Bivariate Extension of Bell Polynomials

Yuanping Zheng and Nadia N. Li School of Mathematics and Statistics Zhoukou Normal University Zhoukou, Henan P. R. China

                    yuanpingzheng0@sohu.com
                   lina20190606@outlook.com

Abstract In this paper, we study bivariate extensions of Bell polynomials and r-Bell polynomials. Some identities related to the r-Stirling numbers and Bell polynomials are presented as special cases.

1 Introduction and motivation

The Stirling number of the second kind is the number of ways to partition a set of n objects into k non-empty subsets (cf. Comtet [2, §5.1]) and denoted by {_{k}^{n}} (see Graham et al. [4, §6.1]). The nth Bell number is the sum of the Stirling numbers of the second kind and denoted by

B_{n} = ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\} , n = 0, 1, ….

The corresponding Bell polynomials are

B_{n}(x) = ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\} x^{k}, n = 0, 1, ….  (1)

Recently, Spivey [6, Eq. 3] discovered the following remarkable formula

B_{m+n} = ∑_{k=0}^{m} ∑_{j=0}^{n} j^{n−k}\binom{n}{k} \begin{cases}m \\ j \end{cases}\} B_{k}  (2)

and gave a short combinatorial proof. Later, Gould and Quaintance [3] extended this Bell number to the following Bell polynomial

B_{m+n}(x) = ∑_{k=0}^{m} ∑_{j=0}^{n} j^{n−k}\binom{n}{k} \begin{cases}m \\ j \end{cases}\} B_{k}(x)x^{j}.  (3)

For any positive integer r, Carlitz [1] introduced the r-Stirling number {_{k}^{n}}_{r}, which is the number of ways to partition a set of n + r objects into k + r non-empty subsets such that the first r elements are in distinct subsets. He also found that the r-Stirling numbers and the Stirling numbers have the following relationship

\begin{cases}n \\ m \end{cases}\}_{r} = ∑_{k=0}^{n} \binom{n}{k} \begin{cases}k \\ m \end{cases}\} r^{n−k}.  (4)

Analogous to the definition of the r-Stirling number, Mező [5] extended the Bell number B_{n} to the r-Bell number

B_{n,r} = ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\}_{r} , n = 0, 1, …

and derived the following identity

B_{m+n,r} = ∑_{k=0}^{m} ∑_{j=0}^{n} (j + r)^{n−k}\binom{n}{k} \begin{cases}m \\ j \end{cases}\}_{r} B_{k}.

The aim of this paper is to generalize further these identities to the bivariate case. In the next section, we give the bivariate extension of Bell polynomials. The bivariate extension of r-Bell polynomials and some examples will be shown in the third section.

2 Bivariate extension of Bell polynomials

Firstly, for any positive integer n, we define the bivariate Bell polynomials by

B_{n}(x, y) = ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\} (x)_{k}y^{k}, n = 0, 1, …,

where the falling factorial is defined by (x)_{0} = 1 and

(x)_{k} = x(x − 1) ⋯ (x − k + 1) for k = 1, 2, ….

Letting y → \frac{y}{x} and then x → ∞, we can see that the bivariate Bell polynomial B_{n}(x, y) reduces to the univariate Bell polynomial in Eq. (1).

The exponential generating function of B_{n}(x, y) is given by the following theorem.

Theorem 1.

∑_{n=0}^{∞} B_{n}(x, y)\frac{t^{n}}{n!} = \{1 + y(e^{t} − 1)\}^{x}.

Proof. Recalling the generating function [2, §5.2] (also see [4, §6.3])

∑_{n=0}^{∞} \begin{cases}n \\ k \end{cases}\} \frac{x^{n}}{n!} = ∑_{n=0}^{∞} S(n, k)\frac{x^{n}}{n!} = \frac{(e^{x} − 1)^{k}}{k!} ,  (5)

we get

∑_{n=0}^{∞} B_{n}(x, y)\frac{t^{n}}{n!} = ∑_{n=0}^{∞} ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\} (x)_{k}y^{k} \frac{t^{n}}{n!}

= ∑_{k=0}^{∞}(x)_{k}y^{k} ∑_{n=k}^{∞} \begin{cases}n \\ k \end{cases}\} \frac{t^{n}}{n!}

= ∑_{k=0}^{∞}(x)_{k}y^{k} \frac{(e^{t} − 1)^{k}}{k!}

= ∑_{k=0}^{∞} \binom{x}{k}\{y(e^{t} − 1)\}^{k}.

By means of the binomial formula [2, §1.6]

(x + y)^{α} = ∑_{k=0}^{∞} \binom{α}{k}x^{α−k}y^{k},  (6)

we have the desired result.

If the exponential generating function f(x) of the sequence {A_{n}} is

∑_{n=0}^{∞} A_{n} \frac{x^{n}}{n!} = f(x),  (7)

then it is routine to get the generating function of the double sequence {A_{m+n}}

∑_{i=0}^{∞} ∑_{j=0}^{∞} A_{i+j} \frac{x^{i}}{i!} \frac{y^{j}}{j!} = f(x + y).  (8)

From this, we can prove the following theorem. Theorem 2. Let m, n ∈ ℕ. We have

B_{m+n}(x, y) = ∑_{k=0}^{m} ∑_{i=0}^{n} k^{n−i}\binom{n}{i} \begin{cases}m \\ k \end{cases}\} B_{i}(x − k, y)(x)_{k}y^{k}.

Proof. In view of Eqs. (7) and (8), we directly obtain the following bivariate exponential generating function of Bell polynomials {B_{m+n}(x, y)}

∑_{m=0}^{∞} ∑_{n=0}^{∞} B_{m+n}(x, y)\frac{u^{m}}{m!} \frac{v^{n}}{n!} = \{1 + y(e^{u+v} − 1)\}^{x}.  (9)

The right-hand side of Eq. (9) can be reformulated to

{1 + y(e^{u+v} − 1)}^{x} = {1 + y(e^{v} − 1) + ye^{v}(e^{u} − 1)}^{x}

= ∑_{k=0}^{∞} \binom{x}{k}\{1 + y(e^{v} − 1)\}^{x−k}\{ye^{v}(e^{u} − 1)\}^{k},

where we have employed the binomial theorem in Eq. (6). According to \binom{x}{k} = \frac{(x)_{k}}{k!} and Theorem 1, the last expression can be transformed into

\{1 + y(e^{u+v} − 1)\}^{x} = ∑_{k=0}^{∞} \frac{(x)_{k}}{k!} ∑_{i=0}^{∞} B_{i}(x − k, y)\frac{v^{i}}{i!} y^{k}e^{kv}(e^{u} − 1)^{k}

= ∑_{k=0}^{∞}(x)_{k}y^{k} \frac{(e^{u} − 1)^{k}}{k!} ∑_{i=0}^{∞} B_{i}(x − k, y)\frac{v^{i}}{i!} ∑_{j=0}^{∞} \frac{k^{j}v^{j}}{j!} .

By means of Eq. (5), we have

\{1 + y(e^{u+v} − 1)\}^{x} = ∑_{k=0}^{∞}(x)_{k}y^{k} ∑_{m=k}^{∞} \begin{cases}m \\ k \end{cases}\} \frac{u^{m}}{m!} ∑_{i=0}^{∞} B_{i}(x − k, y)\frac{v^{i}}{i!} ∑_{j=0}^{∞} \frac{k^{j}v^{j}}{j!} .

Letting i + j = n and then changing the summation order with respect to k and m, the last expression can be reformulated to

\{1 + y(e^{u+v} − 1)\}^{x} = ∑_{m=0}^{∞} ∑_{n=0}^{∞} \frac{u^{m}}{m!} \frac{v^{n}}{n!} ∑_{k=0}^{m} ∑_{i=0}^{n} k^{n−i}(x)_{k}y^{k}\binom{n}{i} \begin{cases}m \\ k \end{cases}\} B_{i}(x − k, y).

Now extracting the coefficient of \frac{u^{m}}{m!} \frac{v^{n}}{n!} across Eq. (9), we get the desired result.

Replacing y by \frac{y}{x} and then letting x → ∞ in Theorem 2, we recover Eq. (3) for the Bell polynomials of single variable.

3 Bivariate extension of r-Bell polynomials

In this section, we firstly define the following bivariate r-Bell polynomials

B_{n,r}(x, y) = ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\}_{r}(x)_{k}y^{k}, n = 0, 1, ….

Theorem 3. Let r be any positive integer. We have

∑_{n=0}^{∞} B_{n,r}(x, y)\frac{t^{n}}{n!} = e^{rt}\{1 + y(e^{t} − 1)\}^{x}.

Proof. The univariate generating function of B_{n,r}(x, y) can be written as

∑_{n=0}^{∞} B_{n,r}(x, y)\frac{t^{n}}{n!} = ∑_{n=0}^{∞} ∑_{k=0}^{n} \begin{cases}n \\ k \end{cases}\}_{r} (x)_{k}y^{k} \frac{t^{n}}{n!}

= ∑_{k=0}^{∞}(x)_{k}y^{k} ∑_{n=k}^{∞} \begin{cases}n \\ k \end{cases}\}_{r} \frac{t^{n}}{n!}.

In view of Eq. (5), we have

∑_{n=0}^{∞} B_{n,r}(x, y)\frac{t^{n}}{n!} = ∑_{k=0}^{∞}(x)_{k}y^{k} \frac{e^{rt}(e^{t} − 1)^{k}}{k!}

= e^{rt} ∑_{k=0}^{∞} \binom{x}{k}\{y(e^{t} − 1)\}^{k}.

Applying the binomial theorem (Eq. (6)) in the right-hand side of the above equation yields the desired result.

Remark 4. The anonymous referee observed that the last identity is equivalent to

B_{n,r}(x, y) = ∑_{j=0}^{n} \binom{n}{j}B_{n−j}(x, y),

which is meaningful for all r, not just the nonnegative integers. It would be interesting to find combinatorial interpretations of B_{n,r}(x, y) (or B_{n,r}(x, 1)) when r is a negative integer.

Theorem 5. For {r, m, n} ∈ ℕ, we have

B_{m+n,r}(x, y) = ∑_{k=0}^{m} ∑_{i=0}^{n} k^{n−i}\binom{n}{i}\begin{cases}m \\ k \end{cases}\}_{r}B_{i,r}(x − k, y)(x)_{k}y^{k}.

Proof. According to Eqs. (7) and (8), we can write the following bivariate generating function directly:

∑_{m,n≥0} B_{m+n,r}(x, y)\frac{u^{m}}{m!} \frac{v^{n}}{n!} = e^{r(u+v)}\{1 + y(e^{(u+v)} − 1)\}^{x}.  (10)

Reformulating the right-hand side of Eq. (10), we have

e^{r(u+v)}{1 + y(e^{u+v} − 1)}^{x} = e^{ru}e^{rv}{1 + y(e^{v} − 1) + ye^{v}(e^{u} − 1)}^{x}

= e^{ru}e^{rv} ∑_{k=0}^{∞} \binom{x}{k}\{1 + y(e^{v} − 1)\}^{x−k}\{ye^{v}(e^{u} − 1)\}^{k}

= ∑_{k=0}^{∞} e^{rv}\{1 + y(e^{v} − 1)\}^{x−k}e^{kv}e^{ru} \frac{(e^{u} − 1)^{k}}{k!} (x)_{k}y^{k}.

With the help of Theorem 3 and Eq. (5), we get from the last expression

e^{r(u+v)}\{1 + y(e^{(u+v)} − 1)\}^{x} = ∑_{j=0}^{∞} ∑_{s=0}^{∞} ∑_{i=0}^{∞} ∑_{l=0}^{∞} \frac{u^{j}}{j!} \frac{u^{s}}{s!} \frac{v^{l}}{l!} \frac{v^{i}}{i!}
× ∑_{k=0}^{j} k^{l}r^{s}B_{i,r}(x − k, y) \begin{cases}j \\ k \end{cases}\} (x)_{k}y^{k}.

Letting i + l = n and s + j = m, then extracting the coefficients of \frac{u^{m}}{m!} \frac{v^{n}}{n!} across Eq. (10), we arrive, after having simplified the result by means of Eq. (4), at the desired identity.

Using the same method as that for Theorem 5, we get another expression for B_{m+n,r}(x, y). Theorem 6. For {r, m, n} ∈ ℕ, we have

B_{m+n,r}(x, y) = ∑_{i=0}^{n} ∑_{k=0}^{m}(k + r)^{n−i}\binom{n}{i} \begin{cases}m \\ k \end{cases}\}_{r} B_{i}(x − k, y)(x)_{k}y^{k}.

When r = 0, the last two theorems reduce to Theorem 2. Proposition 7.

B_{m+n,r}(y) = ∑_{i=0}^{n} ∑_{k=0}^{m} k^{n−i}\binom{n}{i} \begin{cases}m \\ k \end{cases}\}_{r} B_{i,r}(y)y^{k}.

Proof. Let y → \frac{y}{x}, x → ∞ in Theorem 5. Proposition 8.

B_{m+n,r}(y) = ∑_{i=0}^{n} ∑_{k=0}^{m}(k + r)^{n−i}\binom{n}{i}\begin{cases}m \\ k \end{cases}\}_{r}B_{i}(y)y^{k}.

Proof. Let y → \frac{y}{x}, x → ∞ in Theorem 6.

Corollary 9.

B_{m+n,r} = ∑_{i=0}^{n} ∑_{k=0}^{m} k^{n−i}\binom{n}{i} \begin{cases}m \\ k \end{cases}\}_{r} B_{i,r}.

Proof. Let y → \frac{y}{x}, x → ∞ and then set y = 1 in Theorem 5.

Corollary 10. [Mező [5, Theorem 2]]

B_{m+n,r} = ∑_{i=0}^{n} ∑_{k=0}^{m}(k + r)^{n−i}\binom{n}{i} \begin{cases}m \\ k \end{cases}\}_{r} B_{i}.

Proof. Let y → \frac{y}{x}, x → ∞ and then set y = 1 in Theorem 6.

It should be pointed out that Corollary 9 and Corollary 10 can be obtained by setting y = 1 in Propositions 7 and 8, respectively. When r = 0, the last corollary reduces to Eq. (2).

4 Acknowledgments

The authors are sincerely grateful to Professor Wenchang Chu for his valuable comments and suggestions. The corresponding author is partially supported, during this work, by the National Science foundation of China [Youth Grant No. 11601543] and the Research Foundation of Zhoukou Normal University [No. ZKNUB3201701].

References

[1] L. Carlitz, Weighted Stirling numbers of the first and second kind—I, Fibonacci Quart. 18 (1980), 147–162.

[2] L. Comtet, Advanced Combinatorics, Dordrecht-Holland, 1974.

[3] H. W. Gould and J. Quaintance, Implications of Spivey’s Bell number formula, J. Integer Sequences 11 (2008), Article 08.3.7.

[4] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989.

[5] I. Mező, The dual of Spivey’s Bell number formula, J. Integer Sequences 15 (2012), Article 12.2.4.

[6] M. Z. Spivey, A generalized recurrence for Bell numbers, J. Integer Sequences 11 (2008), Article 08.2.5.

2010 Mathematics Subject Classification: Primary 05A19; Secondary 05A10. Keywords: Stirling number of the second kind, r-Stirling number, Bell polynomial.

Received June 28 2019; revised versions received December 3 2019; December 10 2019; December 17 2019. Published in Journal of Integer Sequences, December 27 2019.

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