On Sackin's original proposal: the variance of the leaves' depths as a phylogenetic balance index.

BACKGROUND
The Sackin indexS of a rooted phylogenetic tree, defined as the sum of its leaves' depths, is one of the most popular balance indices in phylogenetics, and Sackin's paper (Syst Zool 21:225-6, 1972) is usually cited as the source for this index. However, what Sackin actually proposed in his paper as a measure of the imbalance of a rooted tree was not the sum of its leaves' depths, but their "variation". This proposal was later implemented as the variance of the leaves' depths by Kirkpatrick and Slatkin in (Evolution 47:1171-81, 1993), where they also posed the problem of finding a closed formula for its expected value under the Yule model. Nowadays, Sackin's original proposal seems to have passed into oblivion in the phylogenetics literature, replaced by the index bearing his name, which, in fact, was introduced a decade later by Sokal.


RESULTS
In this paper we study the properties of the variance of the leaves' depths, V, as a balance index. Firstly, we prove that the rooted trees with n leaves and maximum V value are exactly the combs with n leaves. But although V achieves its minimum value on every space [Formula: see text] of bifurcating rooted phylogenetic trees with n≤183 leaves at the so-called "maximally balanced trees" with n leaves, this property fails for almost every n≥184. We provide then an algorithm that finds the trees in [Formula: see text] with minimum V value in time O(n log(n)). Secondly, we obtain closed formulas for the expected V value of a bifurcating rooted tree with any number n of leaves under the Yule and the uniform models and, as a by-product of the computations leading to these formulas, we also obtain closed formulas for the variance under the uniform model of the Sackin index and the total cophenetic index (Mir et al., Math Biosci 241:125-36, 2013) of a bifurcating rooted tree, as well as of their covariance, thus filling this gap in the literature.


CONCLUSION
The phylogenetics community has been wise in preferring the sum S(T) of the leaves' depths of a phylogenetic tree T over their variance V(T) as a balance index, because the latter does not seem to capture correctly the notion of balance of large bifurcating rooted trees. But it is still a valid and useful shape index.


SN-1 Further background
In this section we gather several definitions and preliminary results that were omitted in the main text's Background section because they are only used in this Supplementary Material.
Given k 2 trees T 1 , . . . , T k , with every T i ∈ T * ni , their root join is the tree T 1 T 2 · · · T k ∈ T * n1+···+n k obtained by connecting the roots of (disjoint copies of) T 1 , . . . , T k to a new common root r; see Fig. 8. In a similar way, given k 2 phylogenetic trees T 1 ∈ T(X 1 ), . . . , T k ∈ T(X k ), with X 1 , . . . , X k pairwise disjoint sets of labels, their root join is the phylogenetic tree T 1 · · · T k ∈ T k j=1 X j obtained by connecting the roots of T 1 , . . . , T k to a new common root. Notice that the shape of the root join of a family of phylogenetic trees is the root join of their shapes, and that the root join of a pair of bifurcating trees is again bifurcating. Figure 8 The tree T 1 · · · T k .
A probabilistic model of bifurcating phylogenetic trees P n is shape invariant when, for every n 1 and for every T, T ∈ BT n , if T and T have the same shape, then P n (T ) = P n (T ). When P n is shape invariant, it induces well-defined probability mappings P X : BT(X) → [0, 1], for every set X of cardinality n, through any shape-preserving bijection BT(X) ↔ BT n induced by a bijection X ↔ [n].
A probabilistic model of bifurcating phylogenetic trees P n is Markovian when there exists a symmetric mapping q : N × N → R, called the conditional split distribution of P n , such that, for every T k ∈ BT(X k ) and T l ∈ BT(X l ), with X k and X l disjoint sets of cardinalities |X k | = k and |X l | = l, P X k ∪X l (T k T l ) = q(k, l)P X k (T k )P X l (T l ).
The formulas (1) and (2) in the main text clearly imply that both the Yule and the uniform models are shape invariant. As to their Markovianity, it is easy to deduce from those formulas that P Y,n (T k T n−k ) = 2 (n − 1) n k P Y,k (T k )P Y,n−k (T n−k ) (11) where C k,n−k := 1 2 Consider the random variables S n and Φ n that take a phylogenetic tree T ∈ BT n and compute S(T ) and Φ(T ), respectively. Their expected values under the Yule model (denoted by E Y ) and the uniform model (denoted by E U ), as well as their variance under the Yule model (denoted by σ 2 Y ), are known:

SN-2 Proof of Proposition 5
Proposition 5 For every T ∈ BT * n , the following conditions are equivalent: (a) T is of type F n . (b) There exists a d 0 ∈ N such that δ T (x) ∈ {d 0 , d 0 + 1} for every x ∈ L(T ). and therefore each |δ T (x) − S(T )| is either (n − l)/n or l/n. If 0 < l < n, these two values are smaller than 1, and if l = 0 or l = n, then all leaves in T have the same depth and therefore |δ T (x) − S(T )| = 0 for every x ∈ L(T ). • If |δ T (x) − S(T )| < 1, then, since δ T (x) ∈ N, δ T (x) ∈ { S(T ) , S(T ) }. As far as (b)⇒(a) goes, we prove it by induction on n. The base case when n = 1 is obvious, because the only tree with one leaf is the fully symmetric tree B 1 . Assume now that (b)⇒(a) is true for every tree in BT * n and let T be a bifurcating tree with n + 1 leaves for which there exists a d 0 ∈ N such that δ T (x) ∈ {d 0 , d 0 + 1} for every x ∈ L(T ); to simplify the discussion, we shall assume that δ(T ) = d 0 + 1. Let x 0 be a leaf in T of depth d 0 + 1 and let T ∈ BT * n be obtained by replacing the cherry that contained x 0 by a leaf of depth d 0 . Then, δ T (x) ∈ {d 0 , d 0 + 1} for every x ∈ L(T ), too. Thus, by the induction hypothesis, T is of type F n . Finally, since T is obtained from a tree of type F n by replacing a leaf of minimum depth, d 0 , by a cherry with leaves of depth one unit larger, T is of type F n+1 .
This completes the proof that conditions (a), (b) and (c) are equivalent.
Let us prove now that if a bifurcating tree T satisfies (d) then it satisfies (b) with d 0 = log 2 (n) , by induction on the depth of the tree. This implication is trivially true when δ(T ) = 0, because the only tree of depth 0 is B 1 . Assume now that the implication is true for every tree of depth at most δ and let T ∈ BT * n be a bifurcating tree of depth δ + 1 that is depth-equivalent to B n . Since (b) is an assertion on the depths of the leaves of T , and ∆(T ) = ∆(B n ), in order to prove that T satisfies (b) we can assume without any loss of generality that T = B n . Let m = log 2 (n) and k = n − 2 m .
Let T 1 and T 2 be the subtrees rooted at the children of the root of T and n 1 and n 2 their respective numbers of leaves, with n 1 n 2 and n = n 1 + n 2 . Since T is maximally balanced, T 1 and T 2 are also maximally balanced and n 1 = n/2 and n 2 = n/2 . Then, since δ(T 1 ), δ(T 2 ) δ = δ(T )−1, by the induction hypothesis we deduce that if, for every i = 1, 2, we set d i = log 2 (n i ) , then δ Ti (x) ∈ {d i , d i + 1} for every x ∈ L(T i ). Now: • If k < 2 m+1 − 1, then d 1 = d 2 = m − 1.
• If k = 2 m+1 − 1, then n 1 = 2 m − 1, and thus d 1 = m − 1, and n 2 = 2 m , and thus d 2 = m; but then, T 2 is fully symmetric, because it is maximally balanced with 2 m leaves, which implies in particular that all its leaves have depth m. Then, in both cases, δ Ti (x) ∈ {m − 1, m} for every x ∈ L(T i ) and i = 1, 2. Since . This is what we wanted to prove, and hence the implication (d)⇒(b) is established.
Finally, we prove the implication (a)⇒(d). Let T ∈ BT * n be of type F n . Since we have already proved that (d)⇒(b)⇒(a), we know that B n is also of type F n . But then, by Remark 3 in the main text, T and B n are depth-equivalent.

SN-3 Proofs of Lemmas 1 and 2
Recall that we denote the numbers of leaves of depths δ(T ) and δ(T ) − 1 of a tree T ∈ BT * n by p 0 (T ) and p 1 (T ), respectively.
, then there does not exist any tree T of type T n;l1,...,lj .
If j = 0, T has only leaves of depths δ and δ − 1 and then it is of type F n . In particular, in this case, if k = 0, T is fully symmetric of depth m, and hence p 1 = 0, while if k > 0, then p 1 = 2 m − k and δ = m + 1. This proves (a) as well as (b) when j = 0. So, we shall assume henceforth that j 1.
To begin with, notice that, in order to complete T to a fully symmetric tree with 2 δ leaves, we must append a cherry to each leaf of depth δ − 1, which adds p 1 new leaves, and we must append a fully symmetric tree of depth l i to each leaf of depth δ − l i , thus adding for each such leaf 2 li − 1 new leaves. This implies that Since we are assuming that j > 0, this implies that n < 2 δ and hence m δ − 1.
Now, on the one hand, if δ = m + 1, (19) implies that In particular, in this case, k + j i=1 (2 li − 1) 2 m because p 1 0. On the other hand, if δ = m + 2, (19) implies that And in this case, since T contains at least 2 leaves of depth δ and j leaves of depths different from δ or δ − 1, it must happen that p 1 n − j − 2, that is Since δ = m + 1 or δ = m + 2, this completes the proof of the statement.

SN-4 Proof of Proposition 6
In this section, and henceforth, p F q denotes the (generalized ) hypergeometric function defined by where (a) 0 = 1 and (a) k := a · (a + 1) · · · (a + k − 1) for k 1. Recall, moreover, that for every n 2 and for every 1 k n − 1, We start by proving two auxiliary lemmas. These lemmas will be used not only in the proof of Proposition 6, but also later in the proofs of the main results in Sections SN-10 to SN-12.
Lemma 3 For every n 2: Proof We consider first the case when 1 m n − 1. In this case We shall compute now this last sum using the lookup algorithm given in [6, p. 36]. Take But now, it is wrong to deduce from the lookup algorithm that and therefore while the index k in our original sum stops a n − m − 1. Therefore, what the lookup algorithm actually implies is that Then, combining (20), (21), (22), and (23), we obtain This proves (b) when 1 m n − 1. Now notice that when m = n 2, and when m > n, Therefore the identity stated in (b) also holds when m n.
Finally, as far as (a) goes, by the symmetry of C k,n−k , we have that from which we deduce that where the second equality is simply (b) for m = 1.
Lemma 4 For every n 2, (b) For every m 1, Proof We consider again first the case when 1 m n − 1. Let us develop our sum of interest: We shall apply again the lookup algorithm to compute this sum. Taking Then, arguing as in the corresponding step in the proof of Lemma 3 we obtain that, by the lookup algorithm and taking into account that (m − n) k = 0 for every k > n − m, Since m n, the first summand inside the parenthesis is equal, by identity http: and, as we established in the proof of Lemma 3 (page 7), m As for the subtrahend, we have So, by (24) and (25), This finishes the proof of (b) in the case when 1 m n − 1. Now, it is clear that when m n and therefore the equality in (b) holds actually for every m 1.
It remains to cover the case when m = 0, i.e., (a). We start again by developing the sum of interest.

Proposition 6
The solution X n of the equation (where r, s 1 and a 1 , . . . , a r , b 1 , . . . , b s ∈ R) with given initial condition X 1 is Proof Consider a sequence of the form This equality will be satisfied if the coefficients a 1 , . . . , a s+1 , b 1 , . . . , b r satisfy that As to (30), it is easy to check that, if l 1, which implies, by induction on l, that, for every l 2, Indeed, the base case l = 2 is a direct consequence of (31), and, as to the inductive step, if for a given l 3 we assume that Therefore, returning back to (30), using (32) we have that This system of linear equations in a 2 , . . . , a s+1 is non singular, and its only solution satisfies Then, the previous formulas can be rewritten as and the solution of the last recurrence is Indeed: and if (33) holds for every h = l + 1, . . . , s + 1, then Then, finally, for every l = 2, . . . , s, Finally, a 1 is obtained by imposing the initial condition

SN-5 Proof of Theorem 1
We start by proving a series of lemmas describing the behaviour of V when we remove a deepest leaf from a rooted tree. To simplify the notations, we shall set henceforth Moreover, given a tree T ∈ T n , we shall denote by x 1 , . . . , x n its leaves ordered in non-decreasing order of depth and we shall set d i := δ T (x i ), for i = 1, . . . , n, so that the elements of ∆(T ) are Since the maximum depth of a tree is reached at least at two sibling leaves, we have that d n−1 = d n and we shall assume without any loss of generality that x n−1 and x n are sibling.
Lemma 5 Let T ∈ T * n be a tree with two leaves of maximum depth forming a cherry. Let T ∈ T * n−1 be the tree obtained by removing both leaves in this cherry, so that the root of the cherry becomes a leaf. Then, Proof For simplicity, we shall denote δ(T ) by δ. With the notations on the leaves x i of T and their depths d i introduced above, we shall assume without any loss of generality that x n−1 and x n are not only sibling, but they form the cherry in the hypothesis. Let T ∈ T * n−1 be the tree obtained by removing from T these two leaves, so that their common parent becomes a new leaf of depth d n−1 − 1 = d n − 1 = δ − 1, which we shall still denote by x n−1 ; cf. Fig. 9. Thus, and then Finally, ... Lemma 6 Let T ∈ T * n be a tree with k leaves of maximum depth forming a k-fan with k 3. Let T ∈ T * n−1 be the tree obtained by removing one leaf from this k-fan. Then, Proof We shall denote again δ(T ) by δ and we shall use again the notations on the leaves x i of T and their depths d i introduced at the beginning of this section. We shall assume without any loss of generality that x n belongs to the k-fan in the hypothesis, and that this is the leaf we remove to obtain T . In this way, since the remaining leaves have the same depths in T and in T , Then and hence, computing W (T ) in terms of W (T ) as in the proof of the previous lemma, we obtain as we claimed.
Lemma 7 Let T ∈ T * n be a tree with two leaves of maximum depth forming a cherry. Then, 2n and the equality holds only when T = K n .
Proof First of all, since (cf. Example 1 in the main text) we have that δ(K n ) − S(K n ) = (n − 1)(n − 2)/(2n), for every n, and therefore the inequality in the statement is an equality for combs. Now, we shall prove the statement by induction on n. The cases when n = 1 or n = 2 are obvious, because then T * n = {K n }, and the case when n = 3 is also obvious, because the only tree in T * 3 with two leaves of maximum depth forming a cherry is K 3 . Let us prove now the inductive step when n 4. To simplify the notations, for every tree T we shall set Let T ∈ T * n , with n 4, be a tree with two leaves of maximum depth forming a cherry. We use again the notations on the leaves x i of T and their depths d i introduced at the beginning of this section, and we assume that the cherry at maximum depth is formed by the leaves x n−1 and x n . Let T ∈ T * n−1 be the tree obtained from T by removing this cherry and replacing it by its root, that becomes a leaf which we shall still denote by x n−1 ; cf. again Fig. 9. On the one hand, we have that δ(T ) = δ(T ) − 1 if T contains only the leaves x n−1 , x n at depth δ(T ), and δ(T ) = δ(T ) if T contains leaves of depth δ(T ) other than x n−1 and x n . On the other hand, by the expression (34) for S(T ) obtained in the proof of Lemma 5, we have that Combining these two facts we obtain that and the equality holds only when T contains no other leaf of depth δ(T ) than x n−1 and x n . Now we must distinguish two cases: (a) If T contains again a cherry at maximum depth, we can apply the induction hypothesis and we obtain that with the equality holding only if T = K n−1 . Then, and the equality holds only if T = K n−1 and T is obtained by replacing a leaf of largest depth by a cherry, that is, when T = K n . (b) Assume now that T has no cherry at maximum depth, and in particular that x n−1 belongs to a k-fan with k 3. Without any loss of generality, assume that this fan is x n−k , . . . , x n−1 and let y be their common parent. In this case, let T ∈ T * n−1 be the tree obtained from T by adding a new node z and replacing the arcs (y, x n−2 ), (y, x n−1 ) by new arcs (y, z), (z, x n−2 ), (z, x n−1 ); see Fig. 10.
x n−2 x n−1 y T Figure 10 The trees T and T in the proof of case (b) of Lemma 7.
The tree T obtained in this way has the cherry (x n−2 , x n−1 ) at maximum depth δ(T ) = δ(T ) + 1, and, since these two leaves increase their depths in T in one unit with respect to T and the other leaves in T maintain their depths in T , we have that S(T ) = S(T ) + 2/(n − 1). Thus, (because we are assuming n 4) and hence where now T ∈ T * n−1 has a cherry at maximum depth and therefore we can apply to it the induction hypothesis: We can now proceed as in the last step in (a): Let us emphasize that in this case T can never be a comb, and the inequality in the statement is strict. This completes the proof of the statement by induction on n.
We are now in position to prove Theorem 1.

Theorem 1
The maximum value of V on T * n is always reached exactly at the combs K n .
Proof Since, for any fixed n, the maximum values on T * n of V and W are reached at the same trees, it will be enough to prove, by induction on n, the following property for every T ∈ T * n : , and the equality holds if, and only if, The cases when n = 1 or n = 2 are obvious, because T * 1 = {K 1 } and T * 2 = {K 2 }. So, we assume henceforth that n 3 and that assertion (35) is true for n − 1.
Let T ∈ T * n . With the notations on the leaves x i of T and their depths d i introduced at the beginning of this section, we must distinguish two cases, depending on whether the leaf x n has only one sibling, forming a cherry, or more than one sibling. We start with this last case.
(A) Let us assume that x n belongs to a k-fan, with k 3; without any loss of generality we consider that the leaves in this fan are x n−k+1 , . . . , x n−1 , x n . Let y be the common parent of this fan. Let T ∈ T * n be the tree obtained from T by adding a new node z and replacing the arcs (y, x n−1 ), (y, x n ) by new arcs (y, z), (z, x n−1 ), (z, x n ). In this way, the leaves x 1 , . . . , x n−2 in T have the same depths as in T and the depths of x n−1 and x n in T are their depths in T plus 1, and therefore δ(T ) = δ(T ) + 1 and S(T ) = S(T ) + 2/n. . . .  Now, let T ∈ T * n−1 be the tree obtained from T by removing from it the leaf x n . It is obvious that T can also be understood as the tree obtained from T by removing the cherry (x n−1 , x n ) and replacing it by a leaf x n−1 . Therefore, by Lemmas 5 and 6 from where we obtain that where this last inequality holds because δ(T ) S(T ) and, by assumption, n 3.
Therefore, if x n belongs to a k-fan, for some k 3, there exists a tree T ∈ T * n with larger W value that contains a leaf of maximum depth belonging to a cherry. This is the case we consider next.
(B) Assume now that x n−1 and x n form a cherry. Let T ∈ T * n−1 be the tree obtained by removing from T this cherry, leaving its root as a leaf. Then: This proves that, when T has a cherry at the bottom, W (T ) W (K n ). Moreover, the equality holds only when both intermediate inequalities are equalities, that is (by Lemma 7 and the induction hypothesis) exactly when T = K n . In summary, we have proved that if T ∈ T * n has some cherry at maximum depth, then W (T ) W (K n ), with the equality holding only when T = K n , and that if T ∈ T * n does not have any cherry at maximum depth, then there exists a tree T 0 ∈ T * n with a cherry at maximum depth such that W (T ) < W (T 0 ) W (K n ). This completes the proof by induction of property (35).

SN-6 Proof of Theorem 2
The goal of this section is to prove the necessary condition on the bifurcating trees with minimum V value stated in the following result.
Theorem 2 If T ∈ BT * n has the minimum value of V , then it is of some type T n;l1,...,lj with 5 l 1 < · · · < l j δ(T ) − 2.
Now, since each one of BT * 1 , BT * 2 , and BT * 3 contains only one tree, it is enough to consider the case n 4. Moreover, the tree in BT * 4 with minimum V value is clearly B 4 , because when n is a power of 2, B n is fully symmetric and hence V (B n ) = 0 is minimum in BT * n . Therefore, we can restrict ourselves to the case n 5.
Lemma 8 Let n 5. If T ∈ BT * n has a leaf of depth 1, then V (T ) is not minimum in BT * n .
Proof Let T ∈ BT * n be a tree with a leaf of depth 1, so that it has the form T = B 1 T 0 with B 1 a tree with a single leaf and T 0 ∈ BT * n−1 . Let T ∈ BT * n be the tree obtained from T 0 by replacing a leaf of smallest depth in it by a cherry (of depth one unit larger). So, if the depths of the leaves in T are 1 < d 2 d 3 · · · d n , the depths of the leaves in T 0 are d 2 − 1 d 3 − 1 · · · d n − 1, and the depths of the leaves in T are then d 2 , d 2 , d 3 − 1, . . . , d n − 1. Then, Then, if we have that W (T ) < W (T ) and therefore V (T ) cannot be minimum in BT * n . Let us check now that this inequality always holds if n 5. To do that, we rephrase it in terms of T 0 .
To begin with, Then S(T 0 ) 2 then we guarantee that inequality (36) holds. But, since T 0 has at least n − 1 4 leaves, all its leaves but at most two of them have depth at least 3, and if it contains two leaves of depth smaller than 3, they have depths 1 and 2 or depth 2 both of them. Therefore, Therefore, inequality (36) holds and hence V (T ) is not minimum in BT * n , as we claimed.
Lemma 9 Let T ∈ BT * n be a bifurcating tree containing a leaf of depth d < δ(T ). Let T d ∈ BT * n be the tree obtained by removing a cherry of depth δ(T ) and replacing a leaf of depth d by a cherry of depth d + 1. Then, Proof Let T ∈ BT * n . With the notations on the leaves x i of T and their depths d i introduced at the beginning of the previous section, we shall assume without any loss of generality that d = d i and that the pair of leaves of depth δ := δ(T ) that are removed from T are x n−1 , x n . Therefore and as we claimed. Proof If V (T ), or, equivalently, W (T ) is minimum on BT * n and T contains a leaf of depth d = δ(T ) − l < δ(T ) − 1, then, with the notations of the last lemma, it must happen that Replacing in this inequality d by δ(T ) − l and solving for l we finally obtain l 2n(δ(T ) − S(T )) + 3n − 1 n − 1 = 3 + 2(n(δ(T ) − S(T )) + 1) n − 1 > 3.
Corollary 2 Let T ∈ BT * n be a bifurcating tree that has a cherry of depth d < δ(T ). Let T * d ∈ BT * n be the tree obtained by removing this cherry, leaving in its place a leaf of depth d − 1, and replacing a leaf of depth δ(T ) by a cherry of depth δ(T ) + 1. Then, Proof T is obtained from T * d by removing a cherry of maximum depth δ(T * d ) = δ(T ) + 1 and replacing a leaf of depth d − 1 by a cherry of depth d. In other words, with the notations of Lemma 9, T = (T * d ) d−1 . Then, by (37), and, by Lemma 9, from where the expression in the statement follows.
Corollary 3 If T ∈ BT * n contains two leaves of the same depth d < δ(T ) − 1, then V (T ) is not minimum in BT * n .
Proof Let T ∈ BT * n and assume that it has two leaves, y 0 and y 1 , of the same depth d < δ(T ) − 1. If δ(T ) + d + 3 − 2 S(T ) 0 then, with the notations of Lemma 9, and therefore V (T ) cannot be minimum in BT * n . So, assume that δ(T ) + d + 3 − 2 S(T ) < 0. In this case, if one of the leaves y 0 or y 1 belongs to a cherry, then, with the notations of Corollary 2, and therefore in this case V (T ) cannot be minimum in BT * n , either. Finally, if y 0 and y 1 do not belong to any cherry, let v 0 and v 1 their respective parents and z 0 and z 1 their respective siblings, which are not leaves. Let T be obtained from T by interchanging y 1 with z 0 : that is, by removing the arcs (v 0 , z 0 ) and (v 1 , y 1 ) and replacing them by arcs (v 0 , y 1 ) and (v 1 , z 0 ) (see Fig. 12). Since δ T (y 1 ) = δ T (y 0 ) = δ T (z 0 ), the resulting tree T is depth-equivalent to T , and in particular V (T ) = V (T ). But T contains the cherry (y 0 , y 1 ) of depth d < δ(T ) − 1 and therefore, as we have just seen, Figure 12 The depth-equivalent trees T and T appearing in the last paragraph of the proof of Corollary 3.
We can summarize the results obtained so far in the following corollary: Corollary 4 If T ∈ BT * n has the minimum value of V , then it is of some type T n;l1,...,lj with 4 l 1 < · · · < l j δ(T ) − 2.
Next lemma finally completes the proof of Theorem 2.
Lemma 10 Let T be a tree of type T n;l1,...,lj with j 1. If V (T ) is minimum in BT * n , then l 1 5.
Proof Let T be a tree of type T n;l1,...,lj , with j 1, such that V (T ) is minimum in BT * n . By the last corollary, we already know that 4 l 1 < · · · < l j δ − 2. Set δ := δ(T ) and m := log 2 (n) , so that n = 2 m + k with 0 k < 2 m .
In the proof of Lemma 2 we saw that Then, if p 0 n/2, we have by Corollary 1 that l 1 3 + 2(n(δ − S(T )) + 1) which implies that l 1 5.
Assume now that p 0 > n/2. In this case, we can also assume that n 32. Indeed, if n < 32, then m 4 and hence, by Lemma 1, δ 6. Thus, since 4 l 1 δ −2 4, it must happen that δ = 6, m = 4, j = 1, and l 1 = 4. So, n = 2 4 +k with k 15 and since δ = 4 + 2 we are in case (c) in the statement of the aforementioned lemma, and therefore p 1 = 3 · 2 4 − k − (2 4 − 1) = 33 − k. Thus, because k 15, which contradicts the assumption that p 0 > n/2. So, in particular, we can assume that T contains at least 16 leaves of depth δ. Assume that T contains a leaf x of depth δ − 4, so that T is of type T n;4,l2,...,lj , and let y its sibling, also of depth δ − 4, and z their common parent. Since T does not contain either any other leaf of depth δ − 4 or leaves of depths δ − 3 or δ − 2, the leaves that descend from y have depths δ −1 or δ. Since T contains at least 16 leaves of depth δ, by pruning and regrafting cherries at maximum depth if necessary, we can assume without any loss of generality that all the leaves that descend from y have depth δ and hence that the subtree of T rooted at y is the fully symmetric tree B 16 .
Let T be now the tree obtained from T by, on the one hand, removing the leaf x, its parent z and the three arcs incident to z, and replacing them by an arc from the parent of z to y (which now becomes of depth δ − 5), and, on the other hand, replacing the subtree B 16 rooted at y by a maximally balanced tree B 17 ; see Fig. 13. From the 17 leaves in T that are descendant of y, 2 have depth δ in T , and the remaining 15 have depths δ − 1, and the rest of leaves in T have the same depth as in T . In particular, δ(T ) = δ and then T = T n;l2,...,lj . Moreover, p 1 (T ) = p 1 + 15. Therefore, using the expression for the variance of a tree of type T n;l 1 ,...,l j given in Lemma 2, we have that Then, V (T ) < V (T ), against the assumption that T had the minimum value of V of BT * n . So, when p 0 > n/2 we also conclude that if T has the minimum value of V of BT * n , l 1 > 4.

SN-7 Proof of Theorem 3
To simplify the language, in this section and in Section SN-13, for every sequence l = (l 1 , . . . , l j ) ∈ N j we shall set Our goal is to prove the following result: Proof Let n = 2 m + k with m = log 2 (n) and 0 k < 2 m , and take a non-empty sequence of indices 5 l 1 < · · · < l j m − 1 such that k 2 m − j i=1 (2 li − 1), so that V (T n;l1,...,lj ) = 1 Then Since A(l) + 2B(l) > 0 because each l i 5 (a fact that we shall use henceforth without any further notice), this implies that So, we have the following fact: Claim 1. If there exist 5 l 1 < · · · < l j m − 1, with j 1, such that Consider now the particular case of tree types T n;l1 , i.e., with j = 1. Set l 1 = x ∈ {5, . . . , m − 1}, so that A(l) = 2 x − x 2 − 1 and B(l) = 2 x − x − 1. Then, Claim 1 implies that if n = 2 m + k and if k belongs to To simplify the notations, let so that this union of intervals can be rewritten as We shall prove that there exists an m 1 ∈ N such that, for every m m 1 , this union of intervals is the interval (F 1 (m − 1), G 1 (5)]. To do that, we use of the next two Claims.
Claim 2. For every m 7 and for every x ∈ {5, . . . , m − 2}, F 1 (x + 1) < F 1 (x) and The decreasing monotonicity of G 1 is clear. As far as that of F 1 goes, we have that again because x 5. This finishes the proof of Claim 2.
Claims 5 and 6 jointly imply that there exists an m 2 ∈ N such that if m m 2 and 3 log 2 (m) x m − 2, the intervals (F 2 (x), G 2 (x)] satisfy that both their left-hand side and right-hand side ends increase with x, and each interval (F 2 (x), G 2 (x)], for x = 3 log 2 (m) , . . . , m − 3, has non-empty overlap with the "next" interval (F 2 (x + 1), G 2 (x + 1)]. Therefore, when m m 2 , So, using Claim 1, we deduce the following claim: There exists an m 2 ∈ N such that, for every m m 2 , if Now, it turns out that if m is large enough, the intervals overlap: more specifically, for large enough m we have that Indeed, on the one hand, the inequality G 2 (m−2) < G 1 (5), that is, 2 m−2 +m+26 < 2 m − 31, holds if m 7. On the other hand, the inequality Therefore, there exists an m 3 ∈ N such that if m m 3 , and setting M = max{m 1 , m 2 , m 3 } we deduce from Claims 4 and 7 that, for every m M , if n ∈ 2 m + F 2 ( log 2 (m 3 ) ), 2 m + G 1 (5) then V (B n ) is not minimal on BT * n . Let us mention incidentally that 2 m + G 1 (5) = 2 m+1 − 31 and in Section SN-13 we shall prove that this end can be replaced optimally by 2 m+1 − 30. This does not change what remains of our argument.
Since F 2 ( log 2 (m 3 ) ) ∼ 4 9 m 3 , we deduce that, for every m M , the cardinality of the set of numbers n ∈ [2 m , 2 m+1 ) such that V (B n ) is minimal on BT * n is in O(m 3 ). Then, for every m M , the fraction of values n ∈ [2, 2 m+1 ) such that V (B n ) is minimal on BT * n is bounded from above by which tends to 0 as m → ∞.

SN-8 Proof of Proposition 1
To begin with, notice that the index S (2) satisfies the following recurrence.
Lemma 11 For every T ∈ T * n , if T = T 1 · · · T k with k 2, then Proof Under the hypothesis in the statement, S(T i ) + n as we claimed. Now, our goal is to prove the following proposition.
Proof When n = 1, both sides of the identity in the statement are equal to 0, and therefore we shall consider henceforth only the case n 2. Recall from equation Since every T ∈ BT n , with n 2, is produced twice by choosing an integer k = 1, . . . , n − 1, a subset X k ⊆ [n] with |X k | = k, a tree T k ∈ BT(X k ), and a tree T n−k ∈ BT(X c k ), and taking T = T k T n−k , equation (11) and Lemma 11 allow us to compute E Y (S (2) n ) from its very definition as follows: (by the shape invariance of P Y , Lemma 11, and identity (11)) In particular, and therefore n ), dividing this last equality by n we obtain and using the formula for E Y (S n ) given in (13), this recurrence becomes Then, since x 1 = 0 and (for a proof, see Lemma 1.(4) in the Supplementary Material of [1]), we have that from where we finally deduce as we claimed.

SN-9 Proof of Proposition 2
Proposition 2 For every n 2, Proof Arguing as in the beginning of the proof of Proposition 1, but using identity (12) instead of (11), we have: as we claimed.

SN-10 Proof of Proposition 3
We establish first a general result on expected values of bifurcating recursive shape indices in the sense of [4] under the uniform model.
Definition A bifurcating recursive shape index is a mapping I that associates to each bifurcating phylogenetic tree a real number R satisfying the following two conditions: (a) It is invariant under tree isomorphisms and relabelings of leaves. Lemma 12 Let I and J be two bifurcating recursive shape indices, and let I n and J n be the random variables that choose a tree T ∈ BT n and compute I(T ) and J(T ), respectively. Then, for every n 2, the expected values of I n J n and I 2 n under the uniform model are Proof To prove (42), we develop E U (I n J n ) as we did with E ( S n ) in the proof of Proposition 2: T n−k I(T n−k )P U,n−k (T n−k ) T n−k J(T n−k )P U,n−k (T n−k ) using the symmetry of C k,n−k , f I (k, n − k), and f J (k, n − k).
As for (43), it is obtained from (42) by taking I = J in it.
Now we can use equation (43) in the last lemma to derive the recurrence in Proposition 3.

Proposition 3 For every n 2,
Proof Applying equation (43) taking as I the Sackin index S, for which we obtain because, by the symmetry of C k,n−k , Simplifying one step further the sum ( * ), we finally obtain The values of the first two sums appearing in the independent term in this recurrence can be computed using Lemmas 3 and 4: As for the third sum, its value is So, the independent term of the expression for E U (S 2 n ) given by equation (44) is This completes the proof of the identity in the statement.

SN-11 Proof of Theorem 6.(b)
In this section we prove the following result.
Theorem Let Φ n be the random variable that takes a tree T ∈ BT n and computes its total cophenetic index Φ(T ). Then, for every n 2, the variance of Φ n under the uniform model is Proof If we apply identity (43) in Lemma 12 taking as I the total cophenetic index Φ, for which we obtain the following recurrence for E U (Φ 2 n ): (a) If k 2 m − 29, the minimum value of V on BT * n is reached exactly at the trees of type F n . (b) If k = 2 m − 30. the minimum value of V on BT * n is reached exactly at the trees of type F n when m 7 and at the trees of type T n;6 if m 8.