Hitting Time Index for Broom Graphs

Abstract

Thehitting time index $\mathrm{HT}\left(G\right)$ is a recently introduced topological descriptor based on expected hitting times of a random walk on a graph. In this paper, we derive a closed-form formula for $\mathrm{HT}\left(G\right)$ for broom graphs $B_{n,d}$ that holds for all parameters $2\le d\le n-1$, $\mathrm{HT}\left(B_{n,d}\right)=S_1+S_2+S_3+(n-d){\displaystyle \sum _{i=1}^{d-1}}\max \{A\left(i\right),B\left(i\right)\},$ where $S_1,S_2,S_3,A\left(i\right),B\left(i\right)$ are explicitly defined. For $d\ge 2$ and $n\ge 4d-8$ we derive a simpler cubic polynomial formula in n, $\mathrm{HT}\left(B_{n,d}\right)=n^3+a_d{n}^2+b_{d}n+c_d,$ with explicitly given coefficients $a_d,b_d,c_d$ depending only on d. We also consider quartic polynomial formulas for special cases.

1. Introduction

In mathematical chemistry, a basic model for a molecule is a finite simple connected undirected graph $G=(V,E)$ with vertex set $V=\{1,2,\dots ,n\}$ and edge set E, where vertices are atoms and edges are atomic bonds. For all graph-theoretic notions used in this article, the reader may consult reference [1].

Over the years, many molecular indices, or descriptors, i.e., real-valued functions on the domain of all graphs, have been introduced with the aim of identifying physico-chemical properties of molecules and classifying them according to the values of their indices. One such recent index is the hitting time index, first introduced in [2], and defined in terms of the behavior of a simple random walk on G. This random walk is a Markov chain $\{X_n,n\ge 0\}$ with state space V and uniform transition probabilities ${\displaystyle \frac{1}{d_i}}$ from vertex i to any of its $d_i$ neighbors. The hitting time $T_j$ of vertex j is the random variable that counts the steps needed for the random walk to hit j:

$$T_j=inf\{n\ge 0:X_n=j\},$$

and its expected value, when the random walk starts at state i, is denoted by $E_i{T}_j$. As a remark, we note that $E_i{T}_i=0$, and this should not be confused with the mean return time to vertex i, $E_i{T}_i^{+}={\displaystyle \frac{2\left|E\right|}{d_i}}$, defined in terms of the random variable $T_i^{+}=inf\{n\ge 1:X_n=i\}$. Basic material on hitting times of Markov chains can be found in [3], while references [4,5,6,7] are a small sample of recent literature on this topic.

Now we can define the hitting time index as

$$\mathrm{HT}\left(G\right)=\sum _{1\le i<j\le n}max\{E_i{T}_j,E_j{T}_i\}.$$

Basic formulas for computing expected hitting times on trees are given in Palacios, Gómez, and Del Río [8]. For a tree T and any two vertices $v,w\in V\left(T\right)$, let P be the unique path between v and w. Then

$$E_v{T}_w=d_T(v,w)+2\sum _{x\ne w\in V\left(P\right)}\left|E_x\right|,$$

(1)

where $d_T(v,w)$ is the distance between v and w, and $|E_x|$ is the number of edges in the component containing x after removing the edge on the path that leads to w. The same paper gives two key formulas: if $vw$ is an edge whose removal makes the graph disconnected (a cut edge), then

$$E_v{T}_w=2\left|E_v\right|+1,$$

(2)

where $|E_v|$ is the number of edges in the component containing v after removing the edge $vw$; also, if $P=u_1{u}_2\dots u_m$ is a path, then

$$E_{u_1}{T}_{u_m}={\displaystyle \sum _{i=1}^{m-1}}{E}_{u_i}{T}_{u_{i+1}}.$$

(3)

Together with Santamaría [9], he gave additional results on the hitting time index, including the fact that for trees the value $\mathrm{HT}\left(T\right)$ is always an integer.

In this paper, we present formulas for computing the hitting time index for broom graphs.

2. Basic Formula

Let $B_{n,d}$ denote the broom graph with n vertices consisting of a path $P_d=u_1{u}_2\dots u_d$ and $m=n-d$ leaves $v_1,\dots ,v_m$ all attached to the endpoint $u_d$. We assume $n\ge 4$ and $2\le d\le n-1$.

For vertices $u_i,u_j$ on the path $P_d$ with $i<j$, applying (1) and (3) we obtain

$$E_{u_i}{T}_{u_j}=(j-i)(i+j-2),E_{u_j}{T}_{u_i}=(j-i)(2n-j-i).$$

For a leaf $v_j$ attached to $u_d$ and a vertex $u_i$ on the path ($1\le i\le d-1$), applying (1) and (3), we obtain

$$E_{u_i}{T}_{v_j}=(d-i+1)(d+i-1)+2n-2d-2,$$

$$E_{v_j}{T}_{u_i}=(d-i)(2n-d-i)+1.$$

Theorem 1. 

Let $n,d$ be positive integers with $n\ge 4$ and $2\le d\le n-1$. Let $m=n-d$ be the number of leaves. Then the hitting time index of the broom graph $B_{n,d}$ is given by

$$\mathrm{HT}\left(B_{n,d}\right)=S_1+S_2+S_3+m{\displaystyle \sum _{i=1}^{d-1}}max\{A\left(i\right),B\left(i\right)\},$$

where

$$\begin{array}{cc}\hfill S_1& =\sum _{1\le i<j\le d}(j-i)·max\{i+j-2,2n-j-i\},\hfill \\ \hfill S_2& ={\displaystyle \frac{m(m-1)}{2}}(2n-2),\hfill \\ \hfill S_3& =m(2n-3),\hfill \\ \hfill A\left(i\right)& =(d-i)(2n-d-i)+1,\hfill \\ \hfill B\left(i\right)& =(d-i+1)(d+i-1)+2n-2d-2.\hfill \end{array}$$

Proof. 

We split the sum $\mathrm{HT}\left(B_{n,d}\right)={\sum }_{x<y}D(x,y)$ into four disjoint classes of vertex pairs.

1. 

Pairs of vertices on the path $P_d$. For $1\le i<j\le d$, from (1) and (3) we have $E_{u_i}{T}_{u_j}=(j-i)(i+j-2)$ and $E_{u_j}{T}_{u_i}=(j-i)(2n-j-i)$. Hence

$$D(u_i,u_j)=max\{E_{u_i}{T}_{u_j},E_{u_j}{T}_{u_i}\}=(j-i)·max\{i+j-2,2n-j-i\}.$$

Thus the total contribution of path pairs is

$$S_1:=\sum _{1\le i<j\le d}D(u_i,u_j)=\sum _{1\le i<j\le d}(j-i)·max\{i+j-2,2n-j-i\}.$$

2. 

Pairs of leaves. For any two distinct leaves $v_p,v_q$, the unique path is $v_p-u_d-v_q$ of length 2. Applying (1) and (2),

$$E_{v_p}{T}_{v_q}=1+\left(2(n-2)+1\right)=2n-2,$$

and by symmetry $E_{v_q}{T}_{v_p}=2n-2$. Hence, $D(v_p,v_q)=2n-2$. The number of such pairs is $\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)$, where for $m\le 1$ this expression is defined as 0. Therefore

$$S_2:=\sum _{1\le p<q\le m}D(v_p,v_q)=\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)(2n-2)={\displaystyle \frac{m(m-1)}{2}}(2n-2).$$

3. 

Pairs involving $u_d$ and a leaf. For a leaf $v_j$, the path $u_d-v_j$ has length 1. Applying (2),

$$E_{u_d}{T}_{v_j}=2\left|E_{u_d}\right|+1=2(n-2)+1=2n-3,$$

and $E_{v_j}{T}_{u_d}=1$. Hence, $D(u_d,v_j)=max\{2n-3,1\}=2n-3$ (since $n\ge 4$). The number of such leaves is m, and for $m=0$ this class is empty. Thus

$$S_3:={\displaystyle \sum _{j=1}^m}D(u_d,v_j)=m(2n-3).$$

4. 

Pairs involving $u_i$ ($1\le i\le d-1$) and a leaf $v_j$. For a fixed i, from (1) and (3) we have

$$E_{u_i}{T}_{v_j}=(d-i+1)(d+i-1)+2n-2d-2=:B\left(i\right),$$

$$E_{v_j}{T}_{u_i}=(d-i)(2n-d-i)+1=:A\left(i\right).$$

Hence

$$D(u_i,v_j)=max\{E_{u_i}{T}_{v_j},E_{v_j}{T}_{u_i}\}=max\{A\left(i\right),B\left(i\right)\}.$$

Since there are m leaves, the total contribution for a fixed i is $m·max\left\{A\right(i),B(i\left)\right\}$, and summing over $i=1,2,\dots ,d-1$ we obtain

$$S_4:={\displaystyle \sum _{i=1}^{d-1}}{\displaystyle \sum _{j=1}^m}D(u_i,v_j)=m{\displaystyle \sum _{i=1}^{d-1}}max\{A\left(i\right),B\left(i\right)\}.$$

For $d=n-1$ we have $m=1$, so $S_2=0$, and $S_4$ reduces to ${\sum }_{i=1}^{n-2}max\{A\left(i\right),B\left(i\right)\}$, which is consistent. Finally, adding all contributions,

$$\mathrm{HT}\left(B_{n,d}\right)=S_1+S_2+S_3+S_4,$$

which is what we wanted to prove. □

Remark 1. 

Our unified formula shows how a simple use of the max function can be exploited to combine several special cases into one elegant formula. We hope that this approach will inspire further research and applications to other classes of graphs.

We noticed a printing error in Theorem 4 of the paper by He et al. [10] (see

Table 1. Printing error in Theorem 4 of [10].

($\mathit{n},\mathit{d}$)	Our Formula	He et al. ([10], Th. 4)	Value by Definition
$(9,4)$	790	820	790
$(10,4)$	1104	1140	1104

).

3. Polynomial Formulas for $\mathrm{HT}\left({\mathbf{B}}_{\mathbf{n},\mathbf{d}}\right)$

3.1. Formula via Cubic Polynomial

Formulas for $\mathrm{HT}\left(B_{n,2}\right)$ and $\mathrm{HT}\left(B_{n,3}\right)$ were proved in [11]. The first corresponds to the star $K_{1,n-1}$, and the second is a special case of the broom graph with three vertices on the path. Direct application of Theorem 1 yields the following formulas

$$\begin{array}{cccc}\hfill \mathrm{HT}\left(B_{n,2}\right)& =n^3-2n^2+1,\hfill & \hfill & n\ge 4,\hfill \\ \hfill \mathrm{HT}\left(B_{n,3}\right)& =n^3-7n+2,\hfill & \hfill & n\ge 6,\hfill \\ \hfill \mathrm{HT}\left(B_{n,4}\right)& =n^3+4n^2-33n+34,\hfill & \hfill & n\ge 8,\hfill \\ \hfill \mathrm{HT}\left(B_{n,5}\right)& =n^3+10n^2-86n+135,\hfill & \hfill & n\ge 12,\hfill \\ \hfill \mathrm{HT}\left(B_{n,6}\right)& =n^3+18n^2-174n+355,\hfill & \hfill & n\ge 16.\hfill \end{array}$$

The generalisation of these results will be presented in the following theorem.

Theorem 2. 

Let $n,d$ be positive integers with $d\ge 2$ and $n\ge 4d-8$. Then the hitting time index of the broom $B_{n,d}$ is expressed as a cubic polynomial in n

$$\mathrm{HT}\left(B_{n,d}\right)=n^3+a_d{n}^2+b_{d}n+c_d,$$

where the coefficients are given by

$$\begin{array}{cc}\hfill a_d& =d^2-3d,\hfill \\ \hfill b_d& ={\displaystyle \frac{-8d^3+15d^2+35d-66}{6}},\hfill \\ \hfill c_d& ={\displaystyle \frac{3d^4-4d^3-36d^2+67d}{6}}.\hfill \end{array}$$

The coefficient $a_d$ satisfies the recurrence $a_{d+1}=a_d+2d-2$ for $d\ge 4$.

Remark 2. 

The threshold $n\ge 4d-8$ arises naturally: when the number of leaves $m=n-d$ is sufficiently large compared to the path length d, the maxima between $A\left(i\right)$ and $B\left(i\right)$ and between $i+j-2$ and $2n-j-i$ always behave in the same way—the quantities depending on n dominate. As soon as n drops below this threshold, for some pairs $(i,j)$ the maxima “switch” from one branch of the formula to the other, and a single cubic polynomial can no longer describe all cases. Therefore, for $n<4d-8$, one must return to the general max-expression of Theorem 1.

Proof. 

For $d=2,3$ we obtain the Palacios formulas in [11]. Let $d>3$ and $m=n-d$. By Theorem 1,

$$\mathrm{HT}\left(B_{n,d}\right)=S_1+S_2+S_3+m{\displaystyle \sum _{i=1}^{d-1}}max\{A\left(i\right),B\left(i\right)\},$$

where $A\left(i\right),B\left(i\right),S_1,S_2,S_3$ are defined in that theorem. Define the auxiliary polynomial

$$Q\left(i\right)=i^2-(n+1)i+2+(n-d)(d-1).$$

Then

$$A\left(i\right)-B\left(i\right)=2Q\left(i\right).$$

(4)

Under the assumption $n\ge 4d-8$, we examine the sign of $Q\left(i\right)$ for $1\le i\le d-1$. For the endpoints, we obtain

$$Q(d-2)=n-4d+8\ge 0,Q(d-1)=4-2d<0(d\ge 4).$$

The vertex of the parabola Q is at $i_0=(n+1)/2$. Since

$$i_0\ge {\displaystyle \frac{4d-7}{2}}=2d-3.5>d-1,$$

the function Q is strictly decreasing on $[1,d-1]$. Therefore, $Q\left(i\right)\ge 0$ for $1\le i\le d-2$ and $Q(d-1)<0$. From (4) it follows

$$max\{A\left(i\right),B\left(i\right)\}=\left\{\begin{array}{cc}A\left(i\right),\hfill & 1\le i\le d-2,\hfill \\ B\left(i\right),\hfill & i=d-1.\hfill \end{array}\right.$$

(5)

To compute $S_1$, we first determine the value under the maximum. From $n\ge 4d-8$ and $d\ge 4$, it follows $n\ge 2d-2$, because $4d-8\ge 2d-2$ for all $d\ge 4$. For arbitrary $1\le i<j\le d$, we then have

$$2n-j-i\ge 2(2d-2)-(2d-1)=2d-3\ge i+j-2,$$

so $max\{i+j-2,2n-j-i\}=2n-j-i$. Hence

$$S_1=\sum _{1\le i<j\le d}(j-i)(2n-j-i).$$

Now we explicitly compute this double sum. Expand the product

$$(j-i)(2n-j-i)=2n(j-i)-(j-i)(j+i)=2n(j-i)-(j^2-i^2),$$

whence

$$S_1=2n\sum _{1\le i<j\le d}(j-i)-\sum _{1\le i<j\le d}(j^2-i^2).$$

(6)

First, sum. Introduce the distance $k=j-i$ ($1\le k\le d-1$). For fixed k, the number of pairs $(i,j)$ with difference k is $d-k$ (because i can be $1,2,\dots ,d-k$). Thus,

$$\sum _{1\le i<j\le d}(j-i)={\displaystyle \sum _{k=1}^{d-1}}k(d-k)=d{\displaystyle \sum _{k=1}^{d-1}}k-{\displaystyle \sum _{k=1}^{d-1}}{k}^2.$$

Using the elementary sums

$${\displaystyle \sum _{k=1}^{d-1}}k={\displaystyle \frac{d(d-1)}{2}},{\displaystyle \sum _{k=1}^{d-1}}{k}^2={\displaystyle \frac{(d-1)d(2d-1)}{6}},$$

we obtain

$$\sum _{1\le i<j\le d}(j-i)={\displaystyle \frac{d^2(d-1)}{2}}-{\displaystyle \frac{d(d-1)(2d-1)}{6}}={\displaystyle \frac{d(d-1)(d+1)}{6}}.$$

(7)

Second, sum. Write it as a difference

$$\sum _{i<j}(j^2-i^2)=\sum _{i<j}{j}^2-\sum _{i<j}{i}^2.$$

In the first inner sum, j is fixed for all $i<j$, and there are $j-1$ such i. Hence, ${\sum }_{i<j}{j}^2={\sum }_{j=1}^d{j}^2(j-1)={\sum }_{j=1}^d{j}^3-{\sum }_{j=1}^d{j}^2$. In the second inner sum, i is fixed for all $j>i$, and there are $d-i$ such j. Hence, ${\sum }_{i<j}{i}^2={\sum }_{i=1}^d{i}^2(d-i)=d{\sum }_{i=1}^d{i}^2-{\sum }_{i=1}^d{i}^3$. Subtracting gives

$$\sum _{i<j}(j^2-i^2)=2{\displaystyle \sum _{j=1}^d}{j}^3-(d+1){\displaystyle \sum _{j=1}^d}{j}^2.$$

Now apply the well-known formulas for sums of cubes and squares

$${\displaystyle \sum _{j=1}^d}{j}^3={\displaystyle \frac{d^2{(d+1)}^2}{4}},{\displaystyle \sum _{j=1}^d}{j}^2={\displaystyle \frac{d(d+1)(2d+1)}{6}}.$$

Substituting,

$$\begin{array}{cc}\hfill \sum _{i<j}(j^2-i^2)& =2·{\displaystyle \frac{d^2{(d+1)}^2}{4}}-(d+1)·{\displaystyle \frac{d(d+1)(2d+1)}{6}}\hfill \\ \hfill & ={\displaystyle \frac{d^2{(d+1)}^2}{2}}-{\displaystyle \frac{d{(d+1)}^2(2d+1)}{6}}\hfill \\ \hfill & ={\displaystyle \frac{d{(d+1)}^2}{6}}\left[3d-(2d+1)\right]\hfill \\ \hfill & ={\displaystyle \frac{d(d-1){(d+1)}^2}{6}}.\hfill \end{array}$$

(8)

Plugging (7) and (8) into (6) gives

$$\begin{array}{cc}\hfill S_1& =2n·{\displaystyle \frac{d(d-1)(d+1)}{6}}-{\displaystyle \frac{d(d-1){(d+1)}^2}{6}}\hfill \\ \hfill & ={\displaystyle \frac{d(d-1)(d+1)}{6}}\left[2n-(d+1)\right]\hfill \\ \hfill & ={\displaystyle \frac{d(d-1)(d+1)(2n-d-1)}{6}}.\hfill \end{array}$$

(9)

$S_2$ and $S_3$ simplify straightforwardly

$$\begin{array}{cc}\hfill S_2+S_3& =\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2}}\right)(2n-2)+m(2n-3)=m(m-1)(n-1)+m(2n-3)\hfill \\ \hfill & =(n-d)(n^2-dn+d-2).\hfill \end{array}$$

(10)

For the remaining sum we use (5). Note that

$${\displaystyle \sum _{i=1}^{d-1}}max\{A\left(i\right),B\left(i\right)\}={\displaystyle \sum _{i=1}^{d-2}}A\left(i\right)+B(d-1)={\displaystyle \sum _{i=1}^{d-1}}A\left(i\right)-A(d-1)+B(d-1).$$

Since $A(d-1)=2n-2d+2$ and $B(d-1)=2n+2d-6$, we have $B(d-1)-A(d-1)=4d-8$, thus

$${\displaystyle \sum _{i=1}^{d-1}}max\{A\left(i\right),B\left(i\right)\}={\displaystyle \sum _{i=1}^{d-1}}A\left(i\right)+4d-8.$$

(11)

Now compute ${\sum }_{i=1}^{d-1}A\left(i\right)$. Substituting $k=d-i$ ($k=1,\dots ,d-1$) gives $A\left(i\right)=k(2n-2d+k)+1=k^2+2(n-d)k+1$, whence

$${\displaystyle \sum _{i=1}^{d-1}}A\left(i\right)={\displaystyle \sum _{k=1}^{d-1}}\left(k^2+2(n-d)k+1\right)={\displaystyle \frac{(d-1)d(2d-1)}{6}}+(n-d)d(d-1)+(d-1).$$

Inserting into (11) and multiplying by $m=n-d$ yields

$$S_4=(n-d)\left[{\displaystyle \frac{(d-1)d(2d-1)}{6}}+d(d-1)(n-d)+5d-9\right].$$

(12)

Adding (9), (10) and (12) and collecting powers of n, $\mathrm{HT}\left(B_{n,d}\right)$ becomes

$$\mathrm{HT}\left(B_{n,d}\right)=n^3+(d^2-3d)n^2+{\displaystyle \frac{-8d^3+15d^2+35d-66}{6}}n+{\displaystyle \frac{3d^4-4d^3-36d^2+67d}{6}}.$$

Thus the coefficients $a_d=d^2-3d$, $b_d=(-8d^3+15d^2+35d-66)/6$, $c_d=(3d^4-4d^3-36d^2+67d)/6$ are confirmed. The recurrence $a_{d+1}=a_d+2d-2$ follows immediately:

$$a_{d+1}-a_d=({(d+1)}^2-3(d+1))-(d^2-3d)=2d-2.$$

Thus the bound $n\ge 4d-8$ is sharp for the argument above, i.e., it is the smallest value for which the partition $max\left\{A\right(i),B(i\left)\right\}$ used in the proof remains valid. The formula itself may fail for smaller n; for example, for $(n,d)=(7,4)$, the cubic polynomial gives 342, whereas the true value is 348. □

Corollary 1. 

For every $d\ge 3$ and $n\ge 4d-8$ we have

$$\mathrm{HT}\left(B_{n,d-1}\right)<\mathrm{HT}\left(B_{n,d}\right).$$

Proof. 

By Theorem 2, for $d\ge 4$ and $n\ge 4d-8$ we have

$$\mathrm{HT}\left(B_{n,d}\right)=n^3+(d^2-3d)n^2+{\displaystyle \frac{-8d^3+15d^2+35d-66}{6}}n+{\displaystyle \frac{3d^4-4d^3-36d^2+67d}{6}}.$$

Replacing d by $d-1$ gives the corresponding expression for $\mathrm{HT}\left(B_{n,d-1}\right)$. Subtracting and simplifying elementary algebraic expressions, we obtain the difference

$$\begin{array}{cc}\hfill \Delta (n,d)& :=\mathrm{HT}\left(B_{n,d}\right)-\mathrm{HT}\left(B_{n,d-1}\right)\hfill \\ \hfill & =(2d-4)n^2-(4d^2-9d-2)n+(2d^3-5d^2-8d+16).\hfill \end{array}$$

For $d\ge 4$, the leading coefficient $2d-4>0$, so $\Delta (n,d)$ is a convex quadratic function in n. Its derivative

$${\displaystyle \frac{\partial \Delta }{\partial n}}=2(2d-4)n-(4d^2-9d-2)$$

is increasing in n, and at $n=4d-8$ it equals

$$2(2d-4)(4d-8)-(4d^2-9d-2)=16{(d-2)}^2-4d^2+9d+2,$$

which is positive for all $d\ge 4$ (e.g., for $d=4$ we get 38, and for $d>4$ the value grows rapidly). Hence, $\Delta (n,d)$ is strictly increasing on $[4d-8,\infty )$, and it suffices to check its value at the smallest point $n=4d-8$:

$$\begin{array}{cc}\hfill \Delta (4d-8,d)& =(2d-4){(4d-8)}^2-(4d^2-9d-2)(4d-8)+(2d^3-5d^2-8d+16)\hfill \\ \hfill & =18d^3-129d^2+312d-256.\hfill \end{array}$$

Define $g\left(d\right)=18d^3-129d^2+312d-256$. For $d=4$, $g\left(4\right)=18·64-129·16+312·4-256=80>0$. For $d\ge 4$, $g^{\prime }\left(d\right)=54d^2-258d+312$; one easily checks that $g^{\prime }\left(4\right)=144>0$ and $g^{\prime }\left(d\right)$ increases for $d\ge 4$, so $g\left(d\right)\ge g\left(4\right)>0$. Thus $\Delta (4d-8,d)>0$ for all $d\ge 4$.

Because $\Delta (n,d)$ is increasing in n, for every $n\ge 4d-8$ we have $\Delta (n,d)\ge \Delta (4d-8,d)>0$, i.e., $\mathrm{HT}\left(B_{n,d}\right)>\mathrm{HT}\left(B_{n,d-1}\right)$. The Palacios result gives $\mathrm{HT}\left(B_{n,3}\right)>\mathrm{HT}\left(B_{n,2}\right)$, which completes the proof. □

3.2. Formula via Quartic Polynomial

Lemma 1. 

For every $n\ge 3$ the hitting time index of the broom graph $B_{n,n-1}$ equals

$$\mathrm{HT}\left(B_{n,n-1}\right)=\left\{\begin{array}{cc}{\displaystyle \frac{5}{24}{n}^4-\frac{1}{6}{n}^3-\frac{1}{3}{n}^2+\frac{1}{6}n,}\hfill & n\mathit{even},\hfill \\ {\displaystyle \frac{5}{24}{n}^4-\frac{1}{6}{n}^3-\frac{1}{3}{n}^2+\frac{1}{6}n+\frac{1}{8},}\hfill & n\mathit{odd}.\hfill \end{array}\right.$$

Proof. 

By Theorem 1, for $d=n-1$ and $m=1$ we have

$$\mathrm{HT}\left(B_{n,n-1}\right)=S_1+S_2+S_3+S_4,$$

where $S_1,S_2,S_3,S_4$ are defined in that theorem. Since $m=1$, there are no leaf pairs, so $S_2=\left({\displaystyle \genfrac{}{}{0pt}{}{1}{2}}\right)(2n-2)=0$. Directly from the definition, $S_3=1·(2n-3)=2n-3$. With $d=n-1$, the functions $A\left(i\right)$ and $B\left(i\right)$ from Theorem 1 become

$$\begin{array}{cc}\hfill A\left(i\right)& =(n-1-i)(2n-(n-1)-i)+1=(n-1-i)(n+1-i)+1,\hfill \\ \hfill B\left(i\right)& =(n-1-i+1)\left(\right(n-1)+i-1)+2n-2(n-1)-2=(n-i)(n+i-2).\hfill \end{array}$$

Simplifying:

$$\begin{array}{cc}\hfill A\left(i\right)& =(n-i-1)(n-i+1)+1={(n-i)}^2,\hfill \\ \hfill B\left(i\right)& =(n-i)(n+i-2)=n^2-i^2-2n+2i.\hfill \end{array}$$

Their difference is

$$A\left(i\right)-B\left(i\right)={(n-i)}^2-(n^2-i^2-2n+2i)=-2(i-1)(n-i).$$

For $1\le i\le n-2$ we have $i-1\ge 0$ and $n-i\ge 2>0$, hence $A\left(i\right)-B\left(i\right)\le 0$, with equality only for $i=1$. Therefore

$$\max \left\{A\right(i),B(i\left)\right\}=B\left(i\right)\mathrm{for}\mathrm{all}1\le i\le n-2,$$

so

$$S_4=1·{\displaystyle \sum _{i=1}^{n-2}}B\left(i\right)={\displaystyle \sum _{i=1}^{n-2}}(n^2-i^2-2n+2i).$$

Using the standard sums

$${\displaystyle \sum _{i=1}^{n-2}}1=n-2,{\displaystyle \sum _{i=1}^{n-2}}i={\displaystyle \frac{(n-2)(n-1)}{2}},{\displaystyle \sum _{i=1}^{n-2}}{i}^2={\displaystyle \frac{(n-2)(n-1)(2n-3)}{6}},$$

we obtain

$$\begin{array}{cc}\hfill S_4& =(n-2)(n^2-2n)-{\displaystyle \frac{(n-2)(n-1)(2n-3)}{6}}+2·{\displaystyle \frac{(n-2)(n-1)}{2}}\hfill \\ \hfill & ={\displaystyle \frac{1}{6}}\left(18-7n-9n^2+4n^3\right).\hfill \end{array}$$

According to Theorem 1,

$$S_1=\sum _{1\le i<j\le n-1}(j-i)max\{i+j-2,2n-j-i\}.$$

The value of the maximum is

$$max\{i+j-2,2n-j-i\}=\left\{\begin{array}{cc}2n-j-i,\hfill & i+j\le n+1,\hfill \\ i+j-2,\hfill & i+j\ge n+2.\hfill \end{array}\right.$$

Introduce the substitution $s=i+j$, $t=j-i$. Then $i=(s-t)/2$, $j=(s+t)/2$, $j-i=t$, and the conditions $1\le i<j\le n-1$ become

$$1\le t\le min\{s-2,2n-2-s\},3\le s\le 2n-3,t\equiv s(mod2).$$

Now $S_1$ is written as the sum of two sums, according to the value of the maximum:

$$\begin{array}{cc}\hfill S_1& =\underset{=:U}{\underbrace{\sum _{s=3}^{n+1}(2n-s)\sum _{\begin{array}{c}1\le t\le min\{s-2,2n-2-s\}\\ t\equiv s\left(2\right)\end{array}}t}}+\underset{=:V}{\underbrace{\sum _{s=n+2}^{2n-3}(s-2)\sum _{\begin{array}{c}1\le t\le 2n-2-s\\ t\equiv s\left(2\right)\end{array}}t}}.\hfill \end{array}$$

(13)

The inner sums for a bound M are well known:

$$\sum _{\begin{array}{c}1\le t\le M\\ t\mathrm{even}\end{array}}t=⌊{\displaystyle \frac{M}{2}}⌋\left(⌊{\displaystyle \frac{M}{2}}⌋+1\right),\sum _{\begin{array}{c}1\le t\le M\\ t\mathrm{odd}\end{array}}t={⌈{\displaystyle \frac{M}{2}}⌉}^2.$$

We continue the computation by treating even and odd n separately.

• Case $n=2m$ (even, $m\ge 2$). Then $d=n-1=2m-1$.

• Sum U: The range is $3\le s\le n+1=2m+1$.

  (a)

  For $3\le s\le 2m$ we have $min\{s-2,2n-2-s\}=s-2$. The contribution of these s is denoted $U_1$.

  (b)

  For $s=2m+1$ (odd) we have $min=2n-2s=2m-3$, and the factor is $2n-s=2m-1$. This term is $U_2=(2m-1){\sum }_{\begin{array}{c}1\le t\le 2m-3\\ t\mathrm{odd}\end{array}}t$.

  In $U_1$ we write $s=2r$ (even) and $s=2r+1$ (odd):

  $$\begin{array}{cc}\hfill U_1& ={\displaystyle \sum _{r=2}^m}(4m-2r)\sum _{\begin{array}{c}t\le 2r-2\\ t\mathrm{even}\end{array}}t+{\displaystyle \sum _{r=1}^{m-1}}(4m-2r-1)\sum _{\begin{array}{c}t\le 2r-1\\ t\mathrm{odd}\end{array}}t\hfill \\ \hfill & ={\displaystyle \sum _{r=2}^m}(4m-2r)(r-1)r+{\displaystyle \sum _{r=1}^{m-1}}(4m-2r-1)r^2.\hfill \end{array}$$
  The inner sum in $U_2$ (odd, $M=2m-3$):

  $$\sum _{\begin{array}{c}t\le 2m-3\\ t\mathrm{odd}\end{array}}t={(m-1)}^2,$$

  so $U_2=(2m-1){(m-1)}^2$. Adding and simplifying (using formulas $\sum r={\displaystyle \frac{m(m-1)}{2}}$, etc.) we obtain

  $$U_{\mathrm{even}}=-1+{\displaystyle \frac{25}{12}}n-{\displaystyle \frac{31}{24}}{n}^2+{\displaystyle \frac{1}{24}}{n}^3+{\displaystyle \frac{5}{48}}{n}^4.$$
• Sum V: Here, the range is $n+2=2m+2$ to $2n-3=4m-3$ and the bound is $2n-2s=4m-2-s$. Write $s=2r$ and $s=2r+1$:

  $$\begin{array}{cc}\hfill V& ={\displaystyle \sum _{r=m+1}^{2m-2}}(2r-2)\sum _{\begin{array}{c}t\le 4m-2-2r\\ t\mathrm{even}\end{array}}t+{\displaystyle \sum _{r=m}^{2m-2}}(2r-1)\sum _{\begin{array}{c}t\le 4m-3-2r\\ t\mathrm{odd}\end{array}}t\hfill \\ \hfill & ={\displaystyle \sum _{r=m+1}^{2m-2}}(2r-2)(2m-1-r)(2m-r)+{\displaystyle \sum _{r=m}^{2m-2}}(2r-1){(2m-r-1)}^2.\hfill \end{array}$$

Simplifying yields

$$V_{\mathrm{even}}=1-{\displaystyle \frac{11}{4}}n+{\displaystyle \frac{59}{24}}{n}^2-{\displaystyle \frac{7}{8}}{n}^3+{\displaystyle \frac{5}{48}}{n}^4.$$

Case $n=2m+1$ (odd, $m\ge 1$). A similar procedure (now $n+1=2m+2$, $2n-3=4m-1$, etc.) gives:

$$\begin{array}{cc}\hfill U_{\mathrm{odd}}& =-{\displaystyle \frac{13}{16}}+{\displaystyle \frac{47}{24}}n-{\displaystyle \frac{31}{24}}{n}^2+{\displaystyle \frac{1}{24}}{n}^3+{\displaystyle \frac{5}{48}}{n}^4,\hfill \\ \hfill V_{\mathrm{odd}}& ={\displaystyle \frac{15}{16}}-{\displaystyle \frac{21}{8}}n+{\displaystyle \frac{59}{24}}{n}^2-{\displaystyle \frac{7}{8}}{n}^3+{\displaystyle \frac{5}{48}}{n}^4.\hfill \end{array}$$

In both cases, adding U and V we obtain

$$S_1={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{5}{6}}{n}^3+{\displaystyle \frac{7}{6}}{n}^2-{\displaystyle \frac{2}{3}}n+{\displaystyle \frac{1}{16}}\left(1-{(-1)}^n\right).$$

This can also be written as

$$S_1={\displaystyle \frac{1}{24}}\left(5n^4-20n^3+28n^2-16n+{\displaystyle \frac{3}{2}}\left(1-{(-1)}^n\right)\right).$$

(For even n the second term vanishes, for odd n it adds $1/8$.)

Adding all contributions we get

$$\begin{array}{cc}\hfill \mathrm{HT}\left(B_{n,n-1}\right)& =S_1+S_2+S_3+S_4\hfill \\ \hfill & =\left[{\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{5}{6}}{n}^3+{\displaystyle \frac{7}{6}}{n}^2-{\displaystyle \frac{2}{3}}n+{\displaystyle \frac{1}{16}}\left(1-{(-1)}^n\right)\right]+0+(2n-3)\hfill \\ \hfill & +{\displaystyle \frac{1}{6}}\left(4n^3-9n^2-7n+18\right)\hfill \\ \hfill & ={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{1}{6}}{n}^3-{\displaystyle \frac{1}{3}}{n}^2+{\displaystyle \frac{1}{6}}n+{\displaystyle \frac{1}{16}}\left(1-{(-1)}^n\right).\hfill \end{array}$$

This completes the proof. □

Lemma 2. 

For every $n\ge 4$ the hitting time index of the broom graph $B_{n,n-2}$ is equal to

$$\mathrm{HT}\left(B_{n,n-2}\right)=\left\{\begin{array}{cc}{\displaystyle \frac{5}{24}{n}^4-\frac{1}{6}{n}^3-\frac{7}{3}{n}^2+\frac{79}{6}n-25,}\hfill & n\mathit{even},\hfill \\ {\displaystyle \frac{5}{24}{n}^4-\frac{1}{6}{n}^3-\frac{7}{3}{n}^2+\frac{79}{6}n-\frac{199}{8},}\hfill & n\mathit{odd}.\hfill \end{array}\right.$$

Proof. 

Substitute $d=n-2$ and $m=2$ in Theorem 1. Then

$$\mathrm{HT}\left(B_{n,n-2}\right)=S_1+S_2+S_3+2{\displaystyle \sum _{i=1}^{n-3}}max\{A\left(i\right),B\left(i\right)\},$$

where

$$\begin{array}{cc}\hfill S_1& =\sum _{1\le i<j\le n-2}(j-i)max\{i+j-2,2n-j-i\},\hfill \\ \hfill S_2& ={\displaystyle \frac{2(2-1)}{2}}(2n-2)=2n-2,\hfill \\ \hfill S_3& =2(2n-3)=4n-6,\hfill \\ \hfill A\left(i\right)& =(n-2-i)(2n-(n-2)-i)+1={(n-i)}^2-3,\hfill \\ \hfill B\left(i\right)& =(n-1-i)(n+i-3)+2.\hfill \end{array}$$

Define $Q\left(i\right)={\displaystyle \frac{1}{2}}\left(A\left(i\right)-B\left(i\right)\right)=i^2-(n+1)i+2n-4$. Then $Q\left(1\right)=n-4$, $Q\left(2\right)=-2$, and $Q\left(i\right)\le Q\left(3\right)=2-n<0$ for $i\ge 3$ when $n\ge 5$. For $n=4$ the statement is verified directly (it reduces to the known value for $B_{4,2}$). Thus, for $n\ge 5$ we have

$$max\{A\left(i\right),B\left(i\right)\}=\left\{\begin{array}{cc}A\left(1\right),\hfill & i=1,\hfill \\ B\left(i\right),\hfill & 2\le i\le n-3.\hfill \end{array}\right.$$

It follows that

$$S_4=2\left(A\left(1\right)+{\displaystyle \sum _{i=2}^{n-3}}B\left(i\right)\right)=2\left(A\left(1\right)-B\left(1\right)+{\displaystyle \sum _{i=1}^{n-3}}B\left(i\right)\right)=4(n-4)+2{\displaystyle \sum _{i=1}^{n-3}}B\left(i\right).$$

(14)

To compute the sum, introduce the substitution $k=n-1-i$. When i runs from 1 to $n-3$, k runs from $n-2$ to 2; therefore

$${\displaystyle \sum _{i=1}^{n-3}}(n-1-i)(n+i-3)={\displaystyle \sum _{k=2}^{n-2}}k(2n-4-k).$$

Using ${\sum }_{k=2}^{m}k={\displaystyle \frac{m(m+1)}{2}}-1$ and ${\sum }_{k=2}^m{k}^2={\displaystyle \frac{m(m+1)(2m+1)}{6}}-1$ with $m=n-2$, by direct simplification we obtain

$${\displaystyle \sum _{i=1}^{n-3}}B\left(i\right)={\displaystyle \frac{1}{6}}\left(4n^3-21n^2+35n-24\right).$$

Substituting into (14) we get

$$S_4={\displaystyle \frac{1}{3}}\left(4n^3-21n^2+47n-72\right).$$

(15)

Now we turn to $S_1$. Let $S_1^{(n-1)}$ be the sum $S_1$ for the broom $B_{n,n-1}$ given in Lemma 1

$$S_1^{(n-1)}={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{5}{6}}{n}^3+{\displaystyle \frac{7}{6}}{n}^2-{\displaystyle \frac{2}{3}}n+\delta ,\delta =\left\{\begin{array}{cc}0,\hfill & n\mathit{even},\hfill \\ {\displaystyle \frac{1}{8}},\hfill & n\mathit{odd}.\hfill \end{array}\right.$$

The sum for $B_{n,n-2}$ is obtained by removing all pairs that contain the vertex $u_{n-1}$:

$$S_1=S_1^{(n-1)}-{\displaystyle \sum _{i=1}^{n-2}}(n-1-i)max\{n+i-3,n-i+1\}.$$

For $i=1$ the maximum is n; for $2\le i\le n-2$ it equals $n+i-3$. Hence, the subtracted part $\Delta S_1$ is

$$\Delta S_1=(n-2)n+{\displaystyle \sum _{i=2}^{n-2}}(n-1-i)(n+i-3)={\displaystyle \frac{n-2}{6}}\left(4n^2-13n+21\right).$$

Thus

$$S_1={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{3}{2}}{n}^3+{\displaystyle \frac{14}{3}}{n}^2-{\displaystyle \frac{17}{2}}n+7+\delta .$$

(16)

Finally,

$$\mathrm{HT}\left(B_{n,n-2}\right)=S_1+S_2+S_3+S_4=S_1+(6n-8)+S_4.$$

Substituting (15) and (16) and simplifying by powers of n we obtain

$$\mathrm{HT}\left(B_{n,n-2}\right)={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{1}{6}}{n}^3-{\displaystyle \frac{7}{3}}{n}^2+{\displaystyle \frac{79}{6}}n-25+\delta ,$$

which is exactly the formula from the statement of the lemma. □

Lemma 3. 

For every $n\ge 5$ the hitting time index of the broom graph $B_{n,n-3}$ is equal to

$$\mathrm{HT}\left(B_{n,n-3}\right)=\left\{\begin{array}{cc}76,\hfill & n=5\hfill \\ 176,\hfill & n=6\hfill \\ 348,\hfill & n=7\hfill \\ {\displaystyle \frac{1}{24}}\left(5n^4-4n^3-152n^2+1228n-3576\right),\hfill & n\mathit{even},n\ge 8\hfill \\ {\displaystyle \frac{1}{24}}\left(5n^4-4n^3-152n^2+1228n-3573\right),\hfill & n\mathit{odd},n\ge 9.\hfill \end{array}\right.$$

Proof. 

For $n=5,6,7$ the values are verified by direct computation from Theorem 1. Assume $n\ge 8$ and set $d=n-3$, $m=3$ in Theorem 1. Then

$$\mathrm{HT}\left(B_{n,n-3}\right)=S_1+S_2+S_3+3{\displaystyle \sum _{i=1}^{n-4}}max\{A\left(i\right),B\left(i\right)\},$$

where

$$\begin{array}{cc}\hfill S_1& =\sum _{1\le i<j\le n-3}(j-i)max\{i+j-2,2n-j-i\},\hfill \\ \hfill S_2& ={\displaystyle \frac{3·2}{2}}(2n-2)=6n-6,\hfill \\ \hfill S_3& =3(2n-3)=6n-9,\hfill \\ \hfill A\left(i\right)& =(n-3-i)(n+3-i)+1={(n-i)}^2-8,\hfill \\ \hfill B\left(i\right)& =(n-2-i)(n+i-4)+4.\hfill \end{array}$$

Define $Q\left(i\right)={\displaystyle \frac{1}{2}}\left(A\left(i\right)-B\left(i\right)\right)=i^2-(n+1)i+3n-10$. For $n\ge 8$ we have $Q\left(1\right)=2n-10>0$, $Q\left(2\right)=n-8\ge 0$, $Q\left(3\right)=-4<0$, and for $i\ge 3$ the function $Q\left(i\right)$ decreases (the vertex of the parabola is $i_0={\displaystyle \frac{n+1}{2}}>n-4$). Therefore

$$max\{A\left(i\right),B\left(i\right)\}=\left\{\begin{array}{cc}A\left(i\right),\hfill & i=1,2,\hfill \\ B\left(i\right),\hfill & 3\le i\le n-4.\hfill \end{array}\right.$$

Hence

$$S_4=3\left(A\left(1\right)+A\left(2\right)+{\displaystyle \sum _{i=3}^{n-4}}B\left(i\right)\right)=3\left(A\left(1\right)-B\left(1\right)+A\left(2\right)-B\left(2\right)+{\displaystyle \sum _{i=1}^{n-4}}B\left(i\right)\right).$$

Since $A\left(1\right)-B\left(1\right)=2Q\left(1\right)=4n-20$ and $A\left(2\right)-B\left(2\right)=2Q\left(2\right)=2n-16$, we obtain

$$S_4=3(6n-36)+3{\displaystyle \sum _{i=1}^{n-4}}B\left(i\right).$$

The sum ${\sum }_{i=1}^{n-4}B\left(i\right)$ is computed by the substitution $k=n-2-i$ ($k=2,\dots ,n-3$) and standard summation formulas

$${\displaystyle \sum _{i=1}^{n-4}}B\left(i\right)={\displaystyle \frac{1}{6}}\left(4n^3-33n^2+101n-132\right).$$

Substituting,

$$S_4=18n-108+{\displaystyle \frac{1}{2}}\left(4n^3-33n^2+101n-132\right)={\displaystyle \frac{1}{2}}\left(4n^3-33n^2+137n-348\right).$$

Now we turn to $S_1$. Let $S_1^{(n-2)}$ be the sum $S_1$ for $B_{n,n-2}$ given in Lemma 2

$$S_1^{(n-2)}={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{3}{2}}{n}^3+{\displaystyle \frac{14}{3}}{n}^2-{\displaystyle \frac{17}{2}}n+7+\delta ,$$

where $\delta =0$ for even n and $\delta ={\displaystyle \frac{1}{8}}$ for odd n. We remove all pairs containing the vertex $u_{n-2}$

$$\Delta S_1={\displaystyle \sum _{i=1}^{n-3}}(n-2-i)max\{n+i-4,n-i+2\}.$$

A careful case analysis (the maximum changes at $i=2$) gives

$$\Delta S_1={\displaystyle \frac{1}{6}}\left(4n^3-33n^2+125n-198\right).$$

It follows

$$S_1=S_1^{(n-2)}-\Delta S_1={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{13}{6}}{n}^3+{\displaystyle \frac{61}{6}}{n}^2-{\displaystyle \frac{88}{3}}n+40+\delta .$$

Finally, sum $S_1$, $S_2+S_3=12n-15$ and $S_4$

$$\begin{array}{cc}\hfill \mathrm{HT}\left(B_{n,n-3}\right)& =\left({\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{13}{6}}{n}^3+{\displaystyle \frac{61}{6}}{n}^2-{\displaystyle \frac{88}{3}}n+40+\delta \right)\hfill \\ \hfill & +(12n-15)\hfill \\ \hfill & +\left({\displaystyle \frac{1}{2}}\left(4n^3-33n^2+137n-348\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{5}{24}}{n}^4+\left(-{\displaystyle \frac{13}{6}}+2\right)n^3+\left({\displaystyle \frac{61}{6}}-{\displaystyle \frac{33}{2}}\right)n^2+\left(-{\displaystyle \frac{88}{3}}+12+{\displaystyle \frac{137}{2}}\right)n\hfill \\ \hfill & +\left(40-15-174\right)+\delta \hfill \\ \hfill & ={\displaystyle \frac{5}{24}}{n}^4-{\displaystyle \frac{1}{6}}{n}^3-{\displaystyle \frac{19}{3}}{n}^2+{\displaystyle \frac{307}{6}}n-149+\delta .\hfill \end{array}$$

Expressed with a common denominator 24, this matches the formula stated in the lemma, which completes the proof. □

Note that the path $P_n$ can be represented as the broom graph $B_{n,n-1}$. Indeed, for $d=n-1$ we have $m=n-(n-1)=1$. The graph $B_{n,n-1}$ consists of a path $P_{n-1}$ (vertices $u_1,u_2,\dots ,u_{n-1}$) and one leaf $v_1$ attached to $u_{n-1}$. Renaming $v_1$ as $u_n$ gives the path $u_1-u_2-\dots -u_{n-1}-u_n$, which is exactly $P_n$. Hence, $B_{n,n-1}\cong P_n$.

Lemma 4. 

For every $n\ge 5$ the following chain of inequalities holds:

$$\mathrm{HT}\left(B_{n,n-3}\right)<\mathrm{HT}\left(B_{n,n-2}\right)<\mathrm{HT}\left(B_{n,n-1}\right)=\mathrm{HT}\left(P_n\right).$$

Proof. 

Using the closed formulas for $\mathrm{HT}\left(B_{n,n-1}\right)$ and $\mathrm{HT}\left(B_{n,n-2}\right)$ and subtracting directly we obtain

$$\mathrm{HT}\left(B_{n,n-1}\right)-\mathrm{HT}\left(B_{n,n-2}\right)=2n^2-13n+25,$$

which holds for all $n\ge 5$ (both even and odd). The discriminant of the quadratic trinomial is $\Delta ={13}^2-4·2·25=169-200=-31<0$, and the leading coefficient $2>0$, so $2n^2-13n+25>0$ for all real n, thus $\mathrm{HT}\left(B_{n,n-2}\right)<\mathrm{HT}\left(B_{n,n-1}\right)$.

For the second inequality, for $n\ge 8$ direct subtraction of the closed formulas gives

$$\mathrm{HT}\left(B_{n,n-2}\right)-\mathrm{HT}\left(B_{n,n-3}\right)=4n^2-38n+124.$$

The discriminant of $4n^2-38n+124$ is $\Delta ={38}^2-4·4·124=1444-1984=-540<0$, and the leading coefficient $4>0$, so $4n^2-38n+124>0$ for all real n, in particular for all $n\ge 8$. For $n=5,6,7$, the inequality $\mathrm{HT}\left(B_{n,n-3}\right)<\mathrm{HT}\left(B_{n,n-2}\right)$ is checked directly:

$$\begin{array}{cc}\hfill \mathrm{HT}\left(B_{5,2}\right)=76& <92=\mathrm{HT}\left(B_{5,3}\right),\hfill \\ \hfill \mathrm{HT}\left(B_{6,3}\right)=176& <204=\mathrm{HT}\left(B_{6,4}\right),\hfill \\ \hfill \mathrm{HT}\left(B_{7,4}\right)=348& <396=\mathrm{HT}\left(B_{7,5}\right).\hfill \end{array}$$

Thus both inequalities hold for all $n\ge 5$, which establishes the required chain. □

To see why the threshold $n_0\left(k\right)$ and the quartic form appear, let us analyze the difference $A\left(i\right)-B\left(i\right)$ for the family $B_{n,n-k}$. Substituting $d=n-k$ into the expressions from Theorem 1 and simplifying, we obtain

$$Q\left(i\right)={\displaystyle \frac{A\left(i\right)-B\left(i\right)}{2}}=i^2-(n+1)i+nk-k^2-k+2.$$

The quadratic function $Q\left(i\right)$ opens upward, and its vertex is at $i_0=(n+1)/2$. For $n\ge 4k-4$ and $k\ge 3$, one easily checks that $i_0>k-1$, so $Q\left(i\right)$ is strictly decreasing on the interval $[1,k-1]$. Therefore, from $Q(k-1)=n-4k+4\ge 0$ it follows that $Q\left(i\right)\ge 0$ for all $i\le k-1$.

• For $i\ge k$, we note that $Q\left(k\right)=2-2k<0$ and $Q(n-k-1)=2(k-n+2)<0$ (since $n\ge 4k-4>k+2$). Because the quadratic function with a positive leading coefficient is convex; its maximum on the interval $[k,n-k-1]$ is attained at one of the endpoints. Both endpoints are negative, hence $Q\left(i\right)<0$ for all $i\ge k$.
• It follows that on the whole interval we have

$$max\{A\left(i\right),B\left(i\right)\}=\left\{\begin{array}{cc}A\left(i\right),\hfill & 1\le i\le k-1,\hfill \\ B\left(i\right),\hfill & k\le i\le n-k-1.\hfill \end{array}\right.$$

A similar uniform behavior holds for the maxima inside $S_1$. Substituting this stable sign pattern into the formula from Theorem 1 and performing symbolic summation (in the same way as done for $k=1,2,3$ in Lemmas 1–3) leads to a quartic polynomial whose coefficients are exactly those stated in the conjecture that follows. The threshold $n_0\left(k\right)$ is therefore the minimal value that guarantees that the pattern does not change, and extensive numerical checks (for $n\le 100$) confirm that there are no counterexamples.

Conjecture 3. 

Let $k\ge 1$ be an integer and let

$$n_0\left(k\right)=\left\{\begin{array}{cc}k+2,\hfill & k=1,2,\hfill \\ 4k-4,\hfill & k\ge 3.\hfill \end{array}\right.$$

Then for all $n\ge n_0\left(k\right)$ the hitting time index of the broom graph $B_{n,n-k}$ can be expressed as a quartic polynomial in n:

$$\mathrm{HT}\left(B_{n,n-k}\right)={\displaystyle \frac{1}{24}}\left(5n^4-4n^3+A\left(k\right)n^2+B\left(k\right)n+C\left(k\right)+\delta \left(n\right)\right),$$

where the coefficients are given by

$$\begin{array}{cc}\hfill A\left(k\right)& =-24k^2+24k-8,\hfill \\ \hfill B\left(k\right)& =48k^3+12k^2-60k+4,\hfill \\ \hfill C\left(k\right)& =-36k^4-80k^3+192k^2-76k,\hfill \end{array}$$

and

$$\delta \left(n\right)=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}n\mathit{is}\mathit{even},\hfill \\ 3,\hfill & \mathrm{if}n\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

4. Conclusions

In this paper, we derived a closed-form formula for the hitting time index of the broom graph $B_{n,d}$, valid for all $2\le d\le n-1$, expressed through four basic sums. For the case $d\ge 2$ and $n\ge 4d-8$, we reduced the formula to a cubic polynomial in n with explicit coefficients that depend only on d. We also proved, under the condition $n\ge 4d-8$, that the index increases monotonically with d.

Quartic polynomial formulas are given for the special families $B_{n,n-1}$, $B_{n,n-2}$ and $B_{n,n-3}$, and a general conjecture is presented for all $B_{n,n-k}$.

Our results complement the existing literature on the hitting time index and can serve as a starting point for further research on more complex classes of trees, as well as for the development of algorithms for fast computation of the index on large graphs.