Some Identities Involving the Fubini Polynomials and Euler Polynomials

Abstract

In this paper, we first introduce a new second-order non-linear recursive polynomials $U_{h,i}\left(x\right)$, and then use these recursive polynomials, the properties of the power series and the combinatorial methods to prove some identities involving the Fubini polynomials, Euler polynomials and Euler numbers.

1. Introduction

For any real number x and y, the two variable Fubini polynomials $F_n(x,y)$ are defined by means of the following (see [1,2])

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

(1)

The first several terms of $F_n(x,y)$ are $F_0(x,y)=1$, $F_1(x,y)=x+y$, $F_2(x,y)=x^2+2xy+2y^2+y$, ⋯. Taking $x=0$, then $F_n(0,y)=F_n\left(y\right)$ (see [1]) are called the Fubini polynomials. If $y=-\frac{1}{2}$, then $F_n\left(x,-\frac{1}{2}\right)=E_n\left(x\right)$, the Euler polynomials, $E_0\left(x\right)=1$, $E_1\left(x\right)=x-\frac{1}{2}$, $E_2\left(x\right)=x^2-x$, and

$$E_n\left(x\right)=\sum _{i=0}^n{(-1)}^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·x^i·E_{n-i}\left(x\right),n=0,1,2,\cdots .$$

If $x=0$, then $E_n\left(0\right)=E_n$ are the famous Euler numbers. $E_0=1$, $E_1=-\frac{1}{2}$, $E_2=0$, $E_3=\frac{1}{4}$, $E_4=0$, $E_5=-\frac{1}{2}$, $E_6=0$, and $E_{2n}=0$ for all positive integer n.

These polynomials appear in combinatorial mathematics and play a very important role in the theory and application of mathematics, thus many number theory and combination experts have studied their properties, and obtained a series of interesting results. For example, Kim and others proved a series of identities related to $F_n(x,y)$ (see [2,3,4]), one of which is

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

T. Kim et al. [5] also studied the properties of the Fubini polynomials $F_n\left(y\right)$, and proved the identity

$$F_n\left(y\right)=\sum _{k=0}^n{S}_2(n,k)k!y^k,(n\ge 0),$$

where $S_2(n,k)$ are the Stirling numbers of the second kind.

Zhao and Chen [6] proved that, for any positive integers n and k, one has the identity

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

(2)

where the summation is taken over all k-dimensional nonnegative integer coordinates $(a_1,a_2,\cdots ,a_h)$ such that $a_1+a_2+\cdots +a_h=n$. The sequence $\left\{C\right(k,i\left)\right\}$ is defined as follows: For any positive integer k and integers $0\le i\le k$, $C(k,0)=1$, $C(k,k)=k!$ and $C(k+1,i+1)=C(k,i+1)+(k+1)C(k,i)$, for all $0\le i<k$.

Some other papers related to Fubini polynomials and Euler numbers can be found elsewhere [7,8,9,10,11,12,13,14,15,16,17,18,19], and we do not repeat them here.

In this paper, as a note of [6], we study a similar calculating problem of Equation (2) for two variable Fubini polynomials $F_n(x,y)$. We also introduce a new second order non-linear recursive polynomials, and then use this polynomials to give a new expression for the summation

$$\begin{array}{c}\hfill W(h,n,x)=\sum _{a_1+a_2+\cdots +a_{h+1}=n}\frac{F_{a_1}(x,y)}{a_1!}·\frac{F_{a_2}(x,y)}{a_2!}\cdots \frac{F_{a_{h+1}}(x,y)}{a_{h+1}!}.\end{array}$$

That is, we prove the following:

Theorem 1.

Let h be a positive integer. Then, for any integer $n\ge 0$, we have the identity

$$\begin{array}{c}\hfill W(h,n,x)=\frac{1}{{(y+1)}^h·h!·n!}·\sum _{k=0}^h\sum _{i=0}^n{U}_{h,k}\left(x\right)·x^i·h^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·F_{n-i+k}(x,y),\end{array}$$

where $U_{h,k}\left(x\right)$ is a second order non-linear recurrence polynomial defined by $U_{h,h}\left(x\right)=1$, and $U_{h+1,0}\left(x\right)=(h+1-x)U_{h,0}\left(x\right)$, and $U_{h+1,k+1}\left(x\right)=(h+1-x)U_{h,k+1}\left(x\right)+U_{h,k}\left(x\right)$ for all integers $h\ge 1$ and k with $0\le k\le h-1$.

It is clear that our theorem is a generalization of Equation (2). If taking $y=-\frac{1}{2}$, $n=0$, $x=0$ and $x=1$ in this theorem, respectively, and noting that $U_{h,0}\left(1\right)=0$, $E_0\left(1\right)=1$ and $E_n\left(1\right)=-E_n$ for all $n\ge 1$, we can deduce the following five corollaries:

Corollary 1.

For any positive integer $h\ge 1$, we have the identity

$$\begin{array}{c}\hfill \sum _{k=0}^h{U}_{h,k}\left(x\right)·E_k\left(x\right)=\frac{h!}{2^h}.\end{array}$$

Corollary 2.

For any positive integer $h\ge 1$ and real x, we have the identity

$$\begin{array}{c}\hfill \sum _{a_1+a_2+\cdots +a_{h+1}=n}\frac{E_{a_1}\left(x\right)}{a_1!}·\frac{E_{a_2}\left(x\right)}{a_2!}\cdots \frac{E_{a_{h+1}}\left(x\right)}{a_{h+1}!}=\frac{2^h}{h!·n!}·\sum _{k=0}^h\sum _{i=0}^n{U}_{h,k}\left(x\right)·x^i·h^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·E_{n-i+k}\left(x\right).\end{array}$$

Corollary 3.

For any positive integer $h\ge 1$, we have the identity

$$\begin{array}{c}\hfill \sum _{a_1+a_2+\cdots +a_{h+1}=n}\frac{E_{a_1}}{a_1!}·\frac{E_{a_2}}{a_2!}\cdots \frac{E_{a_{h+1}}}{a_{h+1}!}=\frac{2^h}{h!·n!}·\sum _{k=0}^h{U}_{h,k}\left(0\right)·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·E_{n+k}.\end{array}$$

Corollary 4.

For any positive integer $h\ge 1$, we have the identity

$$\begin{array}{c}\hfill \frac{h!}{2^h}+\sum _{k=1}^h{U}_{h,k}\left(1\right)·E_k=0.\end{array}$$

From Equation (2) with $y=-\frac{1}{2}$ and Corollary 3 we can deduce the identities $U_{h,i}\left(0\right)=C(h,h-i)$ for all nonnegative integers $0\le i\le h$.

On the other hand, from the definition of $U_{h,k}\left(1\right)$, we can easily prove that the sequence $U_{h,k}\left(1\right)$ are the coefficients of the polynomial ${\displaystyle F\left(x\right)=\prod _{i=1}^{h-1}(x+i)}$. That is,

$$F\left(x\right)=(x+1)(x+2)\cdots (x+h-1)=\sum _{i=0}^{h-1}{U}_{h,i+1}\left(1\right)·x^i.$$

Thus, if $h=p$ is an odd prime, then using the elementary number theory methods we deduce the following:

Corollary 5.

Let p be an odd prime. Then, for any positive integer $2\le k\le p-1$, we have the congruence

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

Taking $h=p$, noting that $U_{p,p}\left(1\right)=1$, $E_1=-\frac{1}{2}$ and $U_{p,1}\left(1\right)=(p-1)!\equiv -1modp$, and then combining Corollaries 4 and 5, we have the following:

Corollary 6.

Let p be an odd prime. Then, we have the congruence

$$\begin{array}{c}\hfill 2E_p+1\equiv 0modp.\end{array}$$

This congruence is also recently obtained by Hou and Shen [12] using the different methods.

2. Several Simple Lemmas

In this section, we give several necessary lemmas in the proof process of our theorem. First, we have the following:

Lemma 1.

Let function $f\left(t\right)=\frac{e^{xt}}{1-y\left(e^t-1\right)}$. Then, for any positive integer h, real numbers x and t, we have the identity

$${(y+1)}^h·h!·f^{h+1}\left(t\right)=e^{hxt}·\sum _{i=0}^h{U}_{h,i}\left(x\right)·f^{\left(i\right)}\left(t\right),$$

where $U_{h,i}\left(x\right)$ is defined as in the theorem, and $f^{\left(h\right)}\left(t\right)$ denotes the h-order derivative of $f\left(t\right)$ with respect to variable t.

Proof. 

We can prove this Lemma 1 by mathematical induction. First, from the properties of the derivative, we have

$$\begin{array}{c}\hfill f^{\prime }\left(t\right)=\frac{x{e}^{xt}}{1-y\left(e^t-1\right)}+\frac{y·e^t·e^{xt}}{{\left(1-y\left(e^t-1\right)\right)}^2}=xf\left(t\right)-f\left(t\right)+\frac{(y+1)·e^{xt}}{{\left(1-y\left(e^t-1\right)\right)}^2}\end{array}$$

or

$$\begin{array}{c}\hfill (y+1)f^2\left(t\right)=e^{xt}\left[f^{\prime }\left(t\right)+(1-x)f\left(t\right)\right]=e^{xt}·\sum _{i=0}^1{U}_{1,i}\left(x\right)·f^{\left(i\right)}\left(t\right).\end{array}$$

That is, Lemma 1 is correct for $h=1$. □

Assuming that Lemma 1 is correct for $1\le h=k$, i.e.,

$$\begin{array}{c}\hfill {(y+1)}^k·k!·f^{k+1}\left(t\right)=e^{kxt}·\sum _{i=0}^k{U}_{k,i}\left(x\right)·f^{\left(i\right)}\left(t\right).\end{array}$$

(3)

Then, from Equation (3) and the definitions of $U_{k,i}\left(x\right)$ and derivative, we have

$$\begin{array}{cc}\hfill & e^{xt}·{(y+1)}^k·(k+1)!·f^k\left(t\right)·f^{\prime }\left(t\right)\hfill \\ =& {(y+1)}^k(k+1)!·f^k\left(t\right)\left((y+1)f^2\left(t\right)+(x-1)·e^{xt}·f\left(t\right)\right)\hfill \\ =& {(y+1)}^{k+1}(k+1)!·f^{k+2}\left(x\right)+(k+1)!·(x-1)·e^{xt}·{(y+1)}^k·f^{k+1}\left(t\right)\hfill \\ =& e^{(k+1)xt}·\left(xk·\sum _{i=0}^k{U}_{k,i}\left(x\right)·f^{\left(i\right)}\left(t\right)+\sum _{i=0}^k{U}_{k,i}\left(x\right)·f^{(i+1)}\left(t\right)\right)\hfill \end{array}$$

or

$$\begin{array}{cc}\hfill & {(y+1)}^{k+1}·(k+1)!·f^{k+2}\left(t\right)=e^{(k+1)xt}·\sum _{i=0}^{k}xk·U_{k,i}\left(x\right)·f^{\left(i\right)}\left(t\right)\hfill \\ \hfill & +e^{(k+1)xt}·\left(\sum _{i=0}^k{U}_{k,i}\left(x\right)·f^{(i+1)}\left(t\right)+\sum _{i=0}^k(k+1)(1-x)U_{k,i}\left(x\right)·f^{\left(i\right)}\left(t\right)\right)\hfill \\ =& e^{(k+1)xt}·\left(\sum _{i=0}^k{U}_{k,i}\left(x\right)·f^{(i+1)}\left(t\right)+\sum _{i=0}^k(k+1-x)U_{k,i}\left(x\right)·f^{\left(i\right)}\left(t\right)\right)\hfill \\ =& e^{(k+1)xt}·\left(\sum _{i=0}^{k-1}{U}_{k,i}\left(x\right)·f^{(i+1)}\left(t\right)+\sum _{i=1}^k(k+1-x)U_{k,i}\left(x\right)·f^{\left(i\right)}\left(t\right)\right)\hfill \\ \hfill & +e^{(k+1)xt}·\left(U_{k,k}\left(x\right)·f^{(k+1)}\left(t\right)+(k+1-x)·U_{k,0}\left(x\right)·f\left(t\right)\right)\hfill \\ =& e^{(k+1)xt}·\left(\sum _{i=0}^{k-1}{U}_{k,i}\left(x\right)·f^{(i+1)}\left(t\right)+\sum _{i=0}^{k-1}(k+1-x)U_{k,i+1}\left(x\right)·f^{(i+1)}\left(t\right)\right)\hfill \\ \hfill & +e^{(k+1)xt}·\left(U_{k+1,k+1}\left(x\right)·f^{(k+1)}\left(t\right)+(k+1-x)·U_{k,0}\left(x\right)·f\left(t\right)\right)\hfill \\ =& e^{(k+1)xt}·\sum _{i=0}^{k-1}{U}_{k+1,i+1}\left(x\right)·f^{(i+1)}\left(t\right)\hfill \\ \hfill & +e^{(k+1)xt}·\left(U_{k+1,k+1}\left(x\right)·f^{(k+1)}\left(t\right)+U_{k+1,0}\left(x\right)·f\left(t\right)\right)\hfill \\ =& e^{(k+1)xt}·\sum _{i=1}^k{U}_{k+1,i}\left(x\right)·f^{\left(i\right)}\left(t\right)\hfill \\ \hfill & +e^{(k+1)xt}·\left(U_{k+1,k+1}\left(x\right)·f^{(k+1)}\left(t\right)+U_{k+1,0}\left(x\right)·f\left(t\right)\right)\hfill \\ =& e^{(k+1)xt}·\sum _{i=0}^{k+1}{U}_{k+1,i}\left(x\right)·f^{\left(i\right)}\left(t\right).\hfill \end{array}$$

which means Lemma 1 is also correct for $h=k+1$.

This proves Lemma 1 by mathematical induction.

Lemma 2.

For any positive integers h and k, we have the power series expansion

$$e^{xht}·f^{\left(k\right)}\left(t\right)=\sum _{n=0}^{\infty }\left(\sum _{i=0}^n{x}^i·h^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·\frac{F_{n-i+k}(x,y)}{n!}\right)·t^n.$$

Proof. 

For any positive integer k, from Equation (1) and the properties of the power series, we have

$$\begin{array}{c}\hfill f^{\left(k\right)}\left(t\right)=\sum _{n=0}^{\infty }(n+k)(n+k-1)\cdots (n+1)·\frac{F_{n+k}(x,y)}{(n+k)!}·t^n=\sum _{n=0}^{\infty }\frac{F_{n+k}(x,y)}{n!}·t^n.\end{array}$$

(4)

On the other hand, we have

$$\begin{array}{c}\hfill e^{xht}=\sum _{n=0}^{\infty }\frac{x^n·h^n}{n!}·t^n.\end{array}$$

(5)

Thus, from Equations (4) and (5) and the multiplicative properties of the power series, we have

$$\begin{array}{cc}\hfill & e^{xht}·f^{\left(k\right)}\left(t\right)=\left(\sum _{n=0}^{\infty }\frac{x^n·h^n}{n!}·t^n\right)\left(\sum _{n=0}^{\infty }\frac{F_{n+k}(x,y)}{n!}·t^n\right)\hfill \\ =& \sum _{n=0}^{\infty }\left(\sum _{i=0}^n{x}^i·h^i·\frac{\left(\genfrac{}{}{0pt}{}{n}{i}\right)}{n!}·F_{n-i+k}(x,y)\right)·t^n.\hfill \end{array}$$

This proves Lemma 2. □

3. Proof of the Theorem

In this section, we complete the proof of our theorem. In fact from Equation (1) and Lemmas 1 and 2, we have

$$\begin{array}{cc}\hfill & {(y+1)}^h·h!·f^{h+1}\left(t\right)={(y+1)}^h·h!·{\left(\sum _{n=0}^{\infty }\frac{F_n(x,y)}{n!}·t^n\right)}^{h+1}\hfill \\ =& {(y+1)}^h·h!·\sum _{n=0}^{\infty }\left(\sum _{a_1+a_2+\cdots +a_{h+1}=n}\frac{F_{a_1}(x,y)}{a_1!}\frac{F_{a_2}(x,y)}{a_2!}\cdots \frac{F_{a_{h+1}}(x,y)}{a_{h+1}!}\right)·t^n\hfill \\ =& \sum _{k=0}^h{U}_{h,k}\left(x\right)·e^{hxt}·f^{\left(k\right)}\left(t\right)\hfill \\ =& \sum _{k=0}^h{U}_{h,k}\left(x\right)\left(\sum _{n=0}^{\infty }\left(\sum _{i=0}^n{x}^i·h^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·\frac{F_{n-i+k}(x,y)}{n!}\right)·t^n\right)\hfill \\ =& \sum _{n=0}^{\infty }\left(\sum _{k=0}^h{U}_{h,k}\left(x\right)\sum _{i=0}^n{x}^i·h^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·\frac{F_{n-i+k}(x,y)}{n!}\right)·t^n.\hfill \end{array}$$

(6)

Comparing the coefficients of the power series in Equation (6), we may immediately deduce the identity

$$\begin{array}{cc}\hfill & {(y+1)}^h·\sum _{a_1+a_2+\cdots +a_{h+1}=n}\frac{F_{a_1}(x,y)}{a_1!}·\frac{F_{a_2}(x,y)}{a_2!}\cdots \frac{F_{a_{h+1}}(x,y)}{a_{h+1}!}\hfill \\ =& \frac{1}{h!·n!}·\sum _{k=0}^h\sum _{i=0}^n{U}_{h,k}\left(x\right)·x^i·h^i·\left(\genfrac{}{}{0pt}{}{n}{i}\right)·F_{n-i+k}(x,y).\hfill \end{array}$$

This completes the proof of our theorem.