Bounding mean orders of sub-$k$-trees of $k$-trees

For a $k$-tree $T$, we prove that the maximum local mean order is attained in a $k$-clique of degree $1$ and that it is not more than twice the global mean order. We also bound the global mean order if $T$ has no $k$-cliques of degree $2$ and prove that for large order, the $k$-star attains the minimum global mean order. These results solve the remaining problems of Stephens and Oellermann [J. Graph Theory 88 (2018), 61-79] concerning the mean order of sub-$k$-trees of $k$-trees.


Introduction
In [10] and [11] Jamison considered the mean number of nodes in subtrees of a given tree.He showed that for trees of order n, the average number of nodes in a subtree of T is at least (n + 2)/3, with this minimum achieved if and only if T is a path.He also showed that the average number of nodes in a subtree containing a root is at least (n + 1)/2 and always exceeds the average over all unrooted subtrees.The mean subtree order in trees was further investigated, e.g. in [3,7,14,20,22], as well as extensions to arbitrary graphs [2,[4][5][6] and the mean order of the connected induced subgraphs of a graph [8,9,[17][18][19].
In [16], Stephens and Oellermann extended the study to k-trees and families of sub-k-trees.A k-tree is a generalization of a tree that has the following recursive construction.
Definition 1 (k-tree).Let k be a fixed positive integer.
1.The complete graph K k is a k-tree.

If T is a k-tree, then so is the graph obtained from T by joining a new vertex to all vertices
of some k-clique of T .
3. There are no other k-trees.
Note that for k = 1 we have the standard recursive construction of trees.A sub-k-tree of a k-tree T is a subgraph that is itself a k-tree.Let S(T ) denote the collection of all sub-k-trees of T and let N (T ) := |S(T )| be the number of sub-k-trees.We denote by R(T ) = X∈S(T ) |X| the total number of vertices in all sub-k-trees (where for a graph G the notation |G| is used throughout to mean the number of vertices in G).The global mean (sub-k-tree) order is µ(T ) = R(T ) N (T ) .The degree of C is the number of (k + 1)-cliques that contain C.
Stephens and Oellermann concluded their study of the mean order of sub-k-trees of k-trees with several open questions.Our main contribution here is to answer three of them.
It was conjectured by Jamison in [10] and proven by Vince and Wang in [20] that for trees of order n without vertices of degree 2-called series-reduced trees-the global mean subtree order is between n 2 and 3n 4 .For k-trees we provide similar asymptotically sharp bounds, answering [16,Problem 6].
These bounds are asymptotically sharp.In particular, for large k, 3n 4 is not an upper bound.For large n, the k-star is the unique extremal k-tree for the lower bound.
Wagner and Wang proved in [21] that the maximum local mean subtree order occurs at a leaf or a vertex of degree 2. We prove an analogous result for k-trees, answering [16,Problem 4].In contrast to the result for trees, it turns out that for k ≥ 2, the maximum can only occur at a k-clique of degree 1, unless T is a k-tree of order k + 2. Theorem 3. Suppose that k ≥ 2. For a k-tree T of order n = k + 2, if a k-clique C maximizes µ(T ; C), then C must be a k-clique of degree 1.For n = k + 2, every k-clique C satisfies µ(T ; C) = k + 1.
Lastly, Jamison [10] conjectured that for a given tree T and any vertex v, the local mean order is at most twice the global mean order of all subtrees in T .Wagner and Wang [21] proved that this is true.Answering [16,Problem 2] affirmatively, we show Theorem 4. The local mean order of the sub-k-trees containing a fixed k-clique C is less than twice the global mean order of all sub-k-trees of T .

Related Results
A total of six questions were posed in [16].Problems 1 and 3 were solved by Luo and Xu [13].Regarding the first problem, Jamison [10] showed that for any tree T and any vertex v of T , the local mean order of subtrees containing v is an upper bound on the global mean order of subtrees of T .Stephens and Oellermann asked about a generalization to k-trees, to which Luo and Xu showed: Theorem 5 ([13]).For any k-tree T of order n with a k-clique C, we have µ(T ; C) ≥ µ(T ) with equality if and only if T ∼ = K k .
For the third problem, it was shown in [10] that paths have the smallest global mean subtree order.For k-trees we have: Theorem 6 ([13]).For any k-tree T of order n, we have µ(T ) ≥ + k with equality if and only if T is a path-type k-tree.
Very recently, Li, Ma, Dong, and Jin [12] gave a partial proof of Theorem 3, showing that the maximum local mean order always occurs at a k-clique of degree 1 or 2, thus also solving [16,Problem 4].They did this by combining the fact that the 1-characteristic trees of adjacent k-cliques can be obtained from each other by a partial Kelmans operation [12,Lem. 4.6], an inequality between local orders of neighboring vertices after performing a partial Kelmans operation [12,Thm. 3.3], as well as [16,Lem. 11] and [21,Thm. 3.2].Theorem 3 also solves Problem 5.4 from [12], asking whether the maximum local mean order can ever occur at a k-clique of degree 2, but not at a k-clique of degree 1.

Outline:
In Section 2, we go over definitions and notation.Theorems 2, 3, 4 are proven in Sections 3, 4, 5 respectively.We additionally address [16,Problem 5], which is a more general question asking what one can say about the local mean order of sub-k-trees containing a fixed r-clique for 1 ≤ r ≤ k.We give a possible direction and partial results in the concluding section.

Notation and Definitions
The global mean sub-k-tree order µ(T ) and the local mean sub-k-tree order µ(T ; C) are defined in the introduction.The local mean order always counts the k vertices from C and it sometimes will be more convenient to work with the average number of additional vertices in a (uniform random) sub-k-tree of T containing C, in which case we use the notation µ Moreover, the number of sub-k-trees not containing C will be denoted by N(T ; C) = N (T ) − N (T ; C) and the total number of vertices in the sub-k-trees that do not contain C will be denoted by A k-leaf or simplicial vertex is a vertex belonging to exactly one (k + 1)-clique of T , i.e., a vertex of degree k.A simplicial k-clique is a k-clique containing a k-leaf.Note that a k-clique of degree 1 is not necessarily simplicial.A major k-clique is a k-clique with degree at least 3. Two k-cliques are adjacent if they share a (k − 1)-clique.
The stem of a k-tree T is the k-tree obtained by deleting all k-leaves from T .
Subclasses of trees generalize to subclasses of k-trees.We will reference two in particular: paths generalize to path-type k-trees that are either isomorphic to K k or K k+1 or have precisely two k-leaves.Note that for every n ≥ k + 3 and k ≥ 3, there are multiple non-isomorphic path-type k-trees of order n.
A k-star is either K k or K k+1 , or it is the unique k-tree with n − k simplicial vertices when Furthermore, the combination of a k-star and a k-path is called a k-broom: take a k-path of a certain length and add some simplicial vertices to a simplicial k-clique of the k-path.In a 1-tree, any two vertices are connected by a unique path.This fact generalizes to k-trees through the construction of the 1-characteristic tree of a k-tree T [13,16].For a k-clique C in T , a perfect elimination ordering of T to C is an ordering v 1 , v 2 , . . ., v n−k of its vertices different from V (C) = {c 1 , . . ., c k } such that each vertex v i is simplicial in the k-tree spanned by C and v j , 1 ≤ j ≤ i.In [16] it is shown that for any v / ∈ C, there is a unique sequence of vertices that along with C induce a path-type k-tree P (C, v) and that form a perfect elimination ordering of P (C, v) to C. It is also proven that T can be written as P (C, v) where the union is taken over all k-leaves v ∈ V .Each k-tree P (C, v) has an associated 1-tree P ′ (C, v) where the vertices consist of a single vertex representing the entire clique C, along with the remaining non-C vertices of P (C, v).The edges are consecutive pairs from the perfect elimination ordering.Taking P ′ (C, v) over all k-leaves v gives us what is called the 1-characteristic tree of T , which we will denote T ′ C .See Figure 2 for an example.

Excluding k-cliques of degree 2
In this section, we prove Theorem 2. For every k-tree T without k-cliques of degree 2, the global mean sub-k-tree order satisfies These bounds are asymptotically sharp.In particular, for large k, 3n 4 is not an upper bound.For large n, the k-star is the unique extremal k-tree for the lower bound.

The lower bound
Among series-reduced trees of order n, for 4 ≤ n ≤ 8 the star attains the maximum mean subtree order, but for n ≥ 11 it attains the minimum mean subtree order.It can be derived from [7,Lem. 12] and an adapted version of [7,Cor. 11] (with 2 replaced by any ε > 0, at the cost of replacing 30 by n ε ) that this is indeed the case for n sufficiently large. 1 In [1], it is shown that for 6 ≤ n ≤ 10, the series-reduced trees attaining the minimum mean subtree order are those presented in Figure 3, and for n ≥ 11, S n is always the unique extremal graph.
In this subsection, we will prove that the above extremal statement generalizes to k-trees without a k-clique of degree 2. 1 We thank John Haslegrave for this remark Figure 3: Series-reduced trees with minimum mean subtree order for 6 ≤ n ≤ 10 First, we prove the following lemma, which states that every k-tree contains a (k + 1)-clique C that plays the role of a centroid in a tree (a vertex or edge whose removal splits the tree into components of size at most n 2 ). Figure 4 is an example of a 2-tree demonstrating why we must take a (k + 1)-clique and not a k-clique.Proof.Take C as in Lemma 7. Let its vertices be {u 1 , u 2 , . . ., u k+1 }.
For every component of T \C, there is a unique vertex u i , 1 ≤ i ≤ k+1, such that the component together with C \ u i forms a k-tree.Now we consider two cases that are handled analogously.
Case 1: There is some u i such that the union of components that form a k-tree when adding In this case, we consider the r ≤ n − k components of T \ C ′ and for every 1 ≤ i ≤ r, we let T i be the union of such a component and C ′ .
We then apply the following claim: Proof.Consider the 1-characteristic tree T ′ C ′ .For every u ∈ V \ C ′ , either all the k-cliques containing u have degree 1, in which case u has degree 1 in T ′ C ′ , or (at least) one of them has degree at least 3 and so does u in T ′ C ′ .Thus T ′ C ′ is series-reduced and thus has at least n−k 2 + 1 leaves.The latter implies that T has at least n−k 2 simplicial vertices.♦ From the claim, there are at least n−k 4 − 1 simplicial vertices not belonging to T i for every 1 ≤ i ≤ r.Observe that given any sub-k-tree of T containing C ′ , we can map it to its sub-k-tree intersection with T i .If two elements of S(T ; C ′ ) differ only in some subset of k-leaves of j =i T j , then they map to the same element of S(T i ; C ′ ).Thus, each element of S(T Using the perfect elimination ordering, every sub-k-tree S in T not containing C can be extended in a minimal way into a sub-k-tree containing C ′ .Furthermore, by considering the 1-characteristic tree T ′ C ′ of C ′ , it is clear that there are no more than k-trees that extend to the same tree.Here we have used Lemma 7. Thus, if we define a map from S(T i ; C ′ ) to S(T i ; C ′ ) using the minimal extension, every element of S(T i ; C ′ ) is mapped to at most n 2 times.Putting the previous two observations together, we have that the number of sub-k-trees containing By summing over all i, we obtain that The result follows now from µ(T ; C ′ ) ≤ n.
Case 2: For every u i the union of components in T \ C that form a k-tree when adding C \ u i has order smaller than n−k 2 .Let T i , 1 ≤ i ≤ k + 1, be the k-trees obtained above when adding C \ u i .By the claim above, for each i there are at least n−k 4 −1 k-leaves not belonging to T i .Thus, the same computations apply and in particular we have that N (T ; Proof of Theorem 2, lower bound.Take a k-clique C ′ which satisfies Lemma 8. Let T ′ C ′ be the 1-characteristic tree of T with respect to C ′ .By [13,Thm. 33] and [7, Lem.12], we conclude that µ(T ; , where i is the number of internal vertices in T ′ C ′ .By Lemma 8, we conclude.For sharpness, observe that if T is not a k-star, we have i ≥ 2 and the lower bound inequality is strict.When T is a k-star, it contains no k-cliques of degree 2 provided that n > k + 2. As computed in [16]

The upper bound
In this subsection, we generalize to k-trees the statement that a series-reduced tree has average subtree order at most 3n 4 by giving a lower bound for the number of k-leaves and proving that k-leaves belong to at most half of the sub-k-trees.This idea was also used in [7].Note that the upper bound is slightly larger than 3n 4 for larger k, which intuitively can be explained by the fact that the smallest sub-k-tree already has k vertices, and more precisely the vertices in the base k-clique will all be major vertices.
Proof of Theorem 2, upper bound.Our upper bound will come from the observation that µ(T ) = v∈T p(v) where p(v) is the fraction of sub-k-trees containing v. We will specifically consider when v is a k-leaf and bound the corresponding terms in the summation.
We first prove that the 1-characteristic tree T ′ C of a k-tree T without k-cliques of degree 2 is a series-reduced tree, for any k-clique C of T .Indeed, given a k-clique C, there is either exactly one vertex adjacent to C or at least 3.As such, the degree of C in T ′ C is not 2.For any other vertex v ∈ T , either it is a leaf in T ′ C , or at least one other vertex has been added to a k-clique C ′ containing v in T .In the latter case, since T has no k-cliques of degree 2, there must be at least two vertices other than v joined to C ′ and thus the degree of v in T ′ C is at least 3. Since T ′ C has at least n−k+1 2 + 1 leaves, T contains at least this many k-leaves (here one must also observe that if C has degree 1 in We can check that f maps k-cliques to k-cliques and is in fact an injection from sub-k-trees containing v to sub-k-trees not containing v. Indeed, because v is a k-leaf, C ′ ∪ {u} is not a (k + 1)-clique and thus not a sub-k-tree.Hence there does not exist a (k + 1)-clique C ′′ for which This implies that every k-leaf belongs to at most half of the sub-k-trees in T .Remembering that there are at least n−k+3 2 k-leaves, the global mean order of T is then For sharpness, let n = 2s + 3 − k for an integer s.We construct T by first constructing a caterpillar T ′ which consists of a path-type k-tree P k+1 s on vertices v 1 , v 2 , . . ., v s , for which k + 1 consecutive vertices form a clique, and adding a k-leaf connected to every k consecutive vertices.To obtain T , we extend T ′ by adding two k-leaves, which are connected to {v 1 , . . ., v k } and {v s−k+1 , . . ., v s } respectively.Note that T has a "stem" of s vertices and the number of For every 1 ≤ i ≤ s − k, there are i sub-k-trees of the stem each of order s + 1 − i.Each of these can be extended by adding any subset of the ℓ + 1 − i neighboring k-leaves, or even one or two more if some of the end-vertices of the stem are involved.

Now we can compute that
The first expression k(n − k) + 1 − ℓ counts the number of simplicial k-cliques different from the ones consisting of k consecutive vertices in the stem. 2 ℓ+2 is the number of sub-k-trees containing the whole stem.The third term counts the number of sub-k-trees containing v 1 or v s but not both and at least k vertices of the stem.The last summation counts the sub-k-trees containing at least k vertices of the stem and none of v 1 and v s .We can also compute R(T ) by summing the total size of the respective sub-k-trees.

The maximum local mean order
In this section, we prove We consider the k-clique in a k-tree T for which the local mean order is greatest.Our aim is to show that its degree cannot be too large, specifically at most 2. To do so, we prove that if a k-clique C has degree at least 2, then there is a neighboring k-clique whose local mean order is not smaller, with strict inequality when C has degree at least 3. From this, it will follow that any k-clique attaining the maximum has degree at most 2 and there is always at least one k-clique with degree 1 in which the maximum is attained.
The proof requires some technical inequalities, which we prove first.We start with a generalization of [21, Lemma 2.1] to the k-tree case as follows.
Lemma 10.For a sub-k-tree T and a k-clique C, we have Equality holds if and only if T is a path-type k-tree and C is a simplicial k-clique.
Proof.The statement is clearly true when T = C, so assume |T | > k.Observe that every sub-k-tree T ′ of T containing C has a vertex v (not belonging to C) which is simplicial within T ′ (consider a perfect elimination order where C is taken as the base k-clique).This implies that T ′ \v is a sub-k-tree containing C as well.
If there exists a sub-k-tree containing C of order ℓ > k, then the above implies that there must also exist a sub-k-tree of order ℓ − 1 containing C. Thus, if we list the sub-k-trees in S(T ; C) from smallest to largest order, we see that as desired.
The equality case is clear, since every sub-k-tree T ′ = C of T must have exactly one k-leaf not belonging to C, and equality is attained when T is a path-type k-tree.
Using the previous lemma, we can now bound the local mean order in terms of the number of sub-k-trees.
Lemma 11.For a k-tree T and one of its k-cliques C, we have The minimum occurs if and only if T is a k-star and C its base k-clique.The maximum is attained exactly when T is a path-type k-tree and C is simplicial.
Proof.The upper bound follows immediately from Lemma 10 by dividing both sides of the inequality by N (T ; C).The equality cases are the same.
To attain the lower bound, equality must hold for (1).This is the case if and only if T is a k-star and C its base k-clique.Indeed, for a k-star of order n with base clique C, we have N (T ; C) = 2 n−k and µ(T ; C) = n+k 2 .In the other direction, every vertex together with C needs to form a k-tree and thus a (k + 1)-clique, which is possible only for a k-star.
We will also make use of the following elementary inequality, which can be considered as the opposite statement of the inequality between the arithmetic mean and geometric mean (AM-GM).Lemma 12. Let x 1 , x 2 , . . ., x n ∈ R ≥1 , and let P = i x i .Then i x i ≤ P + (n − 1).Furthermore, equality is attained if and only if x i = P for some i and x j = 1 for all j = i.
Proof.This can be proven in multiple ways.The most elementary way is to observe that if Repeating this with pairs of elements which are strictly larger than 1 gives the result.Alternatively, one could consider the variables α i = log(x i ) ≥ 0. Since their sum is the fixed constant log P and the exponential function is convex, as a corollary of Karamata's inequality i exp(α i ) is maximized when all except one are equal to 0.
We are now ready to prove Theorem 3. Recall that T can be decomposed into C and ktrees T 1,1 , . . ., T d,k rooted at C 1,1 , . . ., C d,k respectively that are pairwise disjoint except for the vertices of the cliques C i,j .Let B i denote the (k + 1)-clique that contains C as well as C i,1 , . . ., C i,k , and let v i be the vertex of B i \ C. Finally, set N i,j = N (T i,j ; C i,j ) and µ • i,j = µ • (T i,j ; C i,j ).
Proof of Theorem 3. We first observe that ( This is because a sub-k-tree S of T containing C is specified as follows: given 1 ≤ i ≤ d, choose the sub-k-tree intersection of S with T i,1 , . . ., T i,k .There are k j=1 N i,j ways to do this if v i ∈ S and one if v i / ∈ S. We do this independently for each i, resulting in the product above.
Next, we would like to express the local mean order at C in terms of the quantities N i,j and µ • i,j : it is given by To see this, note that we can interpret µ(T ; C) as the expected size of a random sub-k-tree chosen from S(T ; C), which can then be written as the sum of the expected sizes of the intersection with each component T i,j in the decomposition.We have that is the probability that a randomly chosen sub-k-tree of T that contains C also contains v i (by the same reasoning that gave us (2)).Once v i is included, it adds 1 to the number of vertices, and an average total of k j=1 µ • i,j is added from the extensions in T i,1 , . . ., T i,k .Without loss of generality, we can assume that N 1,1 = min i,j N i,j .We want to compare µ • (T ; C) to µ • (T ; C 1,1 ) and prove that µ Observe that this will be enough to prove our claim: no clique with degree greater than 2 can attain the maximum local mean order, and starting from any clique, we may repeatedly apply the inequality above to obtain a sequence of neighboring cliques whose local mean orders are weakly increasing and the last of which is a degree-1 k-clique (more precisely, for the first step replacing C with C ′ := C 1,1 and considering the decomposition {C ′ i,j } and {T ′ i,j } with respect to C ′ , the clique adjacent to C ′ with equal or larger µ • cannot be the original clique C as C will not correspond to min i,j N ′ i,j .So in repeatedly applying the inequality, we will obtain a sequence of distinct cliques which must terminate but can only terminate once we have reached a clique of degree 1.) Let us first express µ • (T ; C 1,1 ) in terms of the N i,j and µ • i,j as well.First, we have The reasoning is similar to (2): there are N 1,1 sub-k-trees that contain C 1,1 , but not the full (k + 1)-clique B 1 , and the remaining product counts sub-k-trees containing B 1 .We also have using the fact that is the probability that a random sub-k-tree containing C 1,1 does not contain B 1 .Let us now take the difference µ • (T ; C 1,1 ) − µ • (T ; C): we have, after some manipulations, . We want to show that µ • (T ; C 1,1 ) − µ • (T ; C) ≥ 0, i.e., that Equivalently, Using the previous computations, we know that So the left-hand side of ( 4) is equal to We can subtract F from both sides and use (3) to replace µ • (T ; C).Taking into account that We first prove (5) for d = 2, in which case it can be rewritten as . Applying Lemma 12 to the numbers Hence we find that it suffices to prove that We note that the left-hand side is strictly increasing in y and the right-hand side is independent of y, which implies that it is sufficient to prove the equality when y = N 1,1 .That is, we want to prove that for every z ≥ y ≥ 1 If y = N 1,1 = 1, (6) reduces to the equality z 2 = z 2 .
If y = N 1,1 ≥ 2, z ≥ y and k ≥ 2 imply that 2 + (k − 1)y + z − k ≤ kz.Together with log 2 y ≥ 1, we conclude that it is sufficient to prove that For k = 2, the difference between the two sides is an increasing function in y, and for y = 2 it reduces to 3z 2 − 2 ≥ 0, which holds.
For k ≥ 3, the inequality is immediate, using that y Once (5) has been verified for d = 2, we can apply induction to prove it for d ≥ 3. Let . We can then rewrite (5) as By the induction hypothesis, we have for every 2 ≤ m ≤ d.Summing over m, we obtain that Since we have g m ≥ 2 for every m, the conclusion now follows as and since f 1 > 0, inequality ( 5) is even strict in the case that d > 2.
We conclude that if ). Equality can only be attained when d = 2. Looking back over the proof of ( 5), we also see that for equality to hold, all N i,j except for N 2,2 (up to renaming) must be equal to 1, and due to Lemma 11, T 2,2 has to be a path-type k-tree.
For n ≥ k + 3, we compute that in a path-type k-tree T the maximum among the degree-1 k-cliques is attained by a central one, which implies that no degree-2 k-cliques can be extremal.Let B 1 be the unique (k + 1)-clique containing C, and assume that T \ B 1 contains components of size a and b.
) , and this is maximized if and only if |a−b| ≤ 1.The latter can also be derived from considering the 1-characteristic tree.This is illustrated in Figure 6 below.We emphasize that the maximizing k-clique of degree 1 is not simplicial.
When n ≤ k + 2, every k-clique has the same local mean sub-k-tree order, and so the (unique) degree-2 k-clique when n = k + 2 is the only case where equality can occur at a k-clique with degree 2. This concludes the proof of Theorem 3. As before, given a k-tree T and a k-clique C in T , we utilize the decomposition of T into C and ).Since a sub-k-tree not containing C needs to be a sub-k-tree of some k-tree T i,j , we have Following the proof for trees, we show Lemma 13.For any k-tree T and k-clique C ∈ T ,

R(T ; C) > N (T ; C).
Proof.Assume to the contrary that there exists a minimum counterexample T .Since the statement is true when T = C, we have |T | > k and we can consider the decomposition as before.
Note that if N i,j = 1, we have that N i,j = 0, and otherwise we have Expanding using (2) and (3), we note that the coefficient of µ • i,j is at least equal to N i,j .As such, it is sufficient to prove the inequality with µ • i,j = 0 and N i,j bounded by kN i,j .
Let f be a function on the positive integers defined by f We now want to prove that ).When increasing a value N 1,j which is at least equal to 2, the left-hand side increases more than the right-hand side.As such, it is sufficient to consider the case where a of the terms N 1,j equal 2, while the other k − a terms equal 1.In this case, the desired inequality holds since k + (k + 1)2 a > k + 2ak for every integer 0 ≤ a ≤ k.
Next, we consider the case d ≥ 2. In this case, when N i,j increases by 1, the left-hand side of ( 7) increases by at least 2k and the right-hand side by at most 2k.When all N i,j are equal to 1, the conclusion follows from k We now bound the local mean order by the global mean order.
Proof of Theorem 4. Let T be a k-tree and C a k-clique in T .We want to prove that We proceed by induction on the number of vertices in T .Note first that the inequality is trivial if |T | ≤ 2k: since the mean is taken over sub-k-trees, which have at least k vertices each, we have µ(T ) ≥ k.On the other hand, we clearly have µ(T ; C) ≤ |T |, and both inequalities hold with equality only if |T | = k.
We thus proceed to the induction step, and assume that |T | > 2k.We have two cases with respect to C.
Case 1: C is simplicial.
Let v be a k-leaf in C, let C ′ denote the clique adjacent to C, and let We want to prove that By the induction hypothesis, we have Multiplying by N +1 N , this is seen to be equivalent to This can be broken up into three terms as follows: Note here that the second term is trivially nonnegative, and the last term trivially positive.Since |T ′ | ≥ 2k, we have N > k (one gets at least k + 1 sub-k-trees containing C ′ by successively adding vertices); hence N 2 − kN + 3N + N > 0. Thus the first term is positive by Lemma 13, completing the induction step in this case.
Case 2: C is not simplicial.
By Theorem 3, we only need to consider the case where C has degree 1.Let v be the unique common neighbor of C, and let C i , 1 ≤ i ≤ k, be the other k-cliques in the (k + 1)-clique spanned by C ∪ {v}.
Let T i , 1 ≤ i ≤ k, be the sub-k-trees rooted at C i (pairwise disjoint except for the vertices of the cliques We can assume without loss of generality that We can now express the local and global mean in a similar way to Case 1.Here N (T ; C) = k i=1 N i + 1, and all the sub-k-trees counted here, except for C, contain v.We have In the remainder of this section, we will omit the bounds in products and sums if they are over the entire range from 1 to k: N i and N i mean k i=1 N i and k i=1 N i respectively.Then We want to show that this expression is positive.By induction, we know that 2 It follows that ( 8) is greater than Multiplying by and observing that this factor is greater than 1, we find that ( 8) is indeed positive if we can prove that In particular, a potential counterexample would have to satisfy To simplify proving Eq. ( 9), we first note that it is sufficient to consider the case where k = j.
Once N i , µ • i and N i are fixed for 1 ≤ i ≤ j, the terms that are dependent on k are ( i≤j N i + 1)k + k on the left, and (1 + i≤j µ • i )(k − j) on the right.The latter since if N i = 1, then T i only consists of C i , and thus i , the increase of the left side is larger than the increase on the right side.Here i≤j N i ≥ i≤j N i is true since the product of j ≥ 2 numbers, each greater than or equal to 2, is at least equal the sum of the same j numbers.Moreover, N i > µ • i follows from Lemma 11.So from now on, we can assume that j = k and all N i are at least equal to 2. By Lemma 13, 9) is a linear inequality in each N i and the coefficient on the right-hand side is always greater than the coefficient on the left-hand side, we can reduce Eq. ( 9) to a sufficient inequality that is only dependent on k, N i and µ This will be assumed in the following.If now all the parameters in (9) are fixed except for one µ • i , we have a linear inequality in µ • i : the quadratic terms stemming from µ • i N i are equal on both sides and cancel.As such, it is sufficient to prove the inequality for the extremal values of µ • i .Here we use the trivial inequality µ • i ≥ 0 as well as the upper bound µ • i ≤ N i −1 2 , which is taken from Lemma 11.So if ( 9) can be proven in the case where For 2 ≤ k ≤ 5, this is achieved by exhaustively checking all 2 k cases that result (using symmetry, there are actually only k+1 cases to consider).See the detailed verifications in https://github.com/StijnCambie/AvSubOrder_ktree.So for the rest of the proof, we assume that k ≥ 6, and we will use the slightly weaker bound µ We distinguish two further cases, depending on the value of µ • 1 .In these cases, we will use the following two inequalities.
Proof.The first inequality, Eq. ( 11), is true if all the N i are equal to 2, since 2 k−1 > 6(k − 1) for every k ≥ 6. Increasing some N i by 1 increases the product by at least 2 k−2 , while the sum increases by only 3.So the inequality holds by a straightforward inductive argument.
Next, we prove Eq. (12).When all N i are equal to 2, it becomes 5 3 (2 k + 1) ≥ 4(k − 2)k.This is easily checked for k ∈ {6, 7}, and for k ≥ 8, the stronger inequality 2 k ≥ 4k 2 can be shown by induction.Now observe that the difference between the left-and right-hand sides is increasing with respect to N 1 , since i≥2 N i > i≥2 N i by the first inequality.It is also increasing in the other N i 's; for example, we can see this is true for N 2 since In the first step, we have applied Eq. ( 11) but replacing N 1 with N 2 .Again, we may conclude using induction.♦ Claim 15.Given µ • 1 = N 1 2 , it is sufficient to consider the case where µ Proof.Starting from any counterexample to (9), we can iteratively change µ • i (considered as variables) for 1 ≤ i ≤ k based on the worst case of the linearization (to 0 if the coefficient on the left-hand side is greater and to N i 2 if the coefficient on the right-hand side is greater) to obtain further counterexamples.
To show that it suffices to consider µ 2 on the left-hand side in Eq. ( 9) is not greater than the coefficient on the right-hand side.This then implies that µ • 2 = N 2 2 is indeed the worst case.For 3 ≤ i ≤ k, we can argue in the same fashion.
Assume for sake of contradiction that the coefficient on the left-hand side is greater.Recall that Thus we must have Using that (recall here that we are assuming without loss of generality that Adding kN 2 to both sides and multiplying both sides by 1 + µ • i results in In the second inequality, we applied (10) which we may do since we began with the assumption that we have a counterexample to (9).After simplification, we get that Since k ≥ 6, this is a clear contradiction to Eq. (11).♦ Having proven Claim 15, we are left with two cases to consider: It is easy to conclude in the former case, except when k = 6 and at least 5 values N i are equal to 2, which has to be handled separately.See https://github.com/StijnCambie/AvSubOrder_ktree/blob/main/2M-mu_j_large_case1.mw for details.The final remaining case is when µ • 1 = 0. We obtain two new inequalities by multiplying Eq. ( 12) with k+1 2 and Eq. ( 11) with (1 + µ • i ) N 1 6 , and use that N 1 = max{N i } and µ • i ≤ N i −1 2 .
5(k + 1) 6 Summing these two inequalities together, we have that Eq. ( 9) holds as a corollary of We remark that by considering a suitable k-broom, one can show that Theorem 4 is sharp, as was also the case for trees.

Conclusion
This paper together with [13,Thm. 18 & 20] answers all of the open questions from [16] except for one, which was stated as rather general and open-ended: Problem 1.For a given r-clique R, 1 ≤ r < k, what is the (local) mean order of all sub-k-trees containing R?
One natural version of this question is to consider the local mean sub-k-tree order over sub-ktrees that contain a fixed vertex.In this direction, we prove the following result, which can be considered as another monotonicity result related to [13,Thm. 23].Luo and Xu [13,Ques. 35] also asked if for a given order, a k-tree attaining the largest global mean sub-k-tree order is necessarily a caterpillar-type k-tree.In contrast with the questions in [16], this question is still open for trees.We prove that the local version (proven in [3,Thm. 3]), which states that for fixed order, the maximum is attained by a broom, is also true for the generalization of k-trees.This is almost immediate by observations from [16].As such, we conclude that the k-tree variants of many results on the average subtree order for trees are now also proven.The analogue of [10,Ques. (7.5)] can be considered as the only question among them where the answer is slightly different for k-trees: in contrast with the case of trees (k = 1), the maximum local mean sub-k-tree order cannot occur in a k-clique with degree 2 when k ≥ 2 (with one small exception).

For
an arbitrary k-clique C of T , let S(T ; C) denote the collection of sub-k-trees containing C and let N (T ; C) := |S(T ; C)|.The local clique number is R(T ; C) = X∈S(T ;C) |X| and the local mean (sub-k-tree) order is µ(T ; C) = R(T ; C) N (T ; C) .

Figure 5 :
Figure 5: Sketch of k-trees with mean sub-k-tree order roughly n 2 and 3n 4 for k = 3

C 0 CFigure 6 : 2 -C 5 4 .
Figure 6: 2-path for which the maximum local mean is attained in a non-simplicial clique C, with its 1-characteristic tree T ′ C

Theorem 16 .Claim 17 .
Let T be a k-tree, k ≥ 2, C a k-clique of T , and v a vertex in C. Then µ(T ; v) < µ(T ; C).Proof.The statement is trivially true if |T | ≤ k + 1.So assume that T with |T | ≥ k + 2 is a minimum counterexample to the statement.Recalling the decomposition into trees T i,j used earlier, note that all sub-k-trees containing v and not C are part of T i,j for some i, j.Without loss of generality, we can assume that µ(T 1,1 ; v) = max i,j µ(T i,j ; v).It suffices to prove that µ(T 1,1 ; v) < µ(T ; C).Taking into account (3), it is sufficient to consider the case where C is a simplicial k-clique of T .Let u be the simplicial vertex of B 1 .Let T ′ = T \{u} andC ′ = B 1 \{u}.We have µ(T ′ ; C ′ ) < µ(T ; C).Proof.Let R = R(T ′ ; C ′ ) and N = N (T ′ ; C ′ ).We now need to prove that µ(T ; C) = 2R+N +k 2N +1 > R N = µ(T ′ ; C ′ ), which is equivalent to N +k > R N .The latter is immediate since N ≥ |T ′ |−(k−1) and R N = µ(T ′ ; C ′ ) ≤ |T ′ |. ♦Since the sub-k-trees containing v are exactly those that contain C, or sub-k-trees of T ′ containing C ′ , or k-cliques within B 1 different from C, we conclude that µ(T ; v) < µ(T ; C).

Proposition 18 .
If a k-tree T of order n and k-clique C of T attain the maximum possible value of µ(T ; C), then T has to be a k-broom with C being one of its simplicial k-cliques.Proof.Let T ′C be the characteristic 1-tree of T with respect to C. Then by[16, Lem.11]µ(T ; C) = µ(T ′ C ; C) + k − 1.Since T ′ C is a tree on n − (k − 1)vertices, by [3, Thm.3], the maximum local mean subtree order is attained by a broom B. Since there is a k-broom T with C being a simplicial k-clique for which T ′ C ∼ = B with C as root, this maximum can be attained.Reversely, if µ(T ′ C ; C) is maximized, then T ′ C is a broom where C is its root (and thus simplicial).