## Abstract

In this paper, we present differential equation for the generating function of the *p*, *q*-Touchard polynomials. An application to ordered partitions of a set is investigated.

Show Summary Details# Differential equations for *p*, *q*-Touchard polynomials

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## Abstract

## 1 Introduction

## 2 Proofs and Applications

## Acknowledgement

## References

## About the article

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### formerly Central European Journal of Mathematics

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Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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In this paper, we present differential equation for the generating function of the *p*, *q*-Touchard polynomials. An application to ordered partitions of a set is investigated.

Keywords: Exponential polynomials; Ordered partitions; Touchard polynomials

It is well known that the *Touchard polynomials* (*exponential polynomials* or Bell polynomials) *T*_{n}(*x*), play an important role in statistics and probability(see [1-14]). They are given by
$${T}_{n}(x)=\sum _{k=0}^{n}S({n}_{;}k){x}^{k}.$$

where *S*(*n*, *k*) is the Stirling number of second kind (see [6], [14]). Moreover, the Touchard polynomials are given by the generating function (see [6], [13])
$${e}^{x({e}^{t}-1)}=\sum _{n=0}^{\mathrm{\infty}}{T}_{n}(x)\frac{{t}^{n}}{n!}.$$

By using the combinatorial definitions of the Stirling and the Bell numbers and the fact that the generating (see [13]) function of the Bell numbers *B*_{n} is given by
$${e}^{{e}^{t}-1}=\sum _{n=0}^{\mathrm{\infty}}{B}_{n}\frac{{t}^{n}}{n!},$$

we have that *B*_{n} = *T*_{n}(1).

The *q*-exponential function is given by (see [6], [14])
$${\mathrm{exp}}_{q}(x)=(1+(1-q)x{)}^{\frac{1}{1-q}},$$(1)

which corresponds to the degenerated exponential function of Carlitz [2] with λ = 1−*q*. Borges [1] showed that this *q*-exponential function has the following expansion in Taylor series
$${\mathrm{exp}}_{q}(x)=1+x+\sum _{k\ge 1}q(2q-1)(3q-2)\dots (kq-(k-1))\frac{{x}^{k+1}}{(k+1)!}.$$(2)

Following [6] and [3], we define the (*p*, *q*)-deformed Touchard polynomials, *T*_{n}(*x*|*p*, *q*), for integer *n* ≥ 0 by means of the generating function
$${T}_{p,q}(x;t)={\mathrm{exp}}_{p}(x({\mathrm{exp}}_{q}(t)-1))=\sum _{n=0}^{\mathrm{\infty}}{T}_{n}(x|p,\phantom{\rule{thickmathspace}{0ex}}q)\frac{{t}^{n}}{n!},$$(3)

where deformed exponential functions exp_{p}(*t*) and exp_{q}(*t*) are given by (1). The first five (*p*, *q*)-deformed Touchard polynomials are
$$\begin{array}{l}{T}_{0}(x|p,\phantom{\rule{thickmathspace}{0ex}}q)=1,\\ {T}_{1}(x|p,\phantom{\rule{thickmathspace}{0ex}}q)=x,\\ {T}_{2}(x|p,\phantom{\rule{thickmathspace}{0ex}}q)=xq+{x}^{2}p,\\ {T}_{3}(x|p,\phantom{\rule{thickmathspace}{0ex}}q)=(2{q}^{2}-q)x+3pq{x}^{2}+(2{p}^{2}-p){x}^{3},\\ {T}_{4}(x|p,\phantom{\rule{thickmathspace}{0ex}}q)=(6{q}^{3}-7{q}^{2}+2q)x+(11p{q}^{2}-4pq){x}^{2}+(12{p}^{2}q-6pq){x}^{3}+(6{p}^{3}-7{p}^{2}+2p){x}^{4}.\end{array}$$

We denote the coefficient of *x*^{j} in *T*_{n}(*x*|*p*, *q*) by *S*_{p, q}(*n*, *i*) (see [3]). For instance, *S*_{p, q}(1,1)=1, *S*_{p,q}(2,1) = *q* and *S*_{p, q}(2,2) = *p*. Clearly, *S*_{p, q}(*n*, 1) = *q*(2*q*−1)(3*q*−2)…((*n*−1)*q*−(*n*−2)) and *S*_{p, q}(*n, n*) = *p*(2*p*− 1) (3*p*−2)…((*n*−1)*p*−(*n*−2)).

Note that Kim at al. [7] studied the linear differential equations arising from the Poisson-Charlier, actuarial, and Meixner polynomials. Also, in [11] the authors considered differential equations arising from the generating function of general modified degenerate Euler numbers. New identities were obtained for Laguerre polynomials [10] and for Bernoulli polynomials of the second kind [5] using linear and non-linear differential equations. Using differential equations, new identities for Changee and Euler polynomials were obtained in [8], for Frobenius-Euler polynomials in [4], and for degenerate Euler numbers in [9].

Here, we study the differential equations arising from the generating function of (*p*, *q*)-deformed Touchard polynomials. In particular, we show the following result.

*Let* *N* ≥ 1. *Then the differential equation*
$${F}^{(N)}=\sum _{i=1}^{N}{S}_{p,q}(N,\phantom{\rule{thickmathspace}{0ex}}i){x}^{i}(1+(1-q)t{)}^{\frac{i}{1-q}-N}{F}^{1-i(1-p)},$$

*has a solution* *F* = *T*_{p, q}(*x*; *t*), *where* ${F}^{(N)}={\displaystyle \frac{{d}^{N}}{d{t}^{N}}{T}_{p,q}(x;t)}$.

Let *F* = *T*_{p, q}(*x*;*t*) and define ${F}^{(N)}={\displaystyle \frac{{d}^{N}}{d{t}^{N}}{T}_{p,q}(x;t)}$, for all *N* > 1. Observe that
$$F={T}_{p,q}(x;t)=(1-(1-p)x+(1-p)x(1+(1-q)t{)}^{\frac{1}{1-q}}{)}^{\frac{1}{1-p}}.$$

Note that $\frac{d}{dt}({F}^{1-p})=(1-p)x(1+(1-q)t{)}^{\frac{1}{1-q}-1}$. Hence, $$\begin{array}{l}{F}^{(1)}=x(1+(1-q)t{)}^{\frac{1}{1-q}-1}{F}^{p},\\ {F}^{(2)}=qx(1+(1-q)t{)}^{\frac{1}{1-q}-2}{F}^{p}+p{x}^{2}(1+(1-q)t{)}^{\frac{2}{1-q}-2}{F}^{1-2(1-p)}.\end{array}$$

Let us assume that $${F}^{(N)}=\sum _{i=1}^{N}{a}_{N,i}(p,\phantom{\rule{thickmathspace}{0ex}}q){x}^{i}(1+(1-q)t{)}^{\frac{i}{1-q}-N}{F}^{1-i(1-p)}.$$

Clearly, *a*_{1,1}(*p*, *q*) = 1, *a*_{2,1}(*p*, *q*) = *q* and *a*_{2,2}(*p*, *q*) = *p*.

By differentiating *F*^{(N)} with respect to *t*, we obtain
$$\begin{array}{ll}{F}^{(N+1)}& =\frac{d}{dt}{F}^{(N)}\\ & =\sum _{i=1}^{N}(i/(1-q)-N)(1-q){a}_{N,i}(p,\phantom{\rule{thickmathspace}{0ex}}q){x}^{i}{(1+(1-q)t)}^{\frac{i}{1-q}-N-1}{F}^{1-i(1-p)}\\ & +\sum _{i=1}^{N}(1/(1-p)-i)(1-p)x{a}_{N,i}(p,\phantom{\rule{thickmathspace}{0ex}}q){x}^{i}{(1+(1-q)t)}^{\frac{i+1}{1-q}-N-1}{F}^{1-(i+1)(1-p)},\end{array}$$

which implies $$\begin{array}{ll}{F}^{(N+1)}& =\sum _{i=1}(i-N(1-q)){a}_{N,i}(p,\phantom{\rule{thickmathspace}{0ex}}q){x}^{j}{(1N+(1-q)t)}^{\frac{i}{1-q}-N-1}{F}^{1-i(1-p)}\\ & +\sum _{i=2}^{N+1}(1-i(1-p)){a}_{N,i-1}(p,\phantom{\rule{thickmathspace}{0ex}}q){x}^{j}{(1+(1-q)t)}^{\frac{i}{1-q}-N-1}{F}^{1-i(1-p)}.\end{array}$$

Note that we assume that ${F}^{(N+1)}={\sum}_{i=1}^{N+1}{a}_{N+1,i}(p,\phantom{\rule{thickmathspace}{0ex}}q){x}^{j}(1+(1-q)t{)}^{\frac{i}{1-q}-N-1}{F}^{1-j(1-p)}$, so by comparing the coefficients of ${x}^{i}(1+(1-q)t{)}^{\frac{i}{1-q}-N-1}{F}^{1-i(1-p)}$, we obtain $${a}_{N+1,i}(p,\phantom{\rule{thickmathspace}{0ex}}q)=(i-N(1-q)){a}_{N,i}(p,\phantom{\rule{thickmathspace}{0ex}}q)+(1-(i-1)(1-p)){a}_{N,i-1}(p,\phantom{\rule{thickmathspace}{0ex}}q)$$(4)

with the initial conditions *a*_{1,1}(*p*, *q*) = 1 and *a*_{N, i}(*p*, *q*) = 0 whenever *N* < *i* or *i* < 0.

*For all* 1 ≤ *i* ≤ *N*, *a*_{N, i}(*p*, *q*) *is the coefficient of* *x*^{j} *in* *T*_{N}(*x*|*p*, *q*).

Thus, by Lemma 2.1, we obtain that $${F}^{(N)}=\sum _{i=1}^{N}{S}_{p,q}(N,\phantom{\rule{thickmathspace}{0ex}}i){x}^{i}N(1+(1-q)t{)}^{\frac{i}{1-q}-N}{F}^{1-i(1-p)},$$

where *S*_{p, q}(*N*, *i*) is the coefficient of *x*^{i} in *T*_{N}(*x*|*p*, *q*). This completes the proof of Theorem 1.1. □

Note that for *p* = 1, Theorem 1.1 gives
$${F}^{(N)}{|}_{p=1}=\sum _{i=1}^{N}{S}_{1,q}N(N,\phantom{\rule{thickmathspace}{0ex}}i){x}^{i}(1+(1-q)t{)}^{\frac{i}{1-q}-N}F,$$

where, by the main result of [12] and (4), we have $${S}_{1,q}(N,\phantom{\rule{thickmathspace}{0ex}}i)=\sum _{j=1}^{i}\frac{{\prod}_{\ell =1}^{N-1}(j+\ell (q-1))}{{\prod}_{\ell =1,\ell \ne j}^{i}(j-\ell )}.$$

These numbers have the following combinatorial interpretation. An *ordered partition* of a set [*n*] = {1, 2,…, *n*} is a set of nonempty disjoint subsequences *B*_{1}/*B*_{2}/…/*B*_{k}, called blocks, whose union is [*n*] (For partitions of a set [*n*], see [13]). Note that we can assume that min *B*_{1} < min *B*_{2} < … < min *B*_{k}. For example, the ordered partition of the set [3] are given by 123, 132, 213, 231, 312, 321, 12/3, 21/3, 13/2, 31/2, 1/23, 1/32, 1/2/3. We denote the set of all ordered partitions of the set [*n*] with *k* blocks by $\mathcal{O}{\mathcal{P}}_{n,k}$. Define $\mathcal{O}{\mathcal{P}}_{n}=\bigcup _{k=0}^{n}\mathcal{O}{\mathcal{P}}_{n,k}$.

Let π be any ordered partition in $\mathcal{O}{\mathcal{P}}_{n,k}$. Let π^{(i)} be the ordered partition that is obtained from π by removing all the elements *i* + 1, *i* + 2, …, *n* from π. An element *i* is called a *closer* if there exists a block *B* in π^{(i}) such that *i* is the rightmost element of *B* and *B* contains at least two elements. Otherwise, it is called an *opener*. For example, if π = 152/437/68 then 8, 7, 2 is closers and 6, 5, 4, 3, 1 openers.

Let *b*_{j}(*N*, *m*) to be the number of ordered partitions of [*N*] with exactly *m* closers and *i* blocks. We define ${b}_{N,i}(q)={\sum}_{m=0}^{N}{b}_{j}(N,\phantom{\rule{thickmathspace}{0ex}}m){q}^{m}$.

*Let* *N* ≥ 1. *Then* *b*_{1,1}(*q*) = 1, *b*_{N, i}(*q*) = 0 *whenever* *i* > *N* *or* *i* < 0, *and*
$${b}_{N+1,i}(q)={b}_{N,i-1}(q)+(i+Nq){b}_{N,i}(q)\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}1\le i\le N+1.$$

By (4) and Lemma 2.2, we obtain the following result.

*For all* 1 ≤ *i* ≤ *N*, *S*_{1,q}(*N*, *i*) = *b*_{N, i}(*q*−1).

Recall that ${T}_{1,q}(x;t)={\displaystyle \sum _{n\ge 0}{T}_{n}(x|1,q)\frac{{t}^{n}}{n!}}$. Clearly, $${F}^{(N)}=\frac{{d}^{N}}{d{t}^{N}}{T}_{1,q}(x;t)=\sum _{n\ge 0}{T}_{n+N}(x|1,q)\frac{{t}^{n}}{n!}.$$

Hence, by Theorem 1.1, we have $$\begin{array}{ll}{F}^{(N)}& =\sum _{i=1}^{N}\left(\sum _{\ell \ge 0}{S}_{1,q}(N,\phantom{\rule{thickmathspace}{0ex}}i){x}^{j}{(\frac{i}{1-q}-N)}_{\ell}{(1-q)}^{\ell}\frac{{t}^{\ell}}{\ell !}\right)\sum _{m\ge 0}{T}_{m}(x|1,\phantom{\rule{thickmathspace}{0ex}}q)\frac{{t}^{m}}{m!}\\ & =\sum _{n\ge 0}\left(\sum _{i=0}^{N}\sum _{\ell =0}^{n}\left(\begin{array}{l}n\\ \ell \end{array}\right){\left(\frac{i}{1-q}-N\right)}_{\ell}{(1-q)}^{\ell}{S}_{1,q}(N,\phantom{\rule{thickmathspace}{0ex}}i){x}^{j}{T}_{n-\ell}(x|1,\phantom{\rule{thickmathspace}{0ex}}q)\right)\frac{{t}^{n}}{n!},\end{array}$$

where $(c{)}_{\ell}=\{c|-1{\}}_{\ell}$. By comparing the coefficients of $\frac{{t}^{n}}{n!}$, we obtain the following result.

*For all* *N* ≥ 1,
$${T}_{n+N}(x|1,\phantom{\rule{thickmathspace}{0ex}}q)=\sum _{i=0}^{N}\sum _{\ell =0}^{n}\left(\begin{array}{l}n\\ \ell \end{array}\right){\left(\frac{i}{1-q}-N\right)}_{\ell}(1-q{)}^{\ell}{S}_{1,q}(N,\phantom{\rule{thickmathspace}{0ex}}i){x}^{j}{T}_{n-\ell}(x|1,\phantom{\rule{thickmathspace}{0ex}}q)\phantom{\rule{thickmathspace}{0ex}}.$$

This research was supported by Kyungpook National University Bokhyeon Research Fund, 2015. Taekyun Kim was appointed as a chair professor at Tianjin polytechnic University by Tianjin city in China from Aug 2015 to Aug 2019. The second author was supported by the Ministry of Science and Technology, Israel.

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**Received**: 2016-07-20

**Accepted**: 2016-10-03

**Published Online**: 2016-11-11

**Published in Print**: 2016-01-01

**Citation Information: **Open Mathematics, Volume 14, Issue 1, Pages 908–912, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0082.

© 2016 Kim *et al*., published by De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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