Binomial permutations over finite fields with even characteristic
Binomial permutations over finite fields with even characteristic
复制标题
具有偶特征的有限域上的二项式排列
DOI：
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发表时间：
2021
期刊：
影响因子：
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通讯作者：
Yan Li
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文献类型：
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作者：
Ziran Tu;Xiangyong Zeng;Yupeng Jiang;Yan Li
In this paper, we study binomials having the form xr(a+x3(q-1))documentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$x^r(a+x^{3(q-1)})$$end{document} over the finite field Fq2documentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$mathbb {F}_{q^2}$$end{document} with q=2mdocumentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$q=2^m$$end{document}, and determine all the r’s and coefficients a’s making them permutations. For even m or odd m with 3∤mdocumentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$3
ot mid m$$end{document}, we prove that the characterization is necessary and sufficient. For the case of odd m and 3∣mdocumentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$3mid m$$end{document}, we prove that the corresponding sufficient condition is also necessary for almost all r’s. Finally we obtain that the proportion of r’s we cannot prove the necessity is only about 140documentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$frac{1}{40}$$end{document}.
In this paper, we study binomials having the form xr(a+x3(q-1))documentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$x^r(a+x^{3(q-1)})$$end{document} over the finite field Fq2documentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$mathbb {F}_{q^2}$$end{document} with q=2mdocumentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$q=2^m$$end{document}, and determine all the r’s and coefficients a’s making them permutations. For even m or odd m with 3∤mdocumentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$3
ot mid m$$end{document}, we prove that the characterization is necessary and sufficient. For the case of odd m and 3∣mdocumentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$3mid m$$end{document}, we prove that the corresponding sufficient condition is also necessary for almost all r’s. Finally we obtain that the proportion of r’s we cannot prove the necessity is only about 140documentclass[12pt]{minimal} usepackage{amsmath} usepackage{wasysym} usepackage{amsfonts} usepackage{amssymb} usepackage{amsbsy} usepackage{mathrsfs} usepackage{upgreek} setlength{oddsidemargin}{-69pt} egin{document}$$frac{1}{40}$$end{document}.