论文标题
在$ n $和$ n^2 $的二进制数字上
On the binary digits of $n$ and $n^2$
论文作者
论文摘要
令$ s（n）$表示整数$ n $的二进制扩展中的数字之和。 Hare，Laishram和Stoll（2011）研究了$ s（n）= s（n^2）= k $的奇数数量，对于给定的整数$ k \ geq 1 $。这些作者无法治疗的其余案例为$ k \ in \ {9,10,11,14,15 \} $。在本文中，我们表明$ k \ in \ {9,10,11 \} $中只有有限数量的解决方案，并评论了解决剩下的两个情况$ k \ in \ in \ {14,15 \} $的困难。一个相关的问题是研究奇数整数的$ s（n^2）= 4 $的解决方案。 Bennett，Bugeaud和Mignotte（2012）证明，只有有限的解决方案，并推测$ n = 13,15,47,111美元是唯一的解决方案。在本文中，我们给出了一种算法，以查找具有固定数字价值的所有解决方案，支持该猜想，并显示$ s（n^2）= 5 $的相关结果。
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 $s(n^2)=5$.
