A new algorithm by embedding structured data for low-rank tensor ring completion

Ruiping Wen; Tingyan Liu; Yalei Pei; Ruiping Wen; Tingyan Liu; Yalei Pei

doi:10.3934/math.2025297

AIMS Mathematics

2025, Volume 10, Issue 3: 6492-6511. doi: 10.3934/math.2025297

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Research article

A new algorithm by embedding structured data for low-rank tensor ring completion

1.
Shanxi Key Laboratory for Intelligent Optimization Computing and Blockchain Technology, Taiyuan Normal University, Jinzhong 030619, Shanxi, China
2.
School of Mathematics and Statistics, Taiyuan Normal University, Jinzhong 030619, Shanxi, China

Received: 28 September 2024 Revised: 05 March 2025 Accepted: 14 March 2025 Published: 24 March 2025
MSC : 15A69, 90C06, 90C47

In this paper, we put up with a new algorithm for tensor completion problems that include missing slices or row/column fibers, where embedding a structured tensor by a multi-way delay-embedding transform (MDT) makes the tensor to be completed have a special structure. The main idea is to employ a tensor completion algorithm based on the tensor ring rank, constructing latent tensor ring factors with a structure that approximates the original tensor starting from the tensor structure. It is also proved that the sequence generated by the new algorithm converges to the optimal solution. Finally, the feasibility of the proposed algorithm is verified by experiments. Compared with other completed algorithms based on tensor ring rank, the completed accuracy is improved, up to 30%.

Keywords:

Citation: Ruiping Wen, Tingyan Liu, Yalei Pei. A new algorithm by embedding structured data for low-rank tensor ring completion[J]. AIMS Mathematics, 2025, 10(3): 6492-6511. doi: 10.3934/math.2025297

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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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