Some Identities Involving Fubini Polynomials

Abstract

Using Hoppe’s formula, we derive two identities that relate the powers and derivatives of the generating function for Fubini polynomials. As applications, we obtain several identities involving Fubini polynomials, including identities for Sums of Products of Fubini polynomials. These results refine and extend those established in two previous papers by other authors. Furthermore, we prove conjectures posed in one of these papers and derive several congruence identities for Fubini numbers.

1. Introduction

The Fubini polynomials $\left\{F_n\left(y\right)\right\}$ are defined by the generating function

$${\displaystyle \frac{1}{1-y(e^t-1)}}=\stackrel{\infty }{\sum _{n=0}}{F}_n\left(y\right)·{\displaystyle \frac{t^n}{n!}}.$$

(1)

$F_n:=F_n\left(1\right)$ are called Fubini numbers. The Fubini polynomials $\left\{F_n\left(y\right)\right\}$ have the following explicit expressions [1]

$$F_n\left(y\right)=\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·k!·y^k,(n\ge 0)$$

(2)

where $\left\{\begin{array}{c}n\\ k\end{array}\right\}$ are the Stirling numbers of the second kind. The Fubini polynomials of two variables $\left\{F_n(y,x)\right\}$ are defined by the generating function

$${\displaystyle \frac{1}{1-y(e^t-1)}}·e^{xt}=\stackrel{\infty }{\sum _{n=0}}{F}_n(y,x)·{\displaystyle \frac{t^n}{n!}}$$

(3)

satisfying the following identity [1,2]

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

(4)

The Fubini polynomials $\left\{F_n^{\left(r\right)}\left(y\right)\right\}$ of order r and the Fubini polynomials $\left\{F_n^{\left(r\right)}(y,x)\right\}$ of order r of two variables are defined by the generating function

$$f^r\left(t\right):={\left[{\displaystyle \frac{1}{1-y(e^t-1)}}\right]}^r=\stackrel{\infty }{\sum _{n=0}}{F}_n^{\left(r\right)}\left(y\right)·{\displaystyle \frac{t^n}{n!}}$$

(5)

$$f^r\left(t\right)·e^{xt}:={\left[{\displaystyle \frac{1}{1-y(e^t-1)}}\right]}^r·e^{xt}=\stackrel{\infty }{\sum _{n=0}}{F}_n^{\left(r\right)}(y,x)·{\displaystyle \frac{t^n}{n!}}$$

(6)

respectively. Kim, T. et al. [3] proved the identity

$$F_n^{\left(r\right)}\left(y\right)=\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·{\left(r\right)}_{\overline{k}}·y^k$$

(7)

where ${\left(r\right)}_{\overline{k}}:=r·(r+1)\cdots (r+k-1)$.

The Euler polynomials $\left\{E_n\left(x\right)\right\}$ are defined by the generating function [4,5]

$${\displaystyle \frac{2}{1+e^t}}·e^{tx}=\stackrel{\infty }{\sum _{n=0}}{E}_n\left(x\right)·{\displaystyle \frac{t^n}{n!}}.$$

(8)

It is clear that if taking $y=-{\displaystyle \frac{1}{2}}$ in (1) and $x=0$ in (8), then from (1) and (8) we can get the Euler number.

$$E_n:=E_n\left(0\right)=F_n(-{\displaystyle \frac{1}{2}}),n\in N.$$

(9)

Fubini numbers and polynomials, along with their many generalizations, play important roles in mathematics, with applications spanning combinatorics, analysis, and number theory. These include combinatorial calculations, the evaluation of infinite series and power sums, and the study of zeta function values. Owing to this broad utility, they have been extensively studied, as referenced in [3,6,7,8,9,10,11,12,13,14].

Zhao, J. and Chen, Z. [12] considered the computational problem of Sums of Products of Fubini polynomials and proved that, for any positive integers n and k, one has the identity

$$\begin{array}{cc}\hfill & \sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}\left(y\right)}{\left(n_1\right)!}}·{\displaystyle \frac{F_{n_2}\left(y\right)}{\left(n_2\right)!}}\cdots {\displaystyle \frac{F_{n_k}\left(y\right)}{\left(n_k\right)!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{(k-1)!{(y+1)}^{k-1}}}·{\displaystyle \frac{1}{n!}}\underset{i=0}{\sum ^{k-1}}C(k-1,i)F_{n+k-1-i}\left(y\right)\hfill \end{array}$$

(10)

where the summation is taken over all k-dimensional nonnegative integer coordinates $(n_1,n_2,\dots ,n_k)$ such that $n_1+n_2+\dots +n_k=n$. 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

$$C(k+1,i+1)=C(k,i+1)+(k+1)C(k,i)$$

(11)

for all $0\le i<k,$ providing $C(k,i)=0,$ if $i>k.$

As applications of the Sums of Products of Fubini polynomials, Zhao, J. and Chen, Z. discussed the congruence identities of Fubini polynomials and Euler numbers and posed the following conjectures.

Conjecture 1.

For odd prime p, we have the congruence

$$C(p-1,i)\equiv 0\left(modp\right)forall1\le i\le p-2.$$

(12)

Conjecture 2.

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

$$F_{n+p-1}\left(y\right)-F_n\left(y\right)\equiv 0\left(modp\right),y\in \mathbb{Z}.$$

(13)

Chen, G. and Chen, L. [13] generalized the results of Zhao, J. and Chen, Z. to the case of two variables, and proved that, for any positive integers n and k, one has the identity

$$\begin{array}{cc}\hfill W(k,n,x)& :=\sum _{n_1+n_2+\cdots +n_{k+1}=n}{\displaystyle \frac{F_{n_1}(y,x)}{n_1!}}·{\displaystyle \frac{F_{n_2}(y,x)}{n_2!}}\cdots {\displaystyle \frac{F_{n_{k+1}}(y,x)}{n_{k+1}!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{{(y+1)}^k·k!·n!}}·\underset{l=0}{\sum ^k}\stackrel{n}{\sum _{i=0}}{U}_{k,l}\left(x\right)·x^i·k^i·\left(\begin{array}{c}n\\ i\end{array}\right)·F_{n-i+l}(y,x)\hfill \end{array}$$

(14)

where $U_{k,l}\left(x\right)$ is a polynomial defined by $U_{k,k}\left(x\right)=1$, $U_{k+1,0}\left(x\right)=(k+1-x)U_{k,0}\left(x\right)$, and

$$U_{k+1,l+1}\left(x\right)=(k+1-x)U_{k,l+1}\left(x\right)+U_{k,l}\left(x\right)$$

(15)

for all integers $k\ge 1$ and l with $0\le l\le k-1$.

However, Zhao, J. and Chen, Z. [12] and Chen, G. and Chen, L. [13] only provided recurrence relations for the coefficients $C(k,i)$ and the coefficient polynomials $U_{k,l}\left(x\right)$, respectively, without giving their explicit expressions.

Motivated by these works, we explore several identities involving Fubini polynomials. We use Hoppe’s formula to derive several identities that relate the powers and derivatives of the generating function for Fubini polynomials. As applications, we obtain several identities involving Fubini polynomials-including identities for Sums of Products of Fubini polynomials, the explicit expressions of the coefficients $C(k,i)$ and the coefficient polynomials $U_{k,l}\left(x\right)$. These results refine and extend those of Zhao, J. and Chen, Z. [12] and Chen, G. and Chen, L. [13]. Furthermore, we prove conjectures mentioned above, and also derive several congruence identities for Fubini numbers.

Our method is different from those of Zhao, J. and Chen, Z. [12] and Chen, G. and Chen, L. [13]. It is worth noting that giving the explicit expressions of $C(k,i)$ and $U_{k,l}\left(x\right)$ naturally provides solutions to the difference equations with initial conditions in (11) and (15), respectively.

We first use the following Hoppe’s formula for the n-th derivative of a composite function [15]

$${\left({\displaystyle \frac{d}{dt}}\right)}^{n}h\left(\phi \left(t\right)\right)=\underset{k=1}{\sum ^n}[h^{\left(k\right)}\left(u\right)·A_{n,k}\left(t\right)]{|}_{u=\phi \left(t\right)}$$

(16)

where

$$A_{n,k}\left(t\right)={\displaystyle \frac{1}{k!}}\underset{l=1}{\sum ^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·{\left(-1\right)}^{k-l}·{\phi }^{k-l}·{\left({\displaystyle \frac{d}{dt}}\right)}^n·{\phi }^l$$

(17)

to derive two identities establishing the relationship between the powers and derivatives of generating functions for Fubini polynomials in the following two theorems.

Theorem 1.

(Main Theorem 1). Let $f\left(t\right):={\displaystyle \frac{1}{1-y(e^t-1)}},$ and then we have

$$f^{\left(n\right)}\left(t\right)=\underset{k=0}{\sum ^n}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^{n+k}·k!·{(y+1)}^k·f^{k+1}\left(t\right).$$

(18)

Theorem 2.

(Main Theorem 2). Let $f\left(t\right):={\displaystyle \frac{1}{1-y(e^t-1)}},$ and then we have

$$f^{k+1}\left(t\right)={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·f^{\left(n\right)}\left(t\right).$$

(19)

where $\left[\begin{array}{c}k+1\\ n+1\end{array}\right]$ are the Stirling numbers of the first kind.

Then we generalize these two theorems to the case of two variables. The results are given in the following two theorems and one corollary.

Theorem 3.

Let $g(t,x):=f\left(t\right)e^{xt}={\displaystyle \frac{1}{1-y(e^t-1)}}{e}^{xt},$$g^{\left(n\right)}(t,x):={\left({\displaystyle \frac{\partial }{\partial t}}\right)}^{n}g(t,x),$ and then we have

$$f^{k+1}\left(t\right)e^{xt}={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}{\left[\begin{array}{c}k+1-x\\ n+1-x\end{array}\right]}_{1-x}·g^{\left(n\right)}(t,x).$$

(20)

where ${\left[\begin{array}{c}k+1-x\\ n+1-x\end{array}\right]}_{1-x}$ are the r-Stirling numbers of the first kind.

Corollary 1.

Let $g(t,x):=f\left(t\right)e^{xt}={\displaystyle \frac{1}{1-y(e^t-1)}}{e}^{xt},$ and then we have

$$f^{k+1}\left(t\right)e^{(K+1)xt}={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}{\left[\begin{array}{c}k+1-x\\ n+1-x\end{array}\right]}_{1-x}·g^{\left(n\right)}(t,x)e^{Kxt}.$$

(21)

Theorem 4.

Let $g(t,x):=f\left(t\right)e^{xt}={\displaystyle \frac{1}{1-y(e^t-1)}}{e}^{xt},$ and then we have

$$g^{\left(n\right)}(t,x)=\underset{k=0}{\sum ^n}{\left\{\begin{array}{c}n+1-x\\ k+1-x\end{array}\right\}}_{1-x}·{(-1)}^{n+k}·k!·{(y+1)}^k·f^{k+1}\left(t\right)e^{xt}$$

(22)

where ${\left\{\begin{array}{c}n+r\\ k+r\end{array}\right\}}_r$ are the r-Stirling numbers of the second kind. Some contents and properties of Stirling numbers can be found in reference [16,17].

Using the above identities for generating functions and the power series expansion method, we can easily obtain the following identities involving Fubini polynomials.

Corollary 2.

We have the following alternative explicit expression for the Fubini polynomials

$$F_n\left(y\right)=\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^{n+k}·k!·{(y+1)}^k$$

(23)

and an identity

$$\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·k!·y^k=\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^{n+k}·k!·{(y+1)}^k=F_n\left(y\right).$$

(24)

Corollary 3.

We have the following identity

$$\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·{(-1)}^k·k!·{(1-x)}^k=\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^{n+k}·k!·x^k,(n\ge 0).$$

(25)

Theorem 5.

Let $f^r\left(t\right):={\left[{\displaystyle \frac{1}{1-y(e^t-1)}}\right]}^r={\displaystyle \sum _{n=0}^{\infty }}{F}_n^{\left(r\right)}\left(y\right)·{\displaystyle \frac{t^n}{n!}},$ then we have

$$F_{n+s}\left(y\right)=\underset{k=0}{\sum ^s}\left\{\begin{array}{c}s+1\\ k+1\end{array}\right\}·{(-1)}^{s+k}·k!·{(y+1)}^k·F_n^{(k+1)}\left(y\right).$$

(26)

Theorem 6.

(Inverse formula). Let $f^r\left(t\right):={\left[{\displaystyle \frac{1}{1-y(e^t-1)}}\right]}^r={\displaystyle \sum _{n=0}^{\infty }}{F}_n^{\left(r\right)}\left(y\right)·{\displaystyle \frac{t^n}{n!}},$ and then we have

$$F_n^{(s+1)}\left(y\right)={\displaystyle \frac{1}{s!{(1+y)}^s}}\underset{k=0}{\sum ^s}\left[\begin{array}{c}s+1\\ k+1\end{array}\right]·F_{n+k}\left(y\right).$$

(27)

Theorem 7.

For Fubini polynomials of two variables, we have

$$F_n^{(s+1)}(y,x)={\displaystyle \frac{1}{s!{(1+y)}^s}}\underset{k=0}{\sum ^s}{\left[\begin{array}{c}s+1-x\\ k+1-x\end{array}\right]}_{1-x}·F_{n+k}(y,x).$$

(28)

Theorem 8.

For Fubini polynomials of two variables, we have

$$F_{n+s}(y,x)=\underset{k=0}{\sum ^s}{\left\{\begin{array}{c}s+1-x\\ k+1-x\end{array}\right\}}_{1-x}·{(-1)}^{s+k}·k!·{(y+1)}^k·F_n^{(k+1)}(y,x).$$

(29)

Using the above identities for generating functions and the power series expansion method, we investigate a computational problem similar to those of Equations (10) and (14), and obtain identities for Sums of Products of one-variable and two-variable Fubini polynomials of order r.

Theorem 9.

For the Sums of Products of Fubini polynomials of order r, we have the identity

$$\begin{array}{cc}\hfill & \sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}\left(y\right)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}\left(y\right)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}\left(y\right)}{n_k!}}={\displaystyle \frac{F_n^{\left(K\right)}\left(y\right)}{n!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{(K-1)!{(y+1)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{i=0}{\sum ^{K-1}}\left[\begin{array}{c}K\\ i+1\end{array}\right]F_{n+i}\left(y\right)={\displaystyle \frac{1}{n!}}·\stackrel{n}{\sum _{l=0}}\left\{\begin{array}{c}n\\ l\end{array}\right\}·{\left(K\right)}_{\overline{l}}·y^l\hfill \end{array}$$

(30)

where $K:=r_1+r_2+\dots +r_k.$

Corollary 4.

For the Sums of Products of Fubini polynomials, we have the identity

$$\begin{array}{cc}\hfill & \sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}\left(y\right)}{\left(n_1\right)!}}·{\displaystyle \frac{F_{n_2}\left(y\right)}{\left(n_2\right)!}}\cdots {\displaystyle \frac{F_{n_k}\left(y\right)}{\left(n_k\right)!}}={\displaystyle \frac{F_n^{\left(k\right)}\left(y\right)}{n!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{(k-1)!{(y+1)}^{k-1}}}·{\displaystyle \frac{1}{n!}}·\underset{i=0}{\sum ^{k-1}}\left[\begin{array}{c}k\\ i+1\end{array}\right]F_{n+i}\left(y\right)={\displaystyle \frac{1}{n!}}·\stackrel{n}{\sum _{l=0}}\left\{\begin{array}{c}n\\ l\end{array}\right\}·{\left(k\right)}_{\overline{l}}·y^l.\hfill \end{array}$$

(31)

Corollary 5.

For the Sums of Products of Fubini numbers of high order, we have the identity

$$\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}}{n_k!}}={\displaystyle \frac{1}{(K-1)!·2^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{i=0}{\sum ^{K-1}}\left[\begin{array}{c}K\\ i+1\end{array}\right]F_{n+i}$$

(32)

where $K:=r_1+r_2+\dots +r_k.$

Theorem 10.

For the Sums of Products of Fubini polynomials of high order of two variables, we have the identity

$$\begin{array}{cc}\hfill & \sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}(y,x)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}(y,x)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}(y,x)}{n_k!}}={\displaystyle \frac{F_n^{\left(K\right)}(y,kx)}{n!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{(K-1)!{(1+y)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{K-1}}\stackrel{n}{\sum _{j=0}}{U}_{K-1,l}\left(x\right)·{\left((k-1)x\right)}^j·\left(\begin{array}{c}n\\ j\end{array}\right)F_{n-j+l}(y,x)\hfill \\ \hfill & ={\displaystyle \frac{1}{(K-1)!{(1+y)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{K-1}}\stackrel{n}{\sum _{j=0}}{U}_{K-1,l}\left(x\right)·{\left((k-1)x\right)}^{n-j}·\left(\begin{array}{c}n\\ j\end{array}\right)F_{j+l}(y,x)\hfill \\ \hfill & ={\displaystyle \frac{1}{(K-1)!{(1+y)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{K-1}}{U}_{(K-1)l}\left(kx\right)·F_{n+l}(y,kx)\hfill \end{array}$$

(33)

where

$$U_{kl}\left(x\right):=\underset{i=l}{\sum ^k}\left[\begin{array}{c}k+1\\ i+1\end{array}\right]·{(-x)}^{i-l}·\left(\begin{array}{c}i\\ l\end{array}\right)={\left[\begin{array}{c}k+1-x\\ l+1-x\end{array}\right]}_{1-x}$$

and $K:=r_1+r_2+\cdots +r_k.$

Corollary 6.

For the Sums of Products of Fubini polynomials of two variables, we have the identity

$$\begin{array}{cc}\hfill & \sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}(y,x)}{\left(n_1\right)!}}·{\displaystyle \frac{F_{n_2}(y,x)}{\left(n_2\right)!}}\cdots {\displaystyle \frac{F_{n_k}(y,x)}{\left(n_k\right)!}}={\displaystyle \frac{F_n^{\left(k\right)}(y,kx)}{n!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{(k-1)!{(1+y)}^{k-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{k-1}}\stackrel{n}{\sum _{j=0}}{U}_{k-1,l}\left(x\right)·{\left((k-1)x\right)}^j·\left(\begin{array}{c}n\\ j\end{array}\right)F_{n-j+l}(y,x)\hfill \\ \hfill & ={\displaystyle \frac{1}{(k-1)!{(1+y)}^{k-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{k-1}}\stackrel{n}{\sum _{j=0}}{U}_{k-1,l}\left(x\right)·{\left((k-1)x\right)}^{n-j}·\left(\begin{array}{c}n\\ j\end{array}\right)F_{j+l}(y,x)\hfill \\ \hfill & ={\displaystyle \frac{1}{(k-1)!{(1+y)}^{k-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{k-1}}{U}_{k-1,l}\left(kx\right)·F_{n+l}(y,kx)\hfill \end{array}$$

(34)

where

$$U_{kl}\left(x\right):=\underset{i=l}{\sum ^k}\left[\begin{array}{c}k+1\\ i+1\end{array}\right]·{(-x)}^{i-l}·\left(\begin{array}{c}i\\ l\end{array}\right)={\left[\begin{array}{c}k+1-x\\ l+1-x\end{array}\right]}_{1-x}.$$

Remark 1.

By comparing Equations (10) and (31), it is evident that $C(k-1,i)=\left[\begin{array}{c}k\\ k-i\end{array}\right],$ and thus we obtain $C(k,i)=\left[\begin{array}{c}k+1\\ k+1-i\end{array}\right].$ So Corollary 4 gives the explicit expression of the coefficients $C(k,i)$ in (10), and naturally provides solutions to the difference equations with initial conditions in (11).

Remark 2.

We have provided the explicit expressions of coefficients $C(k,i)$, and our method is based on Hoppe’s formula, not induction. This result improves the result of the paper of Zhao, J. and Chen, Z. [12] and can be used to prove the conjectures mentioned above.

Remark 3.

By comparing Equations (14) and (34), it is evident that Corollary 6 gives the explicit expression of the coefficients polynomials, $U_{k,l}\left(x\right)$ in (14), and naturally provides solutions to the difference equations with initial conditions in (15). Our result improves the result of the paper of Chen, G. and Chen, L. [13]. Difference equations play an important role in research in combinatorics; for some recent work, see [18,19]

Next we prove the previously mentioned two conjectures, which are stated in the following two theorems.

Theorem 11.

For odd prime p, we have the congruence

$$C(p-1,i)\equiv 0\left(modp\right)forall1\le i\le p-2.$$

(35)

Theorem 12.

For any positive integer n and odd prime p, one has the congruence

$$F_{n+p-1}\left(y\right)-F_n\left(y\right)\equiv 0\left(modp\right),y\in \mathbb{Z}.$$

(36)

Finally, from the identities for Sums of Products of Fubini polynomials, we can obtain a congruence identity for Fubini numbers.

Theorem 13.

For odd prime p, we have the congruence identity

$$(K-1)!·2^{K-1}·(F_p^{\left(r_1\right)}+F_p^{\left(r_2\right)}+\cdots +F_p^{\left(r_k\right)})=\underset{i=0}{\sum ^{K-1}}\left[\begin{array}{c}K\\ i+1\end{array}\right]F_{p+i}\left(\mathrm{mod}p\right)$$

(37)

where $K:=r_1+r_2+·+r_k.$

Remark 4.

Using the identities we have obtained, we can further explore the congruence properties of Fubini polynomials. However, due to space limitations, we omit the detailed discussion here.

Remark 5.

We thank a reviewer for pointing out that a proof to Conjecture 1 has been given in [20]. That proof does not provide an explicit expression for $C(k,i)$, but instead uses mathematical induction combined with various properties of Fubini polynomials, resulting in a rather lengthy proof. In contrast, our approach differs distinctly: we do not use mathematical induction, but directly derive Identity (1.27), from which the coefficients $C(k,i)$ can be immediately given explicitly. By the explicit expression for $C(k,i)$, we provide a concise proof of Conjecture 1.

2. Several Lemmas

To prove our theorem, we need the following several lemmas.

Lemma 1.

r-Stirling numbers of the first kind and the second kind satisfy the following properties [16,17]

Property 1.

Let $f_k$ and $g_k$ be two sequences, then

$$f_n=\stackrel{n}{\sum _{k=0}}{\left\{\begin{array}{c}n+r\\ k+r\end{array}\right\}}_r·g_k\iff g_n=\stackrel{n}{\sum _{k=0}}{\left[\begin{array}{c}n+r\\ k+r\end{array}\right]}_r·{(-1)}^{n+k}·f_k.$$

(38)

Property 2.

$${\left\{\begin{array}{c}n+r\\ k+r\end{array}\right\}}_1=\left\{\begin{array}{c}n+r\\ k+r\end{array}\right\},{\left[\begin{array}{c}n+r\\ k+r\end{array}\right]}_1=\left[\begin{array}{c}n+r\\ k+r\end{array}\right].$$

(39)

Lemma 2.

$$\underset{l=k}{\sum ^n}\left\{\begin{array}{c}n\\ l\end{array}\right\}\left(\begin{array}{c}l\\ k\end{array}\right)·{(-1)}^l·l!=\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^n·k!.$$

(40)

Proof. 

From the definitions of Stirling numbers and r-Stirling numbers of the first kind, we have

$$\begin{array}{cc}\hfill \underset{n=0}{\sum ^{\infty }}& {\displaystyle \frac{1}{k!}}\underset{l=k}{\sum ^n}\left\{\begin{array}{c}n\\ l\end{array}\right\}\left(\begin{array}{c}l\\ k\end{array}\right)·{(-1)}^{n+l}·l!·{\displaystyle \frac{t^n}{n!}}\hfill \\ \hfill & ={\displaystyle \frac{1}{k!}}\underset{l=0}{\sum ^{\infty }}\left[\sum _{n\ge l}\left\{\begin{array}{c}n\\ l\end{array}\right\}·{\displaystyle \frac{{(-t)}^n}{n!}}\right]·\left(\begin{array}{c}l\\ k\end{array}\right)·{(-1)}^l·l!\hfill \\ \hfill & ={\displaystyle \frac{1}{k!}}\underset{l=0}{\sum ^{\infty }}{\displaystyle \frac{{(e^{-t}-1)}^l}{l!}}·\left(\begin{array}{c}l\\ k\end{array}\right)·{(-1)}^l·l!\hfill \\ \hfill & ={\displaystyle \frac{1}{k!}}\underset{l=k}{\sum ^{\infty }}{(1-e^{-t})}^l·\left(\begin{array}{c}l\\ k\end{array}\right)={\displaystyle \frac{1}{k!}}{(1-e^{-t})}^k\underset{l=0}{\sum ^{\infty }}{(1-e^{-t})}^l·\left(\begin{array}{c}l+k\\ k\end{array}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{k!}}{(1-e^{-t})}^k\underset{l=0}{\sum ^{\infty }}{(1-e^{-t})}^l·\left(\begin{array}{c}-(k+1)\\ l\end{array}\right)={\displaystyle \frac{1}{k!}}{(1-e^{-t})}^k·e^{(k+1)t}\hfill \\ \hfill & ={\displaystyle \frac{1}{k!}}{(e^t-1)}^k·e^t=\underset{n=0}{\sum ^{\infty }}{\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}}_1·{\displaystyle \frac{t^n}{n!}}=\underset{n=0}{\sum ^{\infty }}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{\displaystyle \frac{t^n}{n!}}\hfill \end{array}$$

where we have used the identity (39). Comparing the coefficients of ${\displaystyle \frac{t^n}{n!}},$ we obtain

$${\displaystyle \frac{1}{k!}}\underset{l=k}{\sum ^n}\left\{\begin{array}{c}n\\ l\end{array}\right\}\left(\begin{array}{c}l\\ k\end{array}\right)·{(-1)}^{n+l}·l!=\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}$$

This completes the proof of the Lemma. □

Lemma 3.

$$\underset{n=l}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·{(-x)}^{n-l}·\left(\begin{array}{c}n\\ l\end{array}\right)={\left[\begin{array}{c}k+1-x\\ l+1-x\end{array}\right]}_{1-x}.$$

(41)

Proof. 

From the definition of r-Stirling numbers of the first kind, we have

$$\begin{array}{cc}\hfill & \underset{k=0}{\sum ^{\infty }}{\left[\begin{array}{c}k+1-x\\ l+1-x\end{array}\right]}_{1-x}{\displaystyle \frac{t^k}{k!}}\hfill \\ \hfill & ={\displaystyle \frac{{[-\ln (1-t)]}^l}{l!}}·{\displaystyle \frac{1}{{(1-t)}^{1-x}}}\hfill \\ \hfill & ={\displaystyle \frac{{[-\ln (1-t)]}^l}{l!}}·{\displaystyle \frac{1}{1-t}}·e^{x\ln (1-t)}\hfill \\ \hfill & ={\displaystyle \frac{{[-\ln (1-t)]}^l}{l!}}·{\displaystyle \frac{1}{1-t}}·\underset{k=0}{\sum ^{\infty }}{(-x)}^k·{\displaystyle \frac{{[-\ln (1-t)]}^k}{k!}}\hfill \\ \hfill & =\underset{k=0}{\sum ^{\infty }}{(-x)}^k·{\displaystyle \frac{{[-\ln (1-t)]}^{k+l}}{(k+l)!}}·{\displaystyle \frac{(k+l)!}{l!·k!}}{\displaystyle \frac{1}{1-t}}\hfill \\ \hfill & =\underset{k=l}{\sum ^{\infty }}{(-x)}^{k-l}·{\displaystyle \frac{{[-\ln (1-t)]}^k}{k!}}·\left(\begin{array}{c}k\\ l\end{array}\right){\displaystyle \frac{1}{1-t}}\hfill \\ \hfill & =\underset{k=l}{\sum ^{\infty }}\underset{n=k}{\sum ^{\infty }}{\left[\begin{array}{c}n+1\\ k+1\end{array}\right]}_1{\displaystyle \frac{t^n}{n!}}·{(-x)}^{k-l}·\left(\begin{array}{c}k\\ l\end{array}\right)\hfill \\ \hfill & \underset{n=l}{=\sum ^{\infty }}\underset{k=l}{\sum ^n}{\left[\begin{array}{c}n+1\\ k+1\end{array}\right]}_1{\displaystyle \frac{t^n}{n!}}·{(-x)}^{k-l}·\left(\begin{array}{c}k\\ l\end{array}\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & \underset{k=l}{=\sum ^{\infty }}\underset{n=l}{\sum ^k}{\left[\begin{array}{c}k+1\\ n+1\end{array}\right]}_1{\displaystyle \frac{t^k}{k!}}·{(-x)}^{n-l}·\left(\begin{array}{c}n\\ l\end{array}\right)\left(exchangen\mathrm{and}k\right)\hfill \\ \hfill & \underset{k=l}{=\sum ^{\infty }}\left[\underset{n=l}{\sum ^k}{\left[\begin{array}{c}k+1\\ n+1\end{array}\right]}_1·{(-x)}^{n-l}·\left(\begin{array}{c}n\\ l\end{array}\right)\right]·{\displaystyle \frac{t^k}{k!}}.\hfill \end{array}$$

Comparing the coefficients of ${\displaystyle \frac{t^n}{n!}}$ and using the identity (39), we obtain

$${\left[\begin{array}{c}k+1-x\\ l+1-x\end{array}\right]}_{1-x}=\underset{n=l}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·{(-x)}^{n-l}·\left(\begin{array}{c}n\\ l\end{array}\right).$$

This completes the proof of the Lemma. □

3. Proofs of the Main Results on Identities Involving Fubini Polynomials

In this section, the proofs of all our main results will be completed.

Proof of Theorem 1 

Let $f\left(t\right)={\displaystyle \frac{1}{1-y\left(e^t-1\right)}}=h\left(\phi \left(t\right)\right),$ where $h\left(u\right)={\displaystyle \frac{1}{1-y\left(u-1\right)}}$, and $u=\phi \left(t\right)=e^t.$ By Hoppe’s formulas (16) and (17) for the n-th derivative of a composite function, we have

$$f^{\left(n\right)}\left(t\right)={\left({\displaystyle \frac{d}{dt}}\right)}^{n}f={\left({\displaystyle \frac{d}{dt}}\right)}^{n}h\left(\phi \left(t\right)\right)=\underset{k=1}{\sum ^n}[h^{\left(k\right)}\left(u\right)·A_{n,k}\left(t\right)]{|}_{u=\phi \left(t\right)}$$

where

$$A_{n,k}\left(t\right)={\displaystyle \frac{1}{k!}}\underset{l=1}{\sum ^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·{\left(-1\right)}^{k-l}·{\phi }^{k-l}·{\left({\displaystyle \frac{d}{dt}}\right)}^n·{\phi }^l$$

we have

$$\begin{array}{ccc}\hfill h^{\left(k\right)}\left(u\right)& =& {\left({\displaystyle \frac{d}{du}}\right)}^k{\displaystyle \frac{1}{1-y\left(u-1\right)}}\hfill \\ & =& \left(-1\right)\left(-1-1\right)\cdots \left(-1-k+1\right){\left[{\displaystyle \frac{1}{1-y\left(u-1\right)}}\right]}^{1+k}·{\left(-y\right)}^k\hfill \\ & =& {\left[{\displaystyle \frac{1}{1-y\left(u-1\right)}}\right]}^{1+k}·k!·y^k\hfill \end{array}$$

we have

$$\begin{array}{ccc}\hfill A_{n,k}\left(t\right)& =& {\displaystyle \frac{1}{k!}}\underset{l=1}{\sum ^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·{\left(-1\right)}^{k-l}·{\left(e^t\right)}^{k-l}·{\left({\displaystyle \frac{d}{dt}}\right)}^n·{\left(e^t\right)}^l\hfill \\ & =& {\displaystyle \frac{1}{k!}}\underset{l=1}{\sum ^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·{\left(-1\right)}^{k-l}·e^{\left(k-l\right)t}·e^{lt}·{\left(l\right)}^n\hfill \\ & =& {\displaystyle \frac{1}{k!}}\underset{l=0}{\sum ^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·{\left(-1\right)}^{k-l}·l^n·e^{kt}\left(n\ge 1\right)\hfill \\ & =& \left\{\begin{array}{c}n\\ k\end{array}\right\}·e^{kt}=\left\{\begin{array}{c}n\\ k\end{array}\right\}·u^k\hfill \end{array}$$

where we have used the following identity (closed formula)

$${\displaystyle \frac{1}{k!}}\underset{l=0}{\sum ^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·{\left(-1\right)}^{k-l}·l^n=\left\{\begin{array}{c}n\\ k\end{array}\right\}.$$

Then we have

$$\begin{array}{ccc}\hfill f^{\left(n\right)}\left(t\right)& =& \underset{k=1}{\sum ^n}{\left[{\displaystyle \frac{1}{1-y\left(u-1\right)}}\right]}^{1+k}·k!·y^k·\left\{\begin{array}{c}n\\ k\end{array}\right\}·u^k\hfill \\ & =& {\displaystyle \frac{1}{1-y\left(u-1\right)}}·\underset{k=1}{\sum ^n}\left\{\begin{array}{c}n\\ k\end{array}\right\}·k!·{\left[{\displaystyle \frac{yu}{1-y\left(u-1\right)}}\right]}^k\hfill \\ & =& {\displaystyle \frac{1}{1-y\left(u-1\right)}}·\stackrel{n}{\sum _{k=1}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·k!·{\left[{\displaystyle \frac{1+y}{1-y\left(u-1\right)}}-1\right]}^k\hfill \\ & =& {\displaystyle \frac{1}{1-y\left(u-1\right)}}·\stackrel{n}{\sum _{k=1}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·k!·\stackrel{n}{\sum _{l=0}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right){\left({\displaystyle \frac{1+y}{1-y\left(u-1\right)}}\right)}^l·{\left(-1\right)}^{k-l}\hfill \\ & =& {\displaystyle \frac{1}{1-y\left(u-1\right)}}·\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·k!·\stackrel{n}{\sum _{l=0}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right){\left({\displaystyle \frac{1+y}{1-y\left(u-1\right)}}\right)}^l·{\left(-1\right)}^{k-l}\hfill \\ & =& {\displaystyle \frac{1}{1-y\left(u-1\right)}}·\stackrel{n}{\sum _{l=0}}\left[\stackrel{n}{\sum _{k=l}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·k!·{\left(-1\right)}^k\right]·{\left[{\displaystyle \frac{1+y}{1-y\left(u-1\right)}}\right]}^l·{\left(-1\right)}^l\hfill \\ & =& {\displaystyle \frac{1}{1-y\left(u-1\right)}}·\stackrel{n}{\sum _{l=0}}\left\{\begin{array}{c}n+1\\ l+1\end{array}\right\}·l!·{\left(-1\right)}^n·{\left[{\displaystyle \frac{1+y}{1-y\left(u-1\right)}}\right]}^l·{\left(-1\right)}^l\hfill \end{array}$$

where we have used the identity $\left\{\begin{array}{c}n\\ 0\end{array}\right\}=0,$$(n>0)$ and the identity (40) in Lemma 2

$$\stackrel{n}{\sum _{k=l}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·\left({\displaystyle \genfrac{}{}{0pt}{}{k}{l}}\right)·k!·{\left(-1\right)}^k=\left\{\begin{array}{c}n+1\\ l+1\end{array}\right\}·l!·{\left(-1\right)}^n.$$

Then we have

$$\begin{array}{ccc}\hfill f^{\left(n\right)}\left(t\right)& =& \stackrel{n}{\sum _{l=0}}\left\{\begin{array}{c}n+1\\ l+1\end{array}\right\}·l!·{\left(-1\right)}^{n+l}·{\left(1+y\right)}^l·{\left[{\displaystyle \frac{1}{1-y\left(e^t-1\right)}}\right]}^{1+l}\left(changeletters\right)\hfill \\ & =& \stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·k!·{\left(-1\right)}^{n+k}·{\left(1+y\right)}^k·{\left[{\displaystyle \frac{1}{1-y\left(e^t-1\right)}}\right]}^{1+k}\hfill \\ & =& \stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·k!·{\left(-1\right)}^{n+k}·{\left(1+y\right)}^k·f^{k+1}\left(t\right).\hfill \end{array}$$

This completes the proof of our theorem. □

Proof of Theorem 2. 

Let f_n$={(-1)}^n·f^{\left(n\right)}\left(t\right),$ and g_k$={(-1)}^k·k!·{(y+1)}^k·f^{k+1}\left(t\right).$ By Theorem 1, we have

$$f_n=\underset{k=0}{\sum ^n}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·g_k.$$

From Property (39) and Property (38) with $r=1$ in Lemma 1, we obtain

$$g_n=\underset{k=0}{\sum ^n}\left[\begin{array}{c}n+1\\ k+1\end{array}\right]·{(-1)}^{n+k}·f_k$$

i.e., we have

$${(-1)}^n·n!·{(y+1)}^n·f^{n+1}\left(t\right)=\underset{k=0}{\sum ^n}\left[\begin{array}{c}n+1\\ k+1\end{array}\right]·{(-1)}^{n+k}·{(-1)}^k·f^{\left(k\right)}\left(t\right).$$

The theorem follows immediately. This completes the proof of our theorem. □

Proof of Theorem 3. 

Since $f\left(t\right)=g(t,x)·e^{-xt},$ it follows that

$$\begin{array}{cc}\hfill f^{\left(n\right)}\left(t\right)& ={\left[g(t,x)·e^{-xt}\right]}^{\left(n\right)}=\underset{l=0}{\sum ^n}{g}^{\left(l\right)}(t,x)·{\left(e^{-xt}\right)}^{(n-l)}·\left(\begin{array}{c}n\\ l\end{array}\right)\hfill \\ \hfill & =\underset{l=0}{\sum ^n}{g}^{\left(l\right)}(t,x)·e^{-xt}·{(-x)}^{(n-l)}·\left(\begin{array}{c}n\\ l\end{array}\right).\hfill \end{array}$$

From Theorem 2, we obtain

$$f^{k+1}\left(t\right)={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·f^{\left(n\right)}\left(t\right).$$

It follows that

$$\begin{array}{cc}\hfill f^{k+1}\left(t\right)& ={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·\underset{l=0}{\sum ^n}{g}^{\left(l\right)}(t,x)·e^{-xt}·{(-x)}^{(n-l)}·\left(\begin{array}{c}n\\ l\end{array}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{l=0}{\sum ^k}\underset{n=l}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·g^{\left(l\right)}(t,x)·e^{-xt}·{(-x)}^{(n-l)}·\left(\begin{array}{c}n\\ l\end{array}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{l=0}{\sum ^k}\left[\underset{n=l}{\sum ^k}\left[\begin{array}{c}k+1\\ n+1\end{array}\right]·{(-x)}^{(n-l)}·\left(\begin{array}{c}n\\ l\end{array}\right)\right]·g^{\left(l\right)}(t,x)·e^{-xt}\hfill \\ \hfill & ={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}\left[\underset{l=n}{\sum ^k}\left[\begin{array}{c}k+1\\ l+1\end{array}\right]·{(-x)}^{(l-n)}·\left(\begin{array}{c}l\\ n\end{array}\right)\right]·g^{\left(n\right)}(t,x)·e^{-xt}\left(exchangenandl\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}{U}_{kn}\left(x\right)·g^{\left(n\right)}(t,x)·e^{-xt}\hfill \end{array}$$

where $U_{kn}\left(x\right):={\displaystyle \sum _{l=n}^k}\left[\begin{array}{c}k+1\\ l+1\end{array}\right]·{(-x)}^{(l-n)}·\left(\begin{array}{c}l\\ n\end{array}\right)={\left[\begin{array}{c}k+1-x\\ n+1-x\end{array}\right]}_{1-x}.$ In the last equality, we have applied Lemma 3. Then we obtain

$$f^{k+1}\left(t\right)·e^{xt}={\displaystyle \frac{1}{k!{(1+y)}^k}}\underset{n=0}{\sum ^k}{U}_{kn}\left(x\right)·g^{\left(n\right)}(t,x).$$

The proof of the theorem is completed. □

Proof of Corollary 1. 

The corollary follows immediately from Theorem 3. □

Proof of Theorem 4. 

The theorem follows immediately from Theorem 3 and Equation (38) in Lemma 1, as in the proof of Theorem 2. □

Proof of the Corollary 2. 

By Theorem 1, we have

$$\begin{array}{cc}\hfill F_n\left(y\right)& =f^{\left(n\right)}\left(t\right){{|}_{t=0}=\underset{k=0}{\sum ^n}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^{n+k}·k!·{(y+1)}^k·f^{k+1}\left(t\right)|}_{t=0}\hfill \\ \hfill & =\stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n+1\\ k+1\end{array}\right\}·{(-1)}^{n+k}·k!·{(y+1)}^k,(n\ge 0).\hfill \end{array}$$

From the explicit expressions (2), the proof of the corollary is completed. □

Proof of Corollary 3. 

Let $y=x-1$ in (24), then the corollary follows immediately. □

Proof of Theorem 5. 

From the definition of $f\left(t\right)$ and the properties of the power series we obtain $f^{\left(s\right)}\left(t\right)={\displaystyle \sum _{n=0}^{\infty }}{F}_{n+s}\left(y\right)·{\displaystyle \frac{t^n}{n!}},$ and on the other hand, by formula (18) in Theorem 1, then we obtain

$$\begin{array}{cc}\hfill f^{\left(s\right)}\left(t\right)& =\sum _{k=0}^s\left\{\begin{array}{c}s+1\\ k+1\end{array}\right\}·{(-1)}^{s+k}·k!·{(y+1)}^k·f^{k+1}\left(t\right)\hfill \\ \hfill & =\underset{k=0}{\sum ^s}\left\{\begin{array}{c}s+1\\ k+1\end{array}\right\}·{(-1)}^{s+k}·k!·{(y+1)}^k·\stackrel{\infty }{\sum _{n=0}}{F}_n^{(k+1)}\left(y\right)·{\displaystyle \frac{t^n}{n!}}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}\left[\underset{k=0}{\sum ^s}\left\{\begin{array}{c}s+1\\ k+1\end{array}\right\}·{(-1)}^{s+k}·k!·{(y+1)}^k·F_n^{(k+1)}\left(y\right)\right]·{\displaystyle \frac{t^n}{n!}}.\hfill \end{array}$$

Comparing the coefficient of ${\displaystyle \frac{t^n}{n!}},$ we arrive at

$$F_{n+s}\left(y\right)=\underset{k=0}{\sum ^s}\left\{\begin{array}{c}s+1\\ k+1\end{array}\right\}·{(-1)}^{s+k}·k!·{(y+1)}^k·F_n^{(k+1)}\left(y\right).$$

This completes the proof of our theorem. □

Proof of Theorem 6. 

By Theorem 2, we have

$$f^{s+1}\left(t\right)={\displaystyle \frac{1}{s!{(1+y)}^s}}\underset{k=0}{\sum ^s}\left[\begin{array}{c}s+1\\ k+1\end{array}\right]·f^{\left(k\right)}\left(t\right).$$

Since

$$f^{s+1}\left(t\right)=\stackrel{\infty }{\sum _{n=0}}{F}_n^{(s+1)}\left(y\right)·{\displaystyle \frac{t^n}{n!}}$$

and

$$f^{\left(k\right)}\left(t\right)=\stackrel{\infty }{\sum _{n=0}}{F}_{n+k}\left(y\right)·{\displaystyle \frac{t^n}{n!}}$$

thus we obtain

$$\begin{array}{cc}\hfill \stackrel{\infty }{\sum _{n=0}}{F}_n^{(s+1)}\left(y\right)·{\displaystyle \frac{t^n}{n!}}& ={\displaystyle \frac{1}{s!{(1+y)}^s}}\underset{k=0}{\sum ^s}\left[\begin{array}{c}s+1\\ k+1\end{array}\right]·\stackrel{\infty }{\sum _{n=0}}{F}_{n+k}\left(y\right)·{\displaystyle \frac{t^n}{n!}}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}{\displaystyle \frac{1}{s!{(1+y)}^s}}\underset{k=0}{\sum ^s}\left[\begin{array}{c}s+1\\ k+1\end{array}\right]·F_{n+k}\left(y\right)·{\displaystyle \frac{t^n}{n!}}.\hfill \end{array}$$

Comparing the coefficient of ${\displaystyle \frac{t^n}{n!}},$ we obtain

$$F_n^{(s+1)}\left(y\right)={\displaystyle \frac{1}{s!{(1+y)}^s}}\underset{k=0}{\sum ^s}\left[\begin{array}{c}s+1\\ k+1\end{array}\right]·F_{n+k}\left(y\right).$$

This completes the proof of the theorem. □

Proof of Theorem 7. 

From Theorem 3, the definitions of $F_n^{(s+1)}(y,x)$ and $F_n(y,x)$, and the properties of the power series, the theorem follows immediately, as in the proof of Theorem 6. □

Proof of Theorem 8. 

From Theorem 4, the definitions of $F_n^{(s+1)}(y,x)$ and $F_n(y,x)$, and the properties of the power series, the theorem follows immediately, as in the proof of Theorem 5. □

Proof of Theorem 9. 

The strategy of the proof is as follows. First, we use the generating function method to establish a relation between the Sums of Products of Fubini polynomials of high order and Fubini polynomials of high order. Then, we apply the relation between Fubini polynomials of high order and Fubini polynomials given in Theorem 6, which will yield the desired relation between the Sums of Products of Fubini polynomials of high order and Fubini polynomials. From the definition of $f\left(t\right)$ and the properties of the power series, we obtain $f^{\left(k\right)}\left(t\right)={\displaystyle \sum _{n=0}^{\infty }}{F}_{n+k}\left(y\right)\cdots {\displaystyle \frac{t^n}{n!}}.$ Let $K:=r_1+r_2+·+r_k,$ and then we have

$$\begin{array}{cc}\hfill f^K\left(t\right)& =f^{r_1}\left(t\right)·f^{r_2}\left(t\right)\cdots f^{r_k}\left(t\right)\hfill \\ \hfill & =\left(\underset{n_1=0}{\sum ^{\infty }}{F}_{n_1}^{\left(r_1\right)}\left(y\right)·{\displaystyle \frac{t^{n_1}}{n_1!}}\right)\left(\underset{n_2=0}{\sum ^{\infty }}{F}_{n_2}^{\left(r_2\right)}\left(y\right)·{\displaystyle \frac{t^{n_2}}{n_2!}}\right)\cdots \left(\underset{n_k=0}{\sum ^{\infty }}{F}_{n_k}^{\left(r_k\right)}\left(y\right)·{\displaystyle \frac{t^{n_k}}{n_k!}}\right)\hfill \\ \hfill & =\left(\underset{n_1=0}{\sum ^{\infty }}\underset{n_2=0}{\sum ^{\infty }}\cdots \underset{n_k=0}{\sum ^{\infty }}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}\left(y\right)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}\left(y\right)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}\left(y\right)}{n_k!}}·t^{n_1+n_2\cdots +n_k}\right)\hfill \\ \hfill & =\underset{n=0}{\sum ^{\infty }}\left(\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}\left(y\right)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}\left(y\right)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}\left(y\right)}{n_k!}}\right)t^n.\hfill \end{array}$$

Thus we have

$$f^K\left(t\right)=\underset{n=0}{\sum ^{\infty }}\left(\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}\left(y\right)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}\left(y\right)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}\left(y\right)}{n_k!}}\right)·t^n$$

and on the other hand, we have

$$f^K\left(t\right)=\stackrel{\infty }{\sum _{n=0}}{F}_n^{\left(K\right)}\left(y\right)·{\displaystyle \frac{t^n}{n!}}$$

Comparing the coefficients of $t^n$ in two identities above, we acquire

$$\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}\left(y\right)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}\left(y\right)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}\left(y\right)}{n_k!}}={\displaystyle \frac{F_n^{\left(K\right)}\left(y\right)}{n!}}.$$

Using Theorem 6 and Equation (7), we acquire the identity

$$\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}\left(y\right)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}\left(y\right)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}\left(y\right)}{n_k!}}={\displaystyle \frac{F_n^{\left(K\right)}\left(y\right)}{n!}}$$

$$\begin{array}{c}\hfill ={\displaystyle \frac{1}{(K-1)!{(y+1)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{i=0}{\sum ^{K-1}}\left[\begin{array}{c}K\\ i+1\end{array}\right]F_{n+i}\left(y\right)={\displaystyle \frac{1}{n!}}·\stackrel{n}{\sum _{l=0}}\left\{\begin{array}{c}n\\ l\end{array}\right\}·{\left(K\right)}_{\overline{l}}·y^l.\end{array}$$

This completes the proof of our theorem. □

Proof of Corollary 4. 

Let $r_1=r_2=\cdots =r_k=1$in Theorem 9, the corollary follows immediately. □

Proof of Corollary 5. 

Let $y=1$in Theorem 9, the corollary follows immediately. □

Proof of Theorem 10. 

The strategy of the proof is as follows. First, we use the generating function method to establish a relation between the Sums of Products of Fubini polynomials of high order of two variables and Fubini polynomials of high order of two variables. Then, we apply the relation between Fubini polynomials of high order of two variables and Fubini polynomials of two variables given in Theorem 7, which will yield the desired relation between the Sums of Products of Fubini polynomials of high order of two variables and Fubini polynomials of two variables. Let $g(t,x)=f\left(t\right)·e^{xt}={\displaystyle \sum _{n=0}^{\infty }}{F}_n(y,x)·{\displaystyle \frac{t^n}{n!}},$ and $K:=r_1+r_2+\cdots +r_k.$ From Equation (6), we have

$$\begin{array}{ccc}\hfill f^K\left(t\right)·e^{kxt}& =& f^{r_1}\left(t\right)·e^{xt}·f^{r_2}\left(t\right)·e^{xt}\cdots f^{r_k}\left(t\right)·e^{xt}\hfill \\ & =& \left(\underset{n_1=0}{\sum ^{\infty }}{F}_{n_1}^{\left(r_1\right)}(y,x)·{\displaystyle \frac{t^{n_1}}{n_1!}}\right)\left(\underset{n_2=0}{\sum ^{\infty }}{F}_{n_2}^{\left(r_2\right)}(y,x)·{\displaystyle \frac{t^{n_2}}{n_2!}}\right)\cdots \left(\underset{n_k=0}{\sum ^{\infty }}{F}_{n_k}^{\left(r_k\right)}(y,x)·{\displaystyle \frac{t^{n_k}}{n_k!}}\right)\hfill \\ & =& \left(\underset{n_1=0}{\sum ^{\infty }}\underset{n_2=0}{\sum ^{\infty }}\cdots \underset{n_k=0}{\sum ^{\infty }}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}(y,x)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}(y,x)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}(y,x)}{n_k!}}·t^{n_1+n_2\cdots +n_k}\right)\hfill \\ & =& \underset{n=0}{\sum ^{\infty }}\left(\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}(y,x)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}(y,x)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}(y,x)}{n_k!}}\right)·t^n\hfill \end{array}$$

(42)

and on the other hand, from Theorem 3 and Equation $g^{\left(l\right)}(t,x)={\displaystyle \sum _{n=0}^{\infty }}{F}_{n+l}(y,x)·{\displaystyle \frac{t^n}{n!}}$, then we have

$$\begin{array}{cc}\hfill f^{m+1}\left(t\right)·e^{sxt}& =f^{m+1}\left(t\right)·e^{xt}·e^{(s-1)xt}\hfill \\ \hfill & ={\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}{U}_{ml}\left(x\right)·g^{\left(l\right)}(t,x)·e^{(s-1)xt}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}\left[{\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}{U}_{ml}\left(x\right)·F_{n+l}(y,x)\right]·{\displaystyle \frac{t^n}{n!}}·e^{(s-1)xt}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}\left[{\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}{U}_{ml}\left(x\right)·F_{n+l}(y,x)\right]·{\displaystyle \frac{t^n}{n!}}·\stackrel{\infty }{\sum _{n=0}}{\left((s-1)x\right)}^n·{\displaystyle \frac{t^n}{n!}}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}\left[{\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}\stackrel{n}{\sum _{j=0}}{U}_{ml}\left(x\right)·{\left((s-1)x\right)}^j·\left(\begin{array}{c}n\\ j\end{array}\right)F_{n-j+l}(y,x)\right]·{\displaystyle \frac{t^n}{n!}}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}\left[{\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}\stackrel{n}{\sum _{j=0}}{U}_{ml}\left(x\right)·{\left((s-1)x\right)}^{n-j}·\left(\begin{array}{c}n\\ j\end{array}\right)F_{j+l}(y,x)\right]·{\displaystyle \frac{t^n}{n!}}\hfill \\ \hfill & =\stackrel{\infty }{\sum _{n=0}}{C}_n^{(m+1)}(y,x,s)·{\displaystyle \frac{t^n}{n!}}\hfill \end{array}$$

where

$$C_n^{(m+1)}(y,x,s):={\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}\stackrel{n}{\sum _{j=0}}{U}_{ml}\left(x\right)·{\left((s-1)x\right)}^{n-j}·\left(\begin{array}{c}n\\ j\end{array}\right)F_{j+l}(y,x),$$

and then we have

$$f^{m+1}\left(t\right)·e^{sxt}=\stackrel{\infty }{\sum _{n=0}}{C}_n^{(m+1)}(y,x,s)·{\displaystyle \frac{t^n}{n!}}$$

(43)

It is clear that

$$f^{m+1}\left(t\right)·e^{sxt}=\stackrel{\infty }{\sum _{n=0}}{F}_n^{(m+1)}(y,sx)·{\displaystyle \frac{t^n}{n!}}.$$

(44)

Comparing the coefficients of ${\displaystyle \frac{t^n}{n!}}$in Equations (43) and (44), and using Theorem 7, we get

$$\begin{array}{cc}\hfill C_n^{(m+1)}(y,x,s)& =F_n^{(m+1)}(y,sx)={\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}{\left[\begin{array}{c}m+1-sx\\ l+1-sx\end{array}\right]}_{1-sx}·F_{n+l}(y,sx)\hfill \\ \hfill & ={\displaystyle \frac{1}{m!{(1+y)}^m}}\underset{l=0}{\sum ^m}\stackrel{n}{\sum _{j=0}}{U}_{ml}\left(x\right)·{\left((s-1)x\right)}^{n-j}·\left(\begin{array}{c}n\\ j\end{array}\right)F_{j+l}(y,x).\hfill \end{array}$$

Changing the characters used for variables, we get

$$f^K\left(t\right)·e^{kxt}=\stackrel{\infty }{\sum _{n=0}}{C}_n^{\left(K\right)}(y,x,k)·{\displaystyle \frac{t^n}{n!}}.$$

(45)

Comparing the coefficients of $t^n$in Equations (42) and (45), and using Theorem 7, we get

$$\sum _{n_1+n_2+\cdots +n_k=n}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}(y,x)}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}(y,x)}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}(y,x)}{n_k!}}={\displaystyle \frac{C_n^{\left(K\right)}(y,x,k)}{n!}}={\displaystyle \frac{F_n^{\left(K\right)}(y,kx)}{n!}}$$

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{1}{(K-1)!{(1+y)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{K-1}}\stackrel{n}{\sum _{j=0}}{U}_{K-1,l}\left(x\right)·{\left((k-1)x\right)}^{n-j}·\left(\begin{array}{c}n\\ j\end{array}\right)F_{j+l}(y,x)\hfill \\ \hfill & ={\displaystyle \frac{1}{(K-1)!{(1+y)}^{K-1}}}·{\displaystyle \frac{1}{n!}}·\underset{l=0}{\sum ^{K-1}}{U}_{K-1,l}\left(kx\right)·F_{n+l}(y,kx).\hfill \end{array}$$

This completes the proof of our theorem. □

Proof of Corollary 6. 

Let $r_1=r_2=\dots =r_k=1$in Theorem 10, the corollary follows immediately. □

Proof of Theorem 11. 

(Proof of the Conjecture 1)

From the identity (10), the identity (31) and the identity

$$\underset{i=0}{\sum ^{k-1}}C(k-1,i)F_{n+k-1-i}\left(y\right)=\underset{i=0}{\sum ^{k-1}}C(k-1,k-1-i)F_{n+i}\left(y\right),$$

it is clear that

$$\begin{array}{ccc}\hfill C(k-1,k-1-i)& =& \left[\begin{array}{c}k\\ i+1\end{array}\right]\hfill \\ \hfill C(k,k-i)& =& \left[\begin{array}{c}k+1\\ i+1\end{array}\right].\hfill \end{array}$$

Thus we have

$$C(k,i)=\left[\begin{array}{c}k+1\\ k-i+1\end{array}\right].$$

Let $k=p-1,$$j=p-i,$ and $1\le i\le p-2$, then we have

$$\begin{array}{cc}\hfill C(p-1,i)& =\left[\begin{array}{c}p\\ p-i\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{c}p\\ j\end{array}\right],2\le j\le p-1\hfill \\ \hfill & \equiv 0\left(\mathrm{mod}p\right)\hfill \end{array}$$

where we have used the result

$$\left[\begin{array}{c}p\\ i\end{array}\right]\equiv 0\left(\mathrm{mod}p\right),i=2,\dots ,p-1$$

which can be proved readily by the recurrence formula [17,21]

$$k·\left[\begin{array}{c}n\\ k\end{array}\right]=\underset{i=k-1}{\sum ^{n-1}}\left(\begin{array}{c}n\\ i\end{array}\right)·(n-i-1)!·\left[\begin{array}{c}i\\ k-1\end{array}\right]$$

(46)

and the Property

$$\left(\begin{array}{c}p\\ i\end{array}\right)\equiv 0\left(\mathrm{mod}p\right),i=1,2,\dots ,p-1.$$

In fact, let $n=p,$$2\le k\le p-1$ in Equation (46), we have $k·\left[\begin{array}{c}p\\ k\end{array}\right]\equiv 0$$\left(\mathrm{mod}p\right),$ thus we have $\left[\begin{array}{c}p\\ k\end{array}\right]\equiv 0$$\left(\mathrm{mod}p\right).$ This completes the proof of our theorem. □

Proof of Theorem 12. 

(Proof of the Conjecture 2)

From the identity (27), we obtain

$$s!·{(1+y)}^s·F_n^{(s+1)}\left(y\right)=\underset{k=0}{\sum ^s}\left[\begin{array}{c}s+1\\ k+1\end{array}\right]·F_{n+k}\left(y\right).$$

Let $s=p-1,$ then we obtain

$$(p-1)!·{(1+y)}^{p-1}·F_n^{\left(p\right)}\left(y\right)=\underset{k=0}{\sum ^{p-1}}\left[\begin{array}{c}p\\ k+1\end{array}\right]·F_{n+k}\left(y\right).$$

From the identity (7) with $r=p$, $\left\{\begin{array}{c}n\\ 0\end{array}\right\}=0,$$(n\ge 1),$ and ${\left(p\right)}_{\overline{k}}=0$$\left(\mathrm{mod}p\right),(k\ge 1),$ we have

$$\begin{array}{ccc}\hfill F_n^{\left(p\right)}\left(y\right)& =& \stackrel{n}{\sum _{k=0}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·{\left(p\right)}_{\overline{k}}·y^k,(n\ge 0)\hfill \\ & =& \stackrel{n}{\sum _{k=1}}\left\{\begin{array}{c}n\\ k\end{array}\right\}·{\left(p\right)}_{\overline{k}}·y^k,(n\ge 1)\hfill \\ & =& 0\left(\mathrm{mod}p\right).\hfill \end{array}$$

Thus we obtain

$$\begin{array}{ccc}\hfill 0& \equiv & \underset{k=0}{\sum ^{p-1}}\left[\begin{array}{c}p\\ k+1\end{array}\right]·F_{n+k}\left(y\right)\left(\mathrm{mod}p\right)\hfill \\ & \equiv & \left[\begin{array}{c}p\\ 1\end{array}\right]·F_n\left(y\right)+\underset{k=1}{\sum ^{p-2}}\left[\begin{array}{c}p\\ k+1\end{array}\right]·F_{n+k}\left(y\right)+\left[\begin{array}{c}p\\ p\end{array}\right]·F_{n+p-1}\left(y\right)\left(\mathrm{mod}p\right)\hfill \\ & \equiv & F_n\left(y\right)+\underset{k=1}{\sum ^{p-2}}\left[\begin{array}{c}p\\ k+1\end{array}\right]·F_{n+k}\left(y\right)+(p-1)!·F_{n+p-1}\left(y\right)\left(\mathrm{mod}p\right)\hfill \\ & \equiv & F_n\left(y\right)-F_{n+p-1}\left(y\right)\left(\mathrm{mod}p\right)\hfill \end{array}$$

where we have used the identities: $\left[\begin{array}{c}p\\ 1\end{array}\right]=1,$$\left[\begin{array}{c}p\\ p\end{array}\right]=(p-1)!,$$\left[\begin{array}{c}p\\ k\end{array}\right]=0\left(\mathrm{mod}p\right)$ ($k=2,\dots ,p-1),$ and Wilson Theorem: $(p-1)!\equiv -1$$\left(\mathrm{mod}p\right).$ The proof of the theorem is completed. □

Proof of Theorem 13. 

Let $n=p$$in$ Corollary 5, we have

$$(K-1)!·2^{K-1}·p!·\sum _{n_1+n_2+\cdots +n_k=p}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}}{n_1!}}·{\displaystyle \frac{F_{n_2}^{\left(r_2\right)}}{n_2!}}\cdots {\displaystyle \frac{F_{n_k}^{\left(r_k\right)}}{n_k!}}=\underset{i=0}{\sum ^{K-1}}\left[\begin{array}{c}K\\ i+1\end{array}\right]F_{p+i}$$

then we can easily obtain

$$\begin{array}{ccc}& & (K-1)!·2^{K-1}·(F_p^{\left(r_1\right)}+F_p^{\left(r_2\right)}+\cdots +F_p^{\left(r_k\right)})\hfill \\ & & +(K-1)!·2^{K-1}·p!·\sum _{\begin{array}{c}{n}_1+n_2+\cdots +n_k=p\\ n_i<p,if1\le i\le k\end{array}}{\displaystyle \frac{F_{n_1}^{\left(r_1\right)}·F_{n_2}^{\left(r_2\right)}\cdots F_{n_k}^{\left(r_k\right)}}{n_1!·n_2!\cdots n_k!}}\hfill \\ & =& (K-1)!·2^{K-1}·(F_p^{\left(r_1\right)}+F_p^{\left(r_2\right)}+\cdots +F_p^{\left(r_k\right)})\left(\mathrm{mod}p\right)=\underset{i=0}{\sum ^{K-1}}\left[\begin{array}{c}K\\ i+1\end{array}\right]F_{p+i}\left(\mathrm{mod}p\right).\hfill \end{array}$$

The proof of the theorem is completed. □

4. Conclusions

In this study, we employ Hoppe’s formula to establish relationships between the powers and derivatives of the generating function for Fubini polynomials. Using these relationships, we derived several novel identities for Fubini polynomials and their sums of products, as well as explicit expressions for the coefficients $C(k,i)$ and the coefficient polynomials $U_{k,i}\left(x\right)$. These results refine and extend prior works, particularly those of Zhao, J. and Chen, Z. [12] and Chen, G. and Chen, L. [13]. Furthermore, we proved the conjectures posed in one of these papers and derived several congruence identities for Fubini numbers. Our approach, distinct from previous methods, offers a fresh perspective on exploring the properties of Fubini polynomials. We anticipate that these findings will contribute to future research in combinatorics and number theory.