Lemma 1. Let G be the graph with degree sequence D(G) = [d₁, d₂, …, d_n]. Suppose there exists a pair (d_p, d_q) with d_p ≥ d_q + 2 and let a new graph G^′ be obtained from G by changing the pair (d_p, d_q) with (d_p − 1, d_q + 1). If a > 1, then SEI_a(G) > SEI_a(G^′).

Proof. Using the Mean value theorem and the definition of the SEI_a index, we have

Lemma 2. Let G be the graph with degree sequence D(G) = [d₁, d₂, …, d_n]. Suppose there exists a pair (d_p, d_q) such that 2 ≤ d_q ≤ d_p ≤ n − 2, and let a new graph G^′ be obtained from G by changing the pair (d_p, d_q) with (d_p + 1, d_q − 1). If a > 1, then SEI_a(G^′) > SEI_a(G).

Proof. Using the mean value theorem and the definition of the SEI_a index, we have

Theorem 1. Let a > 1 and T be a tree of order n. Then,

• (1)

  SEI_a(T) attains its maximum value iff T≅S_n

• (2)

  SEI_a(T) attains its 2^nd maximum value iff T≅S_n−3,1

• (3)

  SEI_a(T) attains its 3^rd maximum value iff T≅S_n−4,2

• (4)

  SEI_a(T) attains its minimum value iff T≅P_n

• (5)

  SEI_a(T) attains its 2^nd minimum value iff T≅P_2,n−2

• (6)

  SEI_a(T) attains its 3^rd minimum value iff T≅P_3,n−3

Proof.

• (1)

  Suppose the degree sequence of T is D(T) = [d₁, d₂, …, d_n]. If T is not isomorphic to S_n, then there exists a pair (d_p, d_q) with 2 ≤ d_q ≤ d_p ≤ n − 2. Let T₁ be a tree obtained from T by changing the pair (d_p, d_q) with (d_p + 1, d_q − 1), then it follows from Lemma 2 that SEI_a(T₁) > SEI_a(T). We repeat the above procedure unless no pair (d_p, d_q) is left with 2 ≤ d_q ≤ d_p ≤ n − 2. In this way, we get a sequence of trees T₁, T₂, …, T_l such that SEI_a(T₁) < SEI_a(T₂) < ⋯<SEI_a(T_l−1) < SEI_a(T_l). Clearly, T_l≅S_n with degree sequence D(T_l) = [n − 1, 1ⁿ⁻¹], and for any tree T not isomorphic to S_n, SEI_a(T) < SEI_a(S_n).

• (2)

  Above construction shows that S_n is obtained from T_l by changing the pair (d_p, d_q) with (d_p + 1, d_q − 1), where 2 ≤ d_q ≤ d_p ≤ n − 2. Observe that D(T_l−1) = [n − 2,2, 1ⁿ⁻²] and T_l−1≅S_n−3,1.

• (3)

  Similarly, the degree sequence D(T_l−2) of T_l−2 has two cases: $$ and $$. Note that $$ can be obtained from $$ by replacing the pair (2,2) of $$ by (3,1). Clearly, $$ and $$. By Lemma 1, we have $$. Hence, T_l−2≅S_n−4,2.

• (4)

  Let T^′ be a tree of order n with degree sequence $$. Suppose T^′ is not isomorphic to P_n; then, there exists a pair $$ with $$. Let $$ be a tree obtained from T^′ by changing the pair $$ with the pair $$. Then, by Lemma 1, $$. Repeat the same operation until there is no pair $$ such that $$ for all p, q. In this way, we get a sequence of trees $$ such that $$ and $$. Hence, for any tree T^′ not isomorphic to P_n, SEI_a(T^′) > SEI_a(P_n).

• (5)

  Note that the degree sequence of P_n is D(P_n) = [2ⁿ⁻², 1²]. Now, P_n is obtained from $$ by changing the pair $$ with the pair $$, where $$. It is clear that the degree sequence of $$ is $$ and $$.

• (6)

  Similarly, $$ has the following cases: $$ or $$. Note that [3,3, 2ⁿ⁻⁶, 1⁴] can be obtained from [4,2, 2ⁿ⁻⁶, 1⁴] by replacing the pair (3,3) by the pair (4,2). Then, by Lemma 1, $$. Hence, $$.

From the above theorem, we get the following corollary.

Corollary 1. (see [7]). Let T be a tree of order n. Then,

Theorem 2. Let T be a chemical tree of order n and n − 2 = 3d₄ + i, where i = 0,1,2. If a > 1, then we have

• (1)

  SEI_a(T) attains its maximum value iff $$

• (2)

  SEI_a(T) attains its 2^nd maximum value iff $$ for $$ for i = 1, and $$ for i = 2

Proof.

• (1)

  Let T be a chemical tree of order n and let D(T) = [d₁, d₂, …, d_n] be its degree sequence. Suppose there exists a pair (d_p, d_q) such that 2 ≤ d_q ≤ d_p ≤ 3. We construct a graph T₁ from T by changing the pair (d_p, d_q) with the pair (d_p + 1, d_q − 1). Then, by Lemma 2, we have SEI_a(T₁) > SEI_a(T). We repeat the above procedure unless no pair (d_p, d_q) is left with 2 ≤ d_q ≤ d_p ≤ 3. In this way, we get a sequence of trees T₁, T₂, …, T_l such that SEI_a(T₁) < SEI_a(T₂) < ⋯<SEI_a(T_l−1) < SEI_a(T_l). Note that T_l has exactly one vertex of degree 2 or degree 3, while the remaining vertices are of degree 4 or degree 1. Let d₁, d₂, d₃,  and d₄ be the vertices of degree 1, degree 2, degree 3, and degree 4, respectively; then,

•  

  From the above equations, we get the following solutions:

  • (1)

    d₂ = d₃ = 0, d₁ = n − d₄, and d₄ = (n − 2/3) if n − 2 ≡ 0(mod3)

  • (2)

    d₂ = 1, d₃ = 0, d₁ = n − d₄ − 1, and d₄ = (n − 3/3) if n − 2 ≡ 1(mod3)

  • (3)

    d₂ = 0, d₃ = 1, d₁ = n − d₄ − 1, and d₄ = (n − 4/3) if n − 2 ≡ 2(mod3)

•  

  The solutions shows that SEI_a(T) attains its maximum value if and only if $$.

• (2)

  If i = 0, then from the proof of (2), it follows that $$. One can see that D(T_l) is obtained by D(T_l−1) by changing the pair (d_p, d_q) with the pair (d_p + 1, d_q − 1), where 2 ≤ d_q ≤ d_p ≤ 3. So, we have $$. If i = 2, then D(T_l−1) has the following two cases: $$ or $$. Note that $$ can be obtained from $$ by replacing the pair (3,1) by the pair (2,2). Then, by Lemma 1, $$. Hence, $$. The case for i = 3 follows in the same way.

From the above theorem, we get the following corollary.

Corollary 2. Let T be a chemical tree of order n and n − 2 = 3d₄ + i, where i = 0,1,2. Then, we have

Theorem 3. Let a > 1 and T be a tree of order n with n ≥ 3 and 2 ≤ d ≤ n − 1. Then,

• (1)

  SEI_a(T) attains its maximum value iff D(T) = [n − d + 1, 2^d−2, 1^n−d+1]

• (2)

  For 3 ≤ d ≤ n − 3, SEI_a(T) attains its 2^nd maximum value iff D(T) = [n − d, 3, 2^d−3, 1^n−d+1]

• (3)

  For d = n − 4, SEI_a(T) attains its third maximum value iff D(T) = [3³, 2ⁿ⁻⁸, 1⁵] and for 3 ≤ d ≤ n − 5, SEI_a(T) attains its 3^rd maximum value iff D(T) = [n − d − 1,4, 2^d−3, 1^n−d+1]

Proof. The case d = 2 is trivial. Suppose that d ≥ 3. Let P_d be a path of length d on T and v ∈ V(T) be a vertex of maximum degree on P_d. If D(T) ≠ [n − d + 1, 2^d−2, 1^n−d+1], then there exists a vertex u ∈ V(T) satisfying

• (a)

  u ∉ P_d and d_u ≥ 2

• (b)

  u ≠ v, u ∉ P_d, and d_u ≥ 3

First, suppose that there exists a vertex u ∉ P_d and d_u ≥ 2. Choose u such that u is adjacent to exactly d_u − 1 pendent vertices. Let the neighbors of u be $$, and let T_u = T − uu₁ − uu₂ − ⋯uu_k + vu₁ + vu₂ + ⋯+vu_k.

Claim. :SEI_a(T) < SEI_a(T_u).

If d_u ≤ d_v in T, then the claim follows from Lemma 2. Suppose d_u > d_v in T and let T^′ be the tree obtained from T by changing the pair (d_v, d_u) with the pair (d_u, d_v) by pendent vertices (of u) transformation. Note that d_u ≤ d_v in T′. Then, it is easy to see that SEI_a(T) ≤ SEI_a(T ^′). Now, the tree T_u can be obtained from the tree T′ by applying the successive transformations and changing the pair (d_v, d_u) with the pair (d_v + 1, d_u − 1). Thus, by Lemma 2, if follows SEI_a(T ^′) < SEI_a(T_u). Hence, for d_u > d_v, we have SEI_a(T) < SEI_a(T_u). This proves our claim.

We repeat the above process until there is no vertex in T satisfying (a). Finally, we obtain a tree T₁ which has no vertex satisfying (a) and SEI_a(T) < SEI_a(T₁).

• (1)

  Next, we suppose that there is a vertex u satisfying (b) for T₁ (instead of T). Let the neighbors of u be $$, where u₁ and u₂ lie on the path P_d. Let $$ for i = 1,2, …, d_u − 2. By Lemma 2, $$. Repeating the above operation, we get a sequence of trees $$ on n vertices and diameter d such that $$. Note that, in $$, there is no pair of distinct vertices with degree greater or equal to three. Clearly, $$. This proves (1).

• (2)

  Suppose 3 ≤ d ≤ n − 3. Since $$ and $$ is obtained from $$ by changing the pair (d_p, d_q) with the pair (d_p + 1, d_q − 1), where n − 3 ≥ d_p ≥ d_q ≥ 3, it is easy to see that $$ has two cases: $$ and $$. Note that $$ can be obtained from $$ by replacing the pair (3,1) by the pair (2,2). Then, by Lemma 1, $$. This proves (2).

• (3)

  In the similar way, we get $$ for 3 ≤ d ≤ n − 5 and $$ for d = n − 4. Since $$ can be obtained from $$ by replacing the pair (4,2) by the pair (3,3), by Lemma 1, $$. It follows that $$ for d = n − 4 and $$. This proves (3).

A vertex v ∈ V(T) is called central vertex if e(v) = r. Every tree T has either one or two central vertices. If d = 2r, then the tree T has just one central vertex, and if d = 2r − 1, then T has two central vertices. If we choose a tree with given radius r and d = 2r − 1, then we get the following result.

Theorem 4. Let a > 1 and T be a tree of order n with n ≥ 3 and 2 ≤ r ≤ (n/2). Then,

• (1)

  SEI_a(T) attains its maximum value iff D(T) = [n − 2r + 2, 2^2r−3, 1^n−2r+2]

• (2)

  For 2 ≤ r, (n − 2/2), SEI_a(T) attains its 2^nd maximum value iff D(T) = [n − 2r + 1,3, 2^2r−4, 1^n−2r+2]

• (3)

  For r = (n − 3/2) ≥ 3, SEI_a(T) attains its 3^rd maximum value iff D(T) = [3³, 2ⁿ⁻⁸, 1⁵] and for 2 ≤ r < (n − 3/2), and SEI_a(T) attains its 3^rd maximum value iff D(T) = [n − 2r, 4, 2^2r−4, 1^n−2r+1]