DTSMA: Dominant Swarm with Adaptive T-distribution Mutation-based Slime Mould Algorithm

Shihong Yin; Qifang Luo; Yanlian Du; Yongquan Zhou; Shihong Yin; Qifang Luo; Yanlian Du; Yongquan Zhou

doi:10.3934/mbe.2022105

Mathematical Biosciences and Engineering

2022, Volume 19, Issue 3: 2240-2285. doi: 10.3934/mbe.2022105

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Research article

DTSMA: Dominant Swarm with Adaptive T-distribution Mutation-based Slime Mould Algorithm

1.
College of Artificial Intelligence, Guangxi University for Nationalities, Nanning 530006, China
2.
Key Laboratory of Guangxi High Schools Complex System and Computational Intelligence, Nanning 530006, China
3.
Guangxi Key Laboratories of Hybrid Computation and IC Design Analysis, Nanning 530006, China
4.
College of Information and Communication Engineering, Hainan University, Haikou 570228, China
5.
State Key Laboratory of Marine Resources Utilization in South China Sea, Hainan University, Haikou 570228, China

Received: 27 October 2021 Revised: 02 December 2021 Accepted: 16 December 2021 Published: 04 January 2022

The slime mould algorithm (SMA) is a metaheuristic algorithm recently proposed, which is inspired by the oscillations of slime mould. Similar to other algorithms, SMA also has some disadvantages such as insufficient balance between exploration and exploitation, and easy to fall into local optimum. This paper, an improved SMA based on dominant swarm with adaptive t-distribution mutation (DTSMA) is proposed. In DTSMA, the dominant swarm is used improved the SMA's convergence speed, and the adaptive t-distribution mutation balances is used enhanced the exploration and exploitation ability. In addition, a new exploitation mechanism is hybridized to increase the diversity of populations. The performances of DTSMA are verified on CEC2019 functions and eight engineering design problems. The results show that for the CEC2019 functions, the DTSMA performances are best; for the engineering problems, DTSMA obtains better results than SMA and many algorithms in the literature when the constraints are satisfied. Furthermore, DTSMA is used to solve the inverse kinematics problem for a 7-DOF robot manipulator. The overall results show that DTSMA has a strong optimization ability. Therefore, the DTSMA is a promising metaheuristic optimization for global optimization problems.

Keywords:

Citation: Shihong Yin, Qifang Luo, Yanlian Du, Yongquan Zhou. DTSMA: Dominant Swarm with Adaptive T-distribution Mutation-based Slime Mould Algorithm[J]. Mathematical Biosciences and Engineering, 2022, 19(3): 2240-2285. doi: 10.3934/mbe.2022105

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Abstract

1. Introduction

L. Carlitz (^[1]) introduced analogues of Bernoulli numbers for the rational function (finite) field $K = \mathbb F_r(T)$ , which are called Bernoulli-Carlitz numbers now. Bernoulli-Carlitz numbers have been studied since then (e.g., see ^[2,3,4,5,6]). According to the notations by Goss ^[7], Bernoulli-Carlitz numbers $BC_n$ are defined by

$\begin{equation} \frac{x}{e_C(x)} = \sum\limits_{n = 0}^\infty\frac{BC_n}{\Pi(n)}x^n\,. \end{equation}$

(1.1)

Here, $e_C(x)$ is the Carlitz exponential defined by

$\begin{equation} e_C(x) = \sum\limits_{i = 0}^\infty\frac{x^{r^i}}{D_i}\,, \end{equation}$

(1.2)

where $D_i = [i][i-1]^r\cdots [1]^{r^{i-1}}$ ( $i\ge 1$ ) with $D_0 = 1$ , and $[i] = T^{r^i}-T$ . The Carlitz factorial $\Pi(i)$ is defined by

$\begin{equation} \Pi(i) = \prod\limits_{j = 0}^m D_j^{c_j} \end{equation}$

(1.3)

for a non-negative integer $i$ with $r$ -ary expansion:

$\begin{equation} i = \sum\limits_{j = 0}^m c_j r^j\quad (0\le c_j \lt r)\,. \end{equation}$

(1.4)

As analogues of the classical Cauchy numbers $c_n$ , Cauchy-Carlitz numbers $CC_n$ (^[8]) are introduced as

$\begin{equation} \frac{x}{\log_C(x)} = \sum\limits_{n = 0}^\infty\frac{CC_n}{\Pi(n)}x^n\,. \end{equation}$

(1.5)

Here, $\log_C(x)$ is the Carlitz logarithm defined by

$\begin{equation} \log_C(x) = \sum\limits_{i = 0}^\infty(-1)^i\frac{x^{r^i}}{L_i}\,, \end{equation}$

(1.6)

where $L_i = [i][i-1]\cdots [1]$ ( $i\ge 1$ ) with $L_0 = 1$ .

In ^[8], Bernoulli-Carlitz numbers and Cauchy-Carlitz numbers are expressed explicitly by using the Stirling-Carlitz numbers of the second kind and of the first kind, respectively. These properties are the extensions that Bernoulli numbers and Cauchy numbers are expressed explicitly by using the Stirling numbers of the second kind and of the first kind, respectively.

On the other hand, for $N\ge 1$ , hypergeometric Bernoulli numbers $B_{N, n}$ (^[9,10,11,12]) are defined by the generating function

$\begin{equation} \frac{1}{{}_1 F_1(1;N+1;x)} = \frac{x^N/N!}{e^x-\sum_{n = 0}^{N-1}x^n/n!} = \sum\limits_{n = 0}^\infty B_{N,n}\frac{x^n}{n!}\,, \end{equation}$

(1.7)

where

${}_1 F_1(a;b;z) = \sum\limits_{n = 0}^\infty\frac{(a)^{(n)}}{(b)^{(n)}}\frac{z^n}{n!}$

is the confluent hypergeometric function with $(x)^{(n)} = x(x+1)\cdots(x+n-1)$ ( $n\ge 1$ ) and $(x)^{(0)} = 1$ . When $N = 1$ , $B_n = B_{1, n}$ are classical Bernoulli numbers defined by

$\frac{x}{e^x-1} = \sum\limits_{n = 0}^\infty B_n\frac{x^n}{n!}\,.$

In addition, hypergeometric Cauchy numbers $c_{N, n}$ (see ^[13]) are defined by

$\begin{equation} \frac{1}{{}_2 F_1(1,N;N+1;-x)} = \frac{(-1)^{N-1}x^N/N}{\log(1+t)-\sum_{n = 1}^{N-1}(-1)^{n-1}x^n/n} = \sum\limits_{n = 0}^\infty c_{N,n}\frac{x^n}{n!}\,, \end{equation}$

(1.8)

where

${}_2 F_1(a,b;c;z) = \sum\limits_{n = 0}^\infty\frac{(a)^{(n)}(b)^{(n)}}{(c)^{(n)}}\frac{z^n}{n!}$

is the Gauss hypergeometric function. When $N = 1$ , $c_n = c_{1, n}$ are classical Cauchy numbers defined by

$\frac{x}{\log(1+x)} = \sum\limits_{n = 0}^\infty c_n\frac{x^n}{n!}\,.$

In ^[14], for $N\ge 0$ , the truncated Bernoulli-Carlitz numbers $BC_{N, n}$ and the truncated Cauchy-Carlitz numbers $CC_{N, n}$ are defined by

$\begin{equation} \frac{x^{r^N}/D_N}{e_C(x)-\sum_{i = 0}^{N-1}x^{r^i}/D_i} = \sum\limits_{n = 0}^\infty\frac{BC_{N,n}}{\Pi(n)}x^n \end{equation}$

(1.9)

and

$\begin{equation} \frac{(-1)^N x^{r^N}/L_N}{\log_C(x)-\sum_{i = 0}^{N-1}(-1)^i x^{r^i}/L_i} = \sum\limits_{n = 0}^\infty\frac{CC_{N,n}}{\Pi(n)}x^n\,, \end{equation}$

(1.10)

respectively. When $N = 0$ , $BC_n = BC_{0, n}$ and $CC_n = CC_{0, n}$ are the original Bernoulli-Carlitz numbers and Cauchy-Carlitz numbers, respectively. These numbers $BC_{N, n}$ and $CC_{N, n}$ in (1.9) and (1.10) in function fields are analogues of hypergeometric Bernoulli numbers in (1.7) and hypergeometric Cauchy numbers in (1.8) in complex numbers, respectively. In ^[15], the truncated Euler polynomials are introduced and studied in complex numbers.

It is known that any real number $\alpha$ can be expressed uniquely as the simple continued fraction expansion:

$\begin{equation} \alpha = a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\ddots}}}\,, \end{equation}$

(1.11)

where $a_0$ is an integer and $a_1, a_2, \dots$ are positive integers. Though the expression is not unique, there exist general continued fraction expansions for real or complex numbers, and in general, analytic functions $f(x)$ :

$\begin{equation} f(x) = a_0(x)+\cfrac{b_1(x)}{a_1(x)+\cfrac{b_2(x)}{a_2(x)+\cfrac{b_3(x)}{a_3(x)+{\atop\ddots}}}}\,, \end{equation}$

(1.12)

where $a_0(x), a_1(x), \dots$ and $b_1(x), b_2(x), \dots$ are polynomials in $x$ . In ^[16,17] several continued fraction expansions for non-exponential Bernoulli numbers are given. For example,

$\begin{equation} \sum\limits_{n = 1}^\infty B_{2 n}(4 x)^n = \cfrac{x}{1+\dfrac{1}{2}+\cfrac{x}{\dfrac{1}{2}+\dfrac{1}{3}+\cfrac{x}{\dfrac{1}{3}+\dfrac{1}{4}+\cfrac{x}{\ddots}}}}\,. \end{equation}$

(1.13)

More general continued fractions expansions for analytic functions are recorded, for example, in ^[18]. In this paper, we shall give expressions for truncated Bernoulli-Carlitz numbers and truncated Cauchy-Carlitz numbers.

In ^[19], the hypergeometric Bernoulli numbers $B_{N, n}$ ( $N\ge 1$ , $n\ge 1$ ) can be expressed as

$B_{N,n} = (-1)^n n!\left| \begin{array}{ccccc} \frac{N!}{(N+1)!}&1&0&&\\ \frac{N!}{(N+2)!}&\frac{N!}{(N+1)!}&&&\\ \vdots&\vdots&\ddots&1&0\\ \frac{N!}{(N+n-1)!}&\frac{N!}{(N+n-2)!}&\cdots&\frac{N!}{(N+1)!}&1\\ \frac{N!}{(N+n)!}&\frac{N!}{(N+n-1)!}&\cdots&\frac{N!}{(N+2)!}&\frac{N!}{(N+1)!} \end{array} \right|\,.$

When $N = 1$ , we have a determinant expression of Bernoulli numbers ([,p.53]). In addition, relations between $B_{N, n}$ and $B_{N-1, n}$ are shown in ^[19].

In ^[21,22], the hypergeometric Cauchy numbers $c_{N, n}$ ( $N\ge 1$ , $n\ge 1$ ) can be expressed as

$c_{N,n} = n!\left| \begin{array}{ccccc} \frac{N}{N+1}&1&0&&\\ \frac{N}{N+2}&\frac{N}{N+1}&&&\\ \vdots&\vdots&\ddots&1&0\\ \frac{N}{N+n-1}&\frac{N}{N+n-2}&\cdots&\frac{N}{N+1}&1\\ \frac{N}{N+n}&\frac{N}{N+n-1}&\cdots&\frac{N}{N+2}&\frac{N}{N+1} \end{array} \right|\,.$

When $N = 1$ , we have a determinant expression of Cauchy numbers ([20,p.50]).

Recently, in (^[23]) the truncated Euler-Carlitz numbers $EC_{N, n}$ ( $N\ge 0$ ), introduced as

$\frac{x^{q^{2 N}}/D_{2 N}}{{\rm Cosh}_C(x)-\sum_{i = 0}^{N-1}x^{q^{2 i}}/D_{2 i}} = \sum\limits_{n = 0}^\infty\frac{EC_{N,n}}{\Pi(n)}x^n\,,$

are shown to have some determinant expressions. When $N = 0$ , $EC_n = EC_{0, n}$ are the Euler-Carlitz numbers, denoted by

$\frac{x}{{\rm Cosh}_C(x)} = \sum\limits_{n = 0}^\infty\frac{EC_{n}}{\Pi(n)}x^n\,,$

where

${\rm Cosh}_C(x) = \sum\limits_{i = 0}^\infty\frac{x^{q^{2 i}}}{D_{2 i}}$

is the Carlitz hyperbolic cosine. This reminds us that the hypergeometric Euler numbers $E_{N, n}$ (^[24]), defined by

$\frac{t^{2 N}/(2 N)!}{\cosh t-\sum_{n = 0}^{N-1}t^{2 n}/(2 n)!} = \sum\limits_{n = 0}^\infty E_{N,n}\frac{x^n}{n!}\,,$

have a determinant expression [,Theorem 2.3] for $N\ge 0$ and $n\ge 1$ ,

$E_{N, 2 n} = (-1)^n(2 n)! \left| \begin{array}{cccc} \frac{(2 N)!}{(2 N+2)!}&1&0&\\ \frac{(2 N)!}{(2 N+4)!}&\ddots&\ddots&0\\ \vdots&&\ddots&1\\ \frac{(2 N)!}{(2 N+2 n)!}&\cdots&\frac{(2 N)!}{(2 N+4)!}&\frac{(2 N)!}{(2 N+2)!} \end{array} \right|\,.$

When $N = 0$ , we have a determinant expression of Euler numbers (cf. [20,p.52]). More general cases are studied in ^[26].

In this paper, we also give similar determinant expressions of truncated Bernoulli-Carlitz numbers and truncated Cauchy-Carlitz numbers as natural extensions of those of hypergeometric numbers.

2. Continued fraction expansions of truncated Bernoulli-Carlitz and Cauchy-Carlitz numbers

Let the $n$ -th convergent of the continued fraction expansion of (1.12) be

$\begin{equation} \frac{P_n(x)}{Q_n(x)} = a_0(x)+\cfrac{b_1(x)}{a_1(x)+\cfrac{b_2(x)}{a_2(x)+{\atop\ddots{\atop+\cfrac{b_n(x)}{a_n(x)}}}}}\,. \end{equation}$

(2.1)

There exist the fundamental recurrence formulas:

$\begin{align} P_n(x)& = a_n(x)P_{n-1}(x)+b_n(x)P_{n-2}(x)\; (n\ge 1),\\ Q_n(x)& = a_n(x)Q_{n-1}(x)+b_n(x)Q_{n-2}(x)\; (n\ge 1), \end{align}$

(2.2)

with $P_{-1}(x) = 1$ , $Q_{-1}(x) = 0$ , $P_0(x) = a_0(x)$ and $Q_0(x) = 1$ .

From the definition in (1.9), truncated Bernoulli-Carlitz numbers satisfy the relation

$\left(D_N\sum\limits_{i = 0}^\infty\frac{x^{r^{N+i}-r^N}}{D_{N+i}}\right)\left(\sum\limits_{n = 0}^\infty\frac{BC_{N,n}}{\Pi(n)}x^n\right) = 1\,.$

Thus,

$P_m(x) = \frac{D_{N+m}}{D_N},\quad Q_m(x) = D_{N+m}\sum\limits_{i = 0}^m\frac{x^{r^{N+i}-r^N}}{D_{N+i}}$

yield that

$Q_m(x)\sum\limits_{n = 0}^\infty\frac{BC_{N,n}}{\Pi(n)}x^n\sim P_m(x)\quad(m\to\infty)\,.$

Notice that the $n$ -th convergent $p_n/q_n$ of the simple continued fraction (1.11) of a real number $\alpha$ yields the approximation property

$|q_n\alpha-p_n| \lt \frac{1}{q_{n+1}}\,.$

Now,

$\frac{P_0(x)}{Q_0(x)} = 1 = \frac{1}{1},\quad \frac{P_1(x)}{Q_1(x)} = 1-\frac{x^{r^{N+1}-r^N}}{D_{N+1}/D_N+x^{r^{N+1}-r^N}}$

and $P_n(x)$ and $Q_n(x)$ ( $n\ge 2$ ) satisfy the recurrence relations

$\begin{align*} P_n(x)& = \left(\frac{D_{N+n}}{D_{N+n-1}}+x^{r^{N+n}-r^{N+n-1}}\right)P_{n-1}(x)-\frac{D_{N+n-1}}{D_{N+n-2}}x^{r^{N+n}-r^{N+n-1}}P_{n-2}(x)\\ Q_n(x)& = \left(\frac{D_{N+n}}{D_{N+n-1}}+x^{r^{N+n}-r^{N+n-1}}\right)Q_{n-1}(x)-\frac{D_{N+n-1}}{D_{N+n-2}}x^{r^{N+n}-r^{N+n-1}}Q_{n-2}(x) \end{align*}$

(They are proved by induction). Since by (2.2) for $n\ge 2$

$a_n(x) = \frac{D_{N+n}}{D_{N+n-1}}+x^{r^{N+n}-r^{N+n-1}}\quad\hbox{and}\quad b_n(x) = -\frac{D_{N+n-1}}{D_{N+n-2}}x^{r^{N+n}-r^{N+n-1}}\,,$

we have the following continued fraction expansion.

Theorem 1.

$\sum\limits_{n = 0}^\infty\frac{BC_{N,n}}{\Pi(n)}x^n = 1-\cfrac{x^{r^{N+1}-r^N}}{\frac{D_{N+1}}{D_N}+x^{r^{N+1}-r^N}-\cfrac{\frac{D_{N+1}}{D_N}x^{r^{N+2}-r^{N+1}}}{\frac{D_{N+2}}{D_{N+1}}+x^{r^{N+2}-r^{N+1}}-\cfrac{\frac{D_{N+2}}{D_{N+1}}x^{r^{N+3}-r^{N+2}}}{\frac{D_{N+3}}{D_{N+2}}+x^{r^{N+3}-r^{N+2}}-{\atop\ddots}}}}\,.$

Put $N = 0$ in Theorem 1 to illustrate a simpler case. Then, we have a continued fraction expansion concerning the original Bernoulli-Carlitz numbers.

Corollary 1.

$\sum\limits_{n = 0}^\infty\frac{BC_{n}}{\Pi(n)}x^n = 1-\cfrac{x^{r-1}}{D_{1}+x^{r-1}-\cfrac{D_{1}x^{r^{2}-r}}{\frac{D_{2}}{D_{1}}+x^{r^{2}-r}-\cfrac{\frac{D_{2}}{D_{1}}x^{r^{3}-r^{2}}}{\frac{D_{3}}{D_{2}}+x^{r^{3}-r^{2}}-{\atop\ddots}}}}\,.$

From the definition in (1.10), truncated Cauchy-Carlitz numbers satisfy the relation

$\left(L_N\sum\limits_{i = 0}^\infty\frac{(-1)^i x^{r^{N+i}-r^N}}{L_{N+i}}\right)\left(\sum\limits_{n = 0}^\infty\frac{CC_{N,n}}{\Pi(n)}x^n\right) = 1\,.$

Thus,

$P_m(x) = \frac{L_{N+m}}{L_N},\quad Q_m(x) = L_{N+m}\sum\limits_{i = 0}^m\frac{(-1)^i x^{r^{N+i}-r^N}}{L_{N+i}}$

yield that

$Q_m(x)\sum\limits_{n = 0}^\infty\frac{CC_{N,n}}{\Pi(n)}x^n\sim P_m(x)\quad(m\to\infty)\,.$

Now,

$\frac{P_0(x)}{Q_0(x)} = 1 = \frac{1}{1},\quad \frac{P_1(x)}{Q_1(x)} = 1+\frac{x^{r^{N+1}-r^N}}{L_{N+1}/L_N-x^{r^{N+1}-r^N}}$

and $P_n(x)$ and $Q_n(x)$ ( $n\ge 2$ ) satisfy the recurrence relations

$\begin{align*} P_n(x)& = \left(\frac{L_{N+n}}{L_{N+n-1}}-x^{r^{N+n}-r^{N+n-1}}\right)P_{n-1}(x)+\frac{L_{N+n-1}}{L_{N+n-2}}x^{r^{N+n}-r^{N+n-1}}P_{n-2}(x)\\ Q_n(x)& = \left(\frac{L_{N+n}}{L_{N+n-1}}-x^{r^{N+n}-r^{N+n-1}}\right)Q_{n-1}(x)+\frac{L_{N+n-1}}{L_{N+n-2}}x^{r^{N+n}-r^{N+n-1}}Q_{n-2}(x)\,. \end{align*}$

Since by (2.2) for $n\ge 2$

$a_n(x) = \frac{L_{N+n}}{L_{N+n-1}}-x^{r^{N+n}+r^{N+n-1}}\quad\hbox{and}\quad b_n(x) = \frac{L_{N+n-1}}{L_{N+n-2}}x^{r^{N+n}-r^{N+n-1}}\,,$

we have the following continued fraction expansion.

Theorem 2.

$\sum\limits_{n = 0}^\infty\frac{CC_{N,n}}{\Pi(n)}x^n = 1+\cfrac{x^{r^{N+1}-r^N}}{\frac{L_{N+1}}{L_N}-x^{r^{N+1}-r^N}+\cfrac{\frac{L_{N+1}}{L_N}x^{r^{N+2}-r^{N+1}}}{\frac{L_{N+2}}{L_{N+1}}-x^{r^{N+2}-r^{N+1}}+\cfrac{\frac{L_{N+2}}{L_{N+1}}x^{r^{N+3}-r^{N+2}}}{\frac{L_{N+3}}{L_{N+2}}-x^{r^{N+3}-r^{N+2}}+{\atop\ddots}}}}\,.$

3. A determinant expression of truncated Cauchy-Carlitz numbers

In ^[14], some expressions of truncated Cauchy-Carlitz numbers have been shown. One of them is for integers $N\ge 0$ and $n\ge 1$ ,

$\begin{equation} CC_{N,n} = \Pi(n)\sum\limits_{k = 1}^{n}(-L_N)^k\sum\limits_{i_1,\dots,i_k\ge 1\atop r^{N+i_1}+\cdots+r^{N+i_k} = n+k r^N}\frac{(-1)^{i_1+\cdots+i_k}}{L_{N+i_1}\cdots L_{N+i_k}} \end{equation}$

(3.1)

[14,Theorem 2].

Now, we give a determinant expression of truncated Cauchy-Carlitz numbers.

Theorem 3. For integers $N\ge 0$ and $n\ge 1$ ,

$CC_{N,n} = \Pi(n)\left| \begin{array}{ccccc} -a_1&1&0&&\\ a_2&-a_1&\ddots&&\\ \vdots&&\ddots&\ddots&0\\ \vdots&&&-a_1&1\\ (-1)^n a_n&\cdots&\cdots&a_2&-a_1 \end{array} \right|\,,$

where

$a_l = \frac{(-1)^{i}L_N\delta_l^\ast}{L_{N+i}}\quad(l\ge 1)$

with

$\begin{equation} \delta_l^\ast = \begin{cases} 1&\mathit{\text{if $l = r^{N+i}-r^N\;(i = 0,1,\dots)$}};\\ 0&\mathit{\text{otherwise}}. \end{cases} \end{equation}$

(3.2)

We need the following Lemma in ^[27] in order to prove Theorem 3.

Lemma 1. Let $\{\alpha_n\}_{n\ge 0}$ be a sequence with $\alpha_0 = 1$ , and $R(j)$ be a function independent of $n$ . Then

$\begin{equation} \alpha_n = \left| \begin{array}{ccccc} R(1)&1&0&&\\ R(2)&R(1)&&&\\ \vdots&\vdots&\ddots&1&0\\ R(n-1)&R(n-2)&\cdots&R(1)&1\\ R(n)&R(n-1)&\cdots&R(2)&R(1) \end{array} \right|\,. \end{equation}$

(3.3)

if and only if

$\begin{equation} \alpha_n = \sum\limits_{j = 1}^n(-1)^{j-1}R(j)\alpha_{n-j}\quad(n\ge 1) \end{equation}$

(3.4)

with $\alpha_0 = 1$ .

Proof of Theorem 3. By the definition (1.10) with (1.6), we have

$\begin{align*} 1& = \left(\sum\limits_{i = 0}^\infty(-1)^i\frac{L_N}{L_{N+i}}\right)x^{r^{N+i}-r^N}\left(\sum\limits_{m = 0}^\infty\frac{CC_m}{\Pi(m)}x^m\right)\\ & = \left(\sum\limits_{l = 0}^\infty a_l x^l\right)\left(\sum\limits_{m = 0}^\infty\frac{CC_m}{\Pi(m)}x^m\right)\\ & = \sum\limits_{n = 0}^\infty\sum\limits_{l = 0}^\infty a_l\frac{CC_{n-l}}{\Pi(n-l)}x^n\,. \end{align*}$

Thus, for $n\ge 1$ , we get

$\sum\limits_{l = 0}^\infty a_l\frac{CC_{n-l}}{\Pi(n-l)} = 0\,.$

By Lemma 1, we have

$\begin{align*} \frac{CC_n}{\Pi(n)}& = -\sum\limits_{l = 1}^n a_l\frac{CC_{n-l}}{\Pi(n-l)}\\ & = \sum\limits_{l = 1}^n(-1)^{l-1}(-1)^l a_l\frac{CC_{n-l}}{\Pi(n-l)}\\ & = \left| \begin{array}{ccccc} -a_1&1&0&&\\ a_2&-a_1&\ddots&&\\ \vdots&&\ddots&\ddots&0\\ \vdots&&&-a_1&1\\ (-1)^n a_n&\cdots&\cdots&a_2&-a_1 \end{array} \right|\,. \end{align*}$

Examples. When $n = r^{N+1}-r^N$ ,

$\begin{align*} \frac{CC_{r^{N+1}-r^N}}{\Pi(r^{N+1}-r^N)}& = \left| \begin{array}{cccc} 0&1&&\\ \vdots&&\ddots&\\ 0&&&1\\ (-1)^{r^{N+1}-r^N}a_{r^{N+1}-r^N}&0&\cdots&0 \end{array}\right|\\ & = (-1)^{r^{N+1}-r^N+1}(-1)^{r^{N+1}-r^N}\frac{(-1)^{2 N+1}L_N}{L_{N+1}}\\ & = \frac{L_N}{L_{N+1}}\,. \end{align*}$

Let $n = r^{N+2}-r^N$ . For simplicity, put

$\begin{align*} \bar a& = (-1)^{r^{N+1}-r^N}(-1)^{2 N+1}\frac{L_N}{L_{N+1}}\,,\\ \hat a& = (-1)^{r^{N+2}-r^N}(-1)^{2 N+2}\frac{L_N}{L_{N+2}}\,. \end{align*}$

Then by expanding at the first column, we have

$\begin{align*} &\frac{CC_{r^{N+1}-r^N}}{\Pi(r^{N+1}-r^N)} = \left| \begin{array}{ccc} \underbrace{ \begin{array}{cccc} 0&1&0&\\ \vdots&&\ddots&\\ 0&&&\\ \bar a&&&\\ 0&\ddots&&\\ \vdots&&&\\ 0&&\ddots&\\ \hat a&0&\cdots&0 \end{array} }_{r^{N+2}-r^{N+1}} & \begin{array}{c} \\\\\\\\\\\\\\\\\bar a \end{array}& \underbrace{ \begin{array}{ccc} &&\\ &&\\ &&\\ &&\\ &&\\ &&\\ &\ddots&0\\ &&1\\ 0&\cdots&0 \end{array} }_{r^{N+1}-r^N-1} \end{array} \right|\\ & = (-1)^{r^{N+1}-r^N+1}\bar a\left| \begin{array}{cccc} \underbrace{ \begin{array}{ccc} 1&&\\ &\ddots&\\ &&1\\ \bar a&&\\ &\ddots&\\ &&\ddots\\ &&\\ && \end{array} }_{r^{N+1}-r^{N}-1} & \underbrace{ \begin{array}{cc} \\\\\\0&1\\&\ddots\\\\\\\\\ddots&\\&\bar a \end{array} }_2& \underbrace{ \begin{array}{ccc} &&\\ &&\\ &&\\ &&\\ &&\\ &&\\ &&\\ &\ddots&1\\ &\cdots&0 \end{array} }_{r^{N+1}-r^N-1} \end{array} \right|\\ &\quad +(-1)^{r^{N+2}-r^N+1}\hat a \left|\begin{array}{ccccccc} 1&0&&&&&\\ &\ddots&&&&&\\ &&&&&&\\ \bar a&&&&&&\\ &\ddots&&&&&\\ &&\ddots&&&\ddots&0\\ &&&\bar a&&&1 \end{array}\right|\,. \end{align*}$

The second term is equal to

$(-1)^{r^{N+2}-r^N+1}(-1)^{r^{N+2}-r^N}\frac{L_N}{L_{N+2}} = -\frac{L_N}{L_{N+2}}\,.$

The first term is

$\begin{align*} &(-1)^{r^{N+1}-r^N+1}\bar a\left| \begin{array}{ccc} \underbrace{ \begin{array}{ccc} 0&1&\\ &&\\ \bar a&&\\ &\ddots&\\ &&\ddots\\ && \end{array} }_{r^{N+2}-2 r^{N+1}+r^N} & \begin{array}{c} \\\\\\\\\\\\\bar a \end{array}& \underbrace{ \begin{array}{ccc} &&\\ &&\\ &&\\ &&\\ &&\\ &&1\\ &&0 \end{array} }_{r^{N+1}-r^N-1} \end{array} \right|\\ & = (-1)^{2(r^{N+1}-r^N+1)}\bar a^2 \left| \begin{array}{cccc} \underbrace{ \begin{array}{ccc} 1&&\\ &\ddots&\\ &&1\\ \bar a&&\\ &\ddots&\\ &&\ddots\\ &&\\ && \end{array} }_{r^{N+1}-r^{N}-1} & \underbrace{ \begin{array}{cc} \\\\\\0&1\\&\ddots\\\\\\\\\ddots&\\&\bar a \end{array} }_2& \underbrace{ \begin{array}{ccc} &&\\ &&\\ &&\\ &&\\ &&\\ &&\\ &&\\ &\ddots&1\\ &\cdots&0 \end{array} }_{r^{N+1}-r^N-1} \end{array} \right|\\ & = (-1)^{r(r^{N+1}-r^N+1)}\bar a^r \underbrace{ \left| \begin{array}{cccc} 0&1&&\\ \vdots&&\ddots&\\ 0&&&1\\ \bar a&0&\cdots&0 \end{array} \right| }_{r^{N+1}-r^N} \\ & = (-1)^{(r+1)(r^{N+1}-r^N+1)}\bar a^{r+1}\\ & = (-1)^{(r+1)(r^{N+1}-r^N+1)}(-1)^{(r^{N+1}-r^N)(r+1)}(-1)^{r+1}\frac{L_N^{r+1}}{L_{N+1}^{r+1}}\\ & = \frac{L_N^{r+1}}{L_{N+1}^{r+1}}\,. \end{align*}$

Therefore,

$\frac{CC_{r^{N+1}-r^N}}{\Pi(r^{N+1}-r^N)} = \frac{L_N^{r+1}}{L_{N+1}^{r+1}}-\frac{L_N}{L_{N+2}}\,.$

From this procedure, it is also clear that $CC_{N, n} = 0$ if $r^{N+1}-r^N\nmid n$ , since all the elements of one column (or row) become zero.

4. A determinant expression of truncated Bernoulli-Carlitz numbers

In ^[14], some expressions of truncated Bernoulli-Carlitz numbers have been shown. One of them is for integers $N\ge 0$ and $n\ge 1$ ,

$\begin{equation} BC_{N,n} = \Pi(n)\sum\limits_{k = 1}^{n}(-D_N)^k\sum\limits_{i_1,\dots,i_k\ge 1\atop r^{N+i_1}+\cdots+r^{N+i_k} = n+k r^N}\frac{1}{D_{N+i_1}\cdots D_{N+i_k}} \end{equation}$

(4.1)

[14,Theorem 1].

Now, we give a determinant expression of truncated Bernoulli-Carlitz numbers.

Theorem 4. For integers $N\ge 0$ and $n\ge 1$ ,

$BC_{N,n} = \Pi(n)\left| \begin{array}{ccccc} -d_1&1&0&&\\ d_2&-d_1&\ddots&&\\ \vdots&&\ddots&\ddots&0\\ \vdots&&&-d_1&1\\ (-1)^n d_n&\cdots&\cdots&d_2&-d_1 \end{array} \right|\,,$

where

$d_l = \frac{D_N\delta_l^\ast}{D_{N+i}}\quad(l\ge 1)$

with $\delta_l^\ast$ as in (3.2).

Proof. The proof is similar to that of Theorem 3, using (1.9) and (1.2).

Example. Let $n = 2(r^{N+1}-r^N)$ . For convenience, put

$\bar d = \frac{D_N}{D_{N+1}}\,.$

Then, we have

$\begin{align*} &\frac{BC_{N,2(r^{N+1}-r^N)}}{\Pi\bigl(2(r^{N+1}-r^N)\bigr)} = \left|\begin{array}{cc} \underbrace{ \begin{array}{ccc} 0&1&\\ \vdots&&\\ 0&&\\ \bar d&&\\ &\ddots&\\ &&\bar d \end{array} }_{r^{N+1}-r^N+1}& \underbrace{ \begin{array}{ccc} &&\\ &&\\ &&\\ &&\\ &&1\\ 0&\cdots&0 \end{array} }_{r^{N+1}-r^N-1} \end{array}\right|\\ & = (-1)^{r^{N+1}-r^N+1}\left| \begin{array}{ccc} \underbrace{ \begin{array}{ccc} 1&&\\ &\ddots&\\ &&1\\ \bar d&&\\ &\ddots&\\ &&\ddots\\ && \end{array} }_{r^{N+1}-r^N-1}& \begin{array}{c} \\\\\\0\\\\\\\\\bar d \end{array}& \underbrace{ \begin{array}{ccc} &&\\ &&\\ &&\\ 1&&\\ &\ddots&\\ &&1\\ &&0 \end{array} }_{r^{N+1}-r^N-1} \end{array} \right|\\ & = (-1)^{r^{N+1}-r^N+1}\bar d \underbrace{ \left| \begin{array}{cccc} 0&1&&\\ &&\ddots&\\ &&&1\\ \bar d&&&0 \end{array} \right| }_{r^{N+1}-r^N} \\ & = (-1)^{2(r^{N+1}-r^N+1)}\bar d^2 \left| \begin{array}{ccc} 1&&\\ &\ddots&\\ &&1 \end{array} \right|\\ & = \frac{D_N^2}{D_{N+1}^2}\,. \end{align*}$

It is also clear that $BC_{N, n} = 0$ if $r^{N+1}-r^N\nmid n$ .

5. Applications by the Trudi's formula

We shall use Trudi's formula to obtain different explicit expressions and inversion relations for the numbers $CC_{N, n}$ and $BC_{N, n}$ .

Lemma 2. For a positive integer $n$ , we have

$\left| \begin{array}{ccccc} a_1 & a_0 & 0 &\cdots & \\ a_2 & a_1 & \ddots & & \vdots \\ \vdots & \vdots & \ddots & \ddots & 0 \\ a_{n-1} & & \cdots &a_1 & a_0 \\ a_n & a_{n-1} & \cdots & a_2 & a_1 \end{array} \right| = \sum\limits_{t_1 + 2t_2 + \cdots +n t_n = n}\binom{t_1+\cdots + t_n}{t_1, \dots,t_n}(-a_0)^{n-t_1-\cdots - t_n}a_1^{t_1}a_2^{t_2}\cdots a_n^{t_n},$

where $\binom{t_1+\cdots + t_n}{t_1, \dots, t_n} = \frac{(t_1+\cdots + t_n)!}{t_1!\cdots t_n!}$ are the multinomial coefficients.

This relation is known as Trudi's formula [,Vol.3,p.214], ^[29] and the case $a_0 = 1$ of this formula is known as Brioschi's formula ^[30], [28,Vol.3,pp.208–209].

In addition, there exists the following inversion formula (see, e.g. ^[27]), which is based upon the relation

$\sum\limits_{k = 0}^n(-1)^{n-k}\alpha_k D(n-k) = 0\quad(n\ge 1)\,.$

Lemma 3. If $\{\alpha_n\}_{n\geq 0}$ is a sequence defined by $\alpha_0 = 1$ and

$\alpha_n = \begin{vmatrix} D(1) & 1 &0 & \\ D(2) & \ddots & \ddots &0 \\ \vdots & \ddots & \ddots & 1\\ D(n) & \cdots & D(2) & D(1) \\ \end{vmatrix}, \ \mathit{\text{then}} \ D(n) = \begin{vmatrix} \alpha_1 & 1 &0 & \\ \alpha_2 & \ddots & \ddots &0 \\ \vdots & \ddots & \ddots & 1\\ \alpha_n & \cdots & \alpha_2 & \alpha_1 \\ \end{vmatrix}\,.$

From Trudi's formula, it is possible to give the combinatorial expression

$\alpha_n = \sum\limits_{t_1+2t_2+\cdots +n t_n = n}\binom{t_1+\cdots+t_n}{t_1, \dots, t_n}(-1)^{n-t_1-\cdots - t_n}D(1)^{t_1}D(2)^{t_2}\cdots D(n)^{t_n}\,.$

By applying these lemmata to Theorem 3 and Theorem 4, we obtain an explicit expression for the truncated Cauchy-Carlitz numbers and the truncated Bernoulli-Carlitz numbers.

Theorem 5. For integers $N\ge 0$ and $n\ge 1$ , we have

$CC_{N,n} = \Pi(n)\sum\limits_{t_1+2 t_2+\cdots+n t_n = n}\binom{t_1+\cdots+t_n}{t_1,\dots,t_n} (-1)^{n-t_2-t_4-\cdots-t_{2\left\lfloor{n/2}\right\rfloor}}a_1^{t_1}\cdots a_n^{t_n}\,,$

where $a_n$ are given in Theorem 3.

Theorem 6. For integers $N\ge 0$ and $n\ge 1$ , we have

$BC_{N,n} = \Pi(n)\sum\limits_{t_1+2 t_2+\cdots+n t_n = n}\binom{t_1+\cdots+t_n}{t_1,\dots,t_n} (-1)^{n-t_2-t_4-\cdots-t_{2\left\lfloor{n/2}\right\rfloor}}d_1^{t_1}\cdots d_n^{t_n}\,,$

where $d_n$ are given in Theorem 4.

By applying the inversion relation in Lemma 3 to Theorem 3 and Theorem 4, we have the following.

Theorem 7. For integers $N\ge 0$ and $n\ge 1$ , we have

$a_n = (-1)^n\left|\begin{array}{ccccc} \frac{CC_{N,1}}{\Pi(1)}&1&0&&\\ \frac{CC_{N,2}}{\Pi(2)}&\frac{CC_{N,1}}{\Pi(1)}&&&\\ \vdots&\vdots&\ddots&1&0\\ \frac{CC_{N,n-1}}{\Pi(n-1)}&\frac{CC_{N,n-2}}{\Pi(n-2)}&\cdots&\frac{CC_{N,1}}{\Pi(1)}&1\\ \frac{CC_{N,n}}{\Pi(n)}&\frac{CC_{N,n-1}}{\Pi(n-1)}&\cdots&\frac{CC_{N,2}}{\Pi(2)}&\frac{CC_{N,1}}{\Pi(1)} \end{array}\right|\,,$

where $a_n$ is given in Theorem 3.

Theorem 8. For integers $N\ge 0$ and $n\ge 1$ , we have

$d_n = (-1)^n\left|\begin{array}{ccccc} \frac{BC_{N,1}}{\Pi(1)}&1&0&&\\ \frac{BC_{N,2}}{\Pi(2)}&\frac{BC_{N,1}}{\Pi(1)}&&&\\ \vdots&\vdots&\ddots&1&0\\ \frac{BC_{N,n-1}}{\Pi(n-1)}&\frac{BC_{N,n-2}}{\Pi(n-2)}&\cdots&\frac{BC_{N,1}}{\Pi(1)}&1\\ \frac{BC_{N,n}}{\Pi(n)}&\frac{BC_{N,n-1}}{\Pi(n-1)}&\cdots&\frac{BC_{N,2}}{\Pi(2)}&\frac{BC_{N,1}}{\Pi(1)} \end{array}\right|\,,$

where $d_n$ is given in Theorem 4.

Acknowledgments

We would like to thank the referees for their valuable comments.

Conflict of interest

The authors declare no conflict of interest.

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