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arXiv 2203.05451math.NTmath.CO
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关于$n$和$n^2$的二进制数

On the binary digits of $n$ and $n^2$

Karam Aloui, Damien Jamet, Hajime Kaneko, Steffen Kopecki, Pierre Popoli, Thomas Stoll

摘要:

令$s(n)$表示整数$n$的二进制扩展中的数字之和。Hare,Laishram and Stoll(2011)研究了奇数的个数,使得对于给定的整数$k\geq 1$,有$s(n)=s(n^2)=k$。这些作者无法治疗的其余病例为$k\in\{9,10,11,14,15\}$。在本文中,我们表明,只有有限数量的解决方案$k\in\{9,10,11\}$和评论的困难,解决剩下的两个案件$k\in\{14,15\}$。一个相关的问题是研究$s(n^2)=4$对奇数的解。Bennett,Bugeaud and Mittette(2012)证明了只有两个解,并证明了$n=13,15,47,111$是唯一的解。本文给出了一个算法,证明了这一猜想,并给出了$s(n^2)=5$的相关结果。

英文摘要:

Let $s(n)$ denote the sum of digits in the binary expansion of the integer $n$. Hare, Laishram and Stoll (2011) studied the number of odd integers such that $s(n)=s(n^2)=k$, for a given integer $k\geq 1$. The remaining cases that could not be treated by theses authors were $k\in\{9,10,11,14,15\}$. In this paper we show that there is only a finite number of solutions for $k\in\{9,10,11\}$ and comment on the difficulties to settle the two remaining cases $k\in\{14,15\}$. A related problem is to study the solutions of $s(n^2)=4$ for odd integers. Bennett, Bugeaud and Mignotte (2012) proved that there are only finitely many solutions and conjectured that $n=13,15,47,111$ are the only solutions. In this paper, we give an algorithm to find all solutions with fixed sum of digits value, supporting this conjecture, as well as show related results for $s(n^2)=5$.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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