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# Open Mathematics

### formerly Central European Journal of Mathematics

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

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# Differential equations for p, q-Touchard polynomials

Taekyun Kim
• Department of Mathematics, College of Science, Tianjin Polytechnic University Tianjin City, 300387, China
• Department of Mathematics, Kwangwoon University Seoul 139-701, South Korea (Republic of)
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• Other articles by this author:
/ Orli Herscovici
/ Toufik Mansour
/ Seog-Hoon Rim
• Corresponding author
• Department of Mathematics Education, Kyungpook National University, Taegu, South Korea (Republic of)
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• Other articles by this author:
Published Online: 2016-11-11 | DOI: https://doi.org/10.1515/math-2016-0082

## 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.

MSC 2010: 05A18; 05A48; llB48; llB47

## 1 Introduction

It is well known that the Touchard polynomials (exponential polynomials or Bell polynomials) Tn(x), play an important role in statistics and probability(see [1-14]). They are given by $Tn(x)=∑k=0nS(n;k)xk.$

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]) $ex(et−1)=∑n=0∞Tn(x)tnn!.$

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 Bn is given by $eet−1=∑n=0∞Bntnn!,$

we have that Bn = Tn(1).

The q-exponential function is given by (see [6], [14]) $expq(x)=(1+(1−q)x)11−q,$(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 $expq(x)=1+x+∑k≥1q(2q−1)(3q−2)…(kq−(k−1))xk+1(k+1)!.$(2)

Following [6] and [3], we define the (p, q)-deformed Touchard polynomials, Tn(x|p, q), for integer n ≥ 0 by means of the generating function $Tp,q(x;t)=expp(x(expq(t)−1))=∑n=0∞Tn(x|p,q)tnn!,$(3)

where deformed exponential functions expp(t) and expq(t) are given by (1). The first five (p, q)-deformed Touchard polynomials are $T0(x|p,q)=1,T1(x|p,q)=x,T2(x|p,q)=xq+x2p,T3(x|p,q)=(2q2−q)x+3pqx2+(2p2−p)x3,T4(x|p,q)=(6q3−7q2+2q)x+(11pq2−4pq)x2+(12p2q−6pq)x3+(6p3−7p2+2p)x4.$

We denote the coefficient of xj in Tn(x|p, q) by Sp, q(n, i) (see [3]). For instance, Sp, q(1,1)=1, Sp,q(2,1) = q and Sp, q(2,2) = p. Clearly, Sp, q(n, 1) = q(2q−1)(3q−2)…((n−1)q−(n−2)) and Sp, q(n, n) = p(2p− 1) (3p−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 $F(N)=∑i=1NSp,q(N,i)xi(1+(1−q)t)i1−q−NF1−i(1−p),$

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

## 2 Proofs and Applications

Let F = Tp, q(x;t) and define ${F}^{\left(N\right)}=\frac{{d}^{N}}{d{t}^{N}}{T}_{p,q}\left(x;t\right)$, for all N > 1. Observe that $F=Tp,q(x;t)=(1−(1−p)x+(1−p)x(1+(1−q)t)11−q)11−p.$

Note that $\frac{d}{dt}\left({F}^{1-p}\right)=\left(1-p\right)x\left(1+\left(1-q\right)t{\right)}^{\frac{1}{1-q}-1}$. Hence, $F(1)=x(1+(1−q)t)11−q−1Fp,F(2)=qx(1+(1−q)t)11−q−2Fp+px2(1+(1−q)t)21−q−2F1−2(1−p).$

Let us assume that $F(N)=∑i=1NaN,i(p,q)xi(1+(1−q)t)i1−q−NF1−i(1−p).$

Clearly, a1,1(p, q) = 1, a2,1(p, q) = q and a2,2(p, q) = p.

By differentiating F(N) with respect to t, we obtain $F(N+1)=ddtF(N)=∑i=1N(i/(1−q)−N)(1−q)aN,i(p,q)xi(1+(1−q)t)i1−q−N−1F1−i(1−p)+∑i=1N(1/(1−p)−i)(1−p)xaN,i(p,q)xi(1+(1−q)t)i+11−q−N−1F1−(i+1)(1−p),$

which implies $F(N+1)=∑i=1(i−N(1−q))aN,i(p,q)xj(1N+(1−q)t)i1−q−N−1F1−i(1−p)+∑i=2N+1(1−i(1−p))aN,i−1(p,q)xj(1+(1−q)t)i1−q−N−1F1−i(1−p).$

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

with the initial conditions a1,1(p, q) = 1 and aN, i(p, q) = 0 whenever N < i or i < 0.

For all 1 ≤ iN, aN, i(p, q) is the coefficient of xj in TN(x|p, q).

Thus, by Lemma 2.1, we obtain that $F(N)=∑i=1NSp,q(N,i)xiN(1+(1−q)t)i1−q−NF1−i(1−p),$

where Sp, q(N, i) is the coefficient of xi in TN(x|p, q). This completes the proof of Theorem 1.1. □

Note that for p = 1, Theorem 1.1 gives $F(N)|p=1=∑i=1NS1,qN(N,i)xi(1+(1−q)t)i1−q−NF,$

where, by the main result of [12] and (4), we have $S1,q(N,i)=∑j=1i∏ℓ=1N−1(j+ℓ(q−1))∏ℓ=1,ℓ≠ji(j−ℓ).$

These numbers have the following combinatorial interpretation. An ordered partition of a set [n] = {1, 2,…, n} is a set of nonempty disjoint subsequences B1/B2/…/Bk, called blocks, whose union is [n] (For partitions of a set [n], see [13]). Note that we can assume that min B1 < min B2 < … < min Bk. 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 bj(N, m) to be the number of ordered partitions of [N] with exactly m closers and i blocks. We define ${b}_{N,i}\left(q\right)={\sum }_{m=0}^{N}{b}_{j}\left(N,\phantom{\rule{thickmathspace}{0ex}}m\right){q}^{m}$.

Let N ≥ 1. Then b1,1(q) = 1, bN, i(q) = 0 whenever i > N or i < 0, and $bN+1,i(q)=bN,i−1(q)+(i+Nq)bN,i(q),1≤i≤N+1.$

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

For all 1 ≤ iN, S1,q(N, i) = bN, i(q−1).

Recall that ${T}_{1,q}\left(x;t\right)=\sum _{n\ge 0}{T}_{n}\left(x|1,q\right)\frac{{t}^{n}}{n!}$. Clearly, $F(N)=dNdtNT1,q(x;t)=∑n≥0Tn+N(x|1,q)tnn!.$

Hence, by Theorem 1.1, we have $F(N)=∑i=1N∑ℓ≥0S1,q(N,i)xj(i1−q−N)ℓ(1−q)ℓtℓℓ!∑m≥0Tm(x|1,q)tmm!=∑n≥0∑i=0N∑ℓ=0nnℓi1−q−Nℓ(1−q)ℓS1,q(N,i)xjTn−ℓ(x|1,q)tnn!,$

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

For all N ≥ 1, $Tn+N(x|1,q)=∑i=0N∑ℓ=0nnℓi1−q−Nℓ(1−q)ℓS1,q(N,i)xjTn−ℓ(x|1,q).$

## Acknowledgement

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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Accepted: 2016-10-03

Published Online: 2016-11-11

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455,

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