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Open Mathematics

formerly Central European Journal of Mathematics

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Volume 14, Issue 1 (Jan 2016)


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:
  • De Gruyter OnlineGoogle Scholar
/ Orli Herscovici / Toufik Mansour / Seog-Hoon Rim
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  • Department of Mathematics Education, Kyungpook National University, Taegu, South Korea (Republic of)
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  • Other articles by this author:
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Published Online: 2016-11-11 | DOI: https://doi.org/10.1515/math-2016-0082


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

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(et1)=n=0Tn(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 eet1=n=0Bntnn!,

we have that Bn = Tn(1).

The q-exponential function is given by (see [6], [14]) expq(x)=(1+(1q)x)11q,(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+k1q(2q1)(3q2)(kq(k1))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=0Tn(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)=(2q2q)x+3pqx2+(2p2p)x3,T4(x|p,q)=(6q37q2+2q)x+(11pq24pq)x2+(12p2q6pq)x3+(6p37p2+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+(1q)t)i1qNF1i(1p),

has a solution F = Tp, q(x; t), where F(N)=dNdtNTp,q(x;t).

2 Proofs and Applications

Let F = Tp, q(x;t) and define F(N)=dNdtNTp,q(x;t), for all N > 1. Observe that F=Tp,q(x;t)=(1(1p)x+(1p)x(1+(1q)t)11q)11p.

Note that ddt(F1p)=(1p)x(1+(1q)t)11q1. Hence, F(1)=x(1+(1q)t)11q1Fp,F(2)=qx(1+(1q)t)11q2Fp+px2(1+(1q)t)21q2F12(1p).

Let us assume that F(N)=i=1NaN,i(p,q)xi(1+(1q)t)i1qNF1i(1p).

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/(1q)N)(1q)aN,i(p,q)xi(1+(1q)t)i1qN1F1i(1p)+i=1N(1/(1p)i)(1p)xaN,i(p,q)xi(1+(1q)t)i+11qN1F1(i+1)(1p),

which implies F(N+1)=i=1(iN(1q))aN,i(p,q)xj(1N+(1q)t)i1qN1F1i(1p)+i=2N+1(1i(1p))aN,i1(p,q)xj(1+(1q)t)i1qN1F1i(1p).

Note that we assume that F(N+1)=i=1N+1aN+1,i(p,q)xj(1+(1q)t)i1qN1F1j(1p), so by comparing the coefficients of xi(1+(1q)t)i1qN1F1i(1p), we obtain aN+1,i(p,q)=(iN(1q))aN,i(p,q)+(1(i1)(1p))aN,i1(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+(1q)t)i1qNF1i(1p),

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+(1q)t)i1qNF,

where, by the main result of [12] and (4), we have S1,q(N,i)=j=1i=1N1(j+(q1))=1,ji(j).

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 OPn,k. Define OPn=k=0nOPn,k.

Let π be any ordered partition in OPn,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 bN,i(q)=m=0Nbj(N,m)qm.

Let N ≥ 1. Then b1,1(q) = 1, bN, i(q) = 0 whenever i > N or i < 0, and bN+1,i(q)=bN,i1(q)+(i+Nq)bN,i(q),1iN+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 T1,q(x;t)=n0Tn(x|1,q)tnn!. Clearly, F(N)=dNdtNT1,q(x;t)=n0Tn+N(x|1,q)tnn!.

Hence, by Theorem 1.1, we have F(N)=i=1N0S1,q(N,i)xj(i1qN)(1q)t!m0Tm(x|1,q)tmm!=n0i=0N=0nni1qN(1q)S1,q(N,i)xjTn(x|1,q)tnn!,

where (c)={c|1}. By comparing the coefficients of tnn!, we obtain the following result.

For all N ≥ 1, Tn+N(x|1,q)=i=0N=0nni1qN(1q)S1,q(N,i)xjTn(x|1,q).


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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About the article

Received: 2016-07-20

Accepted: 2016-10-03

Published Online: 2016-11-11

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2016-0082.

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© 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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