On the Minimum First-Inverse Nirmala Index of Trees

Abstract

Let G be a graph with edge set $E\left(G\right)$. The degree of a vertex w in G is denoted by $d\left(w\right)$. The first-inverse Nirmala (FIN) index and the atom-bond connectivity (ABC) index of the graph G are defined, respectively, as $\mathcal{FIN}\left(G\right)={\sum }_{uv\in E\left(G\right)}\sqrt{{\left(d\left(u\right)\right)}^{-1}+{\left(d\left(v\right)\right)}^{-1}}$ and $\mathcal{ABC}\left(G\right)={\sum }_{uv\in E\left(G\right)}\sqrt{{\left(d\left(u\right)\right)}^{-1}+{\left(d\left(v\right)\right)}^{-1}-2{\left(d\left(u\right)d\left(v\right)\right)}^{-1}}$. The problem of determining the trees that minimize the ABC index over the class ${\mathcal{T}}_n$ of all n-vertex trees required around a decade of research and dozens of publications before it was resolved. In view of the similarity between the definitions of the ABC and FIN indices, one might expect their minimizing trees over ${\mathcal{T}}_n$ to coincide. However, the computational findings of Furtula and Öz published in 2025 for $6\le n\le 20$ demonstrate that this is not the case. The present paper establishes the first theoretical results toward determining the trees that minimize the FIN index over ${\mathcal{T}}_n$. We also extend the computational findings of Furtula and Öz from $6\le n\le 20$ to $6\le n\le 60$.

1. Introduction

Let G be a finite simple graph. The first-inverse Nirmala index, introduced in [1], is a graph invariant, defined for a graph G as follows:

$$\mathcal{FIN}\left(G\right)=\sum _{uv\in E\left(G\right)}\sqrt{{\displaystyle \frac{1}{d\left(u\right)}}+{\displaystyle \frac{1}{d\left(v\right)}}},$$

where $E\left(G\right)$ is the edge set of G, while $d\left(u\right)$ and $d\left(v\right)$ represent the degrees of the vertices u and v of G (sometimes, we use $d_G\left(u\right)$ instead of $d\left(u\right)$ to specify the graph under consideration). It is closely related to the atom-bond connectivity index (for example, see [2,3]), which is defined as

$$\mathcal{ABC}\left(G\right)=\sum _{uv\in E\left(G\right)}\sqrt{{\displaystyle \frac{d\left(u\right)+d\left(v\right)-2}{d\left(u\right)d\left(v\right)}}}.$$

The edge weight defining the FIN index is obtained by deleting the term $-2$ from the numerator of the ABC edge weight. Although this formal similarity provides useful motivation, it does not imply that the two indices have analogous extremal behavior. For example, the ABC weight of an edge incident with a vertex of degree 2 is always $1/\sqrt{2}$, independently of the degree of its other endvertex. By contrast, the corresponding $\mathcal{FIN}$ weight is $\sqrt{1/2+1/d}$, which is strictly decreasing in the degree d of the other endvertex. Consequently, the transformations and extremal structures associated with the two indices need not coincide. The problem of determining the trees that minimize the ABC index over the class ${\mathcal{T}}_n$ of all n-vertex trees required around a decade of research and dozens of publications before it was resolved [4,5]. In view of the similarity between the definitions of the ABC and FIN indices, one might expect their minimizing trees over the class ${\mathcal{T}}_n$ to coincide. However, the computational findings of Furtula and Öz [6] for $6\le n\le 20$ demonstrate that this is not the case. The present paper establishes the first theoretical results toward determining the trees that minimize the FIN index over ${\mathcal{T}}_n$. We also extend the computational findings of Furtula and Öz from $6\le n\le 20$ to $6\le n\le 60$.

The chemical applicability of the FIN index was examined in [7]. Basic properties and extremal questions for the FIN index were studied by Furtula and Öz [6]. Gao [8] determined the extremal molecular trees of a given order for the FIN index, where a molecular tree is a tree of maximum degree at most 4. The minimization problem, without maximum-degree restriction, for the FIN index of trees of a given order appears to require a different description because the maximum degree of a minimizing tree grows with its order [6]. Our aim is to develop structural properties of a tree minimizing the FIN index over the class ${\mathcal{T}}_n$. The starting point is the strict negativity of the second-order mixed partial derivatives of the following function related to the $\mathcal{FIN}$ edge weight: $\varphi (x,y)=\sqrt{x^{-1}+y^{-1}}$. Proposition 1, obtained from a result of Damnjanović and Ranđelović [9] refining Wang’s theorem [10], then gives useful information about trees minimizing the FIN index over ${\mathcal{T}}_n$. We also used several edge transformations to obtain structural properties of the aforementioned extremal n-vertex trees.

A tree minimizing the FIN index among all the members of ${\mathcal{T}}_n$ is called a $\mathcal{FIN}$-minimal n-vertex tree, where $n\ge 5$. We show that every vertex of degree 2 in a $\mathcal{FIN}$-minimal n-vertex tree is adjacent with a vertex of degree 1 and a vertex of degree at least 3. We prove bounds on the maximum degree, on the number of maximum-degree vertices, on the cardinality of the degree set (that is, the set of all different members of the degree sequence), and on the diameter for a $\mathcal{FIN}$-minimal n-vertex tree. In particular, there are at most two vertices of maximum degree in a $\mathcal{FIN}$-minimal n-vertex tree, and if there are two, then they are adjacent.

The rest of the paper is organized as follows. Section 2 gives some definitions and notations and provides consequences of some of the results established in [9]. Section 3 gives results concerning vertices of degree 2 and vertices of degree at least 3 in $\mathcal{FIN}$-minimal n-vertex trees. Section 4 provides results regarding maximum-degree vertices and the maximum degree of a $\mathcal{FIN}$-minimal n-vertex tree. Bounds concerning degree set and diameter of a $\mathcal{FIN}$-minimal n-vertex tree are established in Section 5. Section 6 is concerned with a family of trees of diameter at most four and states a conjecture suggested by the structural results and exhaustive computations. The final section gives concluding remarks.

2. Preliminaries

All graphs considered in this study are finite, simple, and connected. For a vertex v of a graph G, its degree is denoted by $d_G\left(v\right)$, or simply $d\left(v\right)$ when the graph under consideration is clear from the context. We write $n_i\left(G\right)$ for the number of vertices of degree i, $\Delta \left(G\right)$ for the maximum degree, and $\mathcal{D}\left(G\right)=\left\{d\right(v):v\in V(G\left)\right\}$ for the degree set. For a vertex $u\in V\left(G\right)$, let $N_G\left(u\right)=\{v\in V\left(G\right):uv\in E\left(G\right)\}$. Every element of the set $N_G\left(u\right)$ is called a neighbor of u. A vertex of degree 1 (respectively, greater than 2) is pendent (respectively, branching). A pendent path (respectively, internal path) in G is a non-trivial path $v_1{v}_2\dots v_k$ of G such that $min(d_G\left(v_1\right),d_G\left(v_k\right))=1$ (respectively, $min(d_G\left(v_1\right),d_G\left(v_k\right))>2$), $max(d_G\left(v_1\right),d_G\left(v_k\right))>2$, and $d_G\left(v_i\right)=2$ when $2\le i\le k-1$. The distance between two vertices u and v in a graph G is denoted by $d_G(u,v)$. The diameter of G is denoted by $\mathrm{diam}\left(G\right)$.

For positive real numbers $x,y$, we take

$$\varphi (x,y)=\sqrt{x^{-1}+y^{-1}}.$$

(1)

Then, we have

$$\mathcal{FIN}\left(G\right)=\sum _{uv\in E\left(G\right)}\varphi (d_G\left(u\right),d_G\left(v\right)).$$

We note that the mixed second-order partial derivatives of $\varphi$ satisfy

$${\varphi }_{yx}(x,y)={\varphi }_{xy}(x,y)=-{\displaystyle \frac{1}{4x^2{y}^2{(x^{-1}+y^{-1})}^{3/2}}}<0.$$

(2)

Let $\mathbb{N}$ be the set of positive integers. Following the terminology of (strict) positive polarity due to Damnjanović and Ranđelović [9], we say that a real-valued function f defined on $\mathbb{N}\times \mathbb{N}$ satisfies “negative polarity” if for any positive integers $a,b,c,d$ with $a>c$ and $b>d$, the following inequality holds:

$$f(a,b)+f(c,d)\le f(a,d)+f(c,b).$$

(3)

If the strict form of (3) holds for all aforementioned $a,b,c,d,$ then we say that f satisfies “strict negative polarity”. The function f satisfies “(strict) negative polarity” if and only if $-f$ satisfies “(strict) positive polarity”. From (2), we have ${\varphi }_{xy}(x,y)<0$, which implies

$$\varphi (a,b)+\varphi (c,d)<\varphi (a,d)+\varphi (c,b)$$

(4)

for any positive integers $a,b,c,d$ with $a>c$ and $b>d$. Hence, $\varphi$ satisfies strict negative polarity.

The following lemma is an immediate consequence.

Lemma 1.

Suppose that $uv$ and $st$ are edges of a tree. Let $T^{\prime }$ be a tree obtained from T by removing $uv,st$ and inserting $ut,sv$. Then, both T and $T^{\prime }$ have the same degree sequence. If $d_T\left(u\right)>d_T\left(s\right)$ and $d_T\left(v\right)<d_T\left(t\right)$, then $\mathcal{FIN}\left(T\right)>\mathcal{FIN}\left(T^{\prime }\right)$.

Proof. 

We note that the vertices $u,t,s,v$ must be pairwise distinct.

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)=& \varphi (d_T\left(u\right),d_T\left(v\right))+\varphi (d_T\left(s\right),d_T\left(t\right))-\varphi (d_T\left(u\right),d_T\left(t\right))\hfill \\ & -\varphi (d_T\left(v\right),d_T\left(s\right)),\hfill \end{array}$$

(5)

where $\varphi$ is defined via (1). From (5) and (4), it follows that $\mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)>0$, as required.    □

Now, we recall from [9] the following tree construction algorithm for a given degree sequence $(d_1,d_2,\dots ,d_n)$ of a tree such that $d_1\ge d_2\ge \cdots \ge d_n$.

Let $D=(d_1,\dots ,d_n)$ be a nonincreasing tree degree sequence, and let ${\mathcal{T}}_D$ be the family of its tree realizations.

Lemma 2

([9]). Suppose that $f:\mathbb{N}\times \mathbb{N}\to \mathbb{R}$ is a symmetric function having strict positive polarity. Then, a tree $T\in {\mathcal{T}}_D$ maximizes ${\sum }_{xy\in E\left(T\right)}f(d_T\left(x\right),d_T\left(y\right))$ over ${\mathcal{T}}_D$ if and only if T is constructible by Algorithm 1.

Algorithm 1: An algorithm referred in Lemma 2 [9].

(i) Add a new vertex, assign its desired degree value to $d_1$ and assign its availability value to $d_1$ as well. (ii) For $j=2,\dots ,n$, repeat the following steps until an output tree is reached. (1) Add a new vertex u and assign its desired degree and availability values both to $d_j$. (2) Let X be the set of the vertices different from u that have a positive availability. (3) Choose a vertex v from X so that this vertex has the greatest possible desired degree among the vertices from X. (4) Add an edge whose end-vertices are u and v , and decrease the availability of these two vertices by one.

Lemma 3

([9]). Suppose that $f:\mathbb{N}\times \mathbb{N}\to \mathbb{R}$ is a symmetric function having strict positive polarity, and that $T\in {\mathcal{T}}_D$ maximizes ${\sum }_{xy\in E\left(T\right)}f(d_T\left(x\right),d_T\left(y\right))$ over ${\mathcal{T}}_D$. If $u,v\in V\left(T\right)$ and w lies on the unique u–v path in T, then $d_T\left(w\right)\ge min(d_T\left(u\right),d_T\left(v\right)).$

Proposition 1.

Let $D=(d_1,\dots ,d_n)$ be a nonincreasing tree degree sequence, and let ${\mathcal{T}}_D$ be the family of its tree realizations. A tree $T\in {\mathcal{T}}_D$ minimizes the FIN index over ${\mathcal{T}}_D$ if and only if T is constructible by Algorithm 1.

Proof. 

We take $g(x,y)=-\varphi (x,y)$, where $\varphi$ is defined via (1). Then, by (4), g has strict positive polarity, and hence, from Lemma 2 it follows that a tree $T\in {\mathcal{T}}_D$ maximizes the graph invariant ${\sum }_{uv\in E\left(T\right)}g(d_T\left(u\right),d_T\left(v\right))$ over ${\mathcal{T}}_D$ if and only if T is constructible by Algorithm 1. Since

$$\sum _{uv\in E\left(T\right)}g(d_T\left(u\right),d_T\left(v\right))=-\mathcal{FIN}\left(T\right),$$

the desired conclusion follows. □

Lemma 4.

Let T be a $\mathcal{FIN}$-minimal n-vertex tree, let $u,v\in V\left(T\right)$, and let w lie on the unique u–v path. Then, $d_T\left(w\right)\ge min(d_T\left(u\right),d_T\left(v\right))$.

Proof. 

Let D be the degree sequence of T, and let ${\mathcal{T}}_D$ be the family of its tree realizations. Since ${\mathcal{T}}_D$ is a subfamily of the family of all n-vertex trees and T is $\mathcal{FIN}$-minimal, T also minimizes the FIN index over ${\mathcal{T}}_D$. Equivalently, for $g=-\varphi$, the tree T maximizes the graph invariant ${\sum }_{xy\in E\left(T\right)}g(d_T\left(x\right),d_T\left(y\right))$ over ${\mathcal{T}}_D$, where $\varphi$ is defined via (1). By (4), g has strict positive polarity, and hence, from Lemma 3 it follows that $d_T\left(w\right)\ge min(d_T\left(u\right),d_T\left(v\right))$. □

Corollary 1.

Let T be a $\mathcal{FIN}$-minimal n-vertex tree. Then, for every positive integer k not larger than the maximum degree of T, the subgraph $T_{\ge k}$ induced on the set $\{v\in V\left(T\right):d_T\left(v\right)\ge k\}$ is a tree. In particular, if T is rooted at a vertex of maximum degree, then degrees are nonincreasing along every path starting at the root.

Proof. 

If $u,v\in V\left(T_{\ge k}\right)$, then by Lemma 4, every vertex of the u–v path belongs to $V\left(T_{\ge k}\right)$. Hence, $T_{\ge k}$ is connected, and therefore, it is a tree. The second conclusion of the lemma follows also from Lemma 4. □

3. Vertices of Degree Two and the Branching Vertices

We first prove that the only tree without a branching vertex cannot be a $\mathcal{FIN}$-minimal n-vertex tree for every $n\ge 5$.

Lemma 5.

For every $n\ge 5$, the n-vertex path graph $P_n$ is not $\mathcal{FIN}$-minimal.

Proof. 

We take $P_n=v_1{v}_2\cdots v_n$, and form $T=P_n-v_3{v}_4+v_2{v}_4$. Then, we have

$$\mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(P_n\right)={\displaystyle \frac{4}{\sqrt{3}}}+\sqrt{{\displaystyle \frac{5}{6}}}-\sqrt{{\displaystyle \frac{3}{2}}}-2<0,$$

which yields the desired conclusion. □

Lemma 6.

Let T contain a path $uvw$ such that $d_T\left(u\right)=b\ge 3$, $d_T\left(v\right)=2$, and $d_T\left(w\right)=c\ge 2$. Then, $T^{\prime }=T-vw+uw$ is a tree satisfying $\mathcal{FIN}\left(T^{\prime }\right)<\mathcal{FIN}\left(T\right)$.

Proof. 

Here, we have

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)=& \sum _{x\in N_T\left(u\right)\setminus \left\{v\right\}}\left[\varphi (b,d_T\left(x\right))-\varphi (b+1,d_T\left(x\right))\right]\hfill \\ \hfill & +\varphi (b,2)+\varphi (2,c)-\varphi (b+1,1)-\varphi (b+1,c),\hfill \end{array}$$

(6)

where $\varphi$ is defined via (1). Since, by (2), each of the derivatives of $\varphi (b,y)-\varphi (b+1,y)$ and $\varphi (2,y)-\varphi (b+1,y)$ with respect to y is positive for real numbers $y\ge 1$, the aforementioned two functions are increasing in $y\ge 1$. Since $d_T\left(x\right)\ge 1$ and $c\ge 2$, from (6), it follows that

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)\ge & (b-1)\left[\varphi (b,1)-\varphi (b+1,1)\right]\hfill \\ \hfill & +\varphi (b,2)+\varphi (2,2)-\varphi (b+1,1)-\varphi (b+1,2).\hfill \end{array}$$

(7)

Now, we define a function $\psi$ as

$$\psi \left(x\right)=(x-1)\sqrt{1+{\displaystyle \frac{1}{x}}}+\sqrt{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{x}}}(x\ge 3).$$

The right-hand side of (7) is $\psi \left(b\right)-\psi (b+1)+1$, and hence, we have

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)\ge & \psi \left(b\right)-\psi (b+1)+1.\hfill \end{array}$$

(8)

Now, we show that ${\psi }^{\prime }\left(x\right)<1$. Using

$$\sqrt{1+{\displaystyle \frac{1}{x}}}-1={\displaystyle \frac{1}{x\left(\sqrt{1+x^{-1}}+1\right)}},$$

we note that the inequality ${\psi }^{\prime }\left(x\right)<1$ is equivalent to

$$\sqrt{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{x}}}\left((x+1)\sqrt{1+{\displaystyle \frac{1}{x}}}-(x-1)\right)<\sqrt{1+{\displaystyle \frac{1}{x}}}\left(\sqrt{1+{\displaystyle \frac{1}{x}}}+1\right).$$

(9)

We note that both sides of (9) are positive, and after the first squaring the resulting two sides are again positive. Hence, two successive squarings are reversible, and (9) is equivalent to

$$7x^4+8x^3-14x^2-8x-9>0.$$

(10)

For $x\ge 3$, the polynomial on the left-hand side of (10) is positive because its value and the value of its derivative are positive at $x=3$, while its derivative is increasing thereafter. Hence, we conclude that ${\psi }^{\prime }\left(x\right)<1$ for every $x\ge 3$. By the mean value theorem, we have $\psi (b+1)-\psi \left(b\right)<1$. Thus, (8) yields $\mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)>0$, as required. □

Corollary 2.

Let T be a $\mathcal{FIN}$-minimal n-vertex tree, where $n\ge 5$. Then, every vertex of degree 2 in T has one pendent neighbor and one branching neighbor. Consequently, in T,

(i) 

No two vertices of degree 2 are adjacent;

(ii) 

Every pendent path has length at most 2;

(iii) 

Every internal path has length 1.

Proof. 

Let $x\in V\left(T\right)$ be a vertex of degree 2. By Lemma 5, T has at least one branching vertex. We consider a path P from x, through vertices of degree 2, until first reaching a branching vertex u. Let v be the neighbor of u on this path P. Then, $d_T\left(v\right)=2$. If $v\ne x$, then the other neighbor of v has degree 2, and hence, by Lemma 6 there exists an n-vertex tree $T^{\prime }$ satisfying $\mathcal{FIN}\left(T^{\prime }\right)<\mathcal{FIN}\left(T\right)$, which contradicts the minimality of T. If $v=x$ and the other neighbor of v is not pendent, then again by Lemma 6, we obtain a contradiction. Hence, $v=x$ and the other neighbor of x is pendent. This proves that every vertex of degree 2 in T has one pendent neighbor and one branching neighbor. Now, statements (i)–(iii) follows directly from the fact that we just proved. □

Proposition 2.

Let T be a $\mathcal{FIN}$-minimal n-vertex tree, where $n\ge 5$. Let $B\left(T\right)$ be the set of all branching vertices of T. Then, the following statements hold.

(i) 

The induced subgraph $C=T\left[B\right(T\left)\right]$ is a tree.

(ii) 

The tree T is obtained from C by attaching to its vertices only pendent edges, pendent paths of length 2, or both. More precisely, for $v\in V\left(C\right)$, let $c_v=d_C\left(v\right)$, and let $p_v$ and $q_v$ be the numbers of pendent edges and pendent paths of length 2, respectively, attached at v. Then, $d_T\left(v\right)=c_v+p_v+q_v$,

$$n_2\left(T\right)=\sum _{v\in V\left(C\right)}{q}_v,\mathit{and}{n}_1\left(T\right)=\sum _{v\in V\left(C\right)}(p_v+q_v).$$

(11)

(iii) 

The inequalities $n_2\left(T\right)\le n_1\left(T\right)$ and $\mathrm{diam}\left(T\right)\le \mathrm{diam}\left(C\right)+4$ hold. Equality $n_2\left(T\right)=n_1\left(T\right)$ holds if and only if $p_v=0$ for every $v\in V\left(C\right)$.

Proof. 

(i). By Lemma 5, $B\left(T\right)$ is nonempty. If $u,v\in B\left(T\right)$, then no internal vertex of the u–v path P of T can have degree 1 in T, and from Corollary 2 it follows that no such vertex has degree 2 in T. Hence, every vertex of P belongs to $B\left(T\right)$, and thus, C is connected, and therefore, C is a tree.

(ii).

This part follows from Corollary 2.

(iii).

The inequality $n_2\left(T\right)\le n_1\left(T\right)$ follows from (11), where the equality holds if and only if ${\sum }_{v\in V\left(C\right)}{p}_v=0$, or, equivalently, if and only if $p_v=0$ for every $v\in V\left(C\right)$. Finally, the inequality $\mathrm{diam}\left(T\right)\le \mathrm{diam}\left(C\right)+4$ follows from the first sentence of part (ii). □

4. Vertices of Maximum Degree

We start this section with the following lemma concerning an inequality involving $\varphi$.

Lemma 7.

For integers k and ℓ satisfying $k\ge 3$ and $1\le \mathsf{\ell }<k$, let

$$\begin{array}{cc}\hfill {\Psi }_k(\mathsf{\ell })=& 2\varphi (k,k)-\varphi (k,k+1)-\varphi (k+1,k-1)+(k-1)\left[\varphi (k,\mathsf{\ell })-\varphi (k+1,\mathsf{\ell })\right]\hfill \\ \hfill & +(k-2)\left[\varphi (k,\mathsf{\ell })-\varphi (k-1,\mathsf{\ell })\right],\hfill \end{array}$$

where ϕ is defined via (1). Then, ${\Psi }_k(\mathsf{\ell })>0$.

Proof. 

Let $s={\displaystyle \frac{k}{\mathsf{\ell }}}+1$. Then, $s>2$. Multiplying ${\Psi }_k(\mathsf{\ell })$ by $\sqrt{k}$ and simplifying gives

$$\begin{array}{cc}\hfill \sqrt{k}{\Psi }_k(\mathsf{\ell })=& (k-1)\left[\sqrt{s}-\sqrt{s-{\displaystyle \frac{1}{k+1}}}\right]+(k-2)\left[\sqrt{s}-\sqrt{s+{\displaystyle \frac{1}{k-1}}}\right]\hfill \\ \hfill & +2\sqrt{2}-\sqrt{2-{\displaystyle \frac{1}{k+1}}}-\sqrt{{\displaystyle \frac{2}{k^2-1}}+2}.\hfill \end{array}$$

(12)

We note that

$$\sqrt{2}-\sqrt{2-{\displaystyle \frac{1}{k+1}}}={\displaystyle \frac{1/(k+1)}{\sqrt{2}+\sqrt{2-\frac{1}{k+1}}}}>{\displaystyle \frac{1}{2\sqrt{2}(k+1)}}$$

(13)

and

$$\sqrt{2+{\displaystyle \frac{2}{k^2-1}}}-\sqrt{2}={\displaystyle \frac{2/(k^2-1)}{\sqrt{2+\frac{2}{k^2-1}}+\sqrt{2}}}<{\displaystyle \frac{1}{\sqrt{2}(k^2-1)}}.$$

(14)

From (13) and (14), it follows that

$$2\sqrt{2}-\sqrt{2-{\displaystyle \frac{1}{k+1}}}-\sqrt{2+{\displaystyle \frac{2}{k^2-1}}}>{\displaystyle \frac{k-3}{2\sqrt{2}(k^2-1)}}.$$

(15)

Similarly, we have

$$\sqrt{s}-\sqrt{s-{\displaystyle \frac{1}{k+1}}}={\displaystyle \frac{1/(k+1)}{\sqrt{s}+\sqrt{s-\frac{1}{k+1}}}}>{\displaystyle \frac{1}{2(k+1)\sqrt{s}}}$$

(16)

and

$$\sqrt{s}-\sqrt{s+{\displaystyle \frac{1}{k-1}}}=-{\displaystyle \frac{1/(k-1)}{\sqrt{s}+\sqrt{s+\frac{1}{k-1}}}}>-{\displaystyle \frac{1}{2(k-1)\sqrt{s}}}.$$

(17)

From (16) and (17), we have

$$\begin{array}{cc}\hfill & (k-1)\left[\sqrt{s}-\sqrt{s-{\displaystyle \frac{1}{k+1}}}\right]+(k-2)\left[\sqrt{s}-\sqrt{s+{\displaystyle \frac{1}{k-1}}}\right]>-{\displaystyle \frac{k-3}{2(k^2-1)\sqrt{s}}}.\hfill \end{array}$$

(18)

Now, using (15) and (18), we obtain

$$\sqrt{k}{\Psi }_k(\mathsf{\ell })>{\displaystyle \frac{k-3}{2(k^2-1)}}\left({\displaystyle \frac{1}{\sqrt{2}}}-{\displaystyle \frac{1}{\sqrt{s}}}\right)\ge 0.$$

for every $k\ge 3$ because $s>2$. □

Theorem 1.

Every $\mathcal{FIN}$-minimal n-vertex tree T has at most two vertices of maximum degree. If T has two such vertices, then they are adjacent.

Proof. 

The conclusion is certainly true for $n\le 4$. In the rest of the proof, we assume that $n\ge 5$. Let $\Delta =\Delta \left(T\right)$. Then, by Lemma 5, we have $\Delta \ge 3$. From Corollary 1, it follows that the vertices of degree $\Delta$ induce a tree, say $T_{\Delta }$.

We claim that $T_{\Delta }$ has at most two vertices. Contrarily, we suppose that $T_{\Delta }$ has at least three vertices. Let v be a pendent vertex of $T_{\Delta }$. Let $u\in V\left(T_{\Delta }\right)$ be the unique neighbor of v in $T_{\Delta }$, and let $w\ne v$ be another neighbor of u in $T_{\Delta }$. Since $d_T\left(u\right)=d_T\left(v\right)=\Delta$, those edges incident with u and v that do not belong to the path $wuv$ of T may be interchanged without changing the FIN index; more precisely, if $a\in N_T\left(u\right)\setminus \{w,v\}$ and $b\in N_T\left(v\right)\setminus \left\{u\right\}$, then replacing the edges $ua$ and $vb$ by $ub$ and $va$ gives another tree with the same degree sequence and the same FIN index. Keeping all neighbors of u of degree $\Delta$ attached to u, we may therefore interchange the remaining edges from those edges incident with u and v that do not belong to the path $wuv$ of T, if necessary, and write $N_T\left(u\right)\setminus \{w,v\}=\{z_1,\dots ,z_{\Delta -2}\}$ and $N_T\left(v\right)\setminus \left\{u\right\}=\{x,y_1,\dots ,y_{\Delta -2}\}$, provided that, by taking $\mathsf{\ell }=d_T\left(x\right)$, it holds that

$$d_T\left(z_i\right)\ge \mathsf{\ell }\ge d_T\left(y_j\right)\mathrm{for}\mathrm{all}i,j\in \{1,\dots ,\Delta -2\}.$$

(19)

Since v remains pendent in $T_{\Delta }$ after the aforementioned interchanges, every vertex in $N_T\left(v\right)\setminus \left\{u\right\}$ has degree less than $\Delta$. Hence, $1\le \mathsf{\ell }<\Delta$. Let $T^{\prime }=T-vx+ux$. By (2), the function $\varphi (\Delta ,t)-\varphi (\Delta +1,t)$ is increasing in $t\ge 1$, whereas $\varphi (\Delta ,t)-\varphi (\Delta -1,t)$ is decreasing in $t\ge 1$. Consequently, by keeping in mind (19), $1\le \mathsf{\ell }<\Delta$ with $\Delta \ge 3$, and Lemma 7, we obtain

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)=& 2\varphi (\Delta ,\Delta )-\varphi (\Delta ,\Delta +1)-\varphi (\Delta +1,\Delta -1)+\varphi (\Delta ,\mathsf{\ell })-\varphi (\Delta +1,\mathsf{\ell })\hfill \\ \hfill & +{\displaystyle \sum _{i=1}^{\Delta -2}}\left[\varphi (\Delta ,d_T\left(z_i\right))-\varphi (\Delta +1,d_T\left(z_i\right))\right]\hfill \\ \hfill & +{\displaystyle \sum _{j=1}^{\Delta -2}}\left[\varphi (\Delta ,d_T\left(y_j\right))-\varphi (\Delta -1,d_T\left(y_j\right))\right]\hfill \\ \hfill \ge & {\Psi }_{\Delta }(\mathsf{\ell })>0,\hfill \end{array}$$

which contradicts the $\mathcal{FIN}$-minimality of T. Hence, $T_{\Delta }$ has at most two vertices. Consequently, T has at most two vertices of degree $\Delta$. Since $T_{\Delta }$ is connected, any two vertices of degree $\Delta$ (if they exist) in T must be adjacent. □

Lemma 8.

Let T be a $\mathcal{FIN}$-minimal n-vertex tree rooted at a vertex r of maximum degree Δ, where $n\ge 5$. If $u\in V\left(T\right)$ is the parent of a vertex v, then $d_T\left(u\right)>d_T\left(v\right)$, with the only possible exception that $u=r$ and $d_T\left(u\right)=d_T\left(v\right)=\Delta$. Consequently, at most one edge of T has endvertices of equal degree, and such an edge, if it exists, joins the two vertices of maximum degree.

Proof. 

By Corollary 1, the vertex degrees are nonincreasing along every path starting at r. Suppose that $d_T\left(u\right)=d_T\left(v\right)=k$, and choose such an edge $uv$ as far from r as possible. Since u is the parent of v, we have $k\ge 2$, and the case $k=2$ is excluded by Corollary 2. Hence, $k\ge 3$. Also, we note that every child of v has degree less than k, by the choice of $uv$.

Suppose that $u\ne r$, and let w be the parent of u. As in the proof of Theorem 1, branches incident with u and v away from the path $wuv$ may be interchanged without changing the FIN index. Keeping all neighbors of u of degree k attached to u, we may, after such interchanges, if necessary, write $N_T\left(u\right)\setminus \{w,v\}=\{z_1,\dots ,z_{k-2}\}$ and $N_T\left(v\right)\setminus \left\{u\right\}=\{x,y_1,\dots ,y_{k-2}\}$ so that, with $\mathsf{\ell }=d_T\left(x\right)$,

$$d_T\left(z_i\right)\ge \mathsf{\ell }\ge d_T\left(y_j\right)\mathrm{for}\mathrm{all}i,j\in \{1,\dots ,k-2\}.$$

(20)

Since every child of v has degree less than k, we have $1\le \mathsf{\ell }<k$. Let $T^{\prime }=T-vx+ux$. Then, we have

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)=& \varphi (d_T\left(w\right),k)-\varphi (d_T\left(w\right),k+1)\hfill \\ \hfill & +\varphi (k,k)-\varphi (k+1,k-1)+\varphi (k,\mathsf{\ell })-\varphi (k+1,\mathsf{\ell })\hfill \\ \hfill & +{\displaystyle \sum _{i=1}^{k-2}}\left[\varphi (k,d_T\left(z_i\right))-\varphi (k+1,d_T\left(z_i\right))\right]\hfill \\ \hfill & +{\displaystyle \sum _{j=1}^{k-2}}\left[\varphi (k,d_T\left(y_j\right))-\varphi (k-1,d_T\left(y_j\right))\right].\hfill \end{array}$$

(21)

Since the degrees are nonincreasing along rooted paths, we have $d_T\left(w\right)\ge k$. Also, by (2), the function $\varphi (t,k)-\varphi (t,k+1)$ is increasing in $t\ge 1$, whereas $\varphi (k,t)-\varphi (k-1,t)$ is decreasing in $t\ge 1$. Consequently (as in the proof of Theorem 1), by keeping in mind (20), $1\le \mathsf{\ell }<k$ with $k\ge 3$, and Lemma 7, from (21) we obtain $\mathcal{FIN}\left(T\right)-\mathcal{FIN}\left(T^{\prime }\right)\ge {\Psi }_k(\mathsf{\ell })>0$. This contradicts the $\mathcal{FIN}$-minimality of T. Thus, $u=r$.

Therefore, every edge of T whose endvertices have equal degree must be incident with r. Since r has degree $\Delta$ and the degrees are nonincreasing along rooted paths, such an edge must have both endvertices of degree $\Delta$. By Theorem 1, there are at most two vertices of degree $\Delta$, and hence at most one such edge. □

We next establish an upper bound on the maximimum degree of a $\mathcal{FIN}$-minimal n-vertex tree in terms of n. To do so, we need the following lemma first.

Lemma 9.

For real numbers $d\ge 2$, define ${\eta }_d\left(y\right)=\varphi (d-1,y)-\varphi (d,y)$, where ϕ is defined via (1). Then, the following statements hold.

(i) 

The function ${\eta }_d$ is positive, strictly increasing, and strictly concave on $[1,\infty )$.

(ii) 

If ${\alpha }_d=3-{(d-2)}^{-1}$, then, for every integer $d\ge 5$,

$$A_d:=\varphi (d-1,2)+\varphi (2,1)-2\varphi (d,1)+(d-2){\eta }_d\left({\alpha }_d\right)<0.$$

(iii) 

If ${\beta }_d=3-2{(d-3)}^{-1}$, then, for every integer $d\ge 18$,

$$\begin{array}{cc}\hfill B_d:=& 2\varphi (d-1,3)+4\varphi (3,1)-3\varphi (d,2)-3\varphi (2,1)+(d-3){\eta }_d\left({\beta }_d\right)<0.\hfill \end{array}$$

Proof. 

(i). The positivity of ${\eta }_d$ follows from the fact that $\varphi$ is strictly decreasing in each of its variables. Also, by (2), we have ${\eta }_d^{\prime }\left(y\right)={\varphi }_y(d-1,y)-{\varphi }_y(d,y)>0$, and hence, ${\eta }_d$ is strictly increasing. Moreover, the mixed third-order partial derivative ${\varphi }_{xyy}$ satisfies

$${\varphi }_{xyy}(x,y)={\displaystyle \frac{x+4y}{8\sqrt{x}{y}^{3/2}{(x+y)}^{5/2}}}>0,$$

and hence, we have ${\eta }_d^{″}\left(y\right)={\varphi }_{yy}(d-1,y)-{\varphi }_{yy}(d,y)<0$. Thus, ${\eta }_d$ is also strictly concave.

(ii).

We note that

$${\eta }_d\left(y\right)={\displaystyle \frac{1}{d(d-1)\left(\varphi (d-1,y)+\varphi (d,y)\right)}},$$

which gives

$${\eta }_d\left({\alpha }_d\right)<{\displaystyle \frac{1}{2d(d-1)\varphi (d,{\alpha }_d)}}$$

(22)

for $d\ge 6$, because $\varphi (d-1,{\alpha }_d)>\varphi (d,{\alpha }_d)$. Using ${\alpha }_d=(3d-7)/(d-2)$, we have

$${\left[\varphi (d,{\alpha }_d)\right]}^2-{\displaystyle \frac{25d^2}{3{(5d-6)}^2}}={\displaystyle \frac{70d^3-597d^2+1368d-756}{3d(3d-7){(5d-6)}^2}}>0,$$

(23)

for $d\ge 6$. Therefore, from (22) and (23), it follows that

$${\eta }_d\left({\alpha }_d\right)<{\displaystyle \frac{\sqrt{3}(5d-6)}{10d^2(d-1)}}$$

for $d\ge 6$, and hence,

$$A_d<\tilde{A}\left(d\right),$$

(24)

where, for $d\ge 6$,

$$\tilde{A}\left(d\right)=\varphi (d-1,2)+\varphi (2,1)-2\varphi (d,1)+{\displaystyle \frac{\sqrt{3}(d-2)(5d-6)}{10d^2(d-1)}}.$$

For $d\ge 6$, we note that $\varphi (d-1,2)<{\displaystyle \frac{6}{7}}$ and $\varphi (d,1)>1$, and therefore, the derivative function ${\tilde{A}}^{\prime }\left(d\right)$ of $\tilde{A}\left(d\right)$ satisfies

$$\begin{array}{cc}\hfill {\tilde{A}}^{\prime }\left(d\right)=& -{\displaystyle \frac{1}{2{(d-1)}^2\varphi (d-1,2)}}+{\displaystyle \frac{1}{d^2\varphi (d,1)}}-{\displaystyle \frac{\sqrt{3}(5d^3-32d^2+52d-24)}{10d^3{(d-1)}^2}}\hfill \\ \hfill <& -{\displaystyle \frac{7}{12{(d-1)}^2}}+{\displaystyle \frac{1}{d^2}}-{\displaystyle \frac{5d^3-32d^2+52d-24}{6d^3{(d-1)}^2}}\hfill \\ \hfill =& -{\displaystyle \frac{(d-4)(5d^2-20d+12)}{12d^3{(d-1)}^2}}<0,\hfill \end{array}$$

where we used $\sqrt{3}>5/3$ and $5d^3-32d^2+52d-24>0$ (for $d\ge 6$); the latter inequality follows by writing $d=z+6$. Thus, $\tilde{A}$ is strictly decreasing on $[6,\infty )$. Also, we note that $A_5<0$ and $\tilde{A}\left(6\right)<0$. Consequently, from (24), it follows that $A_d<0$ for every integer $d\ge 5$.

(iii).

In the rest of the proof, we assume that $d\ge 18$. Since ${\beta }_d=(3d-11)/(d-3)$, we have

$${\left[\varphi (d,{\beta }_d)\right]}^2-{\displaystyle \frac{25d^2}{3{(5d-8)}^2}}={\displaystyle \frac{35d^3-633d^2+2640d-2112}{3d(3d-11){(5d-8)}^2}}>0.$$

Hence, similar to part (ii), we have

$${\eta }_d\left({\beta }_d\right)<{\displaystyle \frac{\sqrt{3}(5d-8)}{10d^2(d-1)}}.$$

Therefore, it holds that

$$B_d<\tilde{B}\left(d\right),$$

(25)

where

$$\tilde{B}\left(d\right)=2\varphi (d-1,3)+4\varphi (3,1)-3\varphi (d,2)-3\varphi (2,1)+{\displaystyle \frac{\sqrt{3}(d-3)(5d-8)}{10d^2(d-1)}}.$$

For $d\ge 18$, we note that

$${\displaystyle \frac{1}{\varphi (d-1,3)}}>{\displaystyle \frac{19}{12}}\mathrm{and}{\displaystyle \frac{1}{\varphi (d,2)}}<{\displaystyle \frac{17}{12}},$$

and hence, the derivative function ${\tilde{B}}^{\prime }\left(d\right)$ of $\tilde{B}\left(d\right)$ satisfies

$$\begin{array}{cc}\hfill {\tilde{B}}^{\prime }\left(d\right)& =-{\displaystyle \frac{1}{{(d-1)}^2\varphi (d-1,3)}}+{\displaystyle \frac{3}{2d^2\varphi (d,2)}}-{\displaystyle \frac{\sqrt{3}(5d^3-46d^2+95d-48)}{10d^3{(d-1)}^2}}\hfill \\ \hfill & <-{\displaystyle \frac{19}{12{(d-1)}^2}}+{\displaystyle \frac{17}{8d^2}}-{\displaystyle \frac{6(5d^3-46d^2+95d-48)}{35d^3{(d-1)}^2}}\hfill \\ \hfill & ={\displaystyle \frac{-265d^3+3054d^2-11895d+6912}{840d^3{(d-1)}^2}}<0,\hfill \end{array}$$

where we also replaced the number $-\sqrt{3}$ with the larger number $-{\displaystyle \frac{12}{7}}$ in the above inequality (because $5d^3-46d^2+95d-48>0$). Thus, $\tilde{B}$ is strictly decreasing on $[18,\infty )$. We observe that $\tilde{B}\left(18\right)<0$. Therefore, from (25) it follows that $B_d<0$ for every integer $d\ge 18$. □

Theorem 2.

If T is a $\mathcal{FIN}$-minimal n-vertex tree, then

$$\Delta \left(T\right)\le max\left(17,⌊{\displaystyle \frac{n+3}{3}}⌋\right),$$

(26)

where $n\ge 5$. Particularly, for $n\ge 48$, it holds that

$$\Delta \left(T\right)\le ⌊{\displaystyle \frac{n+3}{3}}⌋.$$

Proof. 

Let $\Delta =\Delta \left(T\right)$, and let $u\in V\left(T\right)$ be a vertex with $d_T\left(u\right)=\Delta$. Let $T_1,\dots ,T_{\Delta }$ be the components of $T-u$. For every $i\in \{1,\dots ,\Delta \}$, let $v_i$ be the neighbor of u in $T_i$, and take $s_i=\left|V\left(T_i\right)\right|$ and $a_i=d_T\left(v_i\right)$. Clearly, $a_i\le s_i$ holds for every $i\in \{1,\dots ,\Delta \}$.

Contrarily, we assume that $\Delta \ge 18$ and $\Delta >(n+3)/3$. Then,

$$n\le 3\Delta -4.$$

(27)

Let $c_1$ and $c_2$ denote the numbers of components of $T-u$ of orders 1 and 2, respectively. Then, we have

$$n-1={\displaystyle \sum _{i=1}^{\Delta }}{s}_i\ge c_1+2c_2+3(\Delta -c_1-c_2),$$

which together with (27), yields

$$2c_1+c_2\ge 5.$$

(28)

Firstly, we consider the case where $c_1\ge 2$. Let $v,w\in V\left(T\right)$ be two pendent neighbors of u. Let $T^{\prime }=T-uw+vw$. Without loss of generality, we assume that $v=v_{\Delta -1}$ and $w=v_{\Delta }$. Since $a_i\le s_i$ for every $i\in \{1,\dots ,\Delta -2\}$, we have

$${\displaystyle \sum _{i=1}^{\Delta -2}}{a}_i\le {\displaystyle \sum _{i=1}^{\Delta -2}}{s}_i=n-3\le 3\Delta -7,$$

where we also used (27). Thus, we have

$${\displaystyle \frac{1}{\Delta -2}}{\displaystyle \sum _{i=1}^{\Delta -2}}{a}_i\le {\displaystyle \frac{3\Delta -7}{\Delta -2}}={\alpha }_{\Delta },$$

(29)

where ${\alpha }_{\Delta }$ is defined in Lemma 9(ii). Since ${\eta }_{\Delta }$, defined in Lemma 9, is increasing and concave (see Lemma 9(i)), by using Jensen’s inequality and (29), we have

$${\displaystyle \frac{1}{\Delta -2}}{\displaystyle \sum _{i=1}^{\Delta -2}}{\eta }_{\Delta }\left(a_i\right)\le {\eta }_{\Delta }\left({\displaystyle \frac{1}{\Delta -2}}{\displaystyle \sum _{i=1}^{\Delta -2}}{a}_i\right)\le {\eta }_{\Delta }\left({\alpha }_{\Delta }\right),$$

which yields

$${\displaystyle \sum _{i=1}^{\Delta -2}}{\eta }_{\Delta }\left(a_i\right)\le (\Delta -2){\eta }_{\Delta }\left({\alpha }_{\Delta }\right).$$

(30)

Now, using (30) and Lemma 9(ii), we have

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T^{\prime }\right)-\mathcal{FIN}\left(T\right)=& \varphi (\Delta -1,2)+\varphi (2,1)-2\varphi (\Delta ,1)+{\displaystyle \sum _{i=1}^{\Delta -2}}{\eta }_{\Delta }\left(a_i\right)\hfill \\ \hfill \le & \varphi (\Delta -1,2)+\varphi (2,1)-2\varphi (\Delta ,1)+(\Delta -2){\eta }_{\Delta }\left({\alpha }_{\Delta }\right)\hfill \\ \hfill =& A_{\Delta }<0,\hfill \end{array}$$

a contradiction to the $\mathcal{FIN}$-minimality of T.

Next, we consider the case where $c_1\le 1$. Then, (28) gives $c_2\ge 3$. We choose three components of order 2, say $T_1,T_2,T_3$. For every $i\in \{1,2,3\}$, let $V\left(T_i\right)=\{v_i,w_i\}$, where $d_T\left(v_i\right)=2$ and $d_T\left(w_i\right)=1$. Let $T^{″}=T-\{u{v}_3,v_3{w}_3\}+\{v_1{v}_3,v_2{w}_3\}$. We recall that $a_4,\dots ,a_{\Delta }$ are the degrees in T of the remaining neighbors of u. Since $a_i\le s_i$ for every $i\in \{4,\dots ,\Delta \}$, we have

$${\displaystyle \sum _{i=4}^{\Delta }}{a}_i\le {\displaystyle \sum _{i=4}^{\Delta }}{s}_i=n-7\le 3\Delta -11,$$

where we also used (27). Thus, we have

$${\displaystyle \frac{1}{\Delta -3}}{\displaystyle \sum _{i=4}^{\Delta }}{a}_i\le {\displaystyle \frac{3\Delta -11}{\Delta -3}}={\beta }_{\Delta },$$

(31)

where ${\beta }_{\Delta }$ is defined in Lemma 9(iii). Since ${\eta }_{\Delta }$ is increasing and concave (see Lemma 9(i)), by using Jensen’s inequality and (31), we have

$${\displaystyle \frac{1}{\Delta -3}}{\displaystyle \sum _{i=4}^{\Delta }}{\eta }_{\Delta }\left(a_i\right)\le {\eta }_{\Delta }\left({\displaystyle \frac{1}{\Delta -3}}{\displaystyle \sum _{i=4}^{\Delta }}{a}_i\right)\le {\eta }_{\Delta }\left({\beta }_{\Delta }\right),$$

which yields

$${\displaystyle \sum _{i=4}^{\Delta }}{\eta }_{\Delta }\left(a_i\right)\le (\Delta -3){\eta }_{\Delta }\left({\beta }_{\Delta }\right).$$

(32)

Now, using (32) and Lemma 9(iii), we have

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(T^{″}\right)-\mathcal{FIN}\left(T\right)=& 2\varphi (\Delta -1,3)+4\varphi (3,1)-3\varphi (\Delta ,2)-3\varphi (2,1)+{\displaystyle \sum _{i=4}^{\Delta }}{\eta }_{\Delta }\left(a_i\right)\hfill \\ \hfill \le & 2\varphi (\Delta -1,3)+4\varphi (3,1)-3\varphi (\Delta ,2)-3\varphi (2,1)+(\Delta -3){\eta }_{\Delta }\left({\beta }_{\Delta }\right)\hfill \\ \hfill =& B_{\Delta }<0,\hfill \end{array}$$

again a contradiction to the $\mathcal{FIN}$-minimality of T.

Thus, we have either $\Delta \le 17$ or

$$\Delta \le ⌊{\displaystyle \frac{n+3}{3}}⌋,$$

which proves (26). The other inequality follows from the fact that $⌊(n+3)/3⌋\ge 17$ for every $n\ge 48$. □

5. Bounds Concerning Degree Set and Diameter

We start this section by recalling a known result.

Lemma 10

([11]). Let S be a set of s integers $a_1,a_2,\cdots ,a_s$ such that $1=a_1<a_2<\cdots <a_s$. Then there exists a tree whose degree set is S, and the minimum order of such a tree is ${\sum }_{i=1}^s(a_i-1)+2.$

In the following theorem, using Lemma 10, we first establish an upper bound for the number of elements in a degree set $\mathcal{D}\left(T\right)$ of any nontrivial n-vertex tree T. Then, using Theorem 1, the stronger forms of this upper bound become useful for any $\mathcal{FIN}$-minimal n-vertex tree.

Theorem 3.

Let T be an n-vertex tree, where $n\ge 2$. Let $s=\left|\mathcal{D}\right(T\left)\right|$ and $\Delta =\Delta \left(T\right)$. Then,

$$s\le ⌊{\displaystyle \frac{1+\sqrt{8n-15}}{2}}⌋.$$

(33)

Also, if T has exactly μ vertices of maximum degree $\Delta =\Delta \left(T\right)\ge 2$, then

$$s\le ⌊{\displaystyle \frac{3+\sqrt{8n-8\mu (\Delta -1)-15}}{2}}⌋.$$

(34)

Moreover, if T is a $\mathcal{FIN}$-minimal n-vertex tree and $n\ge 3$, then either

$${\displaystyle s\le ⌊\frac{3+\sqrt{8(n-\Delta )-7}}{2}⌋\mathit{or}s\le ⌊\frac{3+\sqrt{8n-16\Delta +1}}{2}⌋.}$$

Proof. 

Let $\mathcal{D}\left(T\right)=\{a_1,a_2,\cdots ,a_s\}$ such that $1=a_1<a_2<\cdots <a_s$. Since $a_i\ge i$ for every $i\in \{1,\dots ,s\}$, from Lemma 10, it follows that

$$n\ge {\displaystyle \sum _{i=1}^s}(a_i-1)+2\ge {\displaystyle \sum _{i=1}^s}(i-1)+2={\displaystyle \frac{s(s-1)}{2}}+2,$$

which yields (33).

For establishing (34), we suppose that T has exactly $\mu$ vertices of maximum degree $\Delta \ge 2$. If $s=2$, then $\mathcal{D}\left(T\right)=\{1,\Delta \}$, and hence, the identity

$$n_1\left(T\right)=\sum _{\begin{array}{c}v\in V\left(T\right)\\ d_T\left(v\right)\ge 3\end{array}}\left(d_T\left(v\right)-2\right)+2,$$

(35)

yields $n_1\left(T\right)=\mu (\Delta -2)+2,$ which implies $n=n_1\left(T\right)+\mu =\mu (\Delta -1)+2.$ Therefore,

$${\displaystyle \frac{3+\sqrt{8n-8\mu (\Delta -1)-15}}{2}}=2=s,$$

and hence, (34) holds when $s=2$. In what follows, we suppose that $s\ge 3$. Then, we have $\mathcal{D}\left(T\right)=\{1,a_2,\dots ,a_{s-1},\Delta \},$ where $2\le a_2<\cdots <a_{s-1}<\Delta$. We note that $a_i\ge i$ for every $i\in \{2,\dots ,s-1\}$. Since T has $\mu$ vertices of degree $\Delta$ and at least one vertex of each of the degrees $a_2,\dots ,a_{s-1}$, identity (35) gives

$$\begin{array}{cc}\hfill n_1\left(T\right)& \ge \mu (\Delta -2)+{\displaystyle \sum _{i=2}^{s-1}}(a_i-2)+2\ge \mu (\Delta -2)+{\displaystyle \sum _{i=2}^{s-1}}(i-2)+2\hfill \\ \hfill & =\mu (\Delta -2)+{\displaystyle \frac{(s-3)(s-2)}{2}}+2.\hfill \end{array}$$

(36)

On the other hand, T has at least $s-2$ vertices of degrees $a_2,\dots ,a_{s-1}$ and exactly $\mu$ vertices of degree $\Delta$. Hence, T has at least $s+\mu -2$ non-pendent vertices. Consequently, using (36), we have

$$\begin{array}{cc}\hfill n& \ge n_1\left(T\right)+s+\mu -2\ge \mu (\Delta -1)+{\displaystyle \frac{s^2-3s+6}{2}},\hfill \end{array}$$

which implies $s^2-3s+6\le 2n-2\mu (\Delta -1)$, and hence,

$$s\le {\displaystyle \frac{3+\sqrt{8n-8\mu (\Delta -1)-15}}{2}},$$

which yields (34) because s is an integer.

If T is a $\mathcal{FIN}$-minimal n-vertex tree and $n\ge 3$, then from Theorem 1 it follows that $\mu \in \{1,2\}$, and hence, the final two inequalities are obtained from (34). □

Theorem 4.

Let T be a $\mathcal{FIN}$-minimal n-vertex tree, where $n\ge 5$. Let $s=\left|\mathcal{D}\right(T\left)\right|$.

(i) 

If T has a unique vertex of maximum degree, then $\mathrm{diam}\left(T\right)\le 2s-2$.

(ii) 

If T has two vertices of maximum degree, then $\mathrm{diam}\left(T\right)\le 2s-1$.

In both cases, it holds that

$$\mathrm{diam}\left(T\right)\le min\left(⌊{\displaystyle \frac{n}{2}}⌋+1,2⌊{\displaystyle \frac{1+\sqrt{8n-15}}{2}}⌋-1\right).$$

(37)

More sharply, if μ denotes the number of maximum-degree vertices of T and if $\Delta =\Delta \left(T\right)$, then the second term on the right-hand side of (37) may be replaced by

$$\left\{\begin{array}{cc}{\displaystyle 2⌊{\displaystyle \frac{3+\sqrt{8(n-\Delta )-7}}{2}}⌋-2}\hfill & \mathit{when}\mu =1,\hfill \\ {\displaystyle 2⌊{\displaystyle \frac{3+\sqrt{8n-16\Delta +1}}{2}}⌋-1}\hfill & \mathit{when}\mu =2.\hfill \end{array}\right.$$

Proof. 

We root the tree T at a vertex r of maximum degree $\Delta$. By Lemma 8, the vertex degrees are strictly decreasing along every path starting at r, with the only possible exception that the first two vertices of such a path both have degree $\Delta$.

(i).

Suppose that r is the unique vertex of degree $\Delta$ in T. Then, by Lemma 8, the degrees are strictly decreasing along every path starting at r. Since T has exactly s distinct vertex degrees, every such path has length at most $s-1$. Hence, for any two vertices $x,y\in V\left(T\right)$, we have

$$d_T(x,y)\le d_T(x,r)+d_T(r,y)\le 2(s-1)=2s-2,$$

which implies that $\mathrm{diam}\left(T\right)\le 2s-2$, which proves (i).

(ii).

Suppose that T has two vertices of degree $\Delta$. By Theorem 1, they are adjacent. Let $r^{\prime }\in V\left(T\right)$ be the second vertex of degree $\Delta$. Thus, $r^{\prime }$ is a child of r. By Lemma 8, every path starting at r and not passing through $r^{\prime }$ has length at most $s-1$, whereas every path starting at r and passing through $r^{\prime }$ has length at most s. Moreover, every path starting at $r^{\prime }$ and contained in the subtree rooted at $r^{\prime }$ has length at most $s-1$, since the degrees strictly decrease after $r^{\prime }$. Let $x,y\in V\left(T\right)$.

If x and y both belong to the subtree rooted at $r^{\prime }$, then

$$d_T(x,y)\le d_T(x,r^{\prime })+d_T(r^{\prime },y)\le 2s-2.$$

If neither x nor y belongs to the subtree rooted at $r^{\prime }$, then

$$d_T(x,y)\le d_T(x,r)+d_T(r,y)\le 2s-2.$$

Finally, if exactly one of x and y belongs to the subtree rooted at $r^{\prime }$, then

$$d_T(x,y)\le s+(s-1)=2s-1.$$

In every case, we have $d_T(x,y)\le 2s-1$, and hence, it holds that $\mathrm{diam}\left(T\right)\le 2s-1$, which proves (ii).

We next establish the first term on the right-hand side of (37). Let $P:v_0{v}_1\cdots v_d$ be a diametral path of T, where $d=\mathrm{diam}\left(T\right)$. If $d\le 3$, then we have $d\le ⌊n/2⌋+1$ because $n\ge 5$. Suppose now that $d\ge 4$. By Corollary 2, it holds that $d_T\left(v_i\right)\ge 3$ for every $i\in \{2,\dots ,d-2\}$; otherwise, we have $d_T\left(v_i\right)=2$ for some $i\in \{2,\dots ,d-2\}$, and hence, by Corollary 2, either $v_{i-1}$ or $v_{i+1}$ is pendent, which is impossible. Therefore, each of the $d-3$ vertices $v_2,\dots ,v_{d-2}$ has a neighbor outside P. These $d-3$ neighbors are distinct. Hence, we have $n\ge (d+1)+(d-3)=2d-2$, which yields $d\le ⌊n/2⌋+1$. Thus, in all cases, we have proved that

$$d\le ⌊n/2⌋+1$$

(38)

On the other hand, by parts (i) and (ii) of the present theorem, we have $d\le 2s-1$, which together with (33) yields

$$d\le 2⌊{\displaystyle \frac{1+\sqrt{8n-15}}{2}}⌋-1.$$

(39)

Now, from (38) and (39), the desired inequality (37) follows.

Finally, suppose that $\mu$ is the number of vertices of maximum degree $\Delta$ in T. If $\mu =1$, then, by part (i) and (34), we have

$$\mathrm{diam}\left(T\right)\le 2⌊{\displaystyle \frac{3+\sqrt{8(n-\Delta )-7}}{2}}⌋-2.$$

If $\mu =2$, then, by part (ii) and (34), we have

$$\mathrm{diam}\left(T\right)\le 2⌊{\displaystyle \frac{3+\sqrt{8n-16\Delta +1}}{2}}⌋-1.$$

This completes the proof. □

6. A Family of Trees of Diameter at Most Four and a Conjecture

For an integer $p\ge 3$ and positive integers $a_1,\dots ,a_p$, let $T(p;a_1,\dots ,a_p)$ be the tree obtained from a root r by joining r to vertices $u_1,\dots ,u_p$, where $d\left(u_i\right)=a_i$ and all neighbors of $u_i$ other than r are pendent. Thus, ${\sum }_{i=1}^p{a}_i=n-1$, where n is the order of $T(p;a_1,\dots ,a_p)$.

Proposition 3.

For fixed n and p, the value $\mathcal{FIN}\left(T(p;a_1,\dots ,a_p)\right)$ is minimized if and only if

$$|a_i-a_j|\le 1\mathit{for}\mathit{all}i,j\in \{1,\dots ,p\}.$$

(40)

Proof. 

We define a function $h_p$ as $h_p\left(x\right)=\varphi (p,x)+(x-1)\varphi (x,1)$, where $\varphi$ is defined via (1). Then, we have

$$\mathcal{FIN}\left(T(p;a_1,\dots ,a_p)\right)={\displaystyle \sum _{i=1}^p}{h}_p\left(a_i\right).$$

(41)

For $x\ge 2$, we have

$$h_p^{″}\left(x\right)={\displaystyle \frac{3+4x/p}{4x^4{(p^{-1}+x^{-1})}^{3/2}}}-{\displaystyle \frac{5x+3}{4x^4{(1+x^{-1})}^{3/2}}}.$$

(42)

We observe that the first term on the right-hand side of (42) increases in $p\ge 3$, and hence, is smallest at $p=3$. For $p=3$, the inequality $h_p^{″}\left(x\right)>0$ reduces, after squaring positive quantities, to $x(23x^4+105x^3+81x^2-141x-108)>0$, which holds for $x\ge 2$, because the polynomial $23x^4+105x^3+81x^2-141x-108$ and its derivative are positive at $x=2$, and the derivative is increasing thereafter. Therefore, we have $h_p^{″}\left(x\right)>0$ for $p\ge 3$ and $x\ge 2$.

Now, let $z=1/p$ and define

$$g\left(z\right)=\sqrt{z+1}+\sqrt{z+{\displaystyle \frac{1}{3}}}-2\sqrt{z+{\displaystyle \frac{1}{2}}}.$$

We note that g is strictly decreasing on the interval $(0,\infty )$.

Since $p\ge 3$, we have $0<z\le 1/3$. Therefore, it holds that

$$\begin{array}{cc}\hfill h_p\left(1\right)+h_p\left(3\right)-2h_p\left(2\right)& =g\left(z\right)+{\displaystyle \frac{4}{\sqrt{3}}}-\sqrt{6}\ge g\left({\displaystyle \frac{1}{3}}\right)+{\displaystyle \frac{4}{\sqrt{3}}}-\sqrt{6}>0,\hfill \end{array}$$

which yields

$$h_p\left(2\right)-h_p\left(1\right)<h_p\left(3\right)-h_p\left(2\right).$$

(43)

Since the derivative function $h_p^{\prime }$ of $h_p$ is strictly increasing on the interval $[2,\infty )$, for every integer $k\ge 2$, we have

$$\left(h_p(k+2)-h_p(k+1)\right)-\left(h_p(k+1)-h_p\left(k\right)\right)={\int }_k^{k+1}\left(h_p^{\prime }(x+1)-h_p^{\prime }\left(x\right)\right)dx>0.$$

(44)

Thus, from (43) and (44), we conclude that the successive differences

$$h_p(k+1)-h_p\left(k\right),k\ge 1,$$

are strictly increasing, and hence, the sequence $h_p\left(1\right),h_p\left(2\right),\dots$ is strictly discretely convex.

Firstly, we suppose that, for fixed n and p, the value $\mathcal{FIN}\left(T(p;a_1,\dots ,a_p)\right)$ is minimized. Suppose, to the contrary, that $a_i\ge a_j+2$ for some $i,j$. Let $T^{\prime }$ be the tree obtained from $T(p;a_1,\dots ,a_p)$ by transferring one pendent neighbor from $u_i$ to $u_j$. Then, $T^{\prime }$ has the same order as $T(p;a_1,\dots ,a_p)$ and is of the form $T(p;b_1,\dots ,b_p)$, where $b_i=a_i-1$, $b_j=a_j+1$, and $b_t=a_t$ for $t\notin \{i,j\}$. By strict discrete convexity, we have $h_p\left(a_i\right)+h_p\left(a_j\right)>h_p(a_i-1)+h_p(a_j+1)$, and hence, using (41), we have $\mathcal{FIN}\left(T(p;a_1,\dots ,a_p)\right)-\mathcal{FIN}\left(T^{\prime }\right)>0,$ which contradicts the minimality of $\mathcal{FIN}\left(T(p;a_1,\dots ,a_p)\right)$. Consequently, (40) holds.

Conversely, under the condition ${\sum }_{i=1}^p{a}_i=n-1$, there is a unique multiset $\{a_1,\dots ,a_p\}$ of positive integers satisfying (40). Since the set of admissible multisets is finite, a minimizing multiset exists, and, by the first part of the proof, every minimizing multiset must satisfy (40). Hence, this unique multiset is necessarily the minimizing one. □

We write $n-1=qp+q^{\prime },0\le q^{\prime }<p$, and denote by $H_{n,p}$ the tree $T(p;a_1,\dots ,a_p)$ in which $p-q^{\prime }$ neighbors of the root have degree q and the remaining $q^{\prime }$ neighbors have degree $q+1$. By Proposition 3 (and (41)),

$$F_n\left(p\right):=\mathcal{FIN}\left(H_{n,p}\right)=(p-q^{\prime })h_p\left(q\right)+q^{\prime }{h}_p(q+1)$$

(45)

is the minimum $\mathcal{FIN}$ value among all trees of the form $T(p;a_1,\dots ,a_p)$ of order n.

Conjecture 1.

For every $n\ge 5$, every $\mathcal{FIN}$-minimal n-vertex tree is isomorphic to $H_{n,p}$, where p is the unique integer in $\{3,\dots ,n-1\}$ for which

$$F_n\left(p\right)=\underset{3\le t\le n-1}{min}{F}_n\left(t\right).$$

By Proposition 1, for each tree degree sequence D of order n, it is sufficient to consider its realizations constructible by Algorithm 1. Therefore, to determine the $\mathcal{FIN}$-minimal n-vertex trees, the computation considers all tree degree sequences of order n, but for each such degree sequence it is sufficient to examine only the corresponding outputs of Algorithm 1, rather than all of its tree realizations. The exact computation described below shows that, for every integer n with $5\le n\le 60$, there is a unique integer $p_n\in \{3,\dots ,n-1\}$ such that

$$F_n\left(p_n\right)=\underset{3\le p\le n-1}{min}{F}_n\left(p\right).$$

Let $q_n$ and $q_n^{\prime }$ be the uniquely determined integers satisfying

$$n-1=q_n{p}_n+q_n^{\prime },0\le q_n^{\prime }<p_n;$$

equivalently,

$$q_n=⌊{\displaystyle \frac{n-1}{p_n}}⌋\mathrm{and}{q}_n^{\prime }=n-1-q_n{p}_n.$$

By the definition of $H_{n,p}$, the root of $H_{n,p_n}$ has degree $p_n$, exactly $q_n^{\prime }$ of its neighbors have degree $q_n+1$, and the remaining $p_n-q_n^{\prime }$ neighbors have degree $q_n$. Every other vertex is pendent. Consequently, the degree sequence of $H_{n,p_n}$ is

$$\left(p_n,{(q_n+1)}^{\left[q_n^{\prime }\right]},q_n^{[p_n-q_n^{\prime }]},1^{[n-p_n-1]}\right),$$

(46)

where the notation $z^{\left[m\right]}$ means that the degree z occurs m times, and equal entries are combined.

Theorem 5.

For every n with $5\le n\le 60$, the tree $H_{n,p_n}$ is the unique $\mathcal{FIN}$-minimal n-vertex tree. Also,

$$\begin{array}{cc}\hfill \mathcal{FIN}\left(H_{n,p_n}\right)=& q_n^{\prime }\left[\varphi (p_n,q_n+1)+q_n\varphi (q_n+1,1)\right]+(p_n-q_n^{\prime })\left[\varphi (p_n,q_n)+(q_n-1)\varphi (q_n,1)\right],\hfill \end{array}$$

where ϕ is defined via (1).

Proof. 

By Proposition 1, for each tree degree sequence of order n, it is sufficient to consider a realization produced by Algorithm 1. An exact finite comparison, performed in SageMath, of the aforementioned realizations shows that, for every n with $5\le n\le 60$, there is a unique minimizing degree sequence, namely (46); for illustration, the last ten computed minimizing degree sequences are shown in

Table 1. The last ten computed minimizing degree sequences in Theorem 5.

n	Minimizing Degree Sequence	n	Minimizing Degree Sequence
51	$(17,3^{\left[16\right]},2,1^{\left[33\right]})$	56	$(19,3^{\left[17\right]},2^{\left[2\right]},1^{\left[36\right]})$
52	$(18,3^{\left[15\right]},2^{\left[3\right]},1^{\left[33\right]})$	57	$(19,3^{\left[18\right]},2,1^{\left[37\right]})$
53	$(18,3^{\left[16\right]},2^{\left[2\right]},1^{\left[34\right]})$	58	$(19,3^{\left[19\right]},1^{\left[38\right]})$
54	$(18,3^{\left[17\right]},2,1^{\left[35\right]})$	59	$(20,3^{\left[18\right]},2^{\left[2\right]},1^{\left[38\right]})$
55	$(18,3^{\left[18\right]},1^{\left[36\right]})$	60	$(20,3^{\left[19\right]},2,1^{\left[39\right]})$

. For every n with $5\le n\le 60$, the computed values satisfy $q_n=⌊(n-1)/p_n⌋\in \{1,2,3\}$. Moreover, whenever $q_n=3$, we have $q_n^{\prime }=0$. Hence, the degree sequence (46) has a unique maximum degree $p_n$, and every other non-pendent vertex degree is either 2 or 3. The only possible equality $p_n=q_n+1$ occurs for $p_n=3$ and $q_n=2$; in the computed cases, this occurs only for $n=7$, where $q_n^{\prime }=0$.

Let T be a $\mathcal{FIN}$-minimal tree having this degree sequence (46). We root T at its unique vertex r of degree $p_n$. We show that every non-pendent vertex different from r is a neighbor of r.

Firstly, let $u\in V\left(T\right)\setminus \left\{r\right\}$ be a vertex of degree 3. Since r is the unique vertex of maximum degree, we have $p_n>3$. Contrarily, we suppose that u is not adjacent to r. Then, the parent of u is different from r; hence, by Lemma 8, the parent of u has degree greater than 3. However, by (46), every vertex different from r has degree at most 3, a contradiction. Therefore, in T, every vertex of degree 3 different from r is adjacent to r.

Now, let $v\in V\left(T\right)$ be a vertex of degree 2. By Corollary 2, the vertex v has one pendent neighbor and one branching neighbor, say w. Contrarily, we assume that v is not adjacent to r. Then, $w\ne r$. Since $d_T\left(w\right)\ge 3$ and every vertex different from r has degree at most 3, we have $d_T\left(w\right)=3$. Consequently, by the preceding paragraph, w is adjacent to r.

Since the degree sequence (46) of T contains at most $p_n$ non-root, non-pendent vertices and since v is not adjacent to r, the root r has a pendent neighbor, say $r^{\prime }$. Let $T^{\prime }=T-\{r{r}^{\prime },wv\}+\{rv,w{r}^{\prime }\}$. We note that both the trees T and $T^{\prime }$ have the same degree sequence. Also, we note that $d_T\left(r\right)>d_T\left(w\right)$ and $d_T\left(r^{\prime }\right)=1<2=d_T\left(v\right)$. Hence, by Lemma 1, we have $\mathcal{FIN}\left(T^{\prime }\right)<\mathcal{FIN}\left(T\right)$, a contradiction. Therefore, in T, every vertex of degree 2 is also adjacent to r.

Consequently, in T, every non-pendent vertex different from r is a neighbor of r, and every remaining vertex is pendent. Thus, $T=T(p_n;a_1,\dots ,a_{p_n})$ for some positive integers $a_1,\dots ,a_{p_n}$ satisfying

$${\displaystyle \sum _{i=1}^{p_n}}{a}_i=n-1.$$

Since T is $\mathcal{FIN}$-minimal, it minimizes the $\mathcal{FIN}$ index among all trees of the form $T(p_n;a_1,\dots ,a_{p_n})$ of order n. Therefore, by Proposition 3, we have $|a_i-a_j|\le 1$ for all $i,j\in \{1,\dots ,p_n\}$. Since $n-1=q_n{p}_n+q_n^{\prime },0\le q_n^{\prime }<p_n$, it follows that exactly $q_n^{\prime }$ of the integers $a_i$ are equal to $q_n+1$, while the remaining $p_n-q_n^{\prime }$ are equal to $q_n$. Hence, $T\cong H_{n,p_n}$, and it is the unique $\mathcal{FIN}$-minimal n-vertex tree.

Finally, $H_{n,p_n}$ has $q_n^{\prime }$ neighbors of the root of degree $q_n+1$, each incident with $q_n$ pendent edges, and $p_n-q_n^{\prime }$ neighbors of degree $q_n$, each incident with $q_n-1$ pendent edges. Therefore, we obtain the same expression for $\mathcal{FIN}\left(H_{n,p_n}\right)$ as given in the statement. □

7. Concluding Remarks

The structural results proved in this paper substantially reduce the class of possible $\mathcal{FIN}$-minimal n-vertex trees. Particularly, for every fixed tree degree sequence, the realizations minimizing the $\mathcal{FIN}$ index are precisely those constructible by Algorithm 1, see Proposition 1. Every vertex of degree 2 in a $\mathcal{FIN}$-minimal n-vertex tree T has one pendent neighbor and one branching neighbor, and the subgraph induced by the branching vertices of T is a tree from which T is obtained by attaching only pendent edges and pendent paths of length 2, see Corollary 2 and Proposition 2. Moreover, T has at most two vertices of maximum degree, and if there are two, then they are adjacent; when T is rooted at a maximum-degree vertex, the degrees strictly decrease along every path starting at the root, with the only possible exception that the root and one of its neighbors have the same maximum degree, see Theorem 1 and Lemma 8. Also, it holds that $\Delta \left(T\right)\le max\left(17,⌊(n+3)/3⌋\right)$, see Theorem 2. Furthermore, if $s=\left|\mathcal{D}\right(T\left)\right|$, then $\mathrm{diam}\left(T\right)\le 2s-2$ when the maximum-degree vertex is unique and $\mathrm{diam}\left(T\right)\le 2s-1$ when there are two maximum-degree vertices, see Theorem 4. Finally, the exact computations, together with the some obtained structural results, show that, for every $5\le n\le 60$, the tree $H_{n,p_n}$ is the unique $\mathcal{FIN}$-minimal n-vertex tree, see Theorem 5.

The main remaining structural step toward Conjecture 1 is to prove that every non-pendent vertex, other than a suitable root r, is adjacent to r. In view of Theorem 1, it is also natural to ask whether every $\mathcal{FIN}$-minimal n-vertex tree has a unique vertex of maximum degree.