A New Sequence and Its Some Congruence Properties

Abstract

The aim of this paper is to study the congruence properties of a new sequence, which is closely related to Fubini polynomials and Euler numbers, using the elementary method and the properties of the second kind Stirling numbers. As results, we obtain some interesting congruences for it. This solves a problem proposed in a published paper.

1. Introduction

Let $n\ge 0$ be an integer, the famous Fubini polynomials $F_n\left(y\right)$ are defined according to the coefficients of following generating function:

$$\begin{array}{c}\hfill \frac{1}{1-y\left(e^t-1\right)}=\sum _{n=0}^{\infty }\frac{F_n\left(y\right)}{n!}·t^n,\end{array}$$

(1)

where $F_0\left(y\right)=1$, $F_1\left(y\right)=y$, and so on.

These polynomials are closely related to the Stirling numbers and Euler numbers. For example, if $y=-\frac{1}{2}$, then (1) becomes

$$\begin{array}{c}\hfill \frac{2}{1+e^t}=\sum _{n=0}^{\infty }\frac{E_n}{n!}·t^n,\end{array}$$

(2)

where $E_n$ denotes the Euler numbers.

At the same time, the Fubini polynomials with two variables can also be defined by the following identity (see [1,2]):

$$\frac{e^{xt}}{1-y\left(e^t-1\right)}=\sum _{n=0}^{\infty }\frac{F_n(x,y)}{n!}·t^n,$$

and $F_n\left(y\right)=F_n(0,y)$ for all integers $n\ge 0$. Many scholars have studied the properties of $F_n(x,y)$, and have obtained many important works. For example, T. Kim et al. proved a series of identities related to $F_n(x,y)$ (see [2,3]), one of which is

$$F_n(x,y)=\sum _{l=0}^n\left(\genfrac{}{}{0pt}{}{n}{l}\right)x^l·F_{n-l}\left(y\right),n\ge 0.$$

Zhao Jianhong and Chen Zhuoyu [4] studied the computational problem of the sums

$$\begin{array}{c}\hfill \sum _{a_1+a_2+\cdots +a_k=n}\frac{F_{a_1}\left(y\right)}{\left(a_1\right)!}·\frac{F_{a_2}\left(y\right)}{\left(a_2\right)!}\cdots \frac{F_{a_k}\left(y\right)}{\left(a_k\right)!},\end{array}$$

where the summation in the formula above denotes all k-dimension non-negative integer coordinates $(a_1,a_2,\cdots ,a_k)$ such that $a_1+a_2+\cdots +a_k=n$. They proved the identity

$$\begin{array}{cc}\hfill & {\sum }_{a_1+a_2+\cdots +a_k=n}\frac{F_{a_1}\left(y\right)}{\left(a_1\right)!}·\frac{F_{a_2}\left(y\right)}{\left(a_2\right)!}\cdots \frac{F_{a_k}\left(y\right)}{\left(a_k\right)!}\hfill \\ \hfill =& \frac{1}{(k-1)!{(y+1)}^{k-1}}·\frac{1}{n!}{\sum }_{i=0}^{k-1}C(k-1,i)F_{n+k-1-i}\left(y\right),\hfill \end{array}$$

(3)

where the sequence $C(k,i)$ is defined for positive integer k and i with $0\le i\le k$, $C(k,0)=1$, $C(k,k)=k!$ and

$$C(k+1,i+1)=C(k,i+1)+(k+1)C(k,i), \mathrm{for} \mathrm{all} 0\le i<k,$$

providing $C(k,i)=0$, if $i>k$.

For clarity, for $1\le k\le 9$, we list values of $C(k,i)$ in the following Table 1.

Table 1. Values of $C(k,i)$.

$\mathit{C}(\mathit{k},\mathit{i})$	$\mathit{i}=0$	$\mathit{i}=1$	$\mathit{i}=2$	$\mathit{i}=3$	$\mathit{i}=4$	$\mathit{i}=5$	$\mathit{i}=6$	$\mathit{i}=7$	$\mathit{i}=8$	$\mathit{i}=9$
$\mathit{k}=\mathbf{1}$	1	1
$\mathit{k}=\mathbf{2}$	1	3	2
$\mathit{k}=\mathbf{3}$	1	6	11	6
$\mathit{k}=\mathbf{4}$	1	10	35	50	24
$\mathit{k}=\mathbf{5}$	1	15	85	225	274	120
$\mathit{k}=\mathbf{6}$	1	21	175	735	1624	1764	720
$\mathit{k}=\mathbf{7}$	1	28	322	1960	6769	13,132	13,068	5040
$\mathit{k}=\mathbf{8}$	1	36	546	4536	22,449	67,284	118,124	109,584	40,320
$\mathit{k}=\mathbf{9}$	1	45	870	9450	63,273	269,325	723,680	1,172,700	1,026,576	362,880

Meanwhile, Zhao Jianhong and Chen Zhuoyu [4] proposed some conjectures related to the sequence. We believe that this sequence is meaningful because it satisfies some very interesting congruence properties, such as

$$\begin{array}{c}\hfill C(p-2,i)\equiv 1\left(modp\right)\end{array}$$

(4)

for all odd primes p and integers $0\le i\le p-2$. The equivalent conclusion is

$$\begin{array}{c}\hfill C(p-1,i)\equiv 0\left(modp\right)\end{array}$$

(5)

for all odd primes p and positive integers $1\le i\le p-2$. Since some related content can be found in references [5,6,7,8,9,10,11,12,13,14,15], we will not go through all of them here.

The aim of this paper is to prove congruence (5) by applying the elementary method and the properties of the second kind Stirling numbers. That is, we will solve the conjectures in [4], which are listed in the following.

Theorem 1.

Let p be an odd prime. For any integer $1\le i\le p-2$, we have congruence

$$\begin{array}{c}\hfill C(p-1,i)\equiv 0\left(modp\right).\end{array}$$

From this theorem and (3), we can deduce following three corollaries:

Corollary 1.

For any positive integer n and odd prime p, we have

$$\begin{array}{c}\hfill F_{n+p-1}\left(y\right)-F_n\left(y\right)\equiv 0\left(modp\right).\end{array}$$

Corollary 2.

For any positive integer n and odd prime p, we have

$$E_{n+p-1}-E_n\equiv 0\left(modp\right).$$

Corollary 3.

For any odd prime p, we have the congruences

$$2E_p\equiv -1\left(modp\right), 4E_{p+2}\equiv 1\left(modp\right), \mathit{and} 2E_{p+4}\equiv -1\left(modp\right).$$

Note. Since $E_n$ is a rational number, we can denote ${\displaystyle E_n=\frac{U_n}{V_n}}$, where $U_n$ and $V_n$ are integers with $(U_n,V_n)=1$. Based on this, in our paper, the expression $E_n\equiv 0\left(modp\right)$ means $p\mid U_n$, while $p\nmid V_n$.

2. Several Lemmas

Lemma 1.

For any positive integer k, we have the identity

$$\begin{array}{c}\hfill k!y{(y+1)}^{k-1}=\sum _{i=0}^{k-1}C(k-1,i)F_{k-i}\left(y\right).\end{array}$$

Proof. 

Taking $n=1$ in (3), and noting that $F_0\left(y\right)=1$, $F_1\left(y\right)=y$, and the equation $a_1+a_2+\cdots +a_k=1$ holds if and only if one of $a_i$ is 1, others are 0. The number of the solutions of this equation is $\left(\genfrac{}{}{0pt}{}{k}{1}\right)=k$. So, from (3), we have

$$\begin{array}{cc}\hfill & \sum _{a_1+a_2+\cdots +a_k=1}\frac{F_{a_1}\left(y\right)}{\left(a_1\right)!}·\frac{F_{a_2}\left(y\right)}{\left(a_2\right)!}\cdots \frac{F_{a_k}\left(y\right)}{\left(a_k\right)!}=\left(\genfrac{}{}{0pt}{}{k}{1}\right)y=ky\hfill \\ \hfill =& \frac{1}{(k-1)!{(y+1)}^{k-1}}·\sum _{i=0}^{k-1}C(k-1,i)F_{k-i}\left(y\right)\hfill \end{array}$$

or identity

$$\begin{array}{c}\hfill k!y{(y+1)}^{k-1}=\sum _{i=0}^{k-1}C(k-1,i)F_{k-i}\left(y\right),\end{array}$$

which proves Lemma 1. ☐

Lemma 2.

For any positive integer n, we have the identity

$$F_n\left(y\right)=\sum _{k=0}^{n}S(n,k)k!y^k,(n\ge 0),$$

where $S(n,k)$ are the second kind Stirling numbers, which are defined for any integer k, n with $0\le k\le n$ as:

$$S(n,k)=kS(n-1,k)+S(n-1,k-1)$$

where $S(0,0)=1$, $S(n,0)=0$ and $S(0,k)=0$ for $n,k>0$.

Proof. 

See Reference [2]. ☐

Lemma 3.

For any positive integers n and k, we have

$$S(n,k)=\frac{1}{k!}\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)j^n{(-1)}^{k-j}.$$

Proof. 

See Theorem 4.3.12 of [16]. ☐

Lemma 4.

For any odd prime p and positive integer $2\le k\le p-1$, we have the congruence

$$k!S(p,k)\equiv 0\left(modp\right).$$

Proof. 

From the definition and properties of $S(n,k)$, we have $S(n,k)=0$, if $k>n$. For any integers $0\le j\le p-1$, from the famous Fermat’s little theorem, we have the congruence $j^p\equiv j\left(modp\right)$. From this congruence and Lemma 3, we have

$$\begin{array}{c}\hfill k!S(p,k)=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)j^p{(-1)}^{k-j}\equiv \sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)j{(-1)}^{k-j}\equiv k!S(1,k)\equiv 0\left(modp\right),\end{array}$$

if $k\ge 2$. This completes the proof of Lemma 4. ☐

3. Proof of the Theorem

In this section, we will prove Theorem by mathematical induction. Taking $k=p$ in Lemma 1 and noting that $C(p-1,0)=1$ and $C(p-1,p-1)=(p-1)!$, we have:

$$\begin{array}{cc}\hfill & p!y{(y+1)}^{p-1}=\sum _{i=0}^{p-1}C(p-1,i)F_{p-i}\left(y\right)\hfill \\ \hfill =& F_p\left(y\right)+y(p-1)!+\sum _{i=1}^{p-2}C(p-1,i)F_{p-i}\left(y\right).\hfill \end{array}$$

Note that $(p-1)!+1\equiv 0\left(modp\right)$, which implies

$$\begin{array}{c}\hfill F_p\left(y\right)-y+\sum _{i=1}^{p-2}C(p-1,i)F_{p-i}\left(y\right)\equiv 0\left(modp\right).\end{array}$$

(6)

From (6), we have the congruence

$$\begin{array}{c}\hfill y-F_p\left(y\right)\equiv \sum _{i=1}^{p-2}C(p-1,i)F_{p-i}\left(y\right)\left(modp\right).\end{array}$$

(7)

From Lemma 2, we have

$$\begin{array}{c}\hfill F_p\left(y\right)=\sum _{k=0}^{p}S(p,k)k!y^k\end{array}$$

(8)

and

$$\begin{array}{c}\hfill F_p^{(p-1)}\left(0\right)=S(p,p-1)(p-1)!·(p-1)!,\end{array}$$

(9)

where $F_n^{\left(k\right)}\left(y\right)$ denotes the k-order derivative of $F_n\left(y\right)$ for variable y.

$$\begin{array}{c}\hfill F_{p-1}^{(p-1)}\left(0\right)=S(p-1,p-1)(p-1)!·(p-1)!=(p-1)!·(p-1)!.\end{array}$$

(10)

Then, applying Lemma 3 and Lemma 4 and noting that $S(1,p-1)=0$, we have

$$\begin{array}{ccc}\hfill (p-1)!S(p,p-1)& \equiv & {\sum }_{j=0}^{p-1}\left(\genfrac{}{}{0pt}{}{p-1}{j}\right)j^p{(-1)}^{p-1-j}\equiv {\sum }_{j=0}^{p-1}\left(\genfrac{}{}{0pt}{}{p-1}{j}\right)j{(-1)}^{p-1-j}\hfill \\ \hfill & \equiv & (p-1)!S(1,p-1)\equiv 0\left(modp\right).\hfill \end{array}$$

(11)

Combining (7), (9), (10), and (11), we have:

$$\begin{array}{c}\hfill 0\equiv -S(p,p-1)(p-1)!(p-1)!\equiv C(p-1,1)(p-1)!·(p-1)!\left(modp\right)\end{array}$$

(12)

or

$$\begin{array}{c}\hfill C(p-1,1)\equiv 0\left(modp\right).\end{array}$$

(13)

That is, the theorem is true for $i=1$.

Assume that the theorem is true for all $1\le i\le s$. That is,

$$\begin{array}{c}\hfill C(p-1,i)\equiv 0\left(modp\right)\end{array}$$

for $1\le i\le s<p-1$. It is clear that if $s=p-2$, then the theorem is true.

If $1<s<p-2$, then from (7) we have the congruence

$$\begin{array}{c}\hfill y-F_p\left(y\right)\equiv \sum _{i=s+1}^{p-2}C(p-1,i)F_{p-i}\left(y\right)\left(modp\right).\end{array}$$

(14)

In congruence (14), taking the $(p-s-1)$-order derivative with respect to t, then let $y=0$, applying Lemma 2, we have:

$$\begin{array}{cc}\hfill & -S(p,p-s-1)(p-s-1)!·(p-s-1)!\hfill \\ \hfill \equiv & C(p-1,s+1)(p-s-1)!(p-s-1)!\left(modp\right).\hfill \end{array}$$

(15)

Note that $\left((p-s-1)!,p\right)=1$, from Lemma 4 and (15) we have the congruence

$$\begin{array}{c}\hfill C(p-1,s+1)(p-s-1)!\equiv -(p-s-1)!S(p,p-s-1)\equiv 0\left(modp\right),\end{array}$$

which implies

$$\begin{array}{c}\hfill C(p-1,s+1)\equiv 0\left(modp\right).\end{array}$$

That is, the theorem is true for $i=s+1$. Now the proof of the theorem completes by mathematical induction.

Now, we prove Corollary 1. For any integer $n\ge 0$, taking $k=p$ in (3) and noting that

$$\begin{array}{c}\hfill n!\sum _{a_1+a_2+\cdots +a_p=n}\frac{F_{a_1}\left(y\right)}{\left(a_1\right)!}·\frac{F_{a_2}\left(y\right)}{\left(a_2\right)!}\cdots \frac{F_{a_p}\left(y\right)}{\left(a_p\right)!}\equiv 0\left(modp\right),\end{array}$$

we have

$$\begin{array}{c}\hfill \sum _{i=0}^{p-1}C(p-1,i)F_{n+p-1-i}\left(y\right)\equiv 0\left(modp\right).\end{array}$$

(16)

From our theorem, we have

$$\begin{array}{c}\hfill \sum _{i=1}^{p-2}C(p-1,i)F_{n+p-1-i}\left(y\right)\equiv 0\left(modp\right).\end{array}$$

(17)

Note that $C(p-1,0)=1$, $C(p-1,p-1)=(p-1)!$. Combining (16) and (17), we can deduce the congruence

$$\begin{array}{c}\hfill F_{n+p-1}\left(y\right)-F_n\left(y\right)\equiv 0\left(modp\right).\end{array}$$

Now the proof of Corollary 1 completes. Since Corollarys 2 and 3 are the special situation of Corollary 1, we will not prove Corollarys 2 and 3 here.