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The weight distributions of constacyclic codes

The paper was supported by National Natural Science Foundation of China under Grants 11601475,61772015 and Foundation of Science and Technology on Information Assurance Laboratory under Grants KJ-15-009,6142112010202.
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  • Let $\Bbb F_q$ be a finite field with $q$ elements. Suppose that $a, λ∈ \Bbb F_q^*$, $a^n=λ$ with $n|(q-1)$. In this paper, we determine the weight distribution of a class of $λ$-constacyclic codes of length $nm$ with the parity check polynomial $h(x)=(x^m-aξ^{st})(x^m-aξ^{s(t+1)})...(x^m-aξ^{s(t+r-1)})$ and $n>(r-1)m$, where $s,t, r$ are positive integers and $ξ∈ \Bbb F_q$ is a primitive n-th root of unity. Moreover, we give the weight distributions of $λ$-constacyclic codes of length $nm$ explicitly in several cases: (1) $r=1$, $n>1$; (2) $r=2$, $m=2$ and $n>2$; (3) $r=2$, $m=3$ and $n>3$; (4) $r=3$, $m=2$ and $n>4$.

    Mathematics Subject Classification: Primary: 94B05, 11T71; Secondary: 11C08, 11C20.

    Citation:

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  • Table 1.  Weight distribution 

    Weight Frequency
    0 1
    $n-1$ $2n(q-1)$
    $n$ $2(q-1)(q+1-n)$
    $2n-2$ $n^2(q-1)^2$
    $2n-1$ $2n(q-1)^2(q+1-n)$
    $2n$ $(q-1)^2(q+1-n)^2$
     | Show Table
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    Table 2.  Weight distribution 

    Weight Frequency
    0 1
    $n-1$ $3n(q-1)$
    $n$ $3(q-1)(q+1-n)$
    $2n-2$ $3n^2(q-1)^2$
    $2n-1$ $6n(q-1)^2(q+1-n)$
    $2n$ $3(q-1)^2(q+1-n)^2$
    $3n-3$ $n^3(q-1)^3$
    $3n-2$ $3n^2(q-1)^3(q+1-n)$
    $3n-1$ $3n(q-1)^3(q+1-n)^2$
    $3n$ $(q-1)^3(q+1-n)^3$
     | Show Table
    DownLoad: CSV

    Table 3.  Weight distribution 

    Weight Frequency
    0 1
    $n-2$ $n(n-1)(q-1)$
    $n-1$ $2n(q-1)(q-n+2)$
    $n$ $2(q-1)^3-2(n-3)(q-1)^2+(n-2)(n-3)(q-1)$
    $2n-4$ $\frac {n^2(n-1)^2}4 (q-1)^2$
    $2n-3$ $n^2(q-1)^2(n-1)(q-n+2)$
    $2n-2$ $n^2(q-1)^2(q-n+2)^2+n(n-1)(q-1)^2[(q-1)^2-(n-3)(q-1)+C_{n-2}^2]$
    $2n-1$ $2n(q-n+2)(q-1)^4-2n(n-3)(q-n+2)(q-1)^3$
    $+n(n-2)(n-3)(q-n+2)(q-1)^2$
    $2n$ $(q-1)^6+(n-3)^2(q-1)^4+\frac {(n-2)^2(n-3)^2}4 (q-1)^2+(n-2)(n-3)(q-1)^4$
    $-2(n-3)(q-1)^5-(n-2)(n-3)^2(q-1)^3$
     | Show Table
    DownLoad: CSV
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