The largest singletons in weighted set partitions and its applications

Recently, Deutsch and Elizalde studied the largest and the smallest fixed points of permutations. Motivated by their work, we consider the analogous problems in weighted set partitions. Let $A_{n,k}(\mathbf{t})$ denote the total weight of partitions on $[n+1]$ with the largest singleton $\{k+1\}$. In this paper, explicit formulas for $A_{n,k}(\mathbf{t})$ and many combinatorial identities involving $A_{n,k}(\mathbf{t})$ are obtained by umbral operators and combinatorial methods. As applications, we investigate three special cases such as permutations, involutions and labeled forests. Particularly in the permutation case, we derive a surprising identity analogous to the Riordan identity related to tree enumerations, namely, \begin{eqnarray*} \sum_{k=0}^{n}\binom{n}{k}D_{k+1}(n+1)^{n-k}&=&n^{n+1}, \end{eqnarray*} where $D_{k}$ is the $k$-th derangement number or the number of permutations of $\{1,2,\dots, k\}$ with no fixed points.


Introduction
A partition of a set [n] = {1, 2, . . . , n} is a collection π = {B 1 , B 2 , . . . , B r } of nonempty and mutually disjoint subsets of [n], called blocks, whose union is [n]. For a block B, we denote by |B| the size of the block B, that is the number of the elements in the block B. A block B will be called singleton if |B| = 1. If {k} is a singleton of a partition, we denote it by k for short. If |B| = j, we assign a weight t j for B. The weight w(π) of a partition π is defined to be the product of the weight of each block of π.
It is well known that the weight of partitions of [n] with r blocks is the partial Bell polynomial B n,r t 1 , t 2 , . . . [3] on the variables {t j } j≥1 , that is B n,r t 1 , t 2 , . . . = κn(r) n! r 1 !r 2 ! · · · r n ! t 1 1!
where the summation κ n (r) is for all the nonnegative integer solutions of r 1 + r 2 + · · · + r n = r and r 1 + 2r 2 + · · · + nr n = n. And the total weight for partitions of [n] is the complete Bell polynomial Y n (t) = Y n t 1 , t 2 , . . . = n r=0 B n,r t 1 , t 2 , . . . , 1 which has the exponential generating function Let A n,k denote the set of partitions of [n + 1] with the largest singleton k + 1. Let A n,k (t) denote the total weight of partitions in A n,k . Clearly, A n,0 (t) = t 1 Y n 0, t 2 , . . . and A n,n (t) = t 1 Y n t 1 , t 2 , . . . , where Y n 0, t 2 , . . . is the weight of partitions of [n] without singletons.
Recently, Deutsch and Elizalde [4] studied the largest fixed points of permutations, which is the special case when t j = (j − 1)! for j ≥ 1. Later, Sun and Wu [15] considered the largest singletons in set partitions, which is the special case when t j = 1 for j ≥ 1.
In this paper we will investigate the largest singletons in weighted set partitions generally. The next section is devoted to studying the properties of A n,k (t), involving its explicit formulas and many combinatorial identities for A n,k (t). In the third section, we consider the permutation case, i.e., the special case when t j = (j − 1)! for j ≥ 1, and derive a surprising identity analogous to the Riordan identity related to tree enumerations. In the forth section, we study the involution case which is the special case when t 1 = t 2 = 1, t j = 0 for j ≥ 3. In the final section, we focus on the labeled forest case which is the special case when t j = j j−1 for j ≥ 1.

The properties of A n,k (t)
According to the definition of A n,k (t), for any weighted partition π of [n + 1] with the largest singleton k + 1, if k is also a singleton, delete the singleton k + 1 and subtracting one from all the entries large than k + 1, we obtain a partition of [n] with the largest singleton k. This contributes the weight t 1 A n−1,k−1 (t); if k is not a singleton, exchange k and k + 1, we obtain a partition of [n + 1] with the largest singleton k. This contributes the weight A n,k−1 (t).
Then we obtain a recurrence for n, k ≥ 1, with the initial conditions A n,0 (t) = t 1 Y n 0, t 2 , . . . for n ≥ 0.
Lemma 2.1. The bivariate exponential generating function for A n+k,k (t) is given by Clearly, A 0 (t; x) = t 1 e −xt 1 Y(t; x). From (2.1), one can derive that This completes the proof.  [7,10,11] for more information on umbral calculus, to cite only a few). Then, by Lemma 2.1, we have When comparing the coefficient of x n y k n!k! , A n+k,k (t) can be represented umbrally as Let [x n ]f (x) denote the coefficient of x n in the formal power series f (x), we get The case λ = 0 in (2.3), yields the explicit formula for A n+m,m (t).  Proof. Let X denote the set of partitions of [n + m + 1] containing at least the singleton m + 1. Clearly, X has the weight t 1 Y n+m (t). Let X i be the subset of X containing another singleton m + i + 1 for 1 ≤ i ≤ n. Set X i = X − X i , then n i=1 X i is just the set of partitions of [n + m + 1] with the largest singleton m + 1, so n i=1 X i has the weight A n+m,m (t). For any nonempty (n − k)-subset S ∈ [n], i∈S X i is the set of partitions of [n + m + 1] containing at least the number n − k + 1 of singletons m + 1 and m + i + 1 for all i ∈ S, so i∈S X i has the weight t n−k+1 1 Y m+k (t). By the Inclusion-Exclusion principle, we have which proves (2.5). 2 Proof. The case n := n + 1 in (2.5), together with the case λ = 1 in (2.3), yields (2.6). The case n := n + 2 in (2.5), together with the case λ = 2 in (2.3), yields (2.7). And (2.8) can be easily obtained from (2.6) and (2.7). 2 Theorem 2.5. For any integers n, m, k ≥ 0, there holds Proof. Here we provide a combinatorial proof. For any π ∈ A n+m+k,m+k , suppose that π has exactly m − j singletons in {k + 1, . . . , k + m} which contribute the weight t m−j 1 , and there are m j ways to do this. The remainder j elements in {k + 1, . . . , k + m} can not be singletons in π. These j elements can be regarded as the roles that greater than m + k + 1, so the remainder n + k + j + 1 elements can be partitioned with the largest singleton m + k + 1, which contributes the weight A n+k+j,k (t). Thus the total weight of such partitions is m j t m−j 1 A n+k+j,k (t). Summing up all the possible cases yields (2.9). 2 Theorem 2.6. For any integers n, m ≥ 0 and any indeterminant y, there hold which proves (2.10). Similarly, (2.11) can be obtained, but here we provide a combinatorial proof. Let X n,m = n k=0 X n,m,k and X n,m,k denote the set of pairs (π, S) such that • S is an (n − k)-subset of [m + 2, n + m + 1] = {m + 2, . . . , n + m + 1}, and each element of S is colored by t 1 or y − t 1 ; • π is a partition of the set [n + m + 1] − S with the largest singleton m + 1, and each element of [n + m + 1] − S is only colored by 1.
Let Y n,m = n k=0 Y n,m,k and Y n,m,k denote the set of pairs (π, S) such that • S is an (n − k)-subset of [m + 2, n + m + 1] and each element of S is only colored by y − t 1 ; • π is a partition of the set [n + m + 1] − S such that m + 1 must be a singleton, and each element of [n + m + 1] − S is only colored by 1.
The weight of (π, S) is defined to be the product of the weight of π and the color of each element of [n + m + 1]. Clearly, the weights of X n,m and Y n,m are counted respectively by the left and right sides of (2.11). Given any pair (π, S) ∈ X n,m , S can be partitioned into two parts S 1 and S 2 such that each element of S 1 is colored by y −t 1 and each element of S 2 is colored by t 1 . Regard each element of S 2 as a singleton which is weighted by t 1 and colored by 1, together with π, we obtain a partition π 1 of [n + m + 1] − S 1 such that m + 1 is always a singleton. Then the pair (π 1 , S 1 ) lies in Y n,m . Conversely, for any pair (π 1 , S 1 ) ∈ Y n,m , let S denote the union of S 1 and the singletons of π 1 greater than m + 1, then π 1 can be partitioned into two parts π and π ′ such that π is a partition of [n + m + 1] − S with the largest singleton m + 1 and π ′ is the singletons of π 1 greater than m + 1. Regard π ′ as a subset of [m + 2, n + m + 1] in which each element is colored by t 1 , together with S 1 , we obtain an (n − k)-subset of [m + 2, n + m + 1] for some k such that each element of S is colored by t 1 or y − t 1 . Then the pair (π, S) lies in X n,m . Clearly we find a bijection between X n,m and Y n,m , which proves (2.11). 2 The cases y = −1 in (2.10) and y = t 1 in (2.11) lead to Corollary 2.7. For any integers n, m ≥ 0, there hold The case y := yt 1 y+1 in (2.11), together with (2.10) generates the following result which has a combinatorial interpretation. is not empty and m + 1 must be the largest singleton if S is empty; • each element of [m + 2, n + m + 1] − S must not be a singleton.
The weight of (π, S) is defined to be the product of the weight of π and the colors of all elements in [n + m + 1]. Clearly, any (π, S) ∈ X * n,m can be obtained as follows. First choose an (n − k)-subset S of [m + 2, n + m + 1], there are n k ways to do this. Regard each element of S as a singleton with color y. Then color each element of [m + 2, n + m + 1] − S by 1 or y, namely, each element of [m + 2, n + m + 1] − S is colored by y + 1. Now partitioning [n + m + 1] − S such that the largest singleton is m + 1, together with the n − k singletons formed form S, we get the partition π of [n + m + 1] such that m + 1 must be a singleton; Hence the total weight of pairs (π, S) ∈ X * n,m is just the left hand side of (2.12). Similarly, the total weight of pairs (π, S) ∈ Y * n,m is just the right hand side of (2.12) if regarding each element of [m + 2, n + m + 1] − S as the role greater than i k when S is not empty. Now we can construct a bijection ϕ between X * n,m and Y * n,m which preserves the weights. For any (π, S) ∈ X * n,m , let S 1 denote the set of elements of [n + m + 1] with colors y. Clearly, S is a subset of S 1 . Assume that S 1 = {i 1 , i 2 , . . . , i k } for some 0 ≤ k ≤ n in increasing order. If S 1 is the empty set ∅, which implies that S = ∅ and all elements of [n + m + 1] are colored by 1, it is obvious that (π, ∅) ∈ Y * n,m . Then define ϕ(π, ∅) = (π, ∅). If S 1 is not the empty set, exchanging m + 1 and i k in π, we obtain a partition π 1 , it is easily to verify that (π 1 , S 1 ) ∈ Y * n,m and has the same weight as (π, S). Then define ϕ(π, S) = (π 1 , S 1 ). Conversely, for any (π 1 , S 1 ) ∈ Y * n,m , if S 1 = ∅, so π 1 has the largest singleton m + 1, then (π 1 , ∅) ∈ X * n,m and define ϕ −1 (π 1 , ∅) = (π 1 , ∅). If S 1 = ∅, assume that S 1 = {i 1 , i 2 , . . . , i k } for some 1 ≤ k ≤ n in increasing order, let S denote the set of all the elements in S 1 such that each forms a singleton of π 1 . Now exchanging m + 1 and i k in π 1 , we obtain a partition π, it is easy verifiable that (π, S) ∈ X * n,m which has the same weight as (π 1 , S 1 ). Then define ϕ −1 (π 1 , S 1 ) = (π, S).
Clearly, ϕ is indeed a bijection between X * n,m and Y * n,m , which proves (2.12). 2

The special case for permutations
In this section, we consider the special case when t j = (j − 1)! for j ≥ 1. That is to assign a cycle structure to each block of partitions of [n + 1], such partitions with weight t = (0!, 1!, 2!, . . . ) is equivalent to permutations of [n + 1]. Let P n,k = A n,k (t) with t = (0!, 1!, 2!, . . . ), namely, P n,k is the number of permutations of [n + 1] with the largest fixed point k + 1. From (2.5) and (2.9), one has the explicit formulas for P n,k Clearly, P n,n = n! = Y n (0!, 1!, 2!, . . . ) and P n,0 = D n = Y n (0, 1!, 2!, . . . ), where D n is the derangement number of [n], i.e., the number of permutations of [n] without fixed points. See Table 1 for some small values of P n,k .  Table 1. The values of P n,k for n and k up to 6.
In fact {P n,k } n≥k≥0 forms the difference table introduced by Euler, which has been investigated in depth in the derangement theory [2,5,6,9,8]. Chen [1] also gave another two interpretations for P n,k using k-relative derangements on [n] and skew derangements from [n] to {−k + 1, . . . , −1, 0, 1, . . . , n − k} for 0 ≤ k ≤ n. Actually, Chen established a bijection between these two settings. In a forthcoming paper, we find the bijective connections between several combinatorial objects which are counted by the Euler difference table. Recently, Deutsch and Elizalde [4] gave a new interpretation of D n+2 as the sum of the values of the largest fixed points of all non-derangements of length n + 1. Namely, n k=0 (k + 1)P n,k = D n+2 , which is the special case of (2.7) when t = (0!, 1!, 2!, . . . ) and m = 0. From the previous section, one can obtain many interesting properties of P n,k which is left to interested readers. Furthermore, one can also explore some new relations between P n,k and other classical sequences such as Bell numbers or Fibonacci numbers.
To our best knowledge, (3.3) and (3.4) are the new and suprising identities analogous to the Riordan identity above. In a forthcoming paper, using the functional digraph theory, we will give a combinatorial interpretation for a more general identity involving the Riordan identity and (3.4) as special cases.

The special case for involutions
In this section, we consider the special case in detail when t 1 = t 2 = 1 and t j = 0 for j ≥ 3. That is to study partitions of [n + 1] with no blocks of sizes greater than 2, such partitions are equivalent to involutions of [n + 1]. Let Q n,k = A n,k (t) with t = (1, 1, 0, . . . ), namely, Q n,k is the number of involutions of [n + 1] with the largest fixed point k + 1. See Table 2 for some small values of Q n,k . Clearly, Q n,n = I n = Y n (1, 1, 0, . . . ) and Q n,0 = M n = Y n (0, 1, 0, . . . ), where I n is the number of involutions of [n], and M n is the number of involutions of [n] without fixed points. It is well known that I n and M n have the explicit formulas otherwise.
where B(n, j) = n! 2 n−j (n−j)!(2j−n)! is the Bessel number counting all the partitions of [n] into j blocks with the restriction of block sizes ≤ 2.
Then acting L 2 on the two sides of (4.11) leads to (4.13). 2

The special case for labeled forests
In this section, we consider the special case when t j = j j−1 for j ≥ 1. That is to assign a (rooted and labeled) tree structure to each block of partitions of [n + 1], such partitions with weight t = (1 0 , 2 1 , 3 2 , . . . ) are equivalent to labeled forests on [n + 1]. Let L n,k = A n,k (t) with t = (1 0 , 2 1 , 3 2 , . . . ), namely, L n,k is the number of labeled forests on [n + 1] with the largest singleton tree labeled by k + 1. A singleton tree is a labeled tree with exactly one point. Clearly, L n,n = Y n (1 0 , 2 1 , 3 2 , . . . ) = (n + 1) n−1 and L n,0 = Y n (0, 2 1 , 3 2 , . . . ), where L n,0 is also the number of labeled forests on [n] with no singleton trees. See Table 3 for some small values of L n,k .
Define the umbra L = Y t with t = (1 0 , 2 1 , 3 2 , . . . ), then L n,k can be represented umbrally as L n,k = L k (L − 1) n−k , (5.1) Similar to the Section 4, using (5.1) one can derive the corresponding results for L n+k,k , the details are left to readers.