Sum of weighted records in set partitions

TL;DR

This paper derives an explicit formula and asymptotic estimates for the total sum of weighted records in set partitions, linking it to Bell numbers and analyzing its generating function.

Contribution

It introduces a new explicit formula and asymptotic analysis for the sum of weighted records in set partitions, expanding understanding of this combinatorial statistic.

Findings

Derived explicit formula for sum of weighted records

Provided asymptotic estimates related to Bell numbers

Analyzed the generating function for the statistic

Abstract

The purpose of this paper is to find an explicit formula and asymptotic estimate for the total number of sum of weighted records over set partitions of $[n]$ in terms of Bell numbers. For that we study the generating function for the number of set partitions of $[n]$ according to the statistic sum of weighted records.

Equations

$$P_k(x,q)=\sum _{n\ge k}\sum _{\pi \in P_{n,k}}{x}^n{q}^{\mathrm{swrec}(\pi )}.$$

$$P_k(x,q)=\prod _{i=1}^k\frac{x{q}^i{q}^{(k+1-i)(k-i)}}{1-ix{\prod }_{j=i+1}^k{q}^j}.$$

$$\pi =1{\pi }^{(1)}2{\pi }^{(2)}\cdots k{\pi }^{(k)}$$

$$P_k(x,q)=\frac{x{q}^k}{1-kx}{P}_{k-1}(x{q}^k,q)$$

$$\frac{3}{4}(B_{n+3}-B_{n+2})-(n+\frac{7}{4})B_{n+1}-\frac{1}{2}(n+1)B_n.$$

$$B_n\frac{n^3}{r^3}\left(1+\frac{r}{n}\right)\left(1+O(\frac{\log n}{n})\right),$$

$$\frac{d}{dq}{P}_k(x,q){\mid }_{q=1}=\frac{x^k((\genfrac{}{}{0ex}{}{k+1}{2})+2(\genfrac{}{}{0ex}{}{k+1}{3}))}{(1-x)\dots (1-kx)}+\frac{x^{k+1}}{(1-x)\dots (1-kx)}\sum _{i=1}^k\frac{i(i+1+k)(k-i)}{2(1-ix)}.$$

$$\frac{d}{dq}{P}_k(x,q){\mid }_{q=1}=P_k(x,1)\sum _{i=1}^k\underset{q\to 1}{\lim }\left(\frac{\frac{d}{dq}{L}_i(q)}{L_i(q)}\right),$$

$$L_i(q)=\frac{x{q}^i{q}^{(k+1-i)(k-i)}}{1-ix{\prod }_{j=i+1}^k{q}^j}.$$

$$\underset{q\to 1}{\lim }\frac{d}{dq}{L}_i(q)=\frac{x\ell A_{i,k}(x,1)-x{\lim }_{q\to 1}\frac{d}{dq}{A}_{i,k}(x,q)}{(A_{i,k}(x,1){)}^2}.$$

$$\underset{q\to 1}{\lim }\frac{d}{dq}{L}_i(q)=\frac{x\left(2(i+(k+1-i)(k-i))(1-ix)+i(i+1+k)(k-i)x\right)}{2(1-ix{)}^2},$$

$$\underset{q\to 1}{\lim }\left(\frac{\frac{d}{dq}{L}_i(q)}{L_i(q)}\right)=i+(k+1-i)(k-i)+\frac{i(i+1+k)(k-i)x}{2(1-ix)}.$$

$$\frac{d}{dq}{P}_k(x,q){\mid }_{q=1}=\frac{x^k}{(1-x)\dots (1-kx)}\sum _{i=1}^k\left(i+(k+1-i)(k-i)+\frac{i(i+1+k)(k-i)x}{2(1-ix)}\right)$$

$$=\frac{x^k((\genfrac{}{}{0ex}{}{k+1}{2})+2(\genfrac{}{}{0ex}{}{k+1}{3}))}{(1-x)\dots (1-kx)}+\frac{x^{k+1}}{(1-x)\dots (1-kx)}\sum _{i=1}^k\frac{i(i+1+k)(k-i)}{2(1-ix)},$$

$${\tilde{P}}_k(x,u,q)=\sum _{n\ge k}\sum _{\pi \in P_{n,k}}\frac{x^n{u}^k{q}^{\mathrm{swrec}(\pi )}}{n!}.$$

$$\frac{d}{dq}{\tilde{P}}_k(x,u,q){\mid }_{u=q=1}=e^{e^x-1}(\frac{3}{4}{e}^{3x}+\frac{3}{2}{e}^{2x}-\frac{7}{4}{e}^x-x{e}^{2x}-\frac{3}{2}x{e}^x-\frac{1}{2}).$$

$$\sum _{m=1}^k\left(\frac{a_{k,m}}{(y-m{)}^2}+\frac{b_{k,m}}{y-m}\right),$$

$$a_{k,m}=\frac{(-1{)}^{k-m}m(1+k+m)(k-m)}{2(m-1)!(k-m)!},$$

$$b_{k,m}=\frac{(-1{)}^{k-m}\left(k^2(\frac{m}{4}+1)+k(\frac{m^2}{2}+\frac{3m}{4}+1)-(\frac{3m^2}{2}+m)\right)}{(m-1)!(k-m)!}.$$

$$\frac{d}{dq}{P}_k(x,q){\mid }_{q=1}=x^k\prod _{i=1}^k(1-ix{)}^{-1}\left(\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)+2\left(\genfrac{}{}{0ex}{}{k+1}{3}\right)+\sum _{i=1}^k\frac{i(1+k+i)(k-i)x}{2(1-ix)}\right)$$

$$=$$

$$\prod _{i=1}^k(y-i{)}^{-1}\left(\frac{k(k+1)(2k+1)}{6}+\sum _{i=1}^k\frac{i(1+k+i)(k-i)}{2(y-i)}\right).$$

$$\sum _{m=1}^k\left(\frac{a_{k,m}}{(y-m{)}^2}+\frac{b_{k,m}}{y-m}\right).$$

$$(y-m{)}^{-1}\prod _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k(y-m+m-i{)}^{-1}\left(\frac{k(k+1)(2k+1)}{6}+\frac{m(1+k+m)(k-m)}{2(y-m)}+\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\frac{i(1+k+i)(k-i)}{2(y-i)}\right)$$

$$=(y-m{)}^{-1}\prod _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\left((m-i{)}^{-1}(1+\frac{y-m}{m-i}{)}^{-1}\right)\cdot$$

$$\left(\frac{k(k+1)(2k+1)}{6}+\frac{m(1+k+m)(k-m)}{2(y-m)}+\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\frac{i(1+k+i)(k-i)}{2(y-i)}\right).$$

$$(y-m{)}^{-1}\frac{(-1{)}^{k-m}}{(m-1)!(k-m)!}\prod _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\left(1-\frac{y-m}{m-i}+O((y-m{)}^2)\right)$$

$$\cdot \left(\frac{k(k+1)(2k+1)}{6}+\frac{m(1+k+m)(k-m)}{2(y-m)}+\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\frac{i(1+k+i)(k-i)}{2(m-i)}+O(y-m)\right),$$

$$(y-m{)}^{-1}\frac{(-1{)}^{k-m}}{(m-1)!(k-m)!}\left(1-\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\frac{y-m}{m-i}+O((y-m{)}^2)\right)$$

$$\cdot \left(\frac{k(k+1)(2k+1)}{6}+\frac{m(1+k+m)(k-m)}{2(y-m)}+\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\frac{i(1+k+i)(k-i)}{2(m-i)}+O(y-m)\right).$$

$$(y-m{)}^{-1}\frac{(-1{)}^{k-m}}{(m-1)!(k-m)!}$$

$$\cdot \left(\frac{k(k+1)(2k+1)}{6}+\frac{m(1+k+m)(k-m)}{2(y-m)}+\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k\frac{i(1+k+i)(k-i)-m(1+k+m)(k-m)}{2(m-i)}+O(y-m)\right).$$

$$(y-m{)}^{-1}\frac{(-1{)}^{k-m}}{(m-1)!(k-m)!}$$

$$\cdot \left(\frac{m(1+k+m)(k-m)}{2(y-m)}+\frac{k(k+1)(2k+1)}{6}+\frac{1}{2}\sum _{\begin{array}{c}i=1\\ i\ne m\end{array}}^k(i^2+i+im-k^2-k+m+m^2)+O(y-m)\right)$$

Full text

Keywords: Records, Sum of weighted records, Set partitions, Generating functions, Bell numbers and Asymptotic estimate.

1 Introduction

Let $\sigma_{i}$ be an element in the permutation $\sigma=\sigma_{1}\sigma_{2}\cdots\sigma_{\ell}$, we say that $\sigma_{i}$ is a record of position $i$ if $\sigma_{i}>\sigma_{j}$ for all $j=1,2,\cdots,i-1$. The study of records in permutations interested Rényi [7]. More recently another statistic which depends on records have been studied by Kortchemski [5] who defined the statistic srec, where ${\rm srec}(\sigma)$ defined as the sum of the positions of all records of $\sigma$. For example, permutation $\pi=12534$ has 3 records, 1, 2, 5 and ${\rm srec}(\pi)=1+2+3=6$. For relevant papers about records you can see for example [2] and [3]. In this Paper we want to focus on partitions of a set. Recall that a partition $\Pi$ of set $[n]$ of size $k$ (a partition of $[n]$ with exactly $k$ blocks) is a collection $\{B_{1},B_{2},\ldots,B_{k}\}$, where $\emptyset\neq B_{i}\subseteq[n]$ for all $i$ and $B_{i}\bigcap B_{j}=\emptyset$ for $i\neq j$, such that $\bigcup_{i=1}^{k}B_{i}=[n].$ The elements $B_{i}$ are called blocks, and we use the assumption that $B_{1},B_{2},\cdots,B_{k}$ are listed in increasing order of their minimal elements, that is, $minB_{1}<minB_{2}<\cdots<minB_{k}$. The set of all partitions of $[n]$ with exactly $k$ blocks is denoted by $P_{n,k}$ and $|P_{n,k}|=S_{n,k}$, which is known as the Stirling numbers of the second kind [8]. And the set of all partitions of $[n]$ is denoted by $P_{n}$ and $|P_{n}|=\sum_{k=1}^{n}S_{n,k}=B_{n}$, which is the n-th Bell number [8]. Any partition $\Pi$ can be written as $\pi_{1}\pi_{2}\cdots\pi_{n}$, where $i\in B_{\pi_{i}}$ for all $i$, and this form is called the canonical sequential form. For example $\Pi=\{\{12\},\{3\},\{4\}\}$ is a partition of $[4]$, the canonical sequential form is $\pi=1123$. For more details about set partitions we suggest Mansour’s book [6]. The important results about records, obtained by Knopfmacher, Mansour and Wagner [4] which state the asymptotic mean value and variance for the number, and for the sum of positions, of record in all partitions of $[n]$ are central to my study. In this paper, we define a new statistic swrec, where ${\rm swrec}(\pi)$ is the sum of the position of a record in $\pi$ multiplied by the value of the record over all the records in $P_{n}$. We will study this statistic from the point of view of canonical sequential form. For instance, if $\pi=121132$ the ${\rm swrec}(\pi)=1\cdot 1+2\cdot 2+3\cdot 5=20$.

2 Main Results

2.1 The ordinary generating function for the number of set partitions according to the statistic ${\rm swrec}$

Let $P_{k}(x,q)$ be the generating function for the number of partitions of $[n]$ with exactly $k$ blocks according to the statistic ${\rm swrec}$, that is

[TABLE]

Theorem 1

The generating function for the number of partitions of $[n]$ with exactly $k$ blocks according to the statistic ${\rm swrec}$ is given by

[TABLE]

Proof

As we know, a set partition of $[n]$ with exactly $k$ blocks can be presented as canonical sequential form:

[TABLE]

for some $k$, where $\pi^{(j)}$ denotes an arbitrary word over an alphabet $[j]$ including the empty word. Thus, the contribution of $\pi=1\pi^{(1)}2\pi^{(2)}\cdots k$ to the generating function $P_{k}(x,q)$ is $xq^{k}P_{k-1}(xq^{k},q)$ and the contribution of $\pi^{(k)}$ to the generating function $P_{k}(x,q)$ is $\frac{1}{1-kx}$. Therefore, the corresponding generating function satisfies

[TABLE]

By using induction on $k$ together with the initial condition $P_{1}(x,q)=\frac{xq}{1-x}$ we obtain the required result.

$\Box$

2.2 Exact and asymptotic expression for $\sum\limits_{\pi\in P_{n}}{\rm swrec}(\pi)$

In this section, we aim to prove that the total number of the ${\rm swrec}$ over all partitions of $[n]$ is

[TABLE]

And we want to show that asymptotically the total number of the ${\rm swrec}$ over all partitions of $[n]$ is

[TABLE]

where $r$ is the positive root of $re^{r}=n+1$.

For that we need to perform the following steps:

$\bullet$ Firstly, we find the partial derivative of $P_{k}(x,q)$ with respect to $q$ and substitute $q=1$, that is $\frac{d}{dq}P_{k}(x,q)\mid_{q=1}$.

$\bullet$ Secondly, we pass from $\frac{d}{dq}P_{k}(x,q)\mid_{q=1}$ to $\frac{d}{dq}\widetilde{P}_{k}(x,u,q)\mid_{u=q=1}$, where $\widetilde{P}_{k}(x,u,q)$ is the exponential generating function for the number of partitions of $[n]$ with exactly $k$ blocks according to the statistic ${\rm swrec}$.

$\bullet$ Finally, we derive the total number of ${\rm swrec}$ over all partitions of $[n]$, and the asymptotic estimate for the total number of ${\rm swrec}$ over all partitions of $[n]$.

Lemma 2

For all $k\geq 1$,

[TABLE]

Proof

By differentiating (1) with respect to $q$, we obtain

[TABLE]

where

[TABLE]

We have

[TABLE]

Where $A_{i,k}(x,q)=1-ix\prod_{j=i+1}^{k}q^{j}$ and $\ell=(k+1-i)(k-i)+i$. By using the differentiation rules we get $\frac{d}{dq}A_{i,k}(x,q)=-ix\sum_{m=i+1}^{k}mq^{m-1}\prod_{\begin{subarray}{c}j=i+1\\ j\neq m\end{subarray}}^{k}q^{j}$. Therefore,

[TABLE]

which leads to

[TABLE]

Hence, by substituting (5) in (3) we obtain

[TABLE]

as claimed. $\Box$

Now we need to find $[x^{n}]\frac{d}{dq}P_{k}(x,q)\mid_{q=1}$ to obtain the total number of ${\rm swrec}$. We will study the exponential generating function instead of the ordinary generating function. Let $\widetilde{P}_{k}(x,u,q)$ be the exponential generating function for the number of partitions of $[n]$ with exactly $k$ blocks according to the statistic ${\rm swrec}$, that is

[TABLE]

Theorem 3

The partial derivative of $\widetilde{P}_{k}(x,u,q)$ with respect to $q$ at $u=q=1$ is given by,

[TABLE]

Proof

In order to prove the above result we need the following proposition:

Proposition 4

The partial derivative $\frac{d}{dq}P_{k}(x,q)\mid_{\begin{subarray}{c}x=y^{-1}\\ q=1\end{subarray}}$ can be decomposed as

[TABLE]

where

[TABLE]

and

[TABLE]

Proof

We rewrite (2) as

[TABLE]

By replacing $x^{-1}=y$ in the above equation we get

[TABLE]

The above expression decomposed as

[TABLE]

In order to find the coefficients $a_{k,m}$ and $b_{k,m}$, we need to consider the expansion of (8) at $y=m$, as follows:

[TABLE]

Using Taylor series to expand $(1+\frac{y-m}{m-i})^{-1}$ and $\frac{i(1+k+i)(k-i)}{2(y-i)}$ at $y=m$ we get

[TABLE]

which is equivalent to

[TABLE]

We need to simplify the product, and consider the coefficients of $(y-m)^{-1}$ and $(y-m)^{-2}$ as follows:

[TABLE]

By using Maple we compute the term in the summation, which hints

[TABLE]

Hence, by finding the coefficients of of $(y-m)^{-1}$ and $(y-m)^{-2}$ we complete the proof. $\Box$

Now we return to the proof of the theorem. We use (7) for passing to exponential generating function, by substituting

[TABLE]

in

[TABLE]

and $\frac{e^{mx}(mx-1)+1}{m^{2}}$ in $\frac{1}{(y-m)^{2}}$. Moreover, by summing over all $k$ we obtain the generating function

[TABLE]

We need to change the order of the summation as follows:

[TABLE]

By substituting $\ell=k-m$ and rewriting the above result we obtain the following form:

[TABLE]

By evaluating the previous terms in $u=1$ we complete the proof. $\Box$

Theorem 5

The total number of ${\rm swrec}$ taken over all set partitions of $[n]$, is given by

[TABLE]

Proof

In order to find the total number of ${\rm swrec}$, we need to find an explicit formula for the coefficient of $x^{n}$ in the generating function $\frac{d}{dq}\widetilde{P}_{k}(x,u,q)\mid_{u=q=1}$. By Theorem 3

[TABLE]

By differentiating the well known generating function $e^{e^{x}-1}=\sum_{n\geq 0}B_{n}\frac{x^{n}}{n!}$ three times we obtain

[TABLE]

[TABLE]

and

[TABLE]

From the above equations, we can derive that

[TABLE]

and

[TABLE]

Using all these facts together leads to

[TABLE]

Hence the total number of ${\rm swrec}$ is given by

[TABLE]

$\Box$

In order to obtain asymptotic estimate for the moment as well as limiting distribution, we need the fact

[TABLE]

uniformly for $h=O(\log n)$, where $r$ is the positive root of $re^{r}=n+1$. For more details about the asymptotic expansion of Bell numbers see [1]. Therefore, Theorem 5 gives the following corollary.

Corollary 6

Asymptotically, the total number of ${\rm swrec}$ taken over all set partitions of $[n]$, is given by

[TABLE]

Acknowledgement. The research of the author was supported by the Ministry of Science and Technology, Israel.