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.
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
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])
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
we have that Bn = Tn(1).
The q-exponential function is given by (see [6], [14])
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
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
where deformed exponential functions expp(t) and expq(t) are given by (1). The first five (p, q)-deformed Touchard polynomials are
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.
LetN ≥ 1. Then the differential equation
has a solutionF = Tp, q(x; t), where
2 Proofs and Applications
Let F = Tp, q(x;t) and define
Note that
Let us assume that
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
which implies
Note that we assume that
with the initial conditions a1,1(p, q) = 1 and aN, i(p, q) = 0 whenever N < i or i < 0.
For all 1 ≤ i ≤ N, aN, i(p, q) is the coefficient ofxjinTN(x|p, q).
Define
with A0(t) = 1.
Define
with A(x; 0) = 1. Solving the partial differentiate equation for A(x; t), gives that there exists a function f(y) such that
By using the initial condition A(x; 0) = 1, we obtain that
which completes the proof. □
Thus, by Lemma 2.1, we obtain that
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
where, by the main result of [12] and (4), we have
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
Let π be any ordered partition in
Let bj(N, m) to be the number of ordered partitions of [N] with exactly m closers and i blocks. We define
LetN ≥ 1. Thenb1,1(q) = 1, bN, i(q) = 0 wheneveri > Nori < 0, and
Clearly the initial conditions are held. Let π be any ordered partition of [N + 1] with exactly 2 ≤ i ≤ N blocks and let us write an equation for bN + 1, i(q). If the element N + 1 is form a block, then it should be the rightmost block of π and in this case we have a contribution of bN, i−1(q) . Otherwise, the element N + 1 belongs to a block, say Bj with 1 ≤ j ≤ i. In the case N + 1 is rightmost element of Bj we have a contribution bN, i(q), and in the case N + 1 is not rightmost element of Bj we obtain a contribution of |Bj|qbN, i(q). Hence,
as required. □
By (4) and Lemma 2.2, we obtain the following result.
For all 1 ≤ i ≤ N, S1,q(N, i) = bN, i(q−1).
Recall that
Hence, by Theorem 1.1, we have
where
For allN ≥ 1,
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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