# New modular symmetric function and its applications: Modular $s$-Stirling
numbers

Bazeniar Abdelghafour  University Center of Abdelhafid Boussouf, Mila.
Department of Mathematics, University of Mohamed Seddik Benyahia, LMAM
laboratory, Jijel 18000, Algeria<EMAIL_ADDRESS>,  Moussa Ahmia
Department of Mathematics, University of Mohamed Seddik Benyahia, LMAM
laboratory, Jijel 18000, Algeria<EMAIL_ADDRESS>,  José L. Ramírez
Departamento de Matemáticas, Universidad Nacional de Colombia, Bogotá,
Colombia<EMAIL_ADDRESS>http://sites.google.com/site/ramirezrjl and
Diego Villamizar  Department of Mathematics and Systems Analysis, Aalto
University, 00076 Aalto, Finland<EMAIL_ADDRESS>https://sites.google.com/view/dvillami/

###### Abstract.

In this paper, we consider a generalization of the Stirling number sequence of
both kinds by using a specialization of a new family of symmetric functions.
We give combinatorial interpretations for this symmetric functions by means of
weighted lattice path and tilings. We also present some new convolutions
involving the complete and elementary symmetric functions. Additionally, we
introduce different families of set partitions to give combinatorial
interpretations for the modular $s$-Stirling numbers.

###### Key words and phrases:

Symmetric functions, generating functions, Stirling numbers.

###### 2010 Mathematics Subject Classification:

05A15, 05A19

##  1\. Introduction

A symmetric function is homogeneous of degree $k$ if every monomial in it has
total degree $k$. Symmetric functions are ubiquitous in mathematics and
mathematical physics. For example, they appear in elementary algebra (e.g.
Viète’s theorem), representation theories of symmetric groups, and general
linear groups over the complex numbers or finite fields. They are also
important objects to study in algebraic combinatorics.

A _set partition_ of a set $[n]:=\\{1,2,\ldots,n\\}$ is a collection of non-
empty disjoint subsets, called _blocks_ , whose union is $[n]$. Let ${n\brace
k}$ denote the number of set partitions of $[n]$ into $k$ non-empty blocks.
This sequence is called the _Stirling numbers of the second kind_. Similarly,
let ${n\brack k}$ denote the number of permutations of $[n]$ into $k$ cycles.
This sequence is called the _Stirling numbers of the first kind_. The
literature contains several generalizations of Stirling numbers of both kinds;
see for example [14, 15, 4].

Given a set of variables $x_{1},x_{2},\ldots,x_{n}$, the $k$-th _elementary
and complete symmetric polynomials_ are defined, respectively, by

$\displaystyle e_{k}(x_{1},x_{2},\dots,x_{n})$ $\displaystyle=\sum_{1\leq
i_{1}<i_{2}<\cdots<i_{k}\leq n}x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}},\qquad 1\leq
k\leq n,$ $\displaystyle h_{k}(x_{1},x_{2},\dots,x_{n})$
$\displaystyle=\sum_{1\leq i_{1}\leq i_{2}\leq\cdots\leq i_{k}\leq
n}x_{i_{1}}x_{i_{2}}\cdots x_{i_{k}},\qquad k\geq 1,$

with initial conditions
$e_{0}(x_{1},x_{2},\dots,x_{n})=h_{0}(x_{1},x_{2},\dots,x_{n})=1$. Note that
$e_{k}(x_{1},x_{2},\dots,x_{n})=0$ if $k>n$. The generating functions for the
$e_{k}$ and $h_{k}$ are given by the expressions

$\displaystyle\sum_{k=0}^{n}e_{k}(x_{1},x_{2},\dots,x_{n})z^{k}$
$\displaystyle=\prod_{i=1}^{n}(1+x_{i}z),$ $\displaystyle\sum_{k\geq
0}h_{k}(x_{1},x_{2},\dots,x_{n})z^{k}$
$\displaystyle=\prod_{i=1}^{n}\frac{1}{1-x_{i}z}.$

A variety of combinatorial sequences can be obtained as evaluations of the
symmetric polynomials at specific points (cf. [6, 13]). Particularly, the
Stirling numbers of both kinds are given by

$h_{k}(1,2,\dots,n)={n+k\brace k}\quad\text{and}\quad
e_{k}(1,2,\dots,n)={n+1\brack n+1-k}.$

In this work, we introduce an extension of the Stirling numbers of both kinds,
called _$s$ -modular Stirling numbers_, by introducing a new class of
symmetric functions, and considering these new sequences as specializations of
this symmetric function. We give a combinatorial interpretation of these
symmetric functions by using weighted lattice path and tilings. Similar
symmetric functions were studied by Doty and Walker under the name of modular
complete symmetric polynomials [5]. Most recently, Ahmnia and Merca [1]
introduced a variation of these symmetric functions. Independently, Grinberg
[8] and Fu and Mei [7] introduced the same concept under the name of Petrie
symmetric functions and truncated symmetric functions, respectively. Finally,
we use set partitions to give a combinatorial interpretation to the
$s$-modular Stirling numbers. Among other things, we give an interpretation
(probably new) of the Stirling numbers of first kind in terms of set
partitions. We also give a relationship with the Stirling numbers with higher
level. This last sequence was recently studied in the context of special
polynomials [10].

##  2\. Definitions and properties

Let $s\geq 1$ be a positive integer. We define a modular symmetric function by

(1)
$M_{k}^{(s)}(n):=M_{k}^{(s)}(x_{1},\ldots,x_{n})=\sum_{{}_{\begin{subarray}{c}a_{1}+\cdots+a_{n}=k\\\
a_{1},\ldots,a_{n}\equiv(0,1)\mod(s+1)\end{subarray}}}x_{1}^{a_{1}}\cdots
x_{n}^{a_{n}},$

with $M_{k}^{(1)}(n)=h_{k}(n)$ and $M_{k}^{(s)}(0)=\delta_{k,0}$, where
$\delta_{k,0}$ is the Kronecker delta.

For example, for $s=2$ and $n=1$ we have

$M_{0}^{(2)}(1)=1,\ \ M_{1}^{(2)}(1)=x_{1},\ \ M_{2}^{(2)}(1)=0,\ \
M_{3}^{(2)}(1)=x^{3}_{1}\ \ M_{4}^{(2)}(1)=x^{4}_{1},\ \ M_{5}^{(2)}(1)=0.$

For $s=2=n$ we have

$M_{0}^{(2)}(2)=1,\ \ M_{1}^{(2)}(2)=x_{1}+x_{2},\ \
M_{2}^{(2)}(2)=x_{1}x_{2},\ \ M_{3}^{(2)}(2)=x^{3}_{1}+x^{3}_{2}.$

From the definition of $M_{k}^{(s)}(n)$ we have the following theorem.

######  Theorem 2.1.

Let $s$ and $n$ be positive integers. Then

(2) $\sum_{k\geq
0}M_{k}^{(s)}(n)t^{k}=\prod_{i=1}^{n}\frac{1+x_{i}t}{1-\left(x_{i}t\right)^{s+1}}.$

Moreover, the modular symmetric function also satisfies the following
recurrence relations.

######  Theorem 2.2.

Let $s$ and $n$ be positive integers. Then

(3) $\displaystyle M_{k}^{(s)}(n)=\sum_{\begin{subarray}{c}0\leq j\leq k\\\
j\equiv(0,1)\mod(s+1)\end{subarray}}x_{n}^{j}M_{k-j}^{(s)}(n-1),$ (4)
$\displaystyle
M_{k}^{(s)}(n)=x_{n}^{s+1}M_{k-s-1}^{(s)}(n)+x_{n}M_{k-1}^{(s)}(n-1)+M_{k}^{(s)}(n-1),$

for $k\geq s+1$.

###### Proof.

From (1) and the Theorem 2.1 we have

$\displaystyle\sum_{k\geq 0}M_{k}^{(s)}(n)t^{k}$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{n}\frac{1+x_{i}t}{1-\left(x_{i}t\right)^{s+1}}=\prod_{i=1}^{n}\left((1+x_{i}t)\sum_{j\geq
0}\left(x_{i}t\right)^{(s+1)j}\right)$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{n}\left(\sum_{j\geq
0}\left(x_{i}t\right)^{(s+1)j}+\sum_{j\geq
0}\left(x_{i}t\right)^{(s+1)j+1}\right)$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{n}\left(\sum_{\ell\equiv(0,1)\mod(s+1)}\left(x_{i}t\right)^{\ell}\right)$
$\displaystyle=$
$\displaystyle\sum_{j\equiv(0,1)\mod(s+1)}\left(x_{n}t\right)^{j}\prod_{i=1}^{n-1}\left(\sum_{\ell\equiv(0,1)\mod(s+1)}\left(x_{i}t\right)^{\ell}\right)$
$\displaystyle=$
$\displaystyle\sum_{j\equiv(0,1)\mod(s+1)}\left(x_{n}t\right)^{j}\sum_{\ell=0}^{\infty}M_{\ell}^{(s)}(n-1)t^{\ell}$
$\displaystyle=$ $\displaystyle\sum_{k\geq
0}t^{k}\sum_{\begin{subarray}{c}j+\ell=k\\\
j\equiv(0,1)\mod(s+1)\end{subarray}}x_{n}^{j}M_{\ell}^{(s)}(n-1)$
$\displaystyle=$ $\displaystyle\sum_{k\geq
0}t^{k}\sum_{\begin{subarray}{c}0\leq j\leq k\\\
j\equiv(0,1)\mod(s+1)\end{subarray}}x_{n}^{j}M_{k-j}^{(s)}(n-1).$

By comparing the $k$-th coefficient we obtain (3). The relation (4) follows in
a similar manner.∎

Notice that from (3) we have the equality
$h_{k}(n)=\sum_{j=0}^{k}x_{n}^{j}h_{k-j}(n-1)$.

##  3\. Combinatorial interpretation

The goal of this section is to present a combinatorial interpretation for the
modular symmetric functions by means of weighted lattice paths in the plane
${\mathbb{Z}}\times{\mathbb{Z}}$. A _lattice path_ $\Gamma$ in the lattice
plane ${\mathbb{Z}}\times{\mathbb{Z}}$, with steps in a given set
$S\subset{\mathbb{Z}}^{2}$, is a concatenation of directed steps of $S$, that
is $\Gamma=s_{1}s_{2}\cdots s_{\ell}$, where $s_{i}\in S$, for each $1\leq
i\leq\ell$. Let ${\mathcal{P}}_{n,k}$ denote the set of lattice paths from the
point $(0,0)$ to the point $(k,n-1)$, with step set $S=\\{H=(1,0),V=(0,1)\\}$,
such that the horizontal steps are labelled with the weight $x_{i}$, where
$i-1$ is the level of the step. Let ${\mathcal{P}}_{n,k}^{(s)}$ denote the
weighted lattice path in ${\mathcal{P}}_{n,k}$ such that the number of
horizontal steps in each level are congruent to $0$ or $1$ modulo $s+1$. Given
a weighted path $\Gamma$ in ${\mathcal{P}}_{n,k}^{(s)}$, we denote by
$\omega(\Gamma)$ the weight associated to the path $\Gamma$. For example, in
Figure 1 we show a lattice path in ${\mathcal{P}}_{6,12}^{(2)}$ of weight
$x_{2}^{6}x_{4}x_{5}x_{6}^{4}$.

Figure 1. Weighted lattice path in ${\mathcal{P}}_{6,12}^{(2)}$.

From (1) we obtain the following combinatorial interpretation.

######  Theorem 3.1.

Let $k,n$ and $s$ be positive integers and let $x_{1},x_{2},\ldots,x_{n}$ be
independent variables. Then

$M_{k}^{(s)}\left(x_{1},x_{2},\ldots,x_{n}\right)=\sum_{\Gamma\in{\mathcal{P}}_{n,k}^{(s)}}\omega(\Gamma).$

Figure 2 shows the weighted lattice path interpretation for
$M_{3}^{(2)}(x_{1},x_{2},x_{3})=x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{1}x_{2}x_{3}$.

$x_{1}$$x_{1}$$x_{1}$$x_{2}$$x_{2}$$x_{2}$$x_{3}$$x_{3}$$x_{3}$$x_{1}$$x_{2}$$x_{3}$
Figure 2. The four paths associated to $M_{3}^{(2)}(x_{1},x_{2},x_{3})$.

###  3.1. Tiling interpretation

In this section, we use weighted tilings to give an additional combinatorial
interpretation of the modular symmetric function. We define a _weighted
tiling_ as a tiling of a board of length $n$ ($n$-board) by gray and black
squares, such that each black square received the weight $x_{m+1}$, where $m$
is equal to the number of gray squares to the left of that black square in the
tiling. Let ${\mathcal{T}}^{(s)}_{n,k}$ denote the set of weighted tilings of
an $(n+k-1)$-board using exactly $k$ black squares and $n-1$ gray squares,
such that the number of successive black squares is congruent to $0$ or $1$
modulo $s+1$. For a tiling $T$, we denote by $\omega(T)$ the weight of $T$.

For example, in Figure 3 we show a weighted tiling in
${\mathcal{T}}_{6,12}^{(2)}$ of weight $x_{2}^{6}x_{4}x_{5}x_{6}^{4}.$

Figure 3. Weighted tiling in ${\mathcal{T}}_{6,12}^{(2)}$.

There is a bijection between the sets ${\mathcal{P}}^{(s)}_{n,k}$ and
${\mathcal{T}}^{(s)}_{n,k}$. Indeed, each vertical step $V$ is replaced by a
gray square and each horizontal step is replaced by a black square. Since the
bijection between lattice paths and tiling is weight-preserving, we obtain the
following result.

######  Theorem 3.2.

Let $k,n$ and $s$ be positive integers and let $x_{1},x_{2},\ldots,x_{n}$ be
independent variables. Then

$M_{k}^{(s)}\left(x_{1},x_{2},\ldots,x_{n}\right)=\sum_{T\in\mathcal{T}^{(s)}_{n,k}}\omega(T).$

Figure 4 shows the tiling interpretation for
$M_{3}^{(2)}(x_{1},x_{2},x_{3})=x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{1}x_{2}x_{3}$.

Figure 4. The four tilings associated to $M_{3}^{(2)}(x_{1},x_{2},x_{3})$.

In Theorem 3.3 we give a combinatorial expression for the sequence
$M_{k}^{(s)}(\underbrace{1,1,\ldots,1}_{n\text{\ times}})$.

######  Theorem 3.3.

For $n,k\geq 0,s\geq 1$, we have

$M_{k}^{(s)}(\underbrace{1,1,\ldots,1}_{n\
\text{\emph{times}}})=\sum_{j=0}^{\left\lfloor\frac{k}{s+1}\right\rfloor}\binom{n}{k-j(s+1)}\binom{j+n-1}{n-1}.$

###### Proof.

From the combinatorial interpretation
$M_{k}^{(s)}(\overbrace{1,1,\ldots,1}^{n\text{\ times}})$ counts the number of
weighted tilings of a $(n+k-1)$-board using exactly $k$ black squares and
$n-1$ gray squares, such that the number of successive black squares is
congruent to $0$ or $1$ modulo $s+1$. On the other hand, let $j$ be the number
of successive black squares multiples of $s+1$. Notice that $0\leq
j\leq\lfloor k/(s+1)\rfloor$. Then there are $j+n-1$ gray and black blocks
tiles. Such a tiling with $j+n-1$ tiles, exactly $j$ of which are black blocks
of size congruent to $0$ module $s+1$ is $\binom{j+n-1}{j}.$ The remaining
$k-j(s+1)$ black squares can be inserted before to each gray square or to the
end of the tiling. Since there are $n-1$ gray squares we have
$\binom{n}{k-j(s+1)}$ ways to insert the black squares. Hence there are
$\binom{n}{k-j(s+1)}\binom{j+n-1}{j}$ tilings altogether. Summing over all $j$
gives the total number of weighted tiling in $\mathcal{T}^{(s)}_{n,k}$, which
implies the identity. ∎

Notice that we can also give an algebraic proof for the above result. Indeed,
from the generating function given in Theorem 2.1 we have

$\displaystyle\sum_{k\geq 0}M_{k}^{(s)}(1,1,\ldots,1)t^{k}$
$\displaystyle=\frac{(1+t)^{n}}{\left(1-t^{s+1}\right)^{n}}=\sum_{j=0}^{n}\binom{n}{j}t^{j}\sum_{\ell\geq
0}\binom{\ell+n-1}{n-1}t^{\ell(s+1)}$
$\displaystyle=\sum_{j=0}^{n}\sum_{\ell\geq
0}\binom{n}{j}\binom{\ell+n-1}{n-1}t^{j+\ell(s+1)}=\sum_{k\geq
0}t^{k}\sum_{j+\ell(s+1)=k}\binom{n}{j}\binom{\ell+n-1}{n-1}$
$\displaystyle=\sum_{k\geq
0}t^{k}\sum_{\ell=0}^{\left\lfloor\frac{k}{s+1}\right\rfloor}\binom{n}{k-\ell(s+1)}\binom{\ell+n-1}{n-1}.$

By comparing the $k$-th coefficient we obtain the desired result.

##  4\. Modular $s$-Stirling numbers

The Stirling numbers of the second kind ${n\brace k}$ can be determined by the
recurrence relation ${n\brace k}={n-1\brace k-1}+k{n-1\brace k}$, with the
initial conditions ${0\brace 0}=1$ and ${n\brace 0}={0\brace n}=0$ for $n\geq
1$. It is well-known that the ${n\brace k}$ are determined by the identities
$x^{n}=\sum_{k=0}^{n}{n\brace k}x^{\underline{k}},n\geq 0$, where
$x^{\underline{n}}=x(x-1)\cdots(x-(n-1))$ for $n\geq 1$ and
$x^{\underline{0}}=1$ or equivalently by the generating function

(5) $\displaystyle\sum_{n\geq k}{n\brace k}x^{n}$
$\displaystyle=\frac{x^{k}}{(1-x)(1-2x)\cdots(1-kx)},\quad k\geq 0.$

Using (5), it is not difficult to show that the Stirling numbers of the second
kind are the specialization of the complete symmetric function given by

(6) $\displaystyle{n+k\brace n}$
$\displaystyle=h_{k}(1,2,\dots,n)=\sum_{a_{1}+\cdots+a_{n}=k}1^{a_{1}}\cdots
n^{a_{n}}.$

The Equation (6) can be interpreted by considering the following algorithm:

  1. (1)

Start with the partition of $[1]$ given by $\\{1\\}$.

  2. (2)

Take every integer from $2$ to $1+a_{1}$ and put it in the block of $1,$ so
you end up having $\\{1,2,\dots,1+a_{1}\\}.$

  3. (3)

Then you place $2+a_{1}$ in a new block and for every integer in between
$3+a_{1}$ and $3+a_{1}+a_{2}$ you have $2$ options, either you place this
number in the first block or in the second one. You place $3+a_{1}+a_{2}$ in a
new block and so now you will have $3$ options.

  4. (4)

You keep doing this until you have placed $n+k$ elements.

Algorithm 1 Interpretation of (6).

For example, for $n=3$ and $k=5$ the term $1^{2}2^{1}3^{2}$ corresponds to a
partition that looks like $\\{1,2,3,*\\},\ \\{4,*\\},\ \\{6,*\\}$, where $5$
can go in either of the first $2$ blocks and $7,8$ can go in any block (there
are $3$ of them). Giving a total of $2^{1}\cdot 3^{2}$ options.

Notice then that the $a_{i}$ integers have a direct relationship with the
minimal elements in each of the blocks of a partition. To see this, consider
the following construction: let $\Pi(n,k)$ denote the set of partitions of
$[n]$ having $k$ blocks. Suppose $\pi\in\Pi(n,k)$ is represented as
$\pi=B_{1}/B_{2}/\cdots/B_{k}$, where $B_{i}$ denotes the $i$-th block, with
$\min(B_{1})<\min(B_{2})<\cdots<\min(B_{k})$. Call $m_{i}=\min(B_{i})$ and
define the vector of consecutive differences by

$d(\pi):=(d_{1},\dots,d_{k})=(m_{2}-m_{1}-1,m_{3}-m_{2}-1,\dots,m_{k}-m_{k-1}-1,n-m_{k}).$

In the example above, notice that $d_{i}$ corresponds to $a_{i}$ because these
are exactly the number of elements that we have to place in $i$ blocks and so
there are a total of $i^{d_{i}}$ ways to do this. Notice, further, that since
$m_{1}=1,$ we have that $d_{1}+\cdots+d_{k}=n-k$. If we impose the modularity
conditions on the $d_{i}$’s, we get the modular symmetric function defined in
Equation (1).

Notice that (6) can be written as ${n\brace k}=h_{n-k}(1,2,\dots,k).$ From
this last equation and the combinatorial motivation of $d(\pi),$ we introduce
a new kind of Stirling numbers. For all integer $n\geq 0$ and all $k$ with
$0\leq k\leq n$ _the modular $s$-Stirling numbers of the second kind_, denoted
by ${n\brace k}^{(s)}$, are defined by the expression

(7) $\displaystyle{n\brace k}^{(s)}=M_{n-k}^{(s)}(1,2,\dots,k).$

It is clear that for $s=1$ we recover the Stirling numbers of the second kind,
that is, ${n\brace k}^{(1)}={n\brace k}$. From Theorem 2.2 we have the
following recurrence relation:

(8) ${n\brace k}^{(s)}={n-1\brace k-1}^{(s)}+k{n-2\brace
k-1}^{(s)}+k^{s+1}{n-s-1\brace k}^{(s)},$

with the initial conditions ${n\brace 0}^{(s)}=\delta_{0,n}$ and ${0\brace
k}^{(s)}=\delta_{k,0}$. Moreover, we have the following generating function.
For positive integers $n$ and $s$, we have

$\sum_{n\geq k}{n\brace
k}^{(s)}x^{n-k}=\prod_{r=1}^{k}\frac{1+rx^{s}}{1-(rx)^{s+1}}.$

In Theorem 4.1 we give a combinatorial interpretation for the modular
$s$-Stirling numbers.

######  Theorem 4.1.

The number of set partitions $\pi$ in $\Pi(n,k)$, such that the entries in the
vector $d(\pi)$ satisfies $d_{i}\equiv 0,1\pmod{s+1}$ for each $1\leq i\leq k$
is given by the modular $s$-Stirling numbers ${n\brace k}^{(s)}$.

###### Proof.

By imposing the modularity conditions on the vector of consecutive differences
given above and applying the Algorithm described on Page 1, the theorem
follows. ∎

For example, ${5\brace 2}^{(2)}=9$ corresponding to the set partitions

$\displaystyle 1234/5,\quad 1345/2,\quad 134/25,\quad 135/24,\quad 13/245,$
$\displaystyle 145/23,\quad 14/235,\quad 15/234,\quad 1/2345.$

If you restrict the difference vector to have elements of the form
$d_{i}\equiv 0\pmod{s+1},$ then the number of such partitions is given by
$h_{\lfloor\frac{n-k}{s+1}\rfloor}(1^{s+1},\dots,k^{s+1}).$

On the other hand, the (unsigned) _Stirling numbers of the first kind_ ,
satisfies the recurrence relation ${n\brack k}=(n-1){n-1\brack k}+{n-1\brack
k-1}$, with the initial conditions ${0\brack 0}=1$ and ${n\brack 0}={0\brack
n}=0$ for $n\geq 1$. This sequence can also defined as the connection
constants in the polynomial identity

(9) $x(x+1)\cdots(x+(n-1))=\sum_{k=0}^{n}{n\brack k}x^{k}.$

######  Theorem 4.2.

Let $n$ and $k$ be non negative integers and $s>0$. If $r$ is the remainder of
$k$ when divided by $s+1$, then the following equation holds

$\displaystyle{n+k\brace
n}^{(s)}=\sum_{i=0}^{\min\\{\lfloor\frac{n-r}{s+1}\rfloor,\lfloor\frac{k}{s+1}\rfloor\\}}h_{\lfloor\frac{k}{s+1}\rfloor-i}(1^{s+1},\dots,n^{s+1}){n+1\brack
n+1-r-i(s+1)}.$

###### Proof.

From Equation (1) consider the subset of the variables $\\{a_{i}\\}_{i\in[n]}$
such that $a_{j}\equiv 1\pmod{s+1}.$ Call $J\subseteq[n]$ the set of their
subindices in such a way that $j\in J$ if and only if $a_{j}\equiv
1\pmod{s+1}.$ Notice that for each $j\in J$, one must have that
$a_{j}=b_{j}(s+1)+1$ for some $b_{j}\in\mathbb{N}.$ Therefore we have

$\displaystyle{n+k\brace n}^{(s)}$
$\displaystyle=M^{(s)}_{k}(1,\dots,n)=\sum_{\begin{subarray}{c}J\subseteq[n]\\\
|J|\equiv
r\pmod{s+1}\end{subarray}}\,\sum_{\begin{subarray}{c}a_{1}+\cdots+a_{n}=k\\\
a_{j}\equiv 1\pmod{s+1}\\\ \text{for }j\in J\end{subarray}}1^{a_{1}}\cdots
n^{a_{n}}$ $\displaystyle=\sum_{\begin{subarray}{c}J\subseteq[n]\\\ |J|\equiv
r\pmod{s+1}\end{subarray}}\prod_{j\in
J}j\sum_{(b_{1}+\cdots+b_{n}+\lfloor\frac{|J|}{s+1}\rfloor)+\frac{r}{s+1}=\frac{k}{s+1}}1^{(s+1)b_{1}}\cdots
n^{(s+1)b_{n}}$ $\displaystyle=\sum_{\begin{subarray}{c}J\subseteq[n]\\\
|J|\equiv r\pmod{s+1}\end{subarray}}\prod_{j\in
J}j\sum_{b_{1}+\cdots+b_{n}=\lfloor\frac{k}{s+1}\rfloor-\lfloor\frac{|J|}{s+1}\rfloor}1^{(s+1)b_{1}}\cdots
n^{(s+1)b_{n}}$
$\displaystyle=\sum_{i=0}^{\lfloor\frac{n-r}{s+1}\rfloor}\left(\sum_{\begin{subarray}{c}|J|=i(s+1)+r\\\
J\subseteq[n]\end{subarray}}\prod_{j\in
J}j\right)h_{\lfloor\frac{k}{s+1}\rfloor-i}(1^{s+1},\dots,n^{s+1}).$

From (9) the equality follows. ∎

For example, for $n=4,k=8,$ and $s=3$ we have ${4+8\brace 4}^{(s)}=107331$. On
the other hand,

$\sum_{i=0}^{1}h_{2-i}(1^{4},2^{4},3^{4},4^{4}){5\brack 5-4i}\\\
=(1^{4}2^{4}+1^{4}3^{4}+2^{4}3^{4}+1^{4}4^{4}+2^{4}4^{4}+3^{4}4^{4}+1^{8}+2^{8}+3^{8}+4^{8})\cdot
1\\\ +(1^{4}+2^{4}+3^{4}+4^{4})\cdot 24=107331.$

From Theorem 4.2 and by the little Fermat’s theorem $a^{p}\equiv a\pmod{p}$
with $a\in\mathbb{Z}$, we conclude the following interesting congruence.

######  Corollary 4.3.

Let $n$ and $k$ be non negative integers and $p$ a prime number. If $r$ is the
remainder of $k$ when divided by $p$, then the following congruence holds

${n+k\brace
n}^{(p-1)}\equiv\sum_{i=0}^{\min\\{\lfloor\frac{n-r}{p}\rfloor,\lfloor\frac{k}{p}\rfloor\\}}{n+\lfloor\frac{k}{p}\rfloor-i\brace
n}{n+1\brack n+1-(r+i\cdot p)}\pmod{p}.$

###  4.1. The $\ell$-modular symmetric function

Given $0\leq\ell<s+1,$ we can define the _$\ell$ -modular symmetric function_
by

$M_{k}^{(s,\ell)}(x_{1},\dots,x_{n}):=\sum_{\begin{subarray}{c}j_{1}+\cdots+j_{n}=k\\\
j_{1},\cdots,j_{n}\equiv 0,\ell\pmod{s+1}\end{subarray}}x_{1}^{j_{1}}\cdots
x_{n}^{j_{n}}.$

Using the definition above, we can extend the Theorem 4.2 using the _Stirling
numbers of the first kind with higher level_ , defined in [11, 12]. Moreover,
some applications of the Stirling numbers of higher level in special
polynomials can be found in [10, 9].

Let ${\mathfrak{S}}_{n}$ denote the set of permutations of the set $[n]$. We
will assume that permutations are expressed in _standard cycle form_ , i.e.,
minimal elements first within each cycle, with cycles arranged left-to-right
in ascending order of minimal elements. If $n,k\geq 0$, then let
${\mathfrak{S}}_{(n,k)}$ denote the set of permutations of
${\mathfrak{S}}_{n}$ having exactly $k$ cycles. It is clear that
${\mathfrak{S}}_{n}=\cup_{k=0}^{n}{\mathfrak{S}}_{(n,k)}$ and
$|{\mathfrak{S}}_{(n,k)}|={n\brack k}$. Given a permutation $\sigma$ in
${\mathfrak{S}}_{n}$, let $\min(\sigma)$ denote the set of the minimal
elements in each cycle of $\sigma$. For example, if
$\sigma=(1\,4\,5)(2\,3)(6)(7\,9)(8)$, then we have that
$\min(\sigma)=\\{1,2,6,7,8\\}$.

Given a positive integer $s$, let $\left[\\!\\!\left[n\atop
k\right]\\!\\!\right]_{s}$ denote the number of ordered $s$-tuples
$(\sigma_{1},\sigma_{2},\dots,\sigma_{s})\in{\mathfrak{S}}_{(n,k)}\times{\mathfrak{S}}_{(n,k)}\times\cdots\times{\mathfrak{S}}_{(n,k)}={\mathfrak{S}}_{(n,k)}^{s}$,
such that $\min(\sigma_{1})=\min(\sigma_{2})=\cdots=\min(\sigma_{s}).$

The sequence $\left[\\!\\!\left[n\atop k\right]\\!\\!\right]_{s}$ satisfies
the following recurrence relation

(10) $\left[\\!\\!\left[n\atop
k\right]\\!\\!\right]_{s}=\left[\\!\\!\left[n-1\atop
k-1\right]\\!\\!\right]_{s}+(n-1)^{s}\left[\\!\\!\left[n-1\atop
k\right]\\!\\!\right]_{s},$

with the initial conditions $\left[\\!\\!\left[0\atop
0\right]\\!\\!\right]_{s}=1$ and $\left[\\!\\!\left[n\atop
0\right]\\!\\!\right]_{s}=\left[\\!\\!\left[0\atop
n\right]\\!\\!\right]_{s}=0$ hold for $n\geq 1$.

Given integers $n\geq 0$ and $s\geq 1$, let $\Omega_{n,s}(x)$ denote the
polynomials

$\Omega_{n,s}(x):=x(x+1^{s})(x+2^{s})\cdots(x+(n-1)^{s}),\quad\text{with}\quad\Omega_{0,s}(x)=1.$

The Stirling numbers of the first kind with higher level are the connection
constants between the polynomials $(\Omega_{n,s}(x))_{n\geq 0}$ and the
canonical basis $(x^{n})_{n\geq 0}$. Indeed, if $n\geq 0$, then

(11)
$\Omega_{n,s}(x)=x(x+1^{s})(x+2^{s})\cdots(x+(n-1)^{s})=\sum_{k=0}^{n}\left[\\!\\!\left[n\atop
k\right]\\!\\!\right]_{s}x^{k}\,.$

From a similar argument as in Theorem 4.2 and from (11) we can obtain the
following theorem.

######  Theorem 4.4.

Let $n$ and $k$ be non negative integers and $s+1>\ell\geq 0$, such that
$\gcd(\ell,s+1)=1$. If $r$ is the remainder of $k\ell^{-1}$ when divided by
$s+1$, then the following equation holds

$M^{(s,\ell)}_{k}(1,\dots,n)\\\
=\sum_{i=0}^{\min\\{\lfloor\frac{n-r}{s+1}\rfloor,\lfloor\frac{k}{s+1}\rfloor-\lfloor\frac{r\ell}{s+1}\rfloor\\}}h_{\lfloor\frac{k}{s+1}\rfloor-\lfloor\frac{r\ell}{s+1}\rfloor-i\ell}(1^{s+1},\dots,n^{s+1})\left[{n+1\brack
n+1-r-i(s+1)}\right]_{\ell}.$

##  5\. The $s$-elementary symmetric function

The _$s$ -elementary symmetric polynomial_ is defined by the expression

(12) $\displaystyle
E_{k}^{(s)}(n)=\sum_{\begin{subarray}{c}a_{1}+\cdots+a_{n}=k\\\ a_{i}\leq
s\end{subarray}}x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}.$

An equivalent definition of this symmetric polynomial already exists in a
paper by Bazeniar et al. [3]. For further properties of this symmetric
function see [1].

######  Theorem 5.1.

If $s\equiv 1\pmod{2},$ then for every positive integers $n$ and $k$ the
following identity holds

$\sum_{i=0}^{k}(-1)^{i}E_{i}^{(s)}(n)\cdot M^{(s)}_{k-i}(n)=0.$

###### Proof.

The inverse of the generating function in Theorem 2.1 is given by

$\displaystyle\prod_{i=1}^{n}\frac{1-(x_{i}t)^{s+1}}{1+x_{i}t}$
$\displaystyle=\prod_{i=1}^{n}\frac{1-(-x_{i}t)^{s+1}}{1-(-x_{i}t)}=\prod_{i=1}^{n}\left(1-x_{i}t+(-x_{i}t)^{2}+\cdots+(-x_{i}t)^{s}\right).$

In each product, we can create any number in between $1$ and $s$. Hence

$\sum_{k=0}^{n\cdot
s}E_{k}^{(s)}(n)(-t)^{k}=\prod_{i=1}^{n}\frac{1-(x_{i}t)^{s+1}}{1+x_{i}t},$

and the desired identity follows. ∎

We can express the modular symmetric function $M_{k}^{(s)}$ as convolutions
involving the complete and elementary symmetric functions as follows.

######  Theorem 5.2.

Let $k$, $n$ and $s$ be positive integers and let $x_{1},x_{2},\ldots,x_{n}$
be independent variables. Then

$M_{k}^{(s)}(x_{1},x_{2},\ldots,x_{n})=\sum_{j=0}^{\lfloor
k/s+1\rfloor}h_{j}(x_{1}^{s+1},x_{2}^{s+1},\ldots,x_{n}^{s+1})e_{k-(s+1)j}(x_{1},x_{2},\ldots,x_{n}).$

###### Proof.

According to (2), we have

$\displaystyle\sum_{k=0}^{\infty}M_{k}^{(s)}(x_{1},x_{2},\ldots,x_{n})t^{k}$
$\displaystyle\qquad=\left(\prod_{i=1}^{n}\frac{1}{1-(x_{i}t)^{s+1}}\right)\left(\prod_{i=1}^{n}(1+x_{i}t)\right)$
$\displaystyle\qquad=\left(\sum_{j=0}^{\infty}h_{j}(x_{1}^{s+1},x_{2}^{s+1},\ldots,x_{n}^{s+1})(t)^{(s+1)j}\right)\left(\sum_{j=0}^{\infty}e_{j}(x_{1},x_{2},\ldots,x_{n})t^{j}\right)$
$\displaystyle\qquad=\sum_{k=0}^{\infty}\left(\sum_{j=0}^{\lfloor
k/s+1\rfloor}h_{j}(x_{1}^{s+1},x_{2}^{s+1},\ldots,x_{n}^{s+1})e_{k-(s+1)j}(x_{1},x_{2},\ldots,x_{n})\right)t^{k}.$

As required. ∎

Inspired by Theorem 5.2, we provide the following generalization.

######  Theorem 5.3.

Let $k$, $n$ and $s$ be three positive integers and let
$x_{1},x_{2},\ldots,x_{n}$ be independent variables. Then

$h_{k}(x_{1}^{s},x_{2}^{s},\ldots,x_{n}^{s})=\sum_{j=0}^{k(s+1)}(-1)^{j}h_{j}(x_{1},x_{2},\ldots,x_{n})M_{k(s+1)-j}^{(s)}(x_{1},x_{2},\ldots,x_{n})$

and

$e_{k}(x_{1},x_{2},\ldots,x_{n})=\sum_{j=0}^{\lfloor
k/s+1\rfloor}(-1)^{j}e_{j}(x_{1},x_{2},\ldots,x_{n})M_{k-j(s+1)}^{(s)}(x_{1},x_{2},\ldots,x_{n}).$

If $k$ is not congruent to $0$ modulo $s+1$, then

$\sum_{j=0}^{k}(-1)^{j}h_{j}(x_{1},x_{2},\ldots,x_{n})M_{k-j}^{(s)}(x_{1},x_{2},\ldots,x_{n})=0.$

###### Proof.

The relation given in Theorem 2.1 can be rewritten as

$\prod_{i=1}^{n}\frac{1}{1+x_{i}t}\sum_{k=0}^{\infty}M_{k}^{(s)}(x_{1},x_{2},\ldots,x_{n})t^{k}=\prod_{i=1}^{n}\frac{1}{1-(x_{i}t)^{s+1}}$

or

$\prod_{i=1}^{n}(1-(x_{i}t)^{s+1})\sum_{k=0}^{\infty}M_{k}^{(s)}(x_{1},x_{2},\ldots,x_{n})t^{k}=\prod_{i=1}^{n}\big{(}1+(x_{i}t)\big{)}.$

Thus we deduce that

$\displaystyle\sum_{k=0}^{\infty}h_{k}(x_{1}^{s+1},x_{2}^{s+1},\ldots,x_{n}^{s+1})t^{k(s+1)}$
$\displaystyle\qquad=\left(\sum_{k=0}^{\infty}(-1)^{k}h_{k}(x_{1},x_{2},\ldots,x_{n})t^{k}\right)\left(\sum_{k=0}^{\infty}M_{k}^{(s)}(x_{1},x_{2},\ldots,x_{n})t^{k}\right)$

and

$\displaystyle\sum_{k=0}^{\infty}e_{k}(x_{1},x_{2},\ldots,x_{n})t^{k}$
$\displaystyle\qquad=\left(\sum_{k=0}^{\infty}(-1)^{k}e_{k}(x_{1}^{s+1},x_{2}^{s+1},\ldots,x_{n}^{s+1})t^{k(s+1)}\right)\left(\sum_{k=0}^{\infty}M_{k}^{(s)}(x_{1},x_{2},\ldots,x_{n})t^{k}\right).$

The proof follows easily by comparing the coefficients of $t^{ks}$ on both
sides of these equations. ∎

The following result allows us to express a convolution of the modular
symmetric function $M_{k}^{(s)}$ as convolutions involving the complete and
elementary symmetric functions.

######  Theorem 5.4.

Let $k$, $n$ and $s$ be three positive integers and let
$x_{1},x_{2},\ldots,x_{n}$ be independent variables. Then

$\sum_{j=0}^{k}e_{j}(x_{1},x_{2},\ldots,x_{n})h_{k-j}(x_{1},x_{2},\ldots,x_{n})=\sum_{j=0}^{k}M^{(s)}_{j}(x_{1},x_{2},\ldots,x_{n})E^{(s)}_{k-j}(x_{1},x_{2},\ldots,x_{n}).$

We can, now define the modular $s$-Stirling numbers of the first kind by the
following equality

${n+1\brack
k+1}^{(s)}=(n!)^{s}E_{k}^{(s)}\left(1,\frac{1}{2},\dots,\frac{1}{n}\right).$

These numbers were introduced independently by Ahmia et al. [2]. They
interpreted ${n\brack k}^{(s)}$ as the number of $s$-tuple permutations of
$[n]$ having together $k$ cycles. Inspired by the combinatorial interpretation
given in Theorem 4.1 for the modular $s$-Stirling numbers of the second kind,
we give in the following theorem another combinatorial interpretation of
${n\brack k}^{(s)}$.

######  Theorem 5.5.

Let $n$ and $k$ be non negative integers and $s>0$. The $s$-modular Stirling
numbers of the first kind ${n+1\brack k+1}^{(s)}$ count the number of set
partitions $\pi\in\Pi(n(s+1)-k,n)$ such that $d(\pi)=(d_{1},\dots,d_{n})$ has
the property that $d_{i}\leq s$ for every $1\leq i\leq n$.

###### Proof.

From (12) we have the equality

$(n!)^{s}E_{k}^{(s)}\left(1,\frac{1}{2},\dots,\frac{1}{n}\right)=E_{n\cdot
s-k}^{(s)}(1,\dots,n).$

Using the same argument as in Theorem 4.1, we obtain the desired result. ∎

From Theorem 5.5 (taking $s=1$) we obtain a probably new combinatorial
interpretation for the Stirling numbers of the first kind in terms of set
partitions. Indeed, ${n+1\brack k+1}$ enumerates the set partitions in
$\Pi(2n-k,n)$, such that the vector $d(\pi)=(d_{1},\dots,d_{n})$ has the
property that $d_{i}\leq 1$ for every $1\leq i\leq n$. For example,
${3+1\brack 1+1}=11$, the partitions being

$\displaystyle 1/23/45,\quad 1/235/4,\quad 12/3/45,\quad 13/2/45,\quad
12/34/5,\quad 12/35/4,$ $\displaystyle 135/2/4,\quad 15/23/4,\quad
124/3/5,\quad 125/3/4,\quad 13/25/4.$

The modular $s$-Stirling numbers of the first kind satisfy the following
recurrence relation

(13) $\displaystyle{n\brack k}^{(s)}=\sum_{\ell=0}^{s}{n-1\brack
k-(s-\ell)}^{(s)}\cdot(n-1)^{\ell},$

where ${0\brack 1-s}^{(s)}=1,{n\brack k}^{(s)}=0$ if $k<1-s$. See also, Ahmia
et al. [2]. From this relation we can give the following combinatorial
interpretation.

######  Theorem 5.6.

Let $n$ and $k$ be non negative integers and $s>0$. Consider the set of
$s$-tuples of permutations $(\sigma_{1},\sigma_{2},\dots,\sigma_{s})$ such
that $\min(\sigma_{i})\subseteq\min(\sigma_{i-1})$ for all $i>1$ and
$\sum_{j=1}^{s}|\min(\sigma_{j})|=k+s-1.$ Then the number of such elements
equals ${n\brack k}^{(s)}$.

###### Proof.

By the recursion given in (13), one has that

$\displaystyle{n\brack k}^{(s)}$ $\displaystyle=\sum_{\ell=0}^{s}{n-1\brack
k-(s-\ell)}^{(s)}(n-1)^{\ell}=\sum_{\ell=0}^{s}{n-1\brack
k-\ell}^{(s)}(n-1)^{s-\ell}.$

This recurrence corresponds to choosing if the last element of each
permutation, i.e., $n$ is going to be fixed or not. Call $\ell$ the number of
permutations where $n$ is going to be a fixed point. By the condition we
imposed in the tuple, these have to be the first $\ell$ elements of the
$s$-tuple. For the remaining $s-\ell$ elements of the tuple, we have to choose
an element from the remaining $n-1$ to have $n$ as a preimage in each one of
the permutations. We can do this in $(n-1)^{s-\ell}$ ways. This shows the
claim because the initial condition is in $1-s,$ meaning we need
$k-(1-s)=k+s-1$ cycles to fill. ∎

For example, take $n=3$ and $k=4,$ the following correspond to the $15$ tuples
counted by ${3\brack 4}^{(3)}$ having in total $3+4-1=6$ cycles.

$((1)(2)(3),(1)(2,3),(1,2,3)),\quad((1)(2)(3),(1)(2,3),(1,3,2)),$
$((1)(2)(3),(1,2)(3),(1,2,3)),\quad((1)(2)(3),(1,2)(3),(1,3,2)),$
$((1)(2)(3),(1,3)(2),(1,2,3)),\quad((1)(2)(3),(1,3)(2),(1,3,2)),$
$((1)(2,3),(1)(2,3),(1)(2,3)),\quad((1)(2,3),(1)(2,3),(1,3)(2)),$
$((1)(2,3),(1,3)(2),(1)(2,3)),\quad((1)(2,3),(1,3)(2),(1,3)(2)),$
$((1,2)(3),(1,2)(3),(1,2)(3)),\quad((1,3)(2),(1)(2,3),(1)(2,3)),$
$((1,3)(2),(1)(2,3),(1,3)(2)),\quad((1,3)(2),(1,3)(2),(1)(2,3)),$
$((1,3)(2),(1,3)(2),(1,3)(2)).$

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