Several explicit and recursive formulas for generalized Motzkin numbers

Feng Qi; Bai-Ni Guo; Feng Qi; Bai-Ni Guo

doi:10.3934/math.2020091

AIMS Mathematics

2020, Volume 5, Issue 2: 1333-1345. doi: 10.3934/math.2020091

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Research article

Several explicit and recursive formulas for generalized Motzkin numbers

Feng Qi ^1,2,3,
Bai-Ni Guo ^{3
,
,}

1.
College of Mathematics, Inner Mongolia University for Nationalities, Tongliao 028043, China
2.
School of Mathematical Sciences, Tianjin Polytechnic University, Tianjin 300387, China
3.
School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo 454010, China

Received: 06 November 2019 Accepted: 09 January 2020 Published: 20 January 2020
MSC : Primary: 05A15; Secondary: 05A19, 05A20, 11B37, 11B83, 34A05

In the paper, the authors find two explicit formulas and recover a recursive formula for generalized Motzkin numbers. Consequently, the authors deduce two explicit formulas and a recursive formula for the Motzkin numbers, the Catalan numbers, and the restricted hexagonal numbers respectively.

Keywords:

Citation: Feng Qi, Bai-Ni Guo. Several explicit and recursive formulas for generalized Motzkin numbers[J]. AIMS Mathematics, 2020, 5(2): 1333-1345. doi: 10.3934/math.2020091

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Abstract

1. Introduction

The Motzkin numbers $M_n$ enumerate various combinatorial objects. In 1977, Donaghey and Shapiro ^[3] gave fourteen different manifestations of the Motzkin numbers $M_n$ . In particular, the Motzkin numbers $M_n$ give the numbers of paths from $(0, 0)$ to $(n, 0)$ which never dip below the $x$ -axis $y = 0$ and are made up only of the steps $(1, 0)$ , $(1, 1)$ , and $(1, -1)$ .

The first seven Motzkin numbers $M_n$ for $0\le n\le6$ are $1, 1, 2, 4, 9, 21, 51$ . All the Motzkin numbers $M_n$ can be generated by

$\begin{equation*} M(x) = \frac{1-x-\sqrt{1-2x-3x^2}\, }{2x^2} = \frac1{1-x+\sqrt{1-2x-3x^2}\, } = \sum\limits_{k = 0}^\infty M_kx^k. \end{equation*}$

In 2007, Mansour et al ^[12] introduced the $(u, l, d)$ -Motzkin numbers $m_n^{(u, l, d)}$ and obtained [,Theorem 2.1] that $m_n^{(u, l, d)} = m_n^{(1, l, ud)}$ ,

$\begin{equation} M_{u, l, d}(x) = \frac{1-lx-\sqrt{(1-lx)^2-4udx^2}\, }{2udx^2} = \sum\limits_{n = 0}^\infty m_n^{(u, l, d)}x^n, \end{equation}$

(1.1)

and

$\begin{equation} m_n^{(u, l, d)} = l^n\sum\limits_{j = 0}^{n/2}\frac1{j+1}\binom{2j}{j}\binom{n}{2j}\biggl(\frac{ud}{l^2}\biggr)^j. \end{equation}$

(1.2)

From (1.1) and (1.2), it is easy to see that $m_n^{(u, l, d)} = m_n^{(d, l, u)}$ .

In 2014, Sun ^[42] generalized the Motzkin numbers $M_n$ to

$\begin{equation} M_n(a, b) = \sum\limits_{k = 0}^{\lfloor n/2\rfloor}\binom{n}{2k}C_ka^{n-2k}b^k \end{equation}$

(1.3)

for $a, b\in\mathbb{N}$ in terms of the Catalan numbers

$\begin{equation} C_n = \frac1{n+1}\binom{2n}n \end{equation}$

(1.4)

and established the generating function

$\begin{equation} M_{a, b}(x) = \frac{1-ax-\sqrt{(1-ax)^2-4bx^2}\, }{2bx^2} = \frac1{1-ax+\sqrt{(1-ax)^2-4bx^2}\, } = \sum\limits_{k = 0}^\infty M_k(a, b)x^k, \end{equation}$

(1.5)

where $\lfloor\lambda\rfloor$ denotes the floor function defined by the largest integer less than or equal to $\lambda\in\mathbb{R}$ . Wang and Zhang pointed out in ^[43] that

$\begin{equation} M_n(1, 1) = M_n, \quad M_n(2, 1) = C_{n+1}, \quad \text{and} \quad M_n(3, 1) = H_n, \end{equation}$

(1.6)

where $H_n$ denote the restricted hexagonal numbers described by Harary and Read ^[4].

For more information on many results, applications, and generalizations of the Motzkin numbers $M_n$ , please refer to the papers ^{[3,9,10,42,43]} and closely related references therein. For more information on many results, applications, and generalizations of the Catalan numbers $C_n$ , please refer to the monograph ^[5], the newly published papers ^{[11,17,19,26,27,31,36,37,38,40,41]}, the survey articles ^[25,29], and closely related references therein.

Comparing (1.1) with (1.5) reveals that $M_k(a, b)$ and $m_k^{(u, l, d)}$ are equivalent to each other and satisfy

$\begin{equation} M_k(a, b) = m_n^{(1, a, b)} = m_k^{(b, a, 1)} \quad\text{and}\quad m_k^{(u, l, d)} = M_k(l, ud). \end{equation}$

(1.7)

Therefore, it suffices to consider generalized Motzkin numbers $M_k(a, b)$ , rather than the $(u, l, d)$ -Motzkin numbers $m_n^{(u, l, d)}$ , in this paper.

By the second relation in (1.7), one can reformulated the formula (1.2) as

$\begin{equation} M_n(a, b) = a^n\sum\limits_{j = 0}^{\lfloor n/2\rfloor} \frac1{j+1}\binom{2j}{j}\binom{n}{2j}\biggl(\frac{b}{a^2}\biggr)^j. \end{equation}$

(1.8)

Substituting (1.4) into (1.3) recovers (1.8) once again.

In 2015, Wang and Zhang [43,Theorem 1] combinatorially obtained, among other things, the recursive formula

$\begin{equation} M_{n+2}(a, b) = aM_{n+1}(a, b)+b\sum\limits_{\ell = 0}^nM_\ell(a, b)M_{n-\ell}(a, b), \quad n\ge0. \end{equation}$

(1.9)

It is not difficult to see that the function $(1-ax)^2-4bx^2 = \bigl(a^2-4b\bigr)x^2-2ax+1$ is nonnegative if and only if

1. either $b = 0$ and $x\in\mathbb{R}$ ,

2. or $a^2-4b = 0$ , $a\ne0$ , and $x\le\frac1{2a}$ ,

3. or $a^2-4b > 0$ , $b < 0$ , and $x\in\mathbb{R}$ ,

4. or $a^2-4b > 0$ , $b > 0$ , and $x\ge\frac1{a-2\sqrt{b}\, }$ or $x\le\frac{1}{a+2\sqrt{b}\, }$ .

Consequently, the generating function $M_{a, b}(x)$ defined by (1.5) in ^[42] is defined for either $b\le0$ or $a\ge2\sqrt{b}\, > 0$ .

In this paper, we will find two explicit formulas, which are different from (1.8), and recover the recursive formula (1.9) for generalized Motzkin numbers $M_n(a, b)$ . Consequently, we will derive two explicit formula and a recursive formula for the Motzkin numbers $M_n$ , the Catalan numbers $C_n$ , and the restricted hexagonal numbers $H_n$ respectively.

We can state our main results as the following three theorems.

Theorem 1. For $n\ge0$ , we can compute generalized Motzkin numbers $M_n(a, b)$ by

$\begin{equation} M_n(a, b) = \frac1{2b}\biggl(\frac{4b-a^2}{2a}\biggr)^{n+2} \sum\limits_{\ell = 0}^{n+2} \biggl(\frac{2a^2}{4b-a^2}\biggr)^\ell \frac{(2\ell-3)!!}{\ell!}\binom{\ell}{n-\ell+2}, \end{equation}$

(1.10)

where $\binom{p}{q} = 0$ for $q > p\ge0$ and the double factorial of negative odd integers $-(2n+1)$ is

$\begin{equation*} [-(2n+1)]!! = \frac{(-1)^n}{(2n-1)!!} = (-1)^n\frac{2^nn!}{(2n)!}, \quad n = 0, 1, \dotsc. \end{equation*}$

Consequently, we can compute the Motzkin numbers $M_n$ and the restricted hexagonal numbers $H_n$ respectively by

$\begin{equation} M_n = \frac98\biggl(\frac32\biggr)^n\sum\limits_{\ell = 0}^{n+2} \biggl(\frac23\biggr)^\ell \frac{(2\ell-3)!!}{\ell!}\binom{\ell}{n-\ell+2} \end{equation}$

(1.11)

and

$\begin{equation} H_n = (-1)^n\frac{25}{72}\biggl(\frac56\biggr)^n \sum\limits_{\ell = 0}^{n+2} (-1)^\ell\biggl(\frac{18}{5}\biggr)^\ell \frac{(2\ell-3)!!}{\ell!}\binom{\ell}{n-\ell+2}. \end{equation}$

(1.12)

Theorem 2. For $n\ge0$ , we can compute generalized Motzkin numbers $M_n(a, b)$ by

$\begin{equation} M_n(a, b) = -\frac{\bigl(a-2\sqrt{b}\, \bigr)^{n+2}}{2b} \sum\limits_{\ell = 0}^{n+2}\frac{(2\ell-3)!!}{(2\ell)!!} \frac{[2(n-\ell+2)-3]!!}{[2(n-\ell+2)]!!} \Biggl(\frac{a+2\sqrt{b}\, }{a-2\sqrt{b}\, }\Biggr)^{\ell}. \end{equation}$

(1.13)

Consequently, we can compute the Motzkin numbers $M_n$ and the restricted hexagonal numbers $H_n$ respectively by

$\begin{equation*} M_n = \frac{(-1)^{n+1}}{2} \sum\limits_{\ell = 0}^{n+2}(-1)^\ell3^{\ell} \frac{(2\ell-3)!!}{(2\ell)!!} \frac{[2(n-\ell+2)-3]!!}{[2(n-\ell+2)]!!} \end{equation*}$

and

$\begin{equation*} H_n = -\frac1{2} \sum\limits_{\ell = 0}^{n+2}5^{\ell}\frac{(2\ell-3)!!}{(2\ell)!!} \frac{[2(n-\ell+2)-3]!!}{[2(n-\ell+2)]!!}. \end{equation*}$

Theorem 3. For $n\ge0$ , generalized Motzkin numbers $M_n(a, b)$ satisfy

$\begin{equation} M_0(a, b) = 1, \quad M_1(a, b) = a, \end{equation}$

(1.14)

and the recursive formula (1.9). Consequently, for $n\ge0$ , the Motzkin numbers $M_n$ , the Catalan numbers $C_n$ , and the restricted hexagonal numbers $H_n$ meet the recursive formulas

$\begin{gather} M_{n+2} = M_{n+1}+\sum\limits_{\ell = 0}^nM_\ell M_{n-\ell}, \end{gather}$

(1.15)

$\begin{gather} C_{n+2} = 2C_{n+1}+\sum\limits_{\ell = 0}^nC_\ell C_{n-\ell}, \end{gather}$

(1.16)

and

$\begin{equation} H_{n+2} = 3H_{n+1}+\sum\limits_{\ell = 0}^nH_\ell H_{n-\ell} \end{equation}$

(1.17)

respectively.

2. Lemmas

In order to prove the explicit formula (1.10), we need the following lemmas.

Lemma 1 ([1,p. 40,Exercise 5)], [16,Section 2.2,p. 849], [22,p. 94], [34,Lemma 3], and [44,Lemma 2.1]). Let $u(x)$ and $v(x)\ne0$ be two differentiable functions. Let $U_{(n+1)\times1}(x)$ be an $(n+1)\times1$ matrix whose elements $u_{k, 1}(x) = u^{(k-1)}(x)$ for $1\le k\le n+1$ , let $V_{(n+1)\times n}(x)$ be an $(n+1)\times n$ matrix whose elements

$\begin{equation*} v_{i, j}(x) = \begin{cases} \dbinom{i-1}{j-1}v^{(i-j)}(x), & i-j\ge0\\ 0, & i-j \lt 0 \end{cases} \end{equation*}$

for $1\le i\le n+1$ and $1\le j\le n$ , and let $|W_{(n+1)\times(n+1)}(x)|$ denote the determinant of the $(n+1)\times(n+1)$ matrix

$\begin{equation*} W_{(n+1)\times(n+1)}(x) = \begin{pmatrix}U_{(n+1)\times1}(x) & V_{(n+1)\times n}(x)\end{pmatrix}. \end{equation*}$

Then the $n$ th derivative of the ratio $\frac{u(x)}{v(x)}$ can be computed by

$\begin{equation*} \frac{ \text{d}^n}{ \text{d} x^n}\biggl[\frac{u(x)}{v(x)}\biggr] = (-1)^n\frac{\bigl|W_{(n+1)\times(n+1)}(x)\bigr|}{v^{n+1}(x)}. \end{equation*}$

Lemma 2 ([2,p. 134,Theorem A and p. 139,Theorem C]). The Faà di Bruno formula can be described in terms of the Bell polynomials of the second kind

$\begin{equation*} \text{B}_{n, k}(x_1, x_2, \dotsc, x_{n-k+1}) = \sum\limits_{\substack{1\le i\le n-k+1\\ \ell_i\in\{0\}\cup\mathbb{N}\\ \sum\nolimits_{i = 1}^{n-k+1}i\ell_i = n\\ \sum\nolimits_{i = 1}^{n-k+1}\ell_i = k}}\frac{n!}{\prod\nolimits_{i = 1}^{n-k+1}\ell_i!} \prod\limits_{i = 1}^{n-k+1}\Bigl(\frac{x_i}{i!}\Bigr)^{\ell_i} \end{equation*}$

for $n\ge k\ge0$ by

$\begin{equation} \frac{ \text{d}^n}{ \text{d} t^n}[f\circ h(t)] = \sum\limits_{k = 0}^nf^{(k)}(h(t)) \text{B}_{n, k}\bigl(h'(t), h''(t), \dotsc, h^{(n-k+1)}(t)\bigr) \end{equation}$

(2.1)

for $n\ge0$ .

Lemma 3 ([2,p. 135]). The Bell polynomials of the second kind $\text{B}_{n, k}$ satisfy

$\begin{equation} \text{B}_{n, k}\bigl(abx_1, ab^2x_2, \dotsc, ab^{n-k+1}x_{n-k+1}\bigr) = a^kb^n \text{B}_{n, k}(x_1, x_2, \dotsc, x_{n-k+1}) \end{equation}$

(2.2)

for $n\ge k\ge0$ .

Lemma 4. For $n\ge k\ge0$ , we have

$\begin{equation} \text{B}_{n, k}(x, 1, 0, \dotsc, 0) = \frac{(n-k)!}{2^{n-k}}\binom{n}{k}\binom{k}{n-k}x^{2k-n}. \end{equation}$

(2.3)

More generally, for $n\ge k\ge0$ and $\lambda, \alpha\in\mathbb{C}$ , we have

$\begin{equation} \text{B}_{n, k}\Biggl(1, 1-\lambda, (1-\lambda)(1-2\lambda), \dotsc, \prod\limits_{\ell = 0}^{n-k}(1-\ell\lambda)\Biggr) = \frac{(-1)^k}{k!} \sum\limits_{\ell = 0}^k (-1)^{\ell} \binom{k}{\ell} \prod\limits_{q = 0}^{n-1}(\ell-q\lambda) \end{equation}$

(2.4)

or, equivalently,

$\begin{equation} \text{B}_{n, k}(\langle\alpha\rangle_1, \langle\alpha\rangle_2, \dotsc, \langle\alpha\rangle_{n-k+1}) = \frac{(-1)^k}{k!}\sum\limits_{\ell = 0}^{k}(-1)^{\ell}\binom{k}{\ell}\langle\alpha\ell\rangle_n. \end{equation}$

(2.5)

Proof. The formula (2.3) can be found in [24,Theorem 5.1], [35,p. 7,(19)], [39,Section 3], and [44,Lemma 2.5]. The explicit formula (2.4) was first established in [30,Remark 1] and then was applied in [18,Section 2], [20,First proof of Theorem 2], [21,Lemma 2.2], [24,Remark 6.1], [28,Lemma 4], and [32,Lemma 2.6]. The formula (2.5) and the equivalence were presented in [33,Theorems 2.1 and 4.1].

Remark 1. In recent years, there have been some literature such as ^{[6,7,8,13,14,15,23,24,30,35,45,46,47,48]} devoting to deep investigation and extensive applications of the Bell polynomials of the second kind $\text{B}_{n, k}(x_1, x_2, \dotsc, x_{n-k+1})$ . Specially, in the papers ^[13,14], the generalized Dyck paths (namely, various type of Motzkin paths) and the Bell polynomials were connected closely.

3. Proofs of Theorems 1 and 3

We are now in a position to prove our main results.

Proof of Theorem 1. By virtue of (2.1), (2.2), and (2.3), we obtain for $k\ge0$ that

$\begin{multline} \Bigl[\sqrt{(1-ax)^2-4bx^2}\, \Bigr]^{(k+2)} = \sum\limits_{\ell = 0}^{k+2}\biggl\langle\frac12\biggr\rangle_{\ell} \bigl[(1-ax)^2-4bx^2\bigr]^{1/2-\ell}\\ \times \text{B}_{k+2, \ell}\bigl(-2\bigl[a+\bigl(4b-a^2\bigr)x\bigr], 2\bigl(a^2-4b\bigr), 0, \dotsc, 0\bigr)\\ \to\sum\limits_{\ell = 0}^{k+2}\biggl\langle\frac12\biggr\rangle_{\ell} \text{B}_{k+2, \ell}\bigl(-2a, 2\bigl(a^2-4b\bigr), 0, \dotsc, 0\bigr)\\ = \sum\limits_{\ell = 0}^{k+2}\biggl\langle\frac12\biggr\rangle_{\ell} \bigl[2\bigl(a^2-4b\bigr)\bigr]^\ell \text{B}_{k+2, \ell}\biggl(\frac{a}{4b-a^2}, 1, 0, \dotsc, 0\biggr)\\ = \sum\limits_{\ell = 0}^{k+2}\biggl\langle\frac12\biggr\rangle_{\ell} \bigl[2\bigl(a^2-4b\bigr)\bigr]^\ell \frac{(k-\ell+2)!}{2^{k-\ell+2}} \binom{k+2}{\ell}\binom{\ell}{k-\ell+2} \biggl(\frac{a}{4b-a^2}\biggr)^{2\ell-k-2} \end{multline}$

(3.1)

as $x\to0$ , where

$\begin{equation*} \langle x\rangle_n = \begin{cases} x(x-1)\dotsm(x-n+1), & n\ge1\\ 1, & n = 0 \end{cases} \end{equation*}$

denotes the falling factorial of $x\in\mathbb{R}$ .

Letting $u(x) = 1-ax-\sqrt{(1-ax)^2-4bx^2}\,$ and $v(x) = x^2$ in Lemma 1 gives

$\begin{gather*} \frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n} = \frac1{2b}\frac{(-1)^n}{x^{2(n+1)}} \begin{vmatrix} u(x) & \binom00x^2 & 0 &\dotsm & 0 & 0 & 0\\ u'(x) & 2\binom10x & \binom11x^2 &\dotsm & 0 & 0 & 0\\ u''(x) & 2\binom20 & 2\binom21x &\dotsm & 0 & 0 & 0\\ u^{(3)}(x) & 0 & 2\binom31 &\dotsm & 0 & 0 & 0\\ u^{(4)}(x) & 0 & 0 &\dotsm & 0 & 0 & 0\\ \dotsm & \dotsm & \dotsm &\ddots & \dotsm & \dotsm & \dotsm\\ u^{(n-2)}(x) & 0 & 0 &\dotsm & 2\binom{n-2}{n-3}x & \binom{n-2}{n-2}x^2 & 0\\ u^{(n-1)}(x) & 0 & 0 &\dotsm & 2\binom{n-1}{n-3} & 2\binom{n-1}{n-2}x & \binom{n-1}{n-1}x^2\\ u^{(n)}(x) & 0 & 0 &\dotsm & 0 & 2\binom{n}{n-2} & 2\binom{n}{n-1}x \end{vmatrix}\\ = \frac1{2b}\frac{(-1)^n}{x^{2(n+1)}}\left[(-1)^{n}u^{(n)}(x) \begin{vmatrix} \binom00x^2 & 0 & 0 &\dotsm & 0 & 0 & 0\\ 2\binom10x & \binom11x^2 & 0 &\dotsm & 0 & 0 & 0\\ 2\binom20 & 2\binom21x & \binom22x^2 &\dotsm & 0 & 0 & 0\\ 0 & 2\binom31 & 2\binom32x &\dotsm & 0 & 0 & 0\\ 0 & 0 & 2\binom42 &\dotsm & 0 & 0 & 0\\ \dotsm & \dotsm & \dotsm &\ddots & \dotsm & \dotsm & \dotsm\\ 0 & 0 & 0 &\dotsm & 2\binom{n-2}{n-3}x & \binom{n-2}{n-2}x^2 & 0\\ 0 & 0 & 0 &\dotsm & 2\binom{n-1}{n-3} & 2\binom{n-1}{n-2}x & \binom{n-1}{n-1}x^2 \end{vmatrix}\right.\\ \left.+2\binom{n}{n-1}x \begin{vmatrix} u(x) & \binom00x^2 & 0 & 0 & 0 & \dotsm & 0 & 0\\ u'(x) & 2\binom10x & \binom11x^2 & 0 & 0 & \dotsm & 0 & 0\\ u''(x) & 2\binom20 & 2\binom21x & \binom22x^2 & 0 & \dotsm & 0 & 0\\ u^{(3)}(x) & 0 & 2\binom31 & 2\binom32x & \binom33x^2 & \dotsm & 0 & 0\\ u^{(4)}(x) & 0 & 0 & 2\binom42 & 2\binom43x & \dotsm & 0 & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \ddots & \dotsm & \dotsm\\ u^{(n-2)}(x) & 0 & 0 & 0 & 0 & \dotsm & 2\binom{n-2}{n-3}x & \binom{n-2}{n-2}x^2\\ u^{(n-1)}(x) & 0 & 0 & 0 & 0 & \dotsm & 2\binom{n-1}{n-3} & 2\binom{n-1}{n-2}x \end{vmatrix}\right.\\ \left. -2\binom{n}{n-2}\binom{n-1}{n-1}x^2 \begin{vmatrix} u(x) & \binom00x^2 & 0 & 0 &\dotsm & 0 & 0\\ u'(x) & 2\binom10x & \binom11x^2 & 0 &\dotsm & 0 & 0\\ u''(x) & 2\binom20 & 2\binom21x & \binom22x^2 &\dotsm & 0 & 0\\ u^{(3)}(x) & 0 & 2\binom31 & 2\binom32x &\dotsm & 0 & 0\\ u^{(4)}(x) & 0 & 0 & 2\binom42 &\dotsm & 0 & 0\\ \dotsm & \dotsm & \dotsm & \dotsm &\ddots & \dotsm & \dotsm\\ u^{(n-3)}(x) & 0 & 0 & 0 &\dotsm & 2\binom{n-3}{n-4}x & \binom{n-3}{n-3}x^2\\ u^{(n-2)}(x) & 0 & 0 & 0 &\dotsm & 2\binom{n-2}{n-4} & 2\binom{n-2}{n-3}x \end{vmatrix} \right]\\ = \frac1{2b}\frac{u^{(n)}(x)}{x^2} -\frac{2n}{x}\frac1{2b}\frac{(-1)^{n-1}}{x^{2n}} \begin{vmatrix} u(x) & \binom00x^2 & 0 &\dotsm & 0 & 0\\ u'(x) & 2\binom10x & \binom11x^2 &\dotsm & 0 & 0\\ u''(x) & 2\binom20 & 2\binom21x &\dotsm & 0 & 0\\ u^{(3)}(x) & 0 & 2\binom31 &\dotsm & 0 & 0\\ u^{(4)}(x) & 0 & 0 &\dotsm & 0 & 0\\ \dotsm & \dotsm & \dotsm &\ddots & \dotsm & \dotsm\\ u^{(n-2)}(x) & 0 & 0 &\dotsm & 2\binom{n-2}{n-3}x & \binom{n-2}{n-2}x^2\\ u^{(n-1)}(x) & 0 & 0 &\dotsm & 2\binom{n-1}{n-3} & 2\binom{n-1}{n-2}x \end{vmatrix}\\ -n(n-1)\frac1{x^2}\frac1{2b}\frac{(-1)^{n-2}}{x^{2(n-1)}} \begin{vmatrix} u(x) & \binom00x^2 & 0 & 0 &\dotsm & 0 & 0\\ u'(x) & 2\binom10x & \binom11x^2 & 0 &\dotsm & 0 & 0\\ u''(x) & 2\binom20 & 2\binom21x & \binom22x^2 &\dotsm & 0 & 0\\ u^{(3)}(x) & 0 & 2\binom31 & 2\binom32x &\dotsm & 0 & 0\\ u^{(4)}(x) & 0 & 0 & 2\binom42 &\dotsm & 0 & 0\\ \dotsm & \dotsm & \dotsm & \dotsm &\ddots & \dotsm & \dotsm\\ u^{(n-3)}(x) & 0 & 0 & 0 &\dotsm & 2\binom{n-3}{n-4}x & \binom{n-3}{n-3}x^2\\ u^{(n-2)}(x) & 0 & 0 & 0 &\dotsm & 2\binom{n-2}{n-4} & 2\binom{n-2}{n-3}x \end{vmatrix}\\ = \frac1{2b}\frac{u^{(n)}(x)}{x^2}-\frac{2n}x\frac{ \text{d}^{n-1}M_{a, b}(x)}{ \text{d} x^{n-1}} -\frac{n(n-1)}{x^2}\frac{ \text{d}^{n-2}M_{a, b}(x)}{ \text{d} x^{n-2}}\\ = \frac1{x^2}\biggl[\frac{u^{(n)}(x)}{2b}-2nx\frac{ \text{d}^{n-1}M_{a, b}(x)}{ \text{d} x^{n-1}} -n(n-1)\frac{ \text{d}^{n-2}M_{a, b}(x)}{ \text{d} x^{n-2}}\biggr]. \end{gather*}$

Therefore, by L'Hôspital's rule, we have

$\begin{align} \lim\limits_{x\to0}\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n} & = \lim\limits_{x\to0}\biggl\{\frac1{x^2}\biggl[\frac{u^{(n)}(x)}{2b} -2nx\frac{ \text{d}^{n-1}M_{a, b}(x)}{ \text{d} x^{n-1}} -n(n-1)\frac{ \text{d}^{n-2}M_{a, b}(x)}{ \text{d} x^{n-2}}\biggr]\biggr\}\\ & = \lim\limits_{x\to0}\biggl\{\frac1{2x}\biggl[\frac{u^{(n+1)}(x)}{2b} -2nx\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n} -n(n+1)\frac{ \text{d}^{n-1}M_{a, b}(x)}{ \text{d} x^{n-1}}\biggr]\biggr\}\\ & = \frac12\lim\limits_{x\to0}\biggl[\frac{u^{(n+2)}(x)}{2b} -2nx\frac{ \text{d}^{n+1}M_{a, b}(x)}{ \text{d} x^{n+1}} -n(n+3)\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n}\biggr]\\ & = \frac12\biggl[\lim\limits_{x\to0}\frac{u^{(n+2)}(x)}{2b} -n(n+3)\lim\limits_{x\to0}\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n}\biggr] \end{align}$

which is equivalent to

$\begin{equation*} \lim\limits_{x\to0}\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n} = \frac1{(n+1)(n+2)}\lim\limits_{x\to0}\frac{u^{(n+2)}(x)}{2b} = \frac1{2b(n+1)(n+2)}\lim\limits_{x\to0}u^{(n+2)}(x). \end{equation*}$

Considering

$\begin{equation*} \lim\limits_{x\to0}\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n} = n!M_n(a, b), \end{equation*}$

making use of (3.1), and simplifying lead to the explicit formula (1.10).

Letting $(a, b) = (1, 1)$ and $(a, b) = (3, 1)$ respectively in (1.10) and considering the three relations in (1.6) derive (1.11) and (1.12) immediately. The proof of Theorem 1 is complete.

Proof of Theorem 2. From (1.5), it is derived that

$\begin{equation*} \sqrt{(1-ax)^2-4bx^2}\, = 1-ax-2b\sum\limits_{k = 0}^\infty M_k(a, b)x^{k+2}. \end{equation*}$

This implies that

$\begin{equation} M_k(a, b) = -\frac1{2b}\frac1{(k+2)!}\lim\limits_{x\to0} \Bigl[\sqrt{(1-ax)^2-4bx^2}\, \Bigr]^{(k+2)}, \quad k\ge0. \end{equation}$

(3.2)

It is easy to see that

1. when $a^2-4b > 0$ and $x\le\min\Bigl\{\frac{1}{a+2 \sqrt{b}\, }, \frac{1}{a-2 \sqrt{b}\, }\Bigr\} = \frac{1}{a+2\sqrt{b}\, }$ , we have

$\begin{gather*} \Bigl[\sqrt{(1-ax)^2-4bx^2}\, \Bigr]^{(k+2)} = \Biggl[\sqrt{\bigl(a^2-4b\bigr)\biggl(x-\frac{1}{a+2 \sqrt{b}\, }\biggr) \biggl(x-\frac{1}{a-2 \sqrt{b}\, }\biggr)}\, \Biggr]^{(k+2)}\\ = \sqrt{a^2-4b}\, \Biggl(\sqrt{\frac{1}{a+2 \sqrt{b}\, }-x}\, \sqrt{\frac{1}{a-2 \sqrt{b}\, }-x}\, \Biggr)^{(k+2)}\\ = \sqrt{a^2-4b}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\Biggl(\sqrt{\frac{1}{a+2 \sqrt{b}\, }-x}\, \Biggr)^{(\ell)} \Biggl(\sqrt{\frac{1}{a-2 \sqrt{b}\, }-x}\, \Biggr)^{(k-\ell+2)}\\ = (-1)^k\sqrt{a^2-4b}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\biggl\langle\frac12\biggr\rangle_\ell \biggl(\frac{1}{a+2 \sqrt{b}\, }-x\biggr)^{1/2-\ell} \biggl\langle\frac12\biggr\rangle_{k-\ell+2}\biggl(\frac{1}{a-2 \sqrt{b}\, }-x\biggr)^{\ell-k-3/2}\\ \to(-1)^k\sqrt{a^2-4b}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\biggl\langle\frac12\biggr\rangle_\ell \biggl(\frac{1}{a+2 \sqrt{b}\, }\biggr)^{1/2-\ell} \biggl\langle\frac12\biggr\rangle_{k-\ell+2}\biggl(\frac{1}{a-2 \sqrt{b}\, }\biggr)^{\ell-k-3/2}\\ = (k+2)!\bigl(a-2\sqrt{b}\, \bigr)^{k+2} \sum\limits_{\ell = 0}^{k+2}\frac{(2\ell-3)!!}{(2\ell)!!}\frac{[2(k-\ell+2)-3]!!}{[2(k-\ell+2)]!!} \biggl(\frac{a+2 \sqrt{b}\, }{a-2 \sqrt{b}\, }\biggr)^{\ell} \end{gather*}$

as $x\to0$ ;

2. when $a^2-4b < 0$ and

$\begin{equation*} \frac{1}{a+2\sqrt{b}\, } = \max\Bigl\{\frac{1}{a+2 \sqrt{b}\, }, \frac{1}{a-2 \sqrt{b}\, }\Bigr\} \gt x \gt \min\Bigl\{\frac{1}{a+2 \sqrt{b}\, }, \frac{1}{a-2 \sqrt{b}\, }\Bigr\} = \frac{1}{a-2\sqrt{b}\, }, \end{equation*}$

we have

$\begin{gather*} \Bigl[\sqrt{(1-ax)^2-4bx^2}\, \Bigr]^{(k+2)} = \Biggl[\sqrt{\bigl(4b-a^2\bigr)\biggl(\frac{1}{a+2 \sqrt{b}\, }-x\biggr) \biggl(x-\frac{1}{a-2 \sqrt{b}\, }\biggr)}\, \Biggr]^{(k+2)}\\ = \sqrt{4b-a^2}\, \Biggl(\sqrt{\frac{1}{a+2 \sqrt{b}\, }-x}\, \sqrt{x-\frac{1}{a-2 \sqrt{b}\, }}\, \Biggr)^{(k+2)}\\ = \sqrt{4b-a^2}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\Biggl(\sqrt{\frac{1}{a+2 \sqrt{b}\, }-x}\, \Biggr)^{(\ell)} \Biggl(\sqrt{x-\frac{1}{a-2 \sqrt{b}\, }}\, \Biggr)^{(k-\ell+2)}\\ = \sqrt{4b-a^2}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell(-1)^\ell \biggl\langle\frac12\biggr\rangle_{\ell} \biggl(\frac{1}{a+2 \sqrt{b}\, }-x\biggr)^{1/2-\ell} \biggl\langle\frac12\biggr\rangle_{k-\ell+2} \biggl(x-\frac{1}{a-2\sqrt{b}\, }\biggr)^{1/2-(k-\ell+2)}\\ \to\sqrt{4b-a^2}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell(-1)^\ell \biggl\langle\frac12\biggr\rangle_{\ell} \biggl(\frac{1}{a+2 \sqrt{b}\, }\biggr)^{1/2-\ell} \biggl\langle\frac12\biggr\rangle_{k-\ell+2} \biggl(\frac{1}{2\sqrt{b}\, -a}\biggr)^{1/2-(k-\ell+2)}\\ = \bigl(2\sqrt{b}\, -a\bigr)^{k+1}\sqrt{4b-a^2}\, \sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\frac{(2\ell-3)!!}{2^\ell} \biggl(\frac{a+2 \sqrt{b}\, }{2\sqrt{b}\, -a}\biggr)^{\ell-1/2} (-1)^{k-\ell}\frac{[2(k-\ell+2)-3]!!}{2^{k-\ell+2}}\\ = \bigl(2\sqrt{b}\, -a\bigr)^{k+2}\sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\frac{(2\ell-3)!!}{2^\ell} \biggl(\frac{a+2 \sqrt{b}\, }{2\sqrt{b}\, -a}\biggr)^{\ell} (-1)^{k-\ell}\frac{[2(k-\ell+2)-3]!!}{2^{k-\ell+2}}\\ = \bigl(a-2\sqrt{b}\, \bigr)^{k+2}\sum\limits_{\ell = 0}^{k+2}\binom{k+2}\ell\frac{(2\ell-3)!!}{2^\ell} \frac{[2(k-\ell+2)-3]!!}{2^{k-\ell+2}} \biggl(\frac{a+2 \sqrt{b}\, }{a-2\sqrt{b}\, }\biggr)^{\ell} \\ = (k+2)!\bigl(a-2\sqrt{b}\, \bigr)^{k+2}\sum\limits_{\ell = 0}^{k+2}\frac{(2\ell-3)!!}{(2\ell)!!} \frac{[2(k-\ell+2)-3]!!}{[2(k-\ell+2)]!!} \biggl(\frac{a+2 \sqrt{b}\, }{a-2\sqrt{b}\, }\biggr)^{\ell} \end{gather*}$

as $x\to0$ .

By virtue of (3.2), we obtain the formula (1.13) readily.

Letting $(a, b) = (1, 1)$ and $(a, b) = (3, 1)$ respectively in (1.13) and making use of the first and third relations in (1.6) lead to (1.11) and (1.12) immediately. The proof of Theorem 2 is complete.

Proof of Theorem 3. From (1.5), it is derived that

$\begin{equation*} \sqrt{(1-ax)^2-4bx^2}\, = 1-ax-2b\sum\limits_{k = 0}^\infty M_k(a, b)x^{k+2}. \end{equation*}$

Squaring on both sides of the above equation gives

$\begin{gather*} (1-ax)^2-4bx^2\, = 1-2ax+\bigl(a^2-4b\bigr)x^2 = \Biggl[1-ax-2b\sum\limits_{k = 0}^\infty M_k(a, b)x^{k+2}\Biggr]^2\\ = 1+a^2x^2+4b^2\Biggl[\sum\limits_{k = 0}^\infty M_k(a, b)x^{k+2}\Biggr]^2 -2ax-4b\sum\limits_{k = 0}^\infty M_k(a, b)x^{k+2}+4abx\sum\limits_{k = 0}^\infty M_k(a, b)x^{k+2}\\ = 1-2ax+a^2x^2+4b^2x^4\sum\limits_{k = 0}^\infty\Biggl[\sum\limits_{\ell = 0}^kM_\ell(a, b)M_{k-\ell}\Biggr]x^k\\ -4b\sum\limits_{k = 2}^\infty M_{k-2}(a, b)x^k+4ab\sum\limits_{k = 3}^\infty M_{k-3}(a, b)x^k\\ = 1-2ax+a^2x^2-4b\sum\limits_{k = 2}^\infty M_{k-2}(a, b)x^k +4ab\sum\limits_{k = 3}^\infty M_{k-3}(a, b)x^k\\ +4b^2 \sum\limits_{k = 4}^\infty \Biggl[\sum\limits_{\ell = 0}^{k-4}M_\ell(a, b) M_{k-\ell-4}(a, b)\Biggr]x^k\\ = 1-2ax+a^2x^2-4b\bigl[M_0(a, b)x^2+M_1(a, b)x^3\bigr]+4abM_0(a, b)x^3\\ -4b\sum\limits_{k = 4}^\infty M_{k-2}(a, b)x^k+4ab\sum\limits_{k = 4}^\infty M_{k-3}(a, b)x^k +4b^2\sum\limits_{k = 4}^\infty\Biggl[\sum\limits_{\ell = 0}^{k-4}M_\ell(a, b)M_{k-\ell-4}(a, b)\Biggr]x^k\\ = 1-2ax+\bigl[a^2-4bM_0(a, b)\bigr]x^2+4b[aM_0(a, b)-M_1(a, b)]x^3\\ -4b\sum\limits_{k = 4}^\infty \Biggl[M_{k-2}(a, b)-aM_{k-3}(a, b)-b\sum\limits_{\ell = 0}^{k-4}M_\ell(a, b) M_{k-\ell-4}(a, b)\Biggr]x^k \end{gather*}$

which means that

$\begin{equation*} a^2-4b = a^2-4bM_0(a, b), \quad 4b[aM_0(a, b)-M_1(a, b)] = 0, \end{equation*}$

and

$\begin{equation*} M_{k-2}(a, b)-aM_{k-3}(a, b)-b\sum\limits_{\ell = 0}^{k-4}M_\ell(a, b)M_{k-\ell-4}(a, b) = 0, \quad k\ge4. \end{equation*}$

Consequently, the identities in (1.14) and the recursive formula (1.9) follow.

Taking $(a, b) = (1, 1)$ , $(a, b) = (2, 1)$ , and $(a, b) = (3, 1)$ respectively in (1.9) and considering the three relations in (1.6) lead to (1.15), (1.16), and (1.17) immediately. The proof of Theorem 3 is complete.

4. Two more remarks

Remark 2. From the proof of Theorem 1, we can conclude that

$\begin{equation*} x^2\frac{ \text{d}^nM_{a, b}(x)}{ \text{d} x^n}+2nx\frac{ \text{d}^{n-1}M_{a, b}(x)}{ \text{d} x^{n-1}} +n(n-1)\frac{ \text{d}^{n-2}M_{a, b}(x)}{ \text{d} x^{n-2}} = \frac{u^{(n)}(x)}{2b}, \quad n\ge2. \end{equation*}$

This implies that the generating function $M_{a, b}(x)$ expressed in (1.5) is an explicit solution of the linear ordinary differential equations

$\begin{equation*} x^2f^{(n)}(x)+2nxf^{(n-1)}(x)+n(n-1)f^{(n-2)}(x) = F_{n;a, b}(x) \end{equation*}$

for all $n\ge2$ , where, by (2.2) and (2.3) or (2.4),

$\begin{equation*} F_{n;a, b}(x) = \frac{n!\bigl(4b-a^2\bigr)^n}{2^{n+1}b} \frac{\sqrt{(1-ax)^2-4bx^2}\, }{[a+(4b-a^2)x]^n} \sum\limits_{\ell = 1}^n\frac{2^\ell(2\ell-3)!!}{\ell!(4b-a^2)^\ell} \binom{\ell}{n-\ell} \frac{\bigl[a+\bigl(4b-a^2\bigr)x\bigr]^{2\ell}}{[(1-ax)^2-4bx^2]^\ell}. \end{equation*}$

Remark 3. This paper is a continuation of the article ^[49] and a revised version of the preprint ^[28].

Conflict of interest

The authors declare that they have no conflict of interest in this paper.

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AIMS Mathematics

Several explicit and recursive formulas for generalized Motzkin numbers

Related Papers:

Abstract

1. Introduction

2. Lemmas

3. Proofs of Theorems 1 and 3

4. Two more remarks

Conflict of interest

References

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通讯作者: 陈斌, bchen63@163.com

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AIMS Mathematics

Several explicit and recursive formulas for generalized Motzkin numbers

Related Papers:

Abstract

1. Introduction

2. Lemmas

3. Proofs of Theorems 1 and 3

4. Two more remarks

Conflict of interest

References

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