(G^{i}_{6}) > 6$$ $$(i \in\{1,2,3,4,5 \})$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

The proof is completed.â€ƒâ–¡

### Lemma 3.10

For $$G\in\mathcal{B}^{\theta}_{n}(2,3,4)$$, we have $$S_{2}(G)< m(G)+3$$.

### Proof

For $$G\in\mathcal{B}^{\theta}_{n}(2,3,4)$$, its kernel is $$\theta(2,3,4)$$, denoted by $$G^{0}_{7}$$ for short shown in FigureÂ 9. Let $$V(G^{0}_{7})= \{v_{1},\ldots,v_{5}\}$$.

If there exists a vertex $$v_{i}$$ such that $$e(v_{i}) \geq2$$, then by LemmaÂ 2.10, it suffices to consider the cases that G is isomorphic to $$G^{i}_{7}$$ $$(i \in\{1,2,3\})$$ shown in FigureÂ 9.

When G is isomorphic to $$G^{i}_{7}$$ $$(i \in\{1,2,3\})$$ shown in FigureÂ 9, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5} \geq\mu_{3}(H^{i}_{7})+\mu _{4}(H^{i}_{7})+\mu_{5}(H^{i}_{7}) > 6$$, where $$H^{i}_{7}$$ shown in FigureÂ 9 is a subgraph of $$G^{i}_{7}$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$.

Now, it suffices to consider the case that G is isomorphic to $$G_{7}$$ shown in FigureÂ 9. For $$n(G) \leq8$$, it is easy to check that $$S_{2}(G) < m(G)+3$$ by a direct calculation. In the following, we assume that $$n(G) \geq9$$.

If $$\Vert \{n_{1}, n_{2}, \ldots, n_{5}\} \Vert \geq3$$, then G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10.

If $$\Vert \{n_{1},n_{2}, \ldots,n_{5}\} \Vert = 2$$, then we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda _{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. If G is not isomorphic to $$G^{4}_{7}$$ shown in FigureÂ 9, then it contains a subgraph isomorphic to $$H^{i}_{7}$$ $$(i \in\{5,\ldots, 10\})$$ shown in FigureÂ 9. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda_{4}\geq\mu _{3}(G^{i}_{7})+\mu_{4}(G^{i}_{7}) > 5$$ $$(i \in\{5,\ldots, 10\})$$. Therefore $$S_{2}(G)=\lambda_{1} + \lambda_{2}< n+4 = m(G)+3$$. When G is isomorphic to $$G^{4}_{7}$$, we have $$m_{G}(1) \geq n-6$$ by direct calculation and LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$$ be the other six eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda _{4} \geq\mu_{3}(H^{4}_{7})+\mu_{4}(H^{4}_{7})> 4$$, where $$H^{4}_{7}$$ shown in FigureÂ 9 is a subgraph of $$G^{4}_{7}$$. Therefore $$S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

If $$\Vert \{n_{1},n_{2}, \ldots, n_{5}\} \Vert = 1$$, then we have $$m_{G}(1) \geq n-6$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda_{6}=0$$ be the other six eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}\geq\mu_{3}(H^{i}_{7})+\mu_{4}(H^{i}_{7}) > 4$$ $$(i\in\{11, 12, 13\})$$, where $$H^{i}_{7}$$ shown in FigureÂ 9 is a subgraph of G. Therefore $$S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

The proof is completed.â€ƒâ–¡

### Lemma 3.11

Let $$G\in\mathcal{B}^{\theta}_{n}(2,3,3)$$ and $$G^{0}$$ shown in FigureÂ  10 be the kernel of G. If there exists a vertex $$v_{i}$$ of $$G^{0}$$ such that $$e(v_{i})\geq3$$, then $$S_{2}(G)< m(G)+3$$.

### Proof

By LemmaÂ 2.10, it suffices to consider the case that G is isomorphic to $$G_{i}$$ $$(i \in\{8,9,10\})$$ shown in FigureÂ 10.

When G is isomorphic $$G_{i}$$ $$(i \in\{8,9,10\})$$, we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq \cdots\geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of $$G_{i}$$. We have $$\lambda_{1} + \lambda_{2} + \cdots +\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}+\lambda_{5}\geq\mu_{3}(G^{\prime}_{i})+\mu _{4}(G^{\prime}_{i}) +\mu_{5}(G^{\prime}_{i}) > 6$$, where $$G^{\prime}_{i}$$ shown in FigureÂ 10 is the subgraph of $$G_{i}$$. Therefore $$S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

The proof is completed.â€ƒâ–¡

### Lemma 3.12

Let $$G\in\mathcal{B}^{\theta}_{n}(2,3,3)$$ and $$G^{0}$$ shown in FigureÂ  10 be the kernel of G. If $$\max_{i=1}^{i=4} e(v_{i}) = 2$$, then $$S_{2}(G)< m(G)+3$$, where $$v_{i}$$ is the vertices of $$G^{0}$$.

### Proof

By LemmaÂ 2.10, it suffices to consider the case that G is isomorphic to $$G_{11}$$ or $$G_{12}$$ shown in FigureÂ 11. Here, we only prove the case that G is isomorphic to $$G_{11}$$. For the case G is isomorphic to $$G_{12}$$, we can discuss similarly.

When $$\Vert \{n_{1}, n_{2}, n_{3}, n_{4}\} \Vert \geq2$$, G has 4 vertex-disjoint $$P_{2}$$ (or 3 vertex-disjoint $$P_{3}$$) such that each component of $$G-4P_{2}$$ (or $$G-3P_{3}$$) is a tree and the result follows from LemmaÂ 2.10.

When $$\Vert \{n_{1}, n_{2}, n_{3}, n_{4}\} \Vert \leq1$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq \lambda_{2} \geq\cdots\geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda _{4}\geq\mu_{3}(G^{\prime}_{11})+\mu_{4}(G^{\prime }_{11})=3.00+2.00 = 5$$, where $$G^{\prime}_{11}$$ shown in FigureÂ 11 is a subgraph of $$G_{11}$$. Thus $$S_{2}(G_{11}) < m(G_{11})+3$$.â€ƒâ–¡

### Lemma 3.13

Let $$G_{13}$$ be the bicyclic graph of order n shown in FigureÂ  12, where $$a \geq b \geq1$$ and $$a+b+5=n$$. Then $$S_{2}(G_{13}) < m(G_{13})+3$$.

### Proof

For $$n(G_{13}) \leq14$$, it is easy to check that $$S_{2}(G_{13}) < m(G_{13})+3$$ by a direct calculation. In following, we assume that $$n(G_{13}) \geq15$$.

If $$b \geq5$$, then $$G_{13}$$ contains $$G^{\prime}_{13}$$ shown in FigureÂ 12 as a subgraph. Note that $$m_{G_{13}}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots \geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of $$G_{13}$$. We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}\geq\mu_{3}(G^{\prime}_{13})+\mu_{4}(G^{\prime }_{13})=3.197+1.810 > 5$$. Therefore $$S_{2}(G_{13})= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

If $$b = 4$$, then we have $$\lambda_{1}(G_{13})< \triangle+2 =n-4$$ by LemmaÂ 2.3 and $$\lambda_{2}(G_{13})< \lambda _{1}(L_{v_{1}}(G_{13}))=7.95$$ by LemmaÂ 2.7 and direct calculation. It follows that $$S_{2}(G_{13})< n+4= m(G)+3$$.

If $$b = 3$$, then we have $$\lambda_{1}(G_{13})< \triangle+2=n-3$$ by LemmaÂ 2.3 and $$\lambda_{2}(G_{13})< \lambda _{1}(L_{v_{3}}(G_{13}))=6.96$$ by LemmaÂ 2.7 and direct calculation. It follows that $$S_{2}(G_{13})< n+4= m(G)+3$$.

If $$b =2$$, then we have $$\lambda_{1}(G_{13})<(n-4)+\frac{n+3}{n-4}$$ by LemmaÂ 2.2 and $$\lambda_{2}(G_{3})< \lambda _{1}(L_{v_{1}}(G_{13}))=6.005<6.10$$ by LemmaÂ 2.7 and direct calculation. Therefore $$S_{2}(G_{13})< n-4+\frac{n+3}{n-4}+6.10 < n+4= m(G)+3$$.

If $$b =1$$, then we have $$\lambda_{1}(G_{13})<(n-3)+\frac{n+3}{n-4}$$ by LemmaÂ 2.2 and $$\lambda_{2}(G_{13})< \lambda _{1}(L_{v_{1}}(G_{13}))=5.10<5.20$$ by LemmaÂ 2.7 and direct calculation. Thus $$S_{2}(G_{13})< n-3+\frac{n+3}{n-4}+5.10 < n+4= m(G)+3$$.â€ƒâ–¡

### Lemma 3.14

Let $$G_{14}$$ be the bicyclic graph of order n shown in FigureÂ  12, where $$a \geq b \geq1$$ and $$a+b+4=n$$. Then $$S_{2}(G_{14}) < m+3$$.

### Proof

By some elementary calculations, we have $$\phi(G_{14}, x)= x(x-2)(x-1)^{6}g(x)$$, where $$g(x)= x^{4}-(n+6)x^{3}+(6n+ab+9)x^{2}-(9n+2ab+4)x+4n$$. Let $$x_{1}\geq x_{2} \geq x_{3}\geq x_{4}$$ be the roots of $$g(x)=0$$. Then

\begin{aligned}& x_{1}+ x_{2} + x_{3}+ x_{4}=n+6, \end{aligned}
(3.1)
\begin{aligned}& x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}=6n+ab+9, \end{aligned}
(3.2)
\begin{aligned}& x_{2}x_{3}x_{4}+x_{1}x_{3}x_{4}+x_{1}x_{2}x_{4}+x_{1}x_{2}x_{3}=9n+2ab+4. \end{aligned}
(3.3)

If

$$x_{1}+ x_{2}=n+4$$
(3.4)

then, by (3.1), we have

$$x_{3}+ x_{4}=2.$$
(3.5)

From (3.2)-(3.5) follows that

\begin{aligned}& x_{1}x_{2}+x_{3}x_{4}=4n+ab+1, \end{aligned}
(3.6)
\begin{aligned}& (n+4)x_{3}x_{4}+2x_{1}x_{2}=9n+2ab+4. \end{aligned}
(3.7)

By (3.6) and (3.7), we have

$$x_{3}x_{4}=1.$$
(3.8)

Combining (3.5) and (3.8), we have

$$x_{3}=x_{4}=1.$$
(3.9)

Then $$g(1)=-2ab=0$$, which implies that $$b=0$$. Therefore, $$b \geq1$$ implies that $$S_{2}(G_{14}) < m(G_{14})+3$$.â€ƒâ–¡

### Lemma 3.15

Let G be the bicyclic graph of order n shown in FigureÂ  13, where $$a \geq b \geq0$$, $$c \geq d \geq0$$ and $$a+b+c+d+4=n$$. Then $$S_{2}(G) \leq m+3$$ with equality if and only if $$a=n-4$$ and $$b=c=d=0$$ (that is, $$G \cong G_{n+1,n}$$).

### Proof

For $$n(G) \leq12$$, it is easy to check that $$S_{2}(G) \leq m(G)+3$$ with equality if and only if $$G \cong G_{n+1, n}$$ by a direct calculation. In the following, we assume that $$n(G) \geq13$$.

When $$\Vert \{a,b,c,d\} \Vert = 4$$, we have $$m_{G}(1) \geq n-8$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots \geq\lambda_{7} > \lambda_{8}=0$$ be the other eight eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{8} =n+10$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$.

If $$c \geq2$$, then G contains $$H_{1}$$ shown in FigureÂ 12 as a subgraph. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}\geq\mu_{3}(H_{1})+\mu_{4}(H_{1})=3.94+2.14 > 6$$. If $$c=d=1$$ and $$b \geq2$$, then G contains $$H_{2}$$ shown in FigureÂ 12 as a subgraph. By LemmaÂ 2.6 and direct calculations, we have $$\lambda_{3}+\lambda_{4}\geq\mu _{3}(H_{2})+\mu_{4}(H_{2})=3.414+2.593 > 6$$. Therefore $$S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$. Now, we can assume that $$b=c=d=1$$. Under the assumption, G is isomorphic to $$H_{3}$$ shown in FigureÂ 12. We have $$\lambda_{1}(H_{3})< n-2$$ by LemmaÂ 2.3 and $$\lambda _{2}(H_{3})< \lambda_{1}(L_{v_{1}}(H_{3}))=5.23$$ by LemmaÂ 2.7 and direct calculation. Therefore, $$S_{2}(G)< (n-2)+5.23 < n+4 = m(G)+3$$.

When $$\Vert \{a,b,c,d\} \Vert = 3$$, we have $$m_{G}(1) \geq n-7$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots \geq\lambda_{6} > \lambda_{7}=0$$ be the other seven eigenvalues of G. Then we have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{7} =n+9$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. If G is isomorphic to $$G_{13}$$, then $$S_{2}(G) < m(G)+3$$ by LemmaÂ 3.13. Otherwise it contains a subgraph isomorphic to $$H_{i}$$ $$( i \in\{4,5,6,7\})$$ shown in FigureÂ 12. Then we have $$\lambda_{3}+\lambda_{4}\geq\lambda_{3}(H_{i})+\lambda _{i}(H_{i}) > 5$$ $$(i\in\{4,5,6,7\})$$ by LemmaÂ 2.6 and direct calculations. Therefore $$S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$.

When $$\Vert \{a,b,c,d\} \Vert = 2$$, G is isomorphic to $$G_{14}$$ shown in FigureÂ 12 or $$H_{i}$$ $$(i \in\{8,10\})$$ shown in FigureÂ 12. For G is isomorphic to $$G_{14}$$, we have $$S_{2}(G_{14}) < m(G_{14})+3$$ by LemmaÂ 3.14. For G is isomorphic to $$H_{8}$$ (or $$H_{10}$$), we have $$m_{G}(1) \geq n-6$$ by LemmaÂ 2.8. Let $$\lambda_{1} \geq\lambda_{2} \geq\cdots\geq\lambda_{5} > \lambda _{6}=0$$ be the other six eigenvalues of $$H_{8}$$ (or $$H_{10}$$). We have $$\lambda_{1} + \lambda_{2} + \cdots+\lambda_{6} =n+8$$, since $$\sum_{i=1}^{i=n}\mu_{i}=2m(G)$$. Note that $$H_{8}$$ contains $$H_{9}$$ as a subgraph ($$H_{10}$$ contains $$H_{11}$$ or $$H_{12}$$ as a subgraph), where $$H_{i}$$ $$(i= 9, 11, 12)$$ shown in FigureÂ 12. By LemmaÂ 2.6 and direct calculations, we have $$\lambda _{3}+\lambda_{4}\geq\mu_{3}(H_{i})+\mu_{4}(H_{i}) > 4$$ $$(i=9,11,12)$$. Therefore $$S_{2}(G)= \lambda_{1} + \lambda_{2} < n+4 = m(G)+3$$, as required. When $$\Vert \{a,b,c,d\} \Vert = 1$$, G is isomorphic to $$H_{13}$$ or $$G_{n+1, n}$$. We have $$\lambda_{1}(H_{13}) < n$$ by LemmaÂ 2.2 and $$\lambda _{2}(H_{13}) =4$$ by direct calculations. It follows that $$S_{2}(H_{13})< n +4$$. For $$G_{n+1, n}$$, a direct calculation shows that $$S_{2}(G_{n+1, n})=m+3$$.â€ƒâ–¡

From the discussion above, we safely come to the following result.

### Theorem 3.16

For $$G\in\mathcal{B}^{\theta}_{n}(p,q,l)$$, where $$l \geq q \geq p \geq2$$ and at most one of them is 2, we have $$S_{2}(G) \leq m(G)+3$$ and the equality holds if and only if $$G \cong G_{n+1, n}$$.

TheoremÂ 1.3 follows immediately from TheoremsÂ 3.4 andÂ 3.16.

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## Acknowledgements

The authors would like to thank the anonymous referees for their constructive corrections and valuable comments on this paper, which have considerably improved the presentation of this paper. This project is supported by NSF of China (Nos. 11471077, 11301440), the Foundation to the Educational Committee of Fujian (JA13240, JA15381).

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Correspondence to Yirong Zheng.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

YRZ carried out the proofs of main results in the manuscript. AC, JXL and SR participated in the design of the study and drafted the manuscripts. All the authors read and approved the final manuscripts.

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Zheng, Y., Chang, A., Li, J. et al. Bicyclic graphs with maximum sum of the two largest Laplacian eigenvalues. J Inequal Appl 2016, 287 (2016). https://doi.org/10.1186/s13660-016-1235-5

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• DOI: https://doi.org/10.1186/s13660-016-1235-5