1. Introduction

Permutation polynomials and Dickson polynomials are two of the most important topics in the area of finite fields. Let ${\mathbb F}_{q}$ be the finite field of characteristic $p$ with $q$ elements. Let ${\mathbb F}_q[x]$ be the ring of polynomials over ${\mathbb F}_q$ in the indeterminate $x$. If the polynomial $f (x)\in {\mathbb F}_q[x]$ induces a bijective map from ${\mathbb F}_q$ to itself, then $f (x)\in {\mathbb F}_q[x]$ is called a permutation polynomial of ${\mathbb F}_q$. Properties, constructions and applications of permutation polynomials may be found in ^[4], ^[5] and ^[6]. Associated to any integer $n\ge 0$ and a parameter $a\in {\mathbb F}_{q}$, the $n$-th Dickson polynomials of the first kind and of the second kind, denoted by $D_n (x, a)$ and $E_n (x, a)$, are defined for $n\ge 1$ by

$D_n(x, a):=\sum_{i=0}^{[\frac{n}{2}]}\frac{n}{n-i}\binom{n-i}{i}(-a)^ix^{n-2i}$

and

$E_n(x, a):=\sum_{i=0}^{[\frac{n}{2}]}\binom{n-i}{i}(-a)^ix^{n-2i},$

respectively, and $D_0(x, a):=2, E_0(x, a):=1$. It is well known that $D_n (x, 0)$ is a permutation polynomial of ${\mathbb F}_{q}$ if and only if $\gcd (n, q-1)=1$, and if $a\ne 0$, then $D_n (x, a)$ induces a permutation of ${\mathbb F}_{q}$ if and only if $\gcd (n, q^2-1)=1$. There are lots of published results on permutational properties of Dickson polynomial $E_n (x, a)$ of the second kind (see, for example, ^[1]).

The reversed Dickson polynomial of the first kind, denoted by $D_n (a, x)$, was introduced in ^[3] and defined as follows

$D_n(a, x):=\sum_{i=0}^{[\frac{n}{2}]}\frac{n}{n-i}\binom{n-i}{i}(-x)^ia^{n-2i}$

if $n\ge 1$ and $D_0(a, x)=2$, where $[\frac{n}{2}]$ means the largest integer no more than $\frac{n}{2}$. Wang and Yucas ^[7] extended this concept to that of the $n$-th reversed Dickson polynomial of $(k+1)$-th kind $D_{n, k}(a, x)\in {\mathbb F}_{q}[x]$, which is defined for $n\ge 1$ by

$D_{n, k}(a, x):=\sum_{i=0}^{[\frac{n}{2}]}\frac{n-ki}{n-i}\binom{n-i}{i}(-x)^ia^{n-2i}$

(1.1)

and $D_{0, k}(a, x)=2-k$. Some families of permutation polynomials from the revered Dickson polynomials of the first kind were obtained in ^[3]. Hong, Qin and Zhao ^[2] studied the revered Dickson polynomial $E_n (a, x)$ of the second kind that is defined for $n\ge 1$ by

$E_{n}(a, x):=\sum_{i=0}^{[\frac{n}{2}]}\binom{n-i}{i}(-x)^ia^{n-2i}$

and $E_{0}(a, x)=1$. In fact, they gave some necessary conditions for the revered Dickson polynomial $E_n (1, x)$ of the second kind to be a permutation polynomial of ${\mathbb F}_{q}$. Regarding the revered Dickson polynomial $D_{n, 2}(a, x)\in {\mathbb F}_{q}[x]$ of the third kind, from its definition one can derive that

$D_{n, 2}(a, x)=aE_{n-1}(a, x)$

(1.2)

for each $x\in {\mathbb F}_{q}$. Using (1.2), one can deduce immediately from ^[2] the similar results on the permutational behavior of the reversed Dickson polynomial $D_{n, 2}(a, x)$ of the third kind.

In this paper, our main goal is to develop the method presented by Hong, Qin and Zhao in ^[2] to investigate the reversed Dickson polynomial $D_{n, k}(a, x)$ of the $(k+1)$-th kind which is defined by (1.1) if $n\ge 1$ and $D_{0, k}(a, x):=2-k$. For $a\ne 0$, we write $x=y (a-y)$ with an indeterminate $y\ne \frac{a}{2}$. Then one can rewrite $D_{n, k}(a, x)$ as

$D_{n, k}(a, x)=\frac{\big((k-1)a-(k-2)y\big)y^n-\big(a+(k-2)y\big)(a-y)^n}{2y-a}.$

(1.3)

We have

$D_{n, k}\Big(a, \frac{a^2}{4}\Big)=\frac{(kn-k+2)a^n}{2^n}.$

(1.4)

In fact, (1.3) and (1.4) follow from Theorem 2.2 (ⅰ) and Theorem 2.4 (ⅰ) below. It is easy to see that if ${\rm char}({\mathbb F}_{q})=2$, then $D_{n, k}(a, x)=E_n (a, x)$ if $k$ is odd and $D_{n, k}(a, x)=D_n (a, x)$ if $k$ is even. We also find that $D_{n, k}(a, x)=D_{n, k+p}(a, x)$, so we can restrict $p>k$. Thus we always assume $p={\rm char}({\mathbb F}_{q})\ge 3$ in what follows.

The paper is organized as follows. First in section 2, we study the properties of the reversed Dickson polynomial $D_{n, k}(a, x)$ of the $(k+1)$-th kind. Subsequently, in Section 3, we prove a necessary condition for the reversed Dickson polynomial $D_{n, k}(1, x)$ of the $(k+1)$-th kind to be a permutation polynomial of ${\mathbb F}_{q}$ and then introduce an auxiliary polynomial to present a characterization for $D_{n, k}(1, x)$ to be a permutation of ${\mathbb F}_q$. From the Hermite criterion ^[4] one knows that a function $f: {\mathbb F}_{q}\rightarrow {\mathbb F_{q}}$ is a permutation polynomial of ${\mathbb F_{q}}$ if and only if the $i$-th moment

$\sum_{a\in {\mathbb F_{q}}}f(a)^i= {\left\{\begin{array}{rl} 0, & {\rm if} \ 0\le i\le q-2, \\ -1, & {\rm if} \ i=q-1. \end{array} \right.}$

Thus to understand well the permutational behavior of the reversed Dickson polynomial $D_{n, k}(1, x)$ of the ($k+1$)-th kind, we would like to know if the $i$-th moment $\sum_{a\in {\mathbb F}_q}D_{n, k}(1, a)^i$ is computable. We are able to treat with this sum when $i =1$. The final section is devoted to the computation of the first moment $\sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a)$.

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2. Reversed Dickson polynomials of the $(k+1)$-th kind

In this section, we study the properties of the reversed Dickson polynomials $D_{n, k}(a, x)$ of the $(k+1)$-th kind. Clearly, if $a=0$, then

$D_{n, k}(0, x)={\left\{ \begin{array}{rl} 0, & {\rm if} \ n \ {\rm is \ odd}, \\ (-1)^{\frac{n}{2}+1}(k-2)x^{\frac{n}{2}}, & {\rm if} \ n \ {\rm is \ even}. \end{array} \right.}$

Therefore, $D_{n, k}(0, x)$ is a PP (permutation polynomial) of ${\mathbb F_{q}}$ if and only if $n$ is an even integer with $\gcd (\frac{n}{2}, q-1)=1$. In what follows, we always let $a\in {\mathbb F_{q}^*}$. First, we give a basic fact as follows.

Lemma 2.1. ^[4] Let $f (x)\in {\mathbb F_{q}}[x]$. Then $f (x)$ is a PP of ${\mathbb F_{q}}$ if and only if $cf (dx)$ is a PP of ${\mathbb F_{q}}$ for any given $c, d\in {\mathbb F_{q}^*}$.

Then we can deduce the following result.

Theorem 2.2. Let $a, b\in {\mathbb F_{q}^*}$. Then the following are true.

(ⅰ). One has $D_{n, k}(a, x)=\frac{a^n}{b^n}D_{n, k}(b, \frac{b^2}{a^2}x)$.

(ⅱ). We have that $D_{n, k}(a, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$.

Proof. (ⅰ). By the definition of $D_{n, k}(a, x)$, we have

$\frac{a^n}{b^n}D_{n, k}\Big(b, \frac{b^2}{a^2}x\Big) =\frac{a^n}{b^n} \sum_{i=0}^{[\frac{n}{2}]}\frac{n-ki}{n-i}\binom{n-i}{i} (-1)^ib^{n-2i}\frac{b^{2i}}{a^{2i}}x^{i}\\ =\sum_{i=0}^{[\frac{n}{2}]}\frac{n-ki}{n-i}\binom{n-i}{i}(-1)^ia^{n-2i}x^{i}\\ =D_{n, k}(a, x)$

as required. Part (ⅰ) is proved.

(ⅱ). Taking $b=1$ in part (ⅰ), we have

$D_{n, k}(a, x)=a^nD_{n, k}\Big(1, \frac{x}{a^2}\Big).$

It then follows from Lemma 2.1 that $D_{n, k}(a, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$. This completes the proof of part (ⅱ). So Theorem 2.2 is proved.

Theorem 2.2 tells us that to study the permutational behavior of $D_{n, k}(a, x)$ over ${\mathbb F_{q}}$, one only needs to consider that of $D_{n, k}(1, x)$. In the following, we supply several basic properties on the reversed Dickson polynomial $D_{n, k}(1, x)$ of the $(k+1)$-th kind. The following result is given in ^[2].

Lemma 2.3. ^[2] Let $n\ge 0$ be an integer. Then

$D_n(1, x(1-x))=x^n+(1-x)^n$

and

$E_{n}(1, x(1-x))=\frac{x^{n+1}-(1-x)^{n+1}}{2x-1}$

if $x\ne \frac{1}{2}.$

Theorem 2.4. Each of the following is true.

(ⅰ). For any integer $n\ge 0$, we have

$D_{n, k}(1, \frac{1}{4})=\frac{kn-k+2}{2^n}$

and

$D_{n, k}(1, x)=\frac{\big(k-1-(k-2)y\big)y^n-\big(1+(k-2)y\big)(1-y)^n}{2y-1}$

if $x=y (1-y)\ne \frac{1}{4}$.

(ⅱ). If $n_1$ and $n_2$ are positive integers such that $n_1\equiv n_2\pmod{q^2-1}$, then one has $D_{n_1, k}(1, x_0)=D_{n_2, k}(1, x_0)$ for any $x_0\in {\mathbb F_{q}\setminus\{\frac{1}{4}\}}$.

Proof. (ⅰ). First of all, it is easy to see that $D_{0, k}\big (1, \frac{1}{4}\big)=2-k=\frac{k\times 0-k+2}{2^0}$ and $D_{1, k}\big (1, \frac{1}{4}\big)=1=\frac{k\times 1-k+2}{2^1}$. the first identity is true for the cases that $n=0$ and 1. Now let $n\ge 2$. Then one has

$D_{n, k}\Big(1, \frac{1}{4}\Big)=\sum_{i=0}^{[\frac{n}{2}]} \frac{n-ki}{n-i}\binom{n-i}{i}\Big(-\frac{1}{4}\Big)^{i}\\ =\sum_{i=0}^{[\frac{n}{2}]}\frac{n-(k-1)i}{n-i}\binom{n-i}{i} \Big(-\frac{1}{4}\Big)^{i}+\sum_{i=0}^{[\frac{n}{2}]} \frac{-i}{n-i}\binom{n-i}{i}\Big(-\frac{1}{4}\Big)^{i}\\ =D_{n, k-1}\Big(1, \frac{1}{4}\Big)+\frac{1}{4}\sum_{i=0}^{[\frac{n}{2}]-1} \binom{n-2-i}{i}\Big(-\frac{1}{4}\Big)^{i}\\ =D_{n, k-1}\Big(1, \frac{1}{4}\Big)+\frac{1}{4}E_{n-2}\Big(1, \frac{1}{4}\Big),$

which follows from Theorem 2.2 (1) in ^[2] that

$D_{n, k}\Big(1, \frac{1}{4}\Big)=D_{n, 1}\Big(1, \frac{1}{4}\Big)+(k-1) \frac{1}{4}E_{n-2}\Big(1, \frac{1}{4}\Big)\\ =\frac{n+1}{2^n}+\frac{(k-1)n-(k-1)}{2^n}\\ =\frac{kn-k+2}{2^n}$

as desired. So the first identity is proved.

Now we turn our attention to the second identity. Let $x\ne \frac{1}{4}$, then there exists $y\in {\mathbb F_{q^2}\setminus\{\frac{1}{2}\}}$ such that $x=y (1-y)$. So by the definition of the $n$-th reversed Dickson polynomial of the ($k+1$)-th kind, one has

$D_{n, k}(1, y(1-y))=\sum_{i=0}^{[\frac{n}{2}]} \frac{n-ki}{n-i}\binom{n-i}{i}(-y(1-y))^{i}\nonumber\\ =\sum_{i=0}^{[\frac{n}{2}]} \frac{k(n-i)-kn}{n-i}\binom{n-i}{i}(-y(1-y))^{i}\nonumber\\ =k\sum_{i=0}^{[\frac{n}{2}]} \binom{n-i}{i}(-y(1-y))^{i}-(k-1)\sum_{i=0}^{[\frac{n}{2}]} \frac{n}{n-i}\binom{n-i}{i}(-y(1-y))^{i}\nonumber\\ =kE_n(1, y(1-y))-(k-1)D_n(1, y(1-y)).$

(2.1)

But Lemma 2.3 gives us that

$D_n(1, y(1-y))=y^n+(1-y)^n$

(2.2)

and

$E_n(1, y(1-y))=\sum_{i=0}^{n}y^{n-i}(1-y)^i =\frac{y^{n+1}-(1-y)^{n+1}}{2y-1}.$

(2.3)

Thus it follows from (2.1) to (2.3) that

$D_{n, k}(1, x)=D_{n, k}(1, y(1-y))\\ =kE_n(1, y(1-y))-(k-1)D_n(1, y(1-y))\\ =\frac{ky^{n+1}-k(1-y)^{n+1}}{2y-1}-(k-1)\big(y^n+(1-y)^n\big)\\ =\frac{\big(k-1-(k-2)y\big)y^n-\big(1+(k-2)y\big)(1-y)^n}{2y-1}$

as required. So the second identity holds. Part (ⅰ) is proved.

(ⅱ). For each $x_0\in {\mathbb F_{q}\setminus\{\frac{1}{4}\}}$, one can choose an element $y_0\in {\mathbb F_{q^2}\setminus\{\frac{1}{2}\}}$ such that $x_0=y_0(1-y_0)$. Since $n_1\equiv n_2\pmod{q^2-1}$, one has $y_0^{n_1}=y_0^{n_2}$ and $(1-y_0)^{n_1}=(1-y_0)^{n_2}$. It then follows from part (ⅰ) that

$D_{n_1, k}(1, x_0)=D_{n_1, k}(1, y_0(1-y_0))\\ =\frac{\big(k-1-(k-2)y_0\big)y_0^{n_1}-\big(1+(k-2)y_0\big)(1-y_0)^{n_1}}{2y_0-1}\\ =\frac{\big(k-1-(k-2)y_0\big)y_0^{n_2}-\big(1+(k-2)y_0\big)(1-y_0)^{n_2}}{2y_0-1}\\ =D_{n_2, k}(1, x_0)$

as desired. This ends the proof of Theorem 2.4.

Evidently, by Theorem 2.2 (ⅰ) and Theorem 2.4 (ⅰ) one can derive that (1.3) and (1.4) are true.

Proposition 2.5. Let $n\ge 2$ be an integer. Then the recursion

$D_{n, k}(1, x)=D_{n-1, k}(1, x)-xD_{n-2, k}(1, x)$

holds for any $x\in {\mathbb F_{q}}$.

Proof. We consider the following two cases.

C_ASE 1. $x\ne \frac{1}{4}$. For this case, one may let $x=y (1-y)$ with $y\in {\mathbb F_{q^2}}\setminus \{ \frac{1}{2}\}$. Then by Theorem 2.4 (ⅰ), we have

$D_{n-1, k}(1, x)-xD_{n-2, k}(1, x) =D_{n-1, k}(1, y(1-y))-y(1-y)D_{n-2, k}(1, y(1-y))\\ =\frac{\big(k-1-(k-2)y\big)y^{n-1}-\big(1+(k-2)y\big)(1-y)^{n-1}}{2y-1}\\ \ \ \ \ -y(1-y)\frac{\big(k-1-(k-2)y\big)y^{n-2}-\big(1+(k-2)y\big)(1-y)^{n-2}}{2y-1}\\ =\frac{\big(k-1-(k-2)y\big)y^{n}-\big(1+(k-2)y\big)(1-y)^{n}}{2y-1}\\ =D_{n, k}(1, x)$

as required.

C_ASE 2. $x=\frac{1}{4}$. Then by Theorem 2.4 (ⅰ), we have

$D_{n-1, k}\Big(1, \frac{1}{4}\Big)-\frac{1}{4}D_{n-2, k}\Big(1, \frac{1}{4}\Big) =\frac{k(n-1)-k+2}{2^{n-1}}-\frac{1}{4}\frac{k(n-2)-k+2}{2^{n-2}}\\ =\frac{kn-k+2}{2^{n}}\\ =D_{n, k}\Big(1, \frac{1}{4}\Big).$

This concludes the proof of Proposition 2.5.

By Proposition 2.5, we can obtain the generating function of the reversed Dickson polynomial $D_{n, k}(1, x)$ of the ($k+1$)-th kind as follows.

Proposition 2.6. The generating function of $D_{n, k}(1, x)$ is given by

$\sum_{n=0}^{\infty}D_{n, k}(1, x)t^n=\frac{(k-1)t-k+2}{1-t+xt^2}.$

Proof. By the recursion presented in Proposition 2.5, we have

$(1-t+xt^2)\sum_{n=0}^{\infty}D_{n, k}(1, x)t^n =\sum_{n=0}^{\infty}D_{n, k}(1, x)t^n-\sum_{n=0} ^{\infty}D_{n, k}(1, x)t^{n+1}+x\sum_{n=0}^{\infty}D_{n, k}(1, x)t^{n+2}\\ =(k-1)t-k+2+\sum_{n=0}^{\infty}\big(D_{n+2, k} (1, x)-D_{n+1, k}(1, x)+xD_{n, k}(1, x)\big)t^{n+2}\\ =(k-1)t-k+2.$

Thus the desired result follows immediately.

Lemma 2.7. ^[3] Let $x\in{\mathbb F_{q^2}}$. Then $x (1-x)\in{\mathbb F_{q}}$ if and only if $x^q=x$ or $x^q=1-x$.

Let $V$ be defined by

$V:=\{x\in {\mathbb F_{q^2}}: x^q=1-x\}.$

Clearly, ${\mathbb F_{q}}\cap V=\{\frac{1}{2}\}$. Then we obtain a characterization for $D_{n, k}(1, x)$ to be a PP of ${\mathbb F_{q}}$ as follows.

Theorem 2.8. Let $q=p^e$ with $p>3$ being a prime and $e$ being a positive integer. Let

$f:y\mapsto\frac{\big(k-1-(k-2)y\big)y^{n}-\big(1+(k-2)y\big)(1-y)^{n}}{2y-1}$

be a mapping on $({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$. Then $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $f$ is $2$-to-$1$ and $f (y)\ne\frac{kn-k+2}{2^n}$ for any $y\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$.

Proof. First, we show the sufficiency part. Let $f$ be $2$-to-$1$ and $f (y)\ne\frac{kn-k+2}{2^n}$ for any $y\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$. Let $D_{n, k}(1, x_1)=D_{n, k}(1, x_2)$ for $x_1, x_2\in{\mathbb F_{q}}$. To show that $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$, it suffices to show that $x_1=x_2$, which will be done in what follows.

First of all, one can find $y_1, y_2\in{\mathbb F_{q^2}}$ satisfying $x_1=y_1(1-y_1)$ and $x_2=y_2(1-y_2)$. By Lemma 2.7, we know that $y_1, y_2\in{\mathbb F_{q}}\cup V$. We divide the proof into the following two cases.

C_ASE 1. At least one of $x_1$ and $x_2$ is equal to $\frac{1}{4}$. Without loss of any generality, we may let $x_1=\frac{1}{4}$. So by Theorem 2.4 (ⅰ), one derives that

$D_{n, k}(1, x_2)=D_{n, k}(1, x_1)=D_{n, k}\Big(1, \frac{1}{4}\Big)=\frac{kn-k+2}{2^n}.$

(2.4)

We claim that $x_2=\frac{1}{4}$. Assume that $x_2\ne\frac{1}{4}$. Then $y_2\ne \frac{1}{2}$. Since $f (y)\ne\frac{kn-k+2}{2^n}$ for any $y\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$, by Theorem 2.4 (ⅰ), we get that

$D_{n, k}(1, x_2)=\frac{(k-1-(k-2)y_2)y_2^n-(1+(k-2)y_2)(1-y_2)^n}{2y_2-1}=f(y_2)\ne \frac{kn-k+2}{2^n},$

which contradicts to (2.4}). Hence the claim is true, and so we have $x_1=x_2$ as required.

C_ASE 2. Both of $x_1$ and $x_2$ are not equal to $\frac{1}{4}$. Then $y_1\ne \frac{1}{2}$ and $y_2\ne \frac{1}{2}$. Since $D_{n, k}(1, x_1)=D_{n, k}(1, x_2)$, by Theorem 2.4 (ⅰ), one has

$\frac{(k-1-(k-2)y_1)y_1^n-(1+(k-2)y_1)(1-y_1)^n}{2y_1-1} =\frac{(k-1-(k-2)y_2)y_2^n-(1+(k-2)y_2)(1-y_2)^n}{2y_2-1},$

which is equivalent to $f (y_1)=f (y_2)$. However, $f$ is a $2$-to-$1$ mapping on $({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$, and $f (y_2)=f (1-y_2)$ by the definition of $f$. It then follows that $y_1=y_2$ or $y_1=1-y_2$. Thus $x_1=x_2$ as desired. Hence the sufficiency part is proved.

Now we prove the necessity part. Let $D_{n, k}(1, x)$ be a PP of ${\mathbb F_{q}}$. Choose two elements $y_1, y_2\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$ such that $f (y_1)=f (y_2)$, that is,

$\frac{(k-1-(k-2)y_1)y_1^n-(1+(k-2)y_1)(1-y_1)^n}{2y_1-1} =\frac{(k-1-(k-2)y_2)y_2^n-(1+(k-2)y_2)(1-y_2)^n}{2y_2-1}.$

(2.5)

Since $y_1, y_2\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$, it follows from Lemma 2.7 that $y_1(1-y_1)\in{\mathbb F_{q}}$ and $y_2(1-y_2)\in{\mathbb F_{q}}$. So by Theorem 2.4 (ⅰ), (2.5) implies that

$D_{n, k}(1, y_1(1-y_1))=D_{n, k}(1, y_2(1-y_2)).$

Thus $y_1(1-y_1)=y_2(1-y_2)$ since $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$, which infers that $y_1=y_2$ or $y_1=1-y_2$. Since $y_2\ne \frac{1}{2}$, one has $y_2\ne 1-y_2$. Therefore $f$ is a $2$-to-$1$ mapping on $({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$.

Now take $y'\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$. Then from Lemma 2.7 it follows that $y'(1-y')\in{\mathbb F_{q}}$ and

$y'(1-y')\ne \frac{1}{2}\Big(1-\frac{1}{2}\Big).$

Notice that $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$. Hence one has

$D_{n, k}(1, y'(1-y'))\ne D_{n, k}\Big(1, \frac{1}{2}\Big(1-\frac{1}{2}\Big)\Big).$

But Theorem 2.4 (ⅰ) tells us that

$D_{n, k}\Big(1, \frac{1}{2}\Big(1-\frac{1}{2}\Big)\Big)=\frac{kn-k-2}{2^n}.$

Then by Theorem 2.4 (ⅰ) and noting that $y'\ne \frac{1}{2}$, we have

$\frac{\big(k-1-(k-2)y'\big)y'^{n}-\big(1+(k-2)y'\big)(1-y')^{n}}{2y'-1},$

which infers that $f (y')\ne \frac{kn-k-2}{2^n}$ for any $y'\in ({\mathbb F_{q}}\cup V)\setminus \{\frac{1}{2}\}$. So the necessity part is proved.

The proof of Theorem 2.8 is complete.

Now we can use Theorem 2.4 to present an explicit formula for $D_{n, k}(1, x)$ when $n$ is a power of the characteristic $p$. Then we derive the detailed characterization for $D_{n, k}(1, x)$ being a PP of ${\mathbb F}_q$ in this case.

Proposition 2.9. Let $p={\rm char}({\mathbb F_{q}})\ge 3$ and $s\ge 0$ be an integer. Then

$2D_{p^s, k}(1, x)+k-2=k(1-4x)^{\frac{p^s-1}{2}}.$

Proof. We consider the following two cases.

CASE 1. $x\ne \frac{1}{4}$. For this case, putting $x=y (1-y)$ in Theorem 2.4 (ⅰ) gives us that

$D_{p^s, k}(1, x)=D_{p^s, k}(1, y(1-y))\\ =\frac{\big(k-1-(k-2)y\big)y^{p^s}-\big(1+(k-2)y\big)(1-y)^{p^s}}{2y-1}\\ =\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s}}{u}\\ =\frac{1}{2^{p^s+1}u}\Big((k+(2-k)u)(u+1)^{p^s}-(k+(k-2)u)(1-u)^{p^s}\Big)\\ =\frac{1}{2}(ku^{p^s-1}-k+2),$

where $u=2y-1$. So we obtain that

$2D_{p^s, k}(1, x) =k(u^2)^{\frac{p^s-1}{2}}-k+2 =k\big((2y-1)^2\big)^{\frac{p^s-1}{2}}-k+2,$

which infers that

$2D_{p^s, k}(1, x)+k-2=k(1-4x)^{\frac{p^s-1}{2}}$

as desired.

C_ASE 2. $x=\frac{1}{4}$. By Theorem 2.4 (ⅰ), one has

$2D_{p^s, k}\big(1, \frac{1}{4}\big)+k-2=2\times\frac{kp^s-k+2}{2^{p^s}}+k-2=0 =k(1-4\times\frac{1}{4})^{\frac{p^s-1}{2}}$

as required. So Proposition 2.9 is proved.

It is well known that every linear polynomial over ${\mathbb F_{q}}$ is a PP of ${\mathbb F_{q}}$ and that the monomial $x^n$ is a PP of ${\mathbb F_{q}}$ if and only if $\gcd (n, q-1)=1$. Then by Proposition 2.9, we have the following result.

Corollary 2.10. Let $p\ge 3$ be a prime, $q=p^e$ with $e\ge 1$ and $s\ge 0$ be an integer. Then $D_{p^s, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $k\ge 1, p=3$, $s$ is odd and $\gcd (s, e)=1$.

Proof. First assume that $D_{p^s, k}(1, x)$ is a PP of ${\mathbb F_{p^e}}$. It then follows from Proposition 2.9 that $D_{p^s, k}(1, x)$ is a PP of ${\mathbb F_{p^e}}$ if and only if

$k(1-4x)^{\frac{p^s-1}{2}}$

(2.6)

is a PP of ${\mathbb F_{p^e}}$. Clearly, $k\ge 1$ and $s>0$ in this case. Suppose $p>3$, then (2.6) is a PP of ${\mathbb F_{p^e}}$ if and only if

$\gcd\Big(\frac{p^s-1}{2}, p^e-1\Big)=1.$

This is impossible since $\frac{p-1}{2}|\gcd\big (\frac{p^s-1}{2}, q-1\big)$ implies that

$\gcd\Big(\frac{p^s-1}{2}, q-1\Big)\ge\frac{p-1}{2}>1.$

So $p=3, k\ge 1$ and $s>0$ in what following. Now Suppose $s>0$ is even, then it is easy to see that $2|\gcd (\frac{3^s-1}{2}, 3^e-1)$ which is a contradiction. This means that $s$ must be an odd integer and then so is $\frac{3^s-1}{2}$. Thus we have that (2.6) is a PP of ${\mathbb F_{p^e}}$ if and only if

$\gcd\Big(\frac{3^s-1}{2}, 3^e-1\Big)=\frac{1}{2}\gcd\big(3^s-1, 3^e-1\big)=\frac{1}{2}(3^{\gcd(s, e)}-1)=1,$

which is equivalent to that $s$ is odd and $\gcd (s, e)=1$. So Corollary 2.10 is proved.

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3. A necessary condition for $D_{n, k}(1, x)$ to be permutational and an auxiliary polynomial

In this section, we study a necessary condition on $n$ for $D_{n, k}(1, x)$ to be a PP of ${\mathbb F_{q}}$. On one hand, it is easy to check that

$D_{0, k}(1, 0)=2-k, D_{n, k}(1, 0)=1$

for any $n\ge 1$ and $D_{0, k}(1, 1)=2-k, D_{1, k}(1, 1)=1$. On the other hand, Proposition 2.5 tells us that

$D_{n+2, k}(1, 1)=D_{n+1, k}(1, 1)-D_{n, k}(1, 1)$

for $n\ge 0$. Then one can easily show that the sequence $\{D_{n, k}(1, 1)|n\in \mathbb{N}\}$ is periodic with the smallest positive periods 6. In fact, one has

$D_{n, k}(1, 1)={\left\{ \begin{array}{rl} 2-k, & {\rm if} \ n\equiv 0\pmod{6}, \\ 1, & {\rm if} \ n\equiv 1\pmod{6}, \\ k-1, & {\rm if} \ n\equiv 2\pmod{6}, \\ k-2, & {\rm if} \ n\equiv 3\pmod{6}, \\ -1, & {\rm if} \ n\equiv 4\pmod{6}, \\ 1-k, & {\rm if} \ n\equiv 5\pmod{6} \end{array} \right.}$

So we have the following result.

Theorem 3.1. Assume that $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ with $q=p^e$ and $p>3$. Then $n\not\equiv 1\pmod{6}$.

Proof. Let $D_{n, k}(1, x)$ be a PP of ${\mathbb F_{q}}$. Then $D_{n, k}(1, 0)$ and $D_{n, k}(1, 1)$ are distinct. Then by the above results, the desired result $n\not\equiv 1 \pmod{6}$ follows immediately.

Let $n, k$ be nonnegative integers. We define the following auxiliary polynomial $p_{n, k}(x)\in \mathbb{Z}[x]$ by

$p_{n, k}(x):=k\sum_{j\ge 0}\binom{n}{2j+1}x^j-(k-2)\sum_{j\ge 0}\binom{n}{2j}x^j$

for $n\ge 1$, and

$p_{0, k}(x):=2^n(2-k).$

Then we have the following relation between $D_{n, k}(1, x)$ and $p_{n, k}(x)$.

Theorem 3.2. Let $p>3$ be a prime and $n\ge 0$ be an integer. Then each of the following is true.

(ⅰ). One has

$D_{n, k}(1, x)=\frac{1}{2^n}p_{n, k}(1-4x).$

(3.1)

(ⅱ). We have that $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $p_{n, k}(x)$ is a PP of ${\mathbb F_{q}}$.

Proof. (ⅰ). Clearly, (3.1) follows from the definitions of $p_{0, k}(x)$ and $D_{0, k}(1, x)$ if $n=0$. Then we assume that $n\ge 1$ in what follows.

First, let $x\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}$. Then there exists $y\in{\mathbb F_{q^2}}\setminus\{\frac{1}{2}\}$ such that $x=y (1-y)$. Let $u=2y-1$. It then follows from Theorem 2.4 (ⅰ) that

$D_{n, k}(1, x)=D_{n, k}(1, y(1-y))\\ =\frac{\big(k-1-(k-2)y\big)y^{n}-\big(1+(k-2)y\big)(1-y)^{n}}{2y-1}\\ =\frac{1}{u}\Big(\frac{-(k-2)u+k}{2}\big(\frac{u+1}{2}\big)^{n}-\frac{(k-2)u+k}{2}\big(\frac{1-u}{2}\big)^{n}\Big)\\ =\frac{1}{2^{n+1}u}\Big(k\big((u+1)^n-(1-u)^n\big)-(k-2)u\big((u+1)^n+(1-u)^n\big)\Big)\\ =\frac{1}{2^{n}}\Bigg ( k\sum_{j\ge 0}\binom{n}{2j+1}x^j-(k-2)\sum_{j\ge 0}\binom{n}{2j}u^{2j}\Bigg)\\ =\frac{1}{2^n}p_{n, k}(u^2)\\ =\frac{1}{2^n}p_{n, k}(1-4y(1-y))\\ =\frac{1}{2^n}p_{n, k}(1-4x)$

as desired. So (3.1) holds in this case.

Consequently, we let $x=\frac{1}{4}$. Then by Theorem 2.4 (ⅰ), we have

$D_{n, k}\Big(1, \frac{1}{4}\Big)=\frac{kn-k+2}{2^n}.$

On the other hand, we can easily check that

$p_{n, k}(0)=kn-k+2.$

Therefore

$D_{n, k}\Big(1, \frac{1}{4}\Big)=\frac{1}{2^n}p_{n, k}(0) =\frac{1}{2^n}p_{n, k}\Big(1-4\times \frac{1}{4}\Big)$

as one desires. So (3.1) is proved.

(ⅱ). Notice that $\frac{1}{2^n}\in{\mathbb F_{q}^*}$ and $1-4x$ is linear. So $D_{n, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $p_{n, k}(x)$ is a PP of ${\mathbb F_{q}}$. This ends the proof of Theorem 3.2.

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4. The first moment $\sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a)$

In this section, we compute the first moment $\sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a)$. By Proposition 2.6, one has

$\sum_{n=0}^{\infty}D_{n, k}(1, x)t^n=\frac{(k-1)t-k+2}{1-t+xt^2} =\frac{(k-1)t-k+2}{1-t}\frac{1}{1-\frac{t^2}{t-1}x} \notag\\ =\frac{(k-1)t-k+2}{1-t}\Big(1+\sum_{m=1}^{q-1}\sum_{\ell=0}^{\infty } \bigg(\frac{t^2}{t-1}\bigg)^{m+\ell (q-1)}x^{m+\ell (q-1)}\Big) \notag\\ \equiv\frac{2t-1}{1-t}\Big(1+\sum_{m=1}^{q-1}\sum_{\ell=0}^{\infty } \bigg(\frac{t^2}{t-1}\bigg)^{m+\ell (q-1)}x^{m}\Big)\pmod{x^q-x} \notag\\ =\frac{(k-1)t-k+2}{1-t}\Big(1+\sum_{m=1}^{q-1} \frac{(\frac{t^2}{t-1})^m}{1-(\frac{t^2}{t-1})^{q-1}}x^m\Big) \notag\\ =\frac{(k-1)t-k+2}{1-t}\Big(1+\sum_{m=1}^{q-1} \frac{(t-1)^{q-1-m}t^{2m}}{(t-1)^{q-1}-t^{2(q-1)}}x^m\Big).$

(4.1)

Moreover, by Theorem 2.4 (ⅱ), it follows that for any $x\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}$, one has

$D_{n_1, k}(1, x)=D_{n_2, k}(1, x)$

when $n_1\equiv n_2\pmod{q^2-1}$. Thus if $x\ne \frac{1}{4}$, one has

$\sum_{n=0}^\infty D_{n, k}(1, x)t^n=1+\sum_{n=1}^{q^2-1} \sum_{\ell=0}^{\infty }D_{n+\ell(q^2-1), k}(1, x)t^{n+\ell(q^2-1)}\notag\\ =1+\sum_{n=1}^{q^2-1}D_{n, k}(1, x)\sum_{\ell=0}^{\infty }t^{n+\ell(q^2-1)}\notag\\ =1+\frac{1}{1-t^{q^2-1}}\sum_{n=1}^{q^2-1}D_{n, k}(1, x)t^{n}.$

(4.2)

Then (4.1) together with (4.2) gives that for any $x\ne \frac{1}{4}$, we have

$\sum_{n=1}^{q^2-1}D_{n, k}(1, x)t^n= \Big(\sum_{n=0}^{\infty}D_{n, k}(1, x)t^n-1\Big)(1-t^{q^2-1})\notag\\ \equiv \Big(\frac{(k-1)t-k+2}{1-t}-1\Big)(1-t^{q^2-1}) +\frac{(1-t^{q^2-1})((k-1)t-k+2)}{1-t}\sum_{m=1}^{q-1} \frac{(t-1)^{q-1-m}t^{2m}}{(t-1)^{q-1}-t^{2(q-1)}}x^m\pmod{x^q-x}\notag\\ =\frac{(kt+1-k)(1-t^{q^2-1})}{1-t}+h(t)\sum_{m=1}^{q-1} (t-1)^{q-1-m}t^{2m}x^m,$

(4.3)

where

$h(t):=\frac{(t^{q^2-1}-1)((k-1)t-k+2)}{(t-1)^q-(t-1)t^{2(q-1)}}.$

Lemma 4.1. ^[4] Let $u_0, u_1, \cdots, u_{q-1}$ be the list of the all elements of ${\mathbb F_{q}}$. Then

$\sum_{i=0}^{q-1}u_i^k={\left\{ \begin{array}{rl} 0, & {\rm if} \ 0\le k\le q-2, \\ -1, & {\rm if} \ k=q-1. \end{array} \right.}$

Now by Theorem 2.4 (ⅰ), Lemma 4.1 and (4.3), we derive that

$\sum_{n=1}^{q^2-1}\sum_{a\in{\mathbb F_{q}}}D_{n, k}(1, a)t^n =\sum_{n=1}^{q^2-1}D_{n, k}\Big(1, \frac{1}{4}\Big)t^n+\sum_{n=1}^{q^2-1} \sum_{a\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}}D_{n, k}(1, a)t^n\notag\\ =\sum_{n=1}^{q^2-1}\frac{kn-k+2}{2^n}t^n+\sum_{a\in{\mathbb F_{q}} \setminus\{\frac{1}{4}\}}\frac{(kt+1-k)(1-t^{q^2-1})}{1-t}+h(t)\sum_{m=1} ^{q-1}(t-1)^{q-1-m}t^{2m}\sum_{a\in{\mathbb F_{q}}\setminus\{\frac{1}{4}\}}a^m\notag\\ =\sum_{n=1}^{q^2-1}\frac{kn-k+2}{2^n}t^n+(q-1)\frac{(kt+1-k)(1-t^{q^2-1})}{1-t} +h(t)\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\sum_{a\in{\mathbb F_{q}}}a^m\notag\\ \ \ \ -h(t)\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\Big(\frac{1}{4}\Big)^m\notag\\ =\sum_{n=1}^{q^2-1}\frac{kn-k+2}{2^n}t^n-\frac{(kt+1-k)(1-t^{q^2-1})}{1-t} -h(t)t^{2(q-1)}-h(t)\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\Big(\frac{1}{4}\Big)^m.$

(4.4)

Since $(t-1)^q=t^q-1$ and $q$ is odd, one has

$h(t)=\frac{(t^{q^2-1}-1)(2t-1)}{(t-1)^q-(t-1)t^{2(q-1)}}\notag\\ =\frac{(t^{q^2-1}-1)(2t-1)}{(1-t^{q-1})(t^q-t^{q-1}-1)} \notag\\ =\frac{(t^{q^2}-t)(2t-1)}{(t-t^{q})(t^q-t^{q-1}-1)}\notag\\ =\frac{(t^{q}-t)^q+t^q-t}{t-t^{q}}\cdot \frac{2t-1}{t^q-t^{q-1}-1}\notag\\ =\frac{(-1-(t-t^q)^{q-1})(2t-1)}{t^q-t^{q-1}-1}\notag\notag\\ =\frac{(2t-1)\sum_{i=0}^{q^2-q}b_it^i}{t^q-t^{q-1}-1},$

(4.5)

where

$\sum_{i=0}^{q^2-q}b_it^i:=-1-(t-t^q)^{q-1}.$

Then by the binomial theorem applied to $(t-t^q)^{q-1}$, we can derive the following expression for the coefficient $b_i$.

Proposition 4.2. For each integer $i$ with $0\le i\le q^2-q$, write $i=\alpha+\beta q$ with $\alpha$ and $\beta$ being integers such that $0\le \alpha, \beta \le q-1$. Then

$b_i={\left\{ \begin{array}{ll} (-1)^{\beta +1}\binom{q-1}{\beta}, & {\it if} \ \alpha +\beta =q-1, \\ -1, & {\it if} \ \alpha =\beta =0, \\ 0, & {\it otherwise}. \end{array} \right.}$

For convenience, let

$a_n:=\sum_{a\in{\mathbb F_{q}}}D_{n, k}(1, a).$

Then by (4.4) and (4.5), we arrive at

$\sum_{n=1}^{q^2-1}\Big(a_n-\frac{kn-k+2}{2^n}\Big)t^n =-\frac{(kt+1-k)(1-t^{q^2-1})}{1-t}-\frac{(2t-1)\sum_{i=0}^{q^2-q}b_it^i}{t^q-t^{q-1}-1} \Big(t^{2(q-1)}+\sum_{m=1}^{q-1}(t-1)^{q-1-m}t^{2m}\Big(\frac{1}{4}\Big)^m\Big),$

which implies that

$(t^q-t^{q-1}-1)\sum_{n=1}^{q^2-1}\Big(a_n-\frac{kn-k+2}{2^n}\Big)t^n\notag\\ =-(t^q-t^{q-1}-1)(kt+1-k)\sum_{i=0}^{q^2-2}t^i-(2t-1)\Big(t^{2(q-1)}+\sum_{k=1}^{q-1} (t-1)^{q-1-k}t^{2k}\Big(\frac{1}{4}\Big)^k\Big)\sum_{i=0}^{q^2-q}b_it^i.$

(4.6)

Let

$\sum_{i=1}^{q^2+q-1}c_it^i$

denote the right-hand side of (4.6) and let

$d_n:=a_n-\frac{kn-k+2}{2^n}$

for each integer $n$ with $1\le n\le q^2-1$. Then (4.6) can be reduced to

$(t^q-t^{q-1}-1)\sum_{n=1}^{q^2-1}d_nt^n=\sum_{i=1}^{q^2+q-1}c_it^i.$

(4.7)

Then by comparing the coefficient of $t^i$ with $1\le i\le q^2+q-1$ of the both sides in (4.7), we derive the following relations:

${\left\{ \begin{array}{ll} c_j=-d_j, & {\rm if} \ 1\le j\le q-1, \\ c_q=-d_1-d_q, & \\ c_{q+j}=d_j-d_{j+1}-d_{q+j}, & {\rm if} \ 1\le j\le q^2-q-1, \\ c_{q^2+j}=d_{q^2-q+j}-d_{q^2-q+j+1}, & {\rm if} \ 0\le j\le q-2, \\ c_{q^2+q-1}=d_{q^2-1}, & \end{array} \right.}$

from which we can deduce that

${\left\{ \begin{array}{ll} d_j=-c_j, & {\rm if} \ 1\le j\le q-1, \\ d_q=c_1-c_q, & \\ d_{\ell q+j}=d_{(\ell -1)q+j}-d_{(\ell-1)q+j+1}-c_{\ell q+j}, & {\rm if} \ 1\le \ell\le q-2\ \ {\rm and}\ \ 1\le j\le q-1, \\ d_{\ell q}=d_{(\ell -1)q}-d_{(\ell-1)q+1}-c_{\ell q}, & {\rm if} \ 2\le \ell\le q-2, \\ d_{q^2-q+j}=\sum_{i=j}^{q-1}c_{q^2+i}, & {\rm if} \ 0\le j\le q-1. \end{array} \right.}$

(4.8)

Finally, (4.8) together with the following identity

$\sum_{a\in {\mathbb F_{q}}}D_{n, k}(1, a)=d_n+\frac{kn-k+2}{2^n}$

shows that the last main result of this paper is true:

Theorem 4.3. Let $c_i$ be the coefficient of $t^i$ in the right-hand side of (4.6) with $i$ being an integer such that $1\le i\le q^2+q-1$. Then we have

$\sum_{a\in {\mathbb F_{q}}}D_{j, k}(1, a)=-c_j+\frac{kj-k+2}{2^j}\ \ if\ \ 1\le j\le q-1, \\ \sum_{a\in {\mathbb F_{q}}}D_{q, k}(1, a)=c_1-c_q-\frac{k-2}{2}, \\ \sum_{a\in {\mathbb F_{q}}}D_{\ell q+j, k}(1, a)=\sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q+j, k}(1, a)- \sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q+j+1, k}(1, a)-c_{\ell q+j}+\frac{k}{2^{\ell +j}}\\ \ \ \ \ if\ \ 1\le \ell\le q-2 \ {\it and} \ 1\le j\le q-1, \\ \sum_{a\in {\mathbb F_{q}}}D_{\ell q, k}(1, a)=\sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q, k}(1, a)- \sum_{a\in {\mathbb F_{q}}}D_{(\ell-1)q+1, k}(1, a)-c_{\ell q}+\frac{k}{2^{\ell}}\ \ if\ \ 2\le \ell\le q-2$

and

$\sum_{a\in {\mathbb F_{q}}}D_{q^2-q+j, k}(1, a) =\sum_{i=j}^{q-1}c_{q^2+i}+\frac{kj-k+2}{2^j}\ \ if\ \ 0\le j\le q-1.$

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Acknowledgement

Cheng was supported partially by the General Project of Department of Education of Sichuan Province 15ZB0434. [2000] Primary 11T06, 11T55, 11C08.

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Conflict of Interest

The author declares no conflicts of interest in this paper.

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