Cluster synchronization of coupled complex-valued neural networks with leakage and time-varying delays in finite-time

N. Jayanthi; R. Santhakumari; Grienggrai Rajchakit; Nattakan Boonsatit; Anuwat Jirawattanapanit; N. Jayanthi; R. Santhakumari; Grienggrai Rajchakit; Nattakan Boonsatit; Anuwat Jirawattanapanit

doi:10.3934/math.2023104

AIMS Mathematics

2023, Volume 8, Issue 1: 2018-2043. doi: 10.3934/math.2023104

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Research article Special Issues

Cluster synchronization of coupled complex-valued neural networks with leakage and time-varying delays in finite-time

1.
Government Arts College, Coimbatore, India
2.
Sri Ramakrishna College of Arts and Science, Coimbatore, India
3.
Department of Mathematics, Faculty of Science, Maejo University, Chiang Mai 50290, Thailand
4.
Faculty of Science and Technology, Rajamangala University of Technology Suvarnabhumi, Thailand
5.
Department of Mathematics, Faculty of Science, Phuket Rajabhat University (PKRU), Thailand

Received: 31 August 2022 Revised: 11 October 2022 Accepted: 14 October 2022 Published: 26 October 2022
MSC : 93D05, 93D40, 34D06, 92B20, 93C43

In cluster synchronization (CS), the constituents (i.e., multiple agents) are grouped into a number of clusters in accordance with a function of nodes pertaining to a network structure. By designing an appropriate algorithm, the cluster can be manipulated to attain synchronization with respect to a certain value or an isolated node. Moreover, the synchronization values among various clusters vary. The main aim of this study is to investigate the asymptotic and CS problem of coupled delayed complex-valued neural network (CCVNN) models along with leakage delay in finite-time (FT). In this paper, we describe several sufficient conditions for asymptotic synchronization by utilizing the Lyapunov theory for differential systems and the Filippov regularization framework for the realization of finite-time synchronization of CCVNNs with leakage delay. We also propose sufficient conditions for CS of the system under scrutiny. A synchronization algorithm is developed to indicate the usefulness of the theoretical results in case studies.

Keywords:

Citation: N. Jayanthi, R. Santhakumari, Grienggrai Rajchakit, Nattakan Boonsatit, Anuwat Jirawattanapanit. Cluster synchronization of coupled complex-valued neural networks with leakage and time-varying delays in finite-time[J]. AIMS Mathematics, 2023, 8(1): 2018-2043. doi: 10.3934/math.2023104

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Abstract

1. Introduction

The key to solving the general quadratic congruence equation is to solve the equation of the form $x^2\equiv a\mod p$ , where $a$ and $p$ are integers, $p > 0$ and $p$ is not divisible by $a$ . For relatively large $p$ , it is impractical to use the Euler criterion to distinguish whether the integer $a$ with $(a, p) = 1$ is quadratic residue of modulo $p$ . In order to study this issue, Legendre has proposed a new tool-Legendre's symbol.

Let $p$ be an odd prime, the quadratic character modulo $p$ is called the Legendre's symbol, which is defined as follows:

$\left(\frac{a}{p}\right)= \begin{cases}1, & \text { if } a \text { is a quadratic residue modulo } p; \\ -1, & \text { if } a \text { is a quadratic non-residue modulo } p ;\\ 0, & \text { if } p \mid a.\end{cases}$

The Legendre's symbol makes it easy for us to calculate the level of quadratic residues. The basic properties of Legendre's symbol can be found in any book on elementary number theory, such as ^[1,2,3].

The properties of Legendre's symbol and quadratic residues play an important role in number theory. Many scholars have studied them and achieved some important results. For examples, see the ^{[4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21]}.

One of the most representative properties of the Legendre's symbol is the quadratic reciprocal law:

Let $p$ and $q$ be two distinct odd primes. Then, (see Theorem 9.8 in ^[1] or Theorems 4–6 in ^[3])

$\left(\frac{p}{q}\right)\cdot\left(\frac{q}{p}\right) = (-1)^{\frac{(p-1)(q-1)}{4}}.$

For any odd prime $p$ with $p\equiv1\; \rm mod\; 4$ there exist two non-zero integers $\alpha\left(p\right)$ and $\beta\left(p\right)$ such that

$\begin{align*} p = \alpha^2\left(p\right)+\beta^2\left(p\right). \end{align*}$

(1)

In fact, the integers $\alpha\left(p\right)$ and $\beta\left(p\right)$ in the (1) can be expressed in terms of Legendre's symbol modulo $p$ (see Theorems 4–11 in ^[3])

$\alpha\left(p\right) = \frac{1}{2}\sum\limits_{a = 1}^{p-1}\left(\frac{a^3+a}{p}\right)\; \; \; \rm and\; \; \; \it \beta\left(\it p\right) = \frac{\rm1}{\rm 2}\sum\limits_{a = \rm1}^{p-\rm1}\left(\frac {\it a^{\rm3}+\it ra}{\it p}\right),$

where $r$ is any integer, and $\left(r, p\right) = 1$ , $\left(\frac{r}{p}\right) = -1$ , $\left(\frac{\ast}{p}\right) = \chi_2$ denote the Legendre's symbol modulo $p$ .

Noting that Legendre's symbol is a special kind of character. For research on character, Han ^[7] studied the sum of a special character $\chi\left(ma+\bar{a}\right)$ , for any integer $m$ with $\left(m, p\right) = 1$ , then

$\left|\sum\limits_{a = 1}^{p-1}\chi\left(ma+\bar{a}\right)\right|^2 = 2p+\left(\frac{m}{p}\right)\sum\limits_{a = 1}^{p-1}\chi(a)\sum\limits_{b = 1}^{p-1}\left(\frac{b(b-1)(a^2b-1)}{p}\right),$

which is a special case of a general polynomial character sums $\sum_{a = N+1}^{N+M}\chi\left(f\left(a\right)\right)$ , where $M$ and $N$ are any positive integers, and $f\left(x\right)$ is a polynomial.

In ^[8], Du and Li introduced a special character sums $C\left(\chi, m, n, c;p\right)$ in the following form:

$C\left(\chi, m, n, c;p\right) = \sum\limits_{a = 0}^{p-1}\sum\limits_{b = 0}^{p-1}\chi\left(a^2+na-b^2-nb+c\right)\cdot e\left(\frac{mb^2-ma^2}{p}\right),$

and studied the asymptotic properties of it. They obtained

$\begin{equation*} \label{test1} \begin{split} \sum\limits_{c = 1}^{p-1}\left|C\left(\chi, m, n, c;p\right)\right|^{2k} = \left\{ \begin{array}{l} &p^{2k+1}+\frac{k^2-3k-2}{2}\cdot p^{2k}+O\left(p^{2k-1}\right), \; \; \; \; \rm if\; \chi\ \rm is\ \rm the\ \rm Legendre\ \rm symbol\ \rm {modulo} \ \it p \rm; \\ &p^{2k+1}+\frac{k^2-3k-2}{2}\cdot p^{2k}+O\left(p^{2k-1/2}\right), \; \; \rm if\; \chi\ \rm is\ \rm a\ \rm complex\ \rm character\ \rm {modulo} \ \it p.\\ \end{array} \right. \end{split} \end{equation*}$

Recently, Yuan and Zhang ^[12] researched the question about the estimation of the mean value of high-powers for a special character sum modulo a prime, let $p$ be an odd prime with $p\equiv1\; \rm mod\; 6$ , then for any integer $k\geq0$ , they have the identity

$S_k\left(p\right) = \frac{1}{3}\cdot\left[d^k+\left(\frac{-d+9b}{2}\right)^k+\left(\frac{-d-9b}{2}\right)^k\right],$

where

$S_k\left(p\right) = \frac{1}{p-1}\sum\limits_{r = 1}^{p-1}A^k\left(r\right),$

$A\left(r\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+r\bar{a}}{p}\right),$

and for any integer $r$ with $(r, p) = 1$ .

More relevant research on special character sums will not be repeated. Inspired by these papers, we have the question: If we replace the special character sums with Legendre's symbol, can we get good results on $p\equiv1\; \rm mod\; 4$ ?

We will convert $\beta\left(p\right)$ to another form based on the properties of complete residues

$\beta\left(p\right) = \frac{1}{2}\sum\limits_{a = 1}^{p-1}\left(\frac{a+n\bar{a}}{p}\right),$

where $\bar{a}$ is the inverse of $a$ modulo $p$ . That is, $\bar{a}$ satisfy the equation $x\cdot a\; \equiv1\mod p$ for any integer $a$ with $(a, p) = 1$ .

For any integer $k\geq0$ , $G\left(n\right)$ and $K_k(p)$ are defined as follows:

$G\left(n\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\; \; \; \rm {and} \; \; \; \it K_{\it k}\left(\it p\right) = \frac{\rm 1}{\it p-\rm 1}\sum\limits_{\it n\rm = 1}^{\it p-\rm 1}\it G^{\it k}\left(\it n\right).$

In this paper, we will use the analytic methods and properties of the classical Gauss sums and Dirichlet character sums to study the computational problem of $K_k\left(p\right)$ for any positive integer $k$ , and give a linear recurrence formulas for $K_k\left(p\right)$ . That is, we will prove the following result.

Theorem 1. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ , then we have

$K_k\left(p\right) = \left(4p+2\right)\cdot K_{k-2}\left(p\right)-8\left(2\alpha^2-p\right)\cdot K_{k-3}\left(p\right)+\left(16\alpha^4-16p\alpha^2+4p-1\right)\cdot K_{k-4}\left(p\right),$

for all integer $k\geq4$ with

$K_0\left(p\right) = 1, \; K_1\left(p\right) = 0, \; K_2\left(p\right) = 2p+1, \; K_3\left(p\right) = -3(4\alpha^2-2p),$

where

$\alpha = \alpha\left(p\right) = \sum\limits_{a = 1}^{\frac{p-1}{2}}\left(\frac{a+\bar{a}}{p}\right).$

Applying the properties of the linear recurrence sequence, we may immediately deduce the following corollaries.

Corollary 1. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then we have

$\frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\frac{1}{1+\sum _{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)} = \frac{16\alpha^2p-28\alpha^2-8p^2+14p} {16\alpha^4-16\alpha^2p+4p-1}.$

Corollary 2. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then we have

$\begin{align*} \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\sum\limits_{m = 0}^{p-1}\left(1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\right)\cdot e\left(\frac{nm^2}{p}\right) = -\sqrt{p}. \end{align*}$

Corollary 3. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then we have

$\frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\sum\limits_{m = 0}^{p-1}\left[1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\right]^2\cdot e\left(\frac{nm^2}{p}\right) = \left(4\alpha^2-2p\right)\cdot\sqrt{p}.$

Corollary 4. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 8$ . Then we have

$\begin{align*} \sum\limits_{n = 1}^{p-1}\left(1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right)\right)\cdot \sum\limits_{m = 0}^{p-1}e\left(\frac{nm^4}{p}\right) = \sqrt{p}\left(-1+B(1)\right)-p, \end{align*}$

where

$B(1) = \sum\limits_{m = 0}^{p-1}e\left(\frac{m^4}{p}\right).$

If we consider such a sequence $F_k\left(p\right)$ as follows: Let $p$ be a prime with $p\equiv1\; \rm mod\; 8$ , $\chi_4$ be any fourth-order character modulo $p$ . For any integer $k\geq0$ , we define the $F_k\left(p\right)$ as

$F_k\left(p\right) = \sum\limits_{n = 1}^{p-1}\frac{1}{G^{k}(n)},$

we have

$\begin{align*} F_{k}\left(p\right) = &\frac{1}{16\alpha^4-16\alpha^2p+4p-1}F_{k-4}\left(p\right)-\frac{(4p+2)}{16\alpha^4-16\alpha^2p+4p-1}F_{k-2}\left(p\right)\\ &+\frac{4(4\alpha^2-2p)}{16\alpha^4-16\alpha^2p+4p-1}F_{k-1}\left(p\right). \end{align*}$

2. Some lemmas

Lemma 1. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then for any fourth-order character $\chi_4 \; \rm{mod} \; \it p$ , we have the identity

$\tau^2(\chi_4)+\tau^2(\bar{\chi_4}) = 2\sqrt{p}\cdot\alpha,$

where

$\tau(\chi_4) = \sum\limits_{a = 1}^{p-1}\chi_4(a)e\left(\frac{a}{p}\right)$

denotes the classical Gauss sums, $e(y) = e^{2\pi iy}, i^2 = -1$ , and $\alpha$ is the same as in the Theorem 1.

Proof. See Lemma 2.2 in ^[9].

Lemma 2. Let $p$ be an odd prime. Then for any non-principal character $\psi$ modulo $p$ , we have the identity

$\tau(\psi^2) = \frac{\psi^2(2)}{\tau(\chi_2)}\cdot\tau(\psi)\cdot\tau(\psi\chi_2),$

where $\chi_2 = \left(\frac{\ast}{p}\right)$ denotes the Legendre's symbol modulo $p$ .

Proof. See Lemma 2 in ^[12].

Lemma 3. Let $p$ be a prime with $p\equiv 1\; \rm mod\; 4$ , then for any integer $n$ with $\left(n, p\right) = 1$ and fourth-order character $\chi_{4}\; \rm mod \; \it p$ , we have the identity

$\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = -1-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})).$

Proof. For any integer $a$ with $(a, p) = 1$ , we have the identity

$1+\chi_{4}(a)+\chi_2(a)+\bar{\chi_{4}}(a) = 4,$

if $a$ satisfies $a\equiv b^4\; \rm mod \; \it p$ for some integer $b$ with $(b, p) = 1$ and

$1+\chi_{4}(a)+\chi_2(a)+\bar{\chi_{4}}(a) = 0,$

otherwise. So from these and the properties of Gauss sums we have

$\begin{align*} \mathop\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = &\sum\limits_{a = 1}^{p-1}\left(\frac{a^2}{p}\right)\left(\frac{a^4+n}{p}\right)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}\chi_2\left(a^4\right)\chi_2\left(a^4+n\right)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}(1+\chi_{4}(a)+\chi_2(a)+\bar{\chi_{4}}(a))\cdot\chi_2(a)\cdot\chi_2(a+n)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}(1+\chi_{4}(na)+\chi_2(na)+\bar{\chi_{4}}(na))\cdot\chi_2(na)\cdot\chi_2(na+n)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}\chi_2(a)\chi_2(a+1)+\sum\limits_{a = 1}^{p-1}\chi_{4}(na)\chi_2(a)\chi_2(a+1) \end{align*}$

(2)

$\begin{align*} &+\sum\limits_{a = 1}^{p-1}\chi_2(na)\chi_2(a)\chi_2(a+1)+\sum\limits_{a = 1}^{p-1}\bar{\chi_{4}}(na)\chi_2(a)\chi_2(a+1)\nonumber\\ = &\sum\limits_{a = 1}^{p-1}\chi_2(1+\bar{a})+\sum\limits_{a = 1}^{p-1}\chi_{4}(na)\chi_2(a)\chi_2(a+1)\nonumber\\ &+\sum\limits_{a = 1}^{p-1}\chi_2(n)\chi_2(a+1)+\sum\limits_{a = 1}^{p-1}\bar{\chi_{4}}(na)\chi_2(a)\chi_2(a+1)\nonumber. \end{align*}$

Noting that for any non-principal character $\chi$ ,

$\sum\limits_{a = 1}^{p-1}\chi(a) = 0$

and

$\sum\limits_{a = 1}^{p-1}\chi(a)\chi(a+1) = \frac{1}{\tau(\bar{\chi})}\sum\limits_{b = 1}^{p-1}\sum\limits_{a = 1}^{p-1}\bar{\chi}(b)\chi(a)e\left(\frac{b(a+1)}{p}\right).$

Then we have

$\sum\limits_{a = 1}^{p-1}\chi_2(1+\bar{a}) = -1, \; \; \; \sum\limits_{a = 1}^{p-1}\chi_2(a+1) = -1,$

$\begin{align*} \sum\limits_{a = 1}^{p-1}\chi_{4}(a)\chi_2(a)\chi_2(a+1)& = \frac{1}{\tau(\chi_2)}\sum\limits_{b = 1}^{p-1}\sum\limits_{a = 1}^{p-1}\chi_2(b)\chi_{4}(a)\chi_2(a)e\left(\frac{b(a+1)}{p}\right)\nonumber\\ & = \frac{1}{\tau(\chi_2)}\sum\limits_{b = 1}^{p-1}\bar{\chi_{4}}(b)e\left(\frac{b}{p}\right)\sum\limits_{a = 1}^{p-1}\chi_{4}(ab)\chi_2(ab)e\left(\frac{ab}{p}\right) \end{align*}$

(3)

$\begin{align*} & = \frac{1}{\tau(\chi_2)}\cdot\tau(\bar{\chi_{4}})\cdot\tau(\chi_{4}\chi_2)\nonumber. \end{align*}$

For any non-principal character $\psi$ , from Lemma 2 we have

$\begin{align*} \tau(\psi^2) = \frac{\psi^2(2)}{\tau(\chi_2)}\cdot\tau(\psi)\cdot\tau(\psi\chi_2). \end{align*}$

(4)

Taking $\psi = \chi_{4}$ , note that

$\tau(\chi_2) = \sqrt{p}, \ \ \tau(\chi_{4})\cdot\tau(\bar{\chi_{4}}) = \chi_{4}(-1)\cdot p,$

from (3) and (4), we have

$\begin{align*} \sum\limits_{a = 1}^{p-1}\chi_{4}(a)\chi_2(a)\chi_2(a+1)& = \frac{\bar{\chi_{4}}^2(2)\cdot\tau(\chi_{4}^2)\cdot\tau(\chi_2)\cdot\tau(\bar{\chi_{4}})}{\tau(\chi_2)\cdot\tau(\chi_{4})}\nonumber\\ & = \frac{\chi_2(2)\cdot\tau(\chi_2)\cdot\tau^2(\bar{\chi_{4}})}{\tau(\chi_{4})\cdot\tau(\bar{\chi_{4}})}\nonumber\\ & = \frac{\chi_2(2)\cdot\sqrt{p}\cdot\tau^2(\bar{\chi_{4}})}{\chi_4(-1)\cdot p} \end{align*}$

(5)

$\begin{align*} & = \frac{\chi_2(2)\cdot\tau^2(\bar{\chi_{4}})}{\chi_4(-1)\cdot\sqrt{p}}.\nonumber \end{align*}$

Similarly, we also have

$\begin{align*} \sum\limits_{a = 1}^{p-1}\bar{\chi_{4}}(a)\chi_2(a)\chi_2(a+1) = \frac{\chi_2(2)\cdot\tau^2(\chi_{4})}{\chi_4(-1)\cdot\sqrt{p}}. \end{align*}$

(6)

Consider the quadratic character modulo $p$ , we have

$\begin{align*} \left(\frac{2}{p}\right) = \chi_2(2) = \begin{cases} \; \; 1, &\rm{ if\; \ p\equiv\pm1\; mod\; 8};\\ -1, &\rm{ if\; \ p\equiv\pm3\; mod\; 8}.\\ \end{cases} \end{align*}$

(7)

And when $p\equiv1\; \rm mod\; 8$ , we have $\chi_{4}(-1) = 1$ ; when $p\equiv5\; \rm mod\; 8$ , we have $\chi_{4}(-1) = -1.$ Combining (2) and (5)–(7) we can deduce that

$\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = -1-\chi_2(n)+\frac{1}{\sqrt{ p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right).$

This prove Lemma 3.

Lemma 4. Let $p$ be an odd prime with $p\equiv1\; \rm mod\; 4$ . Then for any integer $k\geq4$ and $n$ with $(n, p) = 1$ , we have the fourth-order linear recurrence formula

$G^k(n) = (4p+2)\cdot G^{k-2}(n)+8(p-2\alpha^2)\cdot G^{k-3}(n)+[(4\alpha^2-2p)^2-(2p-1)^2]\cdot G^{k-4}(n),$

where

$\alpha = \alpha(p) = \frac{1}{2}\sum\limits_{a = 1}^{p-1}\left(\frac{a^3+a}{p}\right) = \sum\limits_{a = 1}^{\frac{p-1}{2}}\left(\frac{a+\bar{a}}{p}\right),$

$\left(\frac{\ast} {p}\right) = \chi_2$ denotes the Legendre's symbol.

Proof. For $p\equiv1\; \rm mod\; 4$ , any integer $n$ with $\left(n, p\right) = 1$ , and fourth-order character $\chi_{4}$ modulo $p$ , we have the identity

$\chi_{4}^4(n) = \bar{\chi_{4}}^4(n) = \chi_0(n), \ \ \chi_{4}^2(n) = \chi_2(n),$

where $\chi_0$ denotes the principal character modulo $p$ .

According to Lemma 3,

$\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = -1-\chi_2(n)+\frac{1}{\sqrt{ p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right),$

$G(n) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right).\quad\quad\quad\quad\quad\quad\quad\quad\ \$

We have

$\begin{align*} G(n) = -\chi_2(n)+\frac{1}{\sqrt{p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right), \end{align*}$

(8)

$\begin{align*} G^2(n) = &[-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4}))]^2\\ = &1-2\chi_2(n)\cdot\frac{1}{\sqrt{p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right)\\ &+\frac{1}{ p}\cdot\left(\chi_2(n)\cdot\tau^4(\bar{\chi_{4}})+\chi_2(n)\cdot\tau^4(\chi_{4})+2p^2\right)\\ = &1-2\chi_2(n)\cdot\frac{1}{\sqrt{p}}\cdot\left(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4})\right)\\ &+\frac{1}{p}\cdot\left(\chi_2(n)\cdot(\tau^4(\bar{\chi_{4}})+\tau^4(\chi_{4}))+2p^2\right). \end{align*}$

According to Lemma 1, we have

$\left(\tau^2(\chi_{4})+\tau^2(\bar{\chi_{4}})\right)^2 = \tau^4(\bar{\chi_{4}})+\tau^4(\chi_{4})+2p^2 = 4p\alpha^2.$

Therefore, we may immediately deduce

$\begin{align*} G^2(n) = &1-2(\chi_2(n)\cdot\left( G(n)+\chi_2(n)\right)\nonumber\\ &+\frac{1}{p}\left(\chi_2(n)\cdot(\tau^4(\bar{\chi_{4}})+\tau^4(\chi_{4}))+2p^2\right)\nonumber\\ = &1-2\chi_2(n)\cdot\left( G(n)+\chi_2(n)\right) \end{align*}$

(9)

$\begin{align*} &+\frac{1}{p}\cdot\left[\chi_2(n)((\tau^2(\bar{\chi_{4}})+\tau^2(\chi_{4}))^2-2p^2)+2p^2\right]\nonumber\\ = &2p-1-2\chi_2(n)\cdot G(n)+\left(4\alpha^2-2p\right)\cdot\chi_2(n)\nonumber, \end{align*}$

$\begin{align*} \; \; \; \; \; \; \; \; \; \; \; \; \; \; G^3(n) = &[-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\cdot\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\cdot\tau^2(\chi_{4}))]^3\nonumber\\ = &\left(2p-1-2\chi_2(n)\cdot G(n)+\left(4\alpha^2-2p\right)\cdot\chi_2(n)\right)\cdot G(n) \end{align*}$

(10)

$\begin{align*} = &(4\alpha^2-2p)\chi_2(n)\cdot G(n)+(2p+3)G(n)-(4p-2)\chi_2(n) -2(4\alpha^2-2p)\nonumber \end{align*}$

and

$\begin{align*} \left[G^2(n)-(2p-1)\right]^2 = \left[\chi_2(n)\cdot(4\alpha^2-2p)-2\chi_2(n)\cdot G(n)\right]^2, \end{align*}$

which implies that

$\begin{align*} G^4(n) = (4p+2)\cdot G^{2}(n)+8(p-2\alpha^2)\cdot G(n)+[(4\alpha^2-2p)^2-(2p-1)^2]. \end{align*}$

(11)

So for any integer $k\geq4$ , from (8)–(11), we have the fourth-order linear recurrence formula

$\begin{align*} G^k(n)& = G^{k-4}(n)\cdot G^4(n)\\ & = (4p+2)\cdot G^{k-2}(n)+8(p-2\alpha^2)\cdot G^{k-3}(n)+[(4\alpha^2-2p)^2-(2p-1)^2]\cdot G^{k-4}(n). \end{align*}$

This proves Lemma 4.

3. Proof of the theorem

In this section, we will complete the proof of our theorem.

Let $p$ be any prime with $p\equiv 1\; \rm mod\; 4$ , then we have

$\begin{align*} K_0(p) = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^0(n) = \frac{p-1}{p-1} = 1.\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{align*}$

(12)

$\begin{align*} K_1(p)& = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^1(n)\nonumber\\ & = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\left(-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\tau^2(\chi_{4}))\right)\\& = 0, \end{align*}$

(13)

$\begin{align*} \; \; \; \; \; K_2(p)& = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^2(n)\nonumber\\ & = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\left(-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\tau^2(\chi_{4}))\right)^2 \\ & = 2p+1,\end{align*}$

(14)

$\begin{align*} \; \; \; \; \; K_3(p)& = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^3(n)\nonumber\\ & = \frac{1}{p-1}\sum\limits_{n = 1}^{p-1}\left(-\chi_2(n)+\frac{1}{\sqrt{p}}\cdot(\chi_{4}(n)\tau^2(\bar{\chi_{4}})+\bar{\chi_{4}}(n)\tau^2(\chi_{4}))\right)^3 \\ & = -3(4\alpha^2-2p).\end{align*}$

(15)

It is clear that from Lemma 4, if $k\geq4$ , we have

$\begin{align*} \; \; \; \; \; K_k(p) = &\frac{1}{p-1}\sum\limits_{n = 1}^{p-1}G^k(n)\nonumber\\ = &(4p+2)\cdot K_{k-2}(p)-8\left(2\alpha^2-p\right)\cdot K_{k-3}(p) +(16\alpha^4-16p\alpha^2+4p-1)\cdot K_{k-4}(p). \end{align*}$

(16)

Now Theorem 1 follows (12)–(16). Obviously, using Theorem 1 to all negative integers, and that lead to Corollary 1.

This completes the proofs of our all results.

Some notes:

Note 1: In our theorem, know $n$ is an integer, and $(n, p) = 1$ . According to the properties of quadratic residual, $\chi_2(n) = \pm1$ , $\chi_4(n) = \pm1$ .

Note 2: In our theorem, we only discussed the case $p\equiv1\; \rm mod\; 8$ . If $p\equiv3\; \rm mod\; 4$ , then the result is trivial. In fact, in this case, for any integer $n$ with $(n, p) = 1$ , we have the identity

$\begin{align*} G(n)& = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+n\bar{a}^2}{p}\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^4}{p}\right)\cdot\left(\frac{a^4+n}{p}\right)\\ & = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a}{p}\right)\cdot\left(\frac{a+n}{p}\right) = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{a^2+na}{p}\right)\\ & = 1+\sum\limits_{a = 1}^{p-1}\left(\frac{1+n\bar{a}}{p}\right) = \sum\limits_{a = 0}^{p-1}\left(\frac{1+na}{p}\right) = 0. \end{align*}$

Thus, for all prime $p$ with $p\equiv3\; \rm mod\; 4$ and $k\geq1$ , we have $K_k(p) = 0$ .

4. Conclusions

The main result of this paper is Theorem 1. It gives an interesting computational formula for $K_k(p)$ with $p\equiv1\; \rm mod\; 4$ . That is, for any integer $k$ , we have the identity

$K_k(p) = (4p+2)\cdot K_{k-2}(p)-8\left(2\alpha^2-p\right)\cdot K_{k-3}(p)+(16\alpha^4-16p\alpha^2+4p-1)\cdot K_{k-4}(p).$

Thus, the problems of calculating a linear recurrence formula of one kind special character sums modulo a prime are given.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors are grateful to the anonymous referee for very helpful and detailed comments.

This work is supported by the N.S.F. (11971381, 12371007) of China and Shaanxi Fundamental Science Research Project for Mathematics and Physics (22JSY007).

Conflict of interest

The authors declare no conflicts of interest.

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Cluster synchronization of coupled complex-valued neural networks with leakage and time-varying delays in finite-time

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2. Some lemmas

3. Proof of the theorem

4. Conclusions

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AIMS Mathematics

Cluster synchronization of coupled complex-valued neural networks with leakage and time-varying delays in finite-time

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Abstract

1. Introduction

2. Some lemmas

3. Proof of the theorem

4. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

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1. Introduction

2. Some lemmas

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4. Conclusions

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