Weight enumerator of some irreducible cyclic codes

Abstract

In this article, we show explicitly all possible weight enumerators for every irreducible cyclic code of length $n$ over a finite field ${\mathbb {F}}_q$, in the case which each prime divisor of $n$ is also a divisor of $q-1$.

1 Introduction

A code of length $n$ and dimension $k$ over a finite field ${\mathbb {F}}_q$ is a linear $k$-dimensional subspace of ${\mathbb {F}}_q^n$. A $[n,k]_q$-code $\mathcal C$ is called cyclic if it is invariant by the shift permutation, i.e., if $(a_1,a_2,\dots ,a_n)\in \mathcal C$ then the shift $(a_n,a_1,\ldots ,a_{n-1})$ is also in $\mathcal C$. The cyclic code $\mathcal C$ can be viewed as an ideal in the group algebra ${\mathbb {F}}_qC_n$, where $C_n$ is the cyclic group of order $n$. We note that ${\mathbb {F}}_qC_n$ is isomorphic to $\mathcal R_n=\frac{{\mathbb {F}}_q[x]}{\langle x^n-1\rangle }$ and since subspaces of $\mathcal R_n$ are ideals and $\mathcal R_n$ is a principal ideal domain, it follows that each ideal is generated by a polynomial $g(x)\in {\mathcal {R}}_n$, where $g$ is a divisor of $x^n-1$.

Codes generated by a polynomial of the form $\frac{x^n-1}{g(x)}$, where $g$ is an irreducible factor of $x^n-1$, are called minimal cyclic codes. Thus, each minimal cyclic code is associated of natural form with an irreducible factor of $x^n-1$ in ${\mathbb {F}}_q[x]$. An example of minimal cyclic code is the Golay code that was used on the Mariner Jupiter-Saturn Mission (see [7]), the BCH code used in communication systems like VOIP telephones and Reed–Solomon code used in two-dimensional bar codes and storage systems like compact disc players, DVDs, disk drives, etc (see [5, Sects. 5.8 and 5.9]). The advantage of the cyclic codes, with respect to other linear codes, is that they have efficient encoding and decoding algorithms (see [5, Sect. 3.7]).

For each element of $g\in {\mathcal {R}}_n$, $\omega (g)$ is defined as the number of non-zero coefficients of $g$ and is called the Hamming weight of the word $g$. Denote by $A_j$ the number of codewords with weight $i$ and by $d=\min \{i>0| A_i\ne 0\}$ the minimum distance of the code. A $[n,k]_q$-code with minimum distance $d$ will be denoted by $[n,k,d]_q$-code. The sequence $\{A_i\}_{i=0}^n$ is called the weight distribution of the code and $A(z):=\sum _{i=0}^n A_iz^i$ is its weight enumerator. The importance of the weight distribution is that it allows us to measure the probability of non-detecting an error of the code: For instance, the probability of undetecting an error in a binary symmetric channel is $\sum \limits _{i=0}^n A_i p^i(1-p)^{n-i}$, where $p$ is the probability that, when the transmitter sends a binary symbol ($0$ or $1$), the receptor gets the wrong symbol.

The weight distribution of irreducible cyclic codes has been determined for a small number of special cases. For a survey about this subject see [3, 4] and their references.

In this article, we show all the possible weight distributions of length $n$ over a finite field ${\mathbb {F}}_q$ in the case that every prime divisor of $n$ divides $q-1$.

2 Preliminaries

Throughout this article, ${\mathbb {F}}_q$ denotes a finite field of order $q$, where $q$ is a power of a prime, $n$ is a positive integer such that $\text{ gcd }(n,q)=1$, $\theta$ is a generator of the cyclic group ${\mathbb {F}}_q^*$ and $\alpha$ is a generator of the cyclic group ${\mathbb {F}}_{q^2}^*$ such that $\alpha ^{q+1}=\theta$. For each $a\in {\mathbb {F}}_q^*$, $\mathop {\mathrm{ord}}\nolimits _q a$ denotes the minimal positive integer $k$ such that $a^k=1$, for each prime $p$ and each integer $m$, $\nu _p(m)$ denotes the maximal power of $p$ that divides $m$ and $\mathop {\mathrm{rad}}(m)$ denotes the radical of $m$, i.e., if $m=p_1^{\alpha _1}p_2^{\alpha _2},\ldots , p_l^{\alpha _l}$ is the factorization of $m$ in prime factors, then $\mathop {\mathrm{rad}}(m)=p_1p_2,\ldots , p_l$. Finally, $a_{_{\div b}}$ denotes the integer $\frac{a}{\gcd (a,b)}$.

Since each irreducible factor of $x^n-1\in {\mathbb {F}}_q[x]$ generates an irreducible cyclic code of length $n$, then a fundamental problem of code theory is to characterize these irreducible factors. The problem of finding a “generic algorithm” to split $x^n-1$ in ${\mathbb {F}}_q[x]$, for any $n$ and $q$, is an open one and only some particular cases are known. Since $x^n-1=\prod _{d|n} \varPhi _d(x)$, where $\varPhi _d(x)$ denotes the $d$-th cyclotomic polynomial (see [8] Theorem 2.45), it follows that the factorization of $x^n-1$ strongly depends on the factorization of the cyclotomic polynomial that has been studied by several authors (see [6, 9, 11] and [2]).

In particular, a natural question is to find conditions in order to have all the irreducible factors binomials or trinomials. In this direction, some results are the following ones

Lemma 1

[1, Corollary 3.3] Suppose that

• 1. $\mathop {\mathrm{rad}}(n)|(q-1)$ and

• 2. $8\not \mid n$ or $q\not \equiv 3 \pmod 4$.

Then the factorization of $x^n-1$ in irreducible factors of ${\mathbb {F}}_q[x]$ is

$$\begin{aligned} \prod _{t|m}{\mathop {\mathop {\prod }\limits _{{1\le u\le \gcd (n,q-1)}}}\limits _{\gcd (u,t)=1}} (x^t-\theta ^{ul}), \end{aligned}$$

where $m=n_{_{\div (q-1)}}$ and $l=(q-1)_{_{\div n}}$. In addition, for each $t$ such that $t|m$, the number of irreducible factors of degree $t$ is $\frac{\varphi (t)}{t}\cdot \gcd (n,q-1)$, where $\varphi$ denotes the Euler Totient function.

Lemma 2

[1, Corollary 3.6] Suppose that

• 1. $\mathop {\mathrm{rad}}(n)|(q-1)$ and

• 2. $8\mid n$ and $q\equiv 3 \pmod 4$.

Then the factorization of $x^n-1$ in irreducible factors of ${\mathbb {F}}_q[x]$ is

$$\begin{aligned} {\mathop {\mathop {\prod }\limits _{t|m'}}\limits _{t\mathrm{odd}}}{\mathop {\mathop {\prod }\limits _{{1\le w\le \gcd (n,q-1)}}} \limits _{\gcd (w,t)=1}} (x^t-\theta ^{wl})\cdot \prod _{t|m'}\prod _{ u\in {\mathcal {S}}_t} \Big (x^{2t}-(\alpha ^{ul'}+\alpha ^{qul'})x^t+\theta ^{ul'}\Big ), \end{aligned}$$

where $m'=n_{_{\div (q^2-1)}}$ and $l=(q-1)_{_{\div n}}$, $l'=(q^2-1)_{_{\div n}}$, $r=\min \{\nu _2(\frac{n}{2}),\nu _2(q+1)\}$ and ${\mathcal {S}}_t$ is the set

$$\begin{aligned} \left\{ u\in {\mathbb {N}}\left| \begin{array}{l}{1\le u\le \gcd (n, q^2-1), \gcd (u,t)=1} \\ {2^r\not \mid u \text { and }\ u< \{qu\}_{\gcd (n,q^2-1))}}\end{array}\right. \right\} , \end{aligned}$$

where $\{a\}_b$ denotes the remainder of the division of $a$ by $b$, i.e., it is the number $0\le c<b$ such that $a\equiv c\pmod b$ .

Moreover, for each $t$ odd such that $t|m'$, the number of irreducible binomials of degree $t$ and $2t$ is $\dfrac{\varphi (t)}{t}\cdot \gcd (n,q-1)$ and $\dfrac{\varphi (t)}{2t}\cdot \gcd (n,q-1)$ respectively, and the number irreducible trinomials of degree $2t$ is

$$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{\varphi (t)}{t}\cdot 2^{r-1} \gcd (n,q-1),&{} \quad \text {if } t \text { is even}\\ \dfrac{\varphi (t)}{t}\cdot (2^{r-1}-1) \gcd (n,q-1),&{} \quad \text {if } t \text { is odd}. \end{array}\right. } \end{aligned}$$

3 Weight distribution

Throughout this section, we assume that $\mathop {\mathrm{rad}}(n)$ divides $q-1$ and $m$, $m'$ $l$, $l'$ and $r$ are as in the Lemmas 1 and 2. The following results characterize all the possible cyclic codes of length $n$ over ${\mathbb {F}}_q$ and show explicitly the weight distribution in each case.

Theorem 1

If $8\not \mid n$ or $q\not \equiv 3 \pmod 4$, then every irreducible code of length $n$ over ${\mathbb {F}}_q$ is an $[n, t, \frac{n}{t}]_q$-code where $t$ divides $m$ and its weight enumerator is

$$\begin{aligned} A(z)=\sum _{j=0}^{t} {t\atopwithdelims ()j}(q-1)^j z^{j\frac{n}{t}}=\Big (1+(q-1)z^{\frac{n}{t}}\Big )^t. \end{aligned}$$

Proof

As a consequence of Lemma 1, every irreducible factor of $x^n-1$ is of the form $x^t-a$ where $t|n$ and $a^{n/t}=1$, so every irreducible code $\mathcal C$ of length $n$ is generated by a polynomial of the form

$$\begin{aligned} g(x)=\frac{x^n-1}{x^t-a}=\sum _{j=0}^{n/t-1} a^{\frac{n}{t}-1 -j} x^{t j} \end{aligned}$$

and $\{g(x), xg(x), \dots , x^{t-1}g(x)\}$ is a basis of the ${\mathbb {F}}_q$-linear subspace $\mathcal C$. Thus, every codeword in $\mathcal C$ is of the form $a_0g+a_1xg+\cdots +a_{t -1} x^{t-1}g$, with $a_j\in {\mathbb {F}}_q$ and

$$\begin{aligned} \omega \left( a_0g+a_1xg+\cdots +a_{t -1} x^{t-1}g\right) = \omega (a_0g)+ \omega (a_1xg)+\cdots + \omega \left( a_{t -1} x^{t-1}g\right) . \end{aligned}$$

Since $\omega (g)=\frac{n}{t}$, it follows that

$$\begin{aligned} \omega \left( a_0g+a_1xg+\cdots +a_{t -1} x^{t-1}g\right) =\frac{n}{t} \#\{j| a_j\ne 0\}. \end{aligned}$$

Clearly we have ${A}_k = 0$, for all $k$ that is not divisible by $\frac{n}{t}$. On the other hand, if $k = j\frac{n}{t}$, then exactly $j$ elements of this base have non-zero coefficients in the linear combination and each non-zero coefficient can be chosen of $q-1$ distinct forms. Hence ${A}_k = {t \atopwithdelims ()j} (q-1)^ j.$ Then the weight distribution is

$$\begin{aligned} A_k = {\left\{ \begin{array}{ll} 0, &{} \quad \hbox {if } t \not \mid k \\ {t \atopwithdelims ()j}(q-1)^j, &{} \quad \hbox {if }k = j\frac{n}{t}, \end{array}\right. } \end{aligned}$$

as we want to prove. $\square$

Remark 1

The previous result generalizes Theorem 3 in [10] (see also Theorem 22 in [4]).

Remark 2

As a direct consequence of Lemma 1, for all positive divisor $t$ of $m$, there exist $\frac{\varphi (t)}{t} \gcd (n,q-1)$ irreducible cyclic $[n,t,\frac{n}{t}]_q$-codes.

In order to find the weight distribution in the case that $q\equiv 3 \pmod 4$ and $8|n$, we need some additional lemmas.

Lemma 3

Let $t$ be a positive integer such that $t$ divides $m'$ and assume that $q\equiv 3 \pmod 4$ and $8\not \mid n$. If $x^{2t}-(a+a^q)x^t+a^{q+1} \in {\mathbb {F}}_q[x]$ is an irreducible trinomial, where $a=\alpha ^{ul'}\in {\mathbb {F}}_{q^2}$, and $g(x)$ is the polynomial $\dfrac{x^{n}-1}{x^{2t}-(a + a^q)x^t + a^{q+1}}\in {\mathbb {F}}_q[x]$, then $\nu _2(u)\le r-2$ and

$$\begin{aligned} \omega (g(x)-\lambda x^tg(x))={\left\{ \begin{array}{ll} \frac{n}{t}\left( 1-\frac{1}{2^{r-\nu _2(u)} }\right) ,&{} \quad \text {if }\, \lambda \in \Lambda _u\\ \frac{n}{t},&{} \quad \text {if }\, \lambda \notin \Lambda _u, \end{array}\right. } \end{aligned}$$

where $\Lambda _u=\left\{ \left. \dfrac{a^{i}-a^{qi}}{a^{i+1} - a^{q(i+1)}} \right| i=0,1,\ldots , 2^{r-\nu _2(u)}-2\right\}$.

Proof

Since $x^{2t}-(a+a^q)x^t+a^{q+1}$ is an irreducible trinomial in ${\mathbb {F}}_q[x]$, then $\gcd (t,u)=1$, $2^r\not \mid u$ and $a\ne -a^q$. In particular, $\mathop {\mathrm{ord}}\nolimits _{q^2} a$ does not divide either $q-1$ or $2(q-1)$. Observe that

$$\begin{aligned}\mathop {\mathrm{ord}}\nolimits _{q^2} a&=\frac{q^2-1}{\gcd ( q^2-1, ul')}=\frac{q^2-1}{\gcd \left( q^2-1, u \frac{q^2-1}{\gcd (q^2-1,n)}\right) }\\&=\frac{\gcd (q^2-1, n)}{\gcd (q^2-1,n,u)}\\&=\frac{2^r\gcd (q-1, n)}{\gcd ( 2^r(q-1),n,u)}, \end{aligned}$$

and for each odd prime $p$, we have

$$\begin{aligned} \nu _p\left( \frac{2^r\gcd (q-1, n)}{\gcd ( 2^r(q-1),n,u)}\right) \le \nu _p(\gcd (q-1, n))\le \nu _p(q-1). \end{aligned}$$

(1)

Therefore, $\mathop {\mathrm{ord}}\nolimits _{q^2} a\not \mid 2(q-1)$ implies that $\nu _2(\mathop {\mathrm{ord}}\nolimits _{q^2} a)>\nu _2(2(q-1))=2$, and since

$$\begin{aligned} \nu _2 \left( \frac{2^r\gcd (q-1, n)}{\gcd ( 2^r(q-1),n,u)}\right)&=r+1-\min \Big \{\nu _2(\gcd ( 2^r(q-1),n)),\nu _2(u))\Big \}\\&=r+1-\min \{r+1,\nu _2(u))\}=r+1-\nu _2(u), \end{aligned}$$

we conclude that $\nu _2(u)\le r-2$.

On the other hand

$$\begin{aligned} g(x)&= \frac{x^{n}-1}{x^{2t} - (a + a^q)x^t + a^{q+1}}\\&= \frac{x^{n}-1}{a-a^q}\left( \frac{1}{x^t-a} - \frac{1}{x^t-a^q} \right) \\&= \displaystyle {\sum _{j=1}^{n/t-1}\left( \frac{a^{j}-a^{qj}}{a-a^q}\right) x^{ n-t-tj} },\\ \end{aligned}$$

is a polynomial whose degree is $n-2t$ and every non-zero monomial is such that its degree is divisible by $t$. Now, suppose that there exist $1\le i<j\le \frac{n}{t} -2$ such that the coefficients of the monomials $x^{n-t-jt}$ and $x^{n-t-it}$ in the polynomial $g_{\lambda }:=g(x)-\lambda x^t g(x)$ are simultaneously zero. Then

$$\begin{aligned} \frac{a^{j} - a^{qj}}{a - a^q} = \lambda \frac{a^{j+1} - a^{q(j+1)}}{a - a^q} \quad \hbox {and}\quad \frac{a^{i} - a^{qi}}{a - a^q} = \lambda \frac{a^{i+1} - a^{q(i+1)}}{a - a^q} . \end{aligned}$$

So, in the case of $\lambda \ne 0$, we have

$$\begin{aligned} \lambda = \frac{a^{j}-a^{qj}}{a^{j+1} - a^{q(j+1)}} = \frac{a^{i}-a^{qi}}{a^{i+1} - a^{q(i+1)}}. \end{aligned}$$

This last equality is equivalent to $a^{(q-1)(j-i)} = 1$, i.e., $\mathop {\mathrm{ord}}\nolimits _{q^2} a$ divides $(q-1)(j-i)$. In the case of $\lambda =0$, we obtain that $\mathop {\mathrm{ord}}\nolimits _{q^2} a$ divides $(q-1)j$ and $(q-1)i$ by the same argument. Therefore, we can treat this case as a particular case of the above one making $i = 0$. It follows that $\frac{2^r\gcd (q-1, n)}{\gcd ( 2^r(q-1),n,u)}$ divides $(q-1)(j-i)$.

So, by Eq. (1), the condition $\mathop {\mathrm{ord}}\nolimits _{q^2} a|(q-1)(j-i)$ is equivalent to

$$\begin{aligned} \nu _2 \left( \frac{2^r\gcd (q-1, n)}{\gcd ( 2^r(q-1),n,u)}\right) =r+1-\nu _2(u)\le \nu _2((p-1)(j-i))=1+\nu _2(j-i), \end{aligned}$$

and thus $2^{r-\nu _2(u)}|(j-i)$.

In other words, if the coefficient of the monomial of degree $n-t-it$ is zero, then all the coefficients of the monomials of degree $n-t-jt$ with $j\equiv i\pmod {2^{r-\nu _2(u)}}$ are zero. Thus, if $\lambda \notin \Lambda _u$, then any coefficient of the form $x^{tj}$ is zero and the weight of $g_{\lambda }$ is $\frac{n}{t}$. Otherwise, exactly $\frac{n}{t}\cdot \frac{1}{2^{r-\nu _2(u)}}$ coefficients of the monomials of the form $x^{tj}$ are zero, then the weight of $g_{\lambda }$ is $\frac{n}{t}(1-\frac{1}{2^{r-\nu _2(u)}})$, as we want to prove. $\square$

Corollary 1

Let $g$ be a polynomial in the same condition of Lemma 3. Then

$$\begin{aligned} \# \left\{ (\mu ,\lambda ) \in {\mathbb {F}}_q^2 \left| \omega (\mu g(x) + \lambda x^{t}g(x)) = \dfrac{n}{t}\left( 1- \frac{1}{2^{r-\nu _2(u)}}\right) \right. \right\} = 2^{r-\nu _2(u)}(q-1). \end{aligned}$$

Proof

If $\mu = 0$ and $\lambda \ne 0$, then $\omega (\lambda x^{t}g(x)) = \frac{n}{t}(1-\frac{1}{2^{r-\nu _2(u)}})$ and we have $(q-1)$ ways to choose $\lambda$.

Suppose that $\mu \ne 0$, then $\omega (\mu g(x) + \lambda x^{t}g(x)) = \omega (g(x) + \frac{\lambda }{\mu }x^{t}g(x)),$ i.e., the weight only depends on the quotient $\frac{\lambda }{\mu }$. By Lemma 3, there exist $2^{r-\nu _2(u)}-1$ values of $\frac{\lambda }{\mu }$ such that $g(x) + \frac{\lambda }{\mu } x^tg(x)$ has weight $\frac{n}{t}(1-\frac{1}{2^{r-\nu _2(u)}})$, so we have $(q-1)(2^{r-\nu _2(u)}-1)$ pairs of this type. $\square$

Theorem 2

If $8|n$ and $q\equiv 3 \pmod 4$, then every irreducible code of length $n$ over ${\mathbb {F}}_q$ is one of the following class:

• (a) A $[n, t, \frac{n}{t}]_q$-code, where $4\not \mid t$, $t|m'$ and its weight enumerator is

  $$\begin{aligned} A(z)=\sum _{j=0}^{t} {t \atopwithdelims ()j}(q-1)^j z^{j \frac{n}{t}}=\Big (1+(q-1)z^{\frac{n}{t}}\Big )^t. \end{aligned}$$

• (b) A $[n,2t, d ]_q$-code, where $t|m'$, $d=\frac{n}{t}(1-\frac{1}{2^{r-\nu _2(u)}})$, $0\le u\le r-2$ and its weight enumerator is

  $$\begin{aligned} A(z)=\left( 1+2^{r-\nu _2(u)}(q-1)z^d+(q-1)(q+1-2^{r-\nu _2(u)})z^{\frac{n}{t}}\right) ^t. \end{aligned}$$

  In particular, if $\frac{n}{t2^{r-\nu _2(u)}}\not \mid k$, then $A_k=0$.

Proof

Observe that every irreducible code is generated by a polynomial of the form $\frac{x^n-1}{x^t-a}$, where $a\in {\mathbb {F}}_q$, or a polynomial of the form $g(x)=\frac{x^n-1}{(x^t-a)(x^t-a^q)}$, where $a$ satisfies the condition of Lemma 3. In the first case, the result is the same as Theorem 1. In the second case, each codeword is of the form

$$\begin{aligned} \sum _{j=0}^ {2t-1} \lambda _j x^j g(x)=\sum _{j=0}^{t-1} h_j, \end{aligned}$$

where $h_j=\lambda _j x^jg(x)+\lambda _{t+j} x^{t+j}g(x)$. Since, for $0\le i<j\le t-1$, the polynomial $h_i$ and $h_j$ do not have non-zero monomials of the same degree, it follows that

$$\begin{aligned} \omega \left( \sum _{j=0}^{t-1} h_j\right) =\sum _{j=0}^{t-1} \omega (h_j). \end{aligned}$$

By Lemma 3, $h_j$ has weight $\frac{n}{t}$, $d$ or $0$, for all $j = 0, \dots , t-1$. For each $j=0,1,\dots ,t-1$, there exist $(q^2-1)$ non-zero pairs $(\lambda _j, \lambda _{j+t})$ , and by Corollary 1, we know that there exist $2^{r-\nu _2(u)}(q-1)$ pairs with weight $d$. Therefore, there exist

$$\begin{aligned} q^2-1 -2^{r-\nu _2(u)} (q-1) = (q-1)(q+1-2^{r-\nu _2(u)}) \end{aligned}$$

pairs with weight $\frac{n}{t}$.

So, in order to calculate $A_k$, we need to select the polynomials $h_l$’s which have weight $d=\frac{n}{t}(1-\frac{1}{2^{r-\nu _2(u)}})$ and those ones which have weight $\frac{n}{t}$ in such a way that the total weight is $k$.

If we chose $i$ of the first type and $j$ of the second type, the first $h_l$’s can be chosen by ${t \atopwithdelims ()i}(2^{r-\nu _2(u)}(q-1))^i$ ways and for the other $t-i$ ones, there are ${{t-i} \atopwithdelims ()j}((q-1)(q+1-2^{r-\nu _2{u}}))^j$ ways of choosing $j$ with weight $\frac{n}{t}$. The remaining $h_j$’s have weight zero. Therefore

$$\begin{aligned} A_k = {\mathop {\mathop {\sum }\limits _{k = di + \frac{n}{t}j}}\limits _{0\le i+ j\le t}}{t \atopwithdelims ()i}\left( 2^{r-\nu _2(u)}(q-1)\right) ^i{{t-i} \atopwithdelims ()j}\left( (q-1)(q+1-2^{r-\nu _2(u)})\right) ^j,\\ \end{aligned}$$

and

$$\begin{aligned} A(z)&=\sum _{0\le i+ j\le t}{t \atopwithdelims ()i,j}\left( 2^{r-\nu _2(u)}(q-1)z^d\right) ^i\left( (q+1-2^{r-\nu _2(u)})(q-1)z^{\frac{n}{t}}\right) ^j\\&=\left( 1+2^{r-\nu _2(u)}(q-1)z^d+(q-1)(q+1-2^{r-\nu _2(u)})z^{\frac{n}{t}}\right) ^t. \end{aligned}$$

In particular, the minimum distance is $d$ and every non-zero weight is divisible by $\gcd (d,\frac{n}{t})=\frac{n}{t2^{r-\nu _2(u)}}$. $\square$

Remark 3

As a direct consequence of Lemma 2, for all positive divisor $t$ of $m'$, there exist $2^{r-1-\nu _2(u)}\frac{\varphi (t)}{t} \gcd (n,q-1)$ irreducible cyclic $[n,t,d]_q$-codes if $t$ is odd, and $2^{r-1}\frac{\varphi (t)}{t} \gcd (n,q-1)$ irreducible cyclic $[n,2t,\frac{n}{t}(1-\frac{1}{2^r})]_q$-codes if $t$ is even.

Example 1

Let $q=31$ and $n=288=2^5\times 3$. Then $m'=3$, $l'=10$, $r=4$. If $h(x)$ denotes a irreducible factor of $x^{288}-1$, then $h(x)$ is a binomial of degree $1$, $2$, $3$ or $6$, or a trinomial of degree $2$ or $6$. The irreducible codes generated by $\frac{x^{n}-1}{h(x)}$ (and therefore parity check polynomial is $h$), and its weight enumerators are shown in the following tables

 Codes generated by binomials

$\,[n,t,\frac{n}{t}]_q$-code	$h(x)$	Weight enumerator
$\,[288,1,288]_{31}$	$\begin{array}{c} x + 1\\ x + 5\\ x + 6\\ x + 25\\ x + 26\\ x + 30\\ \end{array}$	$1+30z^{288}$
$\,[288,2,144]_{31}$	$\begin{array}{c} x^2 + 1\\ x^2 + 5\\ x^2 + 25\\ \end{array}$	$(1+30z^{144})^2$
$[288,3,96]_{31}$	$\begin{array}{c}x^3 + 5\\ x^3 + 6\\ x^3 + 25\\ x^3 + 26\\ \end{array}$	$(1+30z^{96})^3$
$\,[288,6,48]_{31}$	$\begin{array}{c} x^6 + 5\\ x^6 + 25\\ \end{array}$	$(1+30z^{48})^6$