arXiv 2005.086512020-05-07math.NTcs.CRcs.ITmath.IT

由连续二次剩余差导出的二元序列

Binary sequences derived from differences of consecutive quadratic residues

Arne Winterhof, Zibi Xiao

摘要：

对于素数$p\ge 5$，令$q_0,q_1,\ldots,q_{(p-3)/2}$是以递增顺序模$p$的二次剩余。本文研究了由$d_n=q_n+q_{n+1}\bmod 2$和$t_n=1$定义的两个$(p-3)/2$ -周期二元序列$(d_n)$和$(t_n)$，如果$q_{n+1}=q_n+1$和$t_n=0$不是，则为$n=0,1,\ldots,(p-5)/2$ .对于这两个序列，我们找到了一些充分条件，达到最大的线性复杂度$(p-3)/2$。研究$(d_n)$的线性复杂度的动机是Caragiu等人的数学。然而，$(d_n)$是不平衡的，我们表明，一个周期的$(d_n)$包含约$1/3$零和$2/3$ 1，如果$p$是足够大的。相比之下，对于任何固定的$s$和足够大的$p$，$(t_n)$不仅基本上平衡，而且长度$s$的所有较长模式在载体序列$(t_n,t_{n+1},\ldots,t_{n+s-1})$、$n=0,1,\ldots,(p-5)/2$中基本上同样经常出现。

英文摘要：

For a prime $p\ge 5$ let $q_0,q_1,\ldots,q_{(p-3)/2}$ be the quadratic residues modulo $p$ in increasing order. We study two $(p-3)/2$-periodic binary sequences $(d_n)$ and $(t_n)$ defined by $d_n=q_n+q_{n+1}\bmod 2$ and $t_n=1$ if $q_{n+1}=q_n+1$ and $t_n=0$ otherwise, $n=0,1,\ldots,(p-5)/2$. For both sequences we find some sufficient conditions for attaining the maximal linear complexity $(p-3)/2$. Studying the linear complexity of $(d_n)$ was motivated by heuristics of Caragiu et al. However, $(d_n)$ is not balanced and we show that a period of $(d_n)$ contains about $1/3$ zeros and $2/3$ ones if $p$ is sufficiently large. In contrast, $(t_n)$ is not only essentially balanced but also all longer patterns of length $s$ appear essentially equally often in the vector sequence $(t_n,t_{n+1},\ldots,t_{n+s-1})$, $n=0,1,\ldots,(p-5)/2$, for any fixed $s$ and sufficiently large $p$.

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摘要：

英文摘要：

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