AI 中文总结
本文证明长度至少18且两字母出现偶数次的光滑二元词必为洗牌平方，并应用于Kolakoski序列，刻画其洗牌平方前缀的条件。
AI 中文摘要
二元词若其位置可划分为两个相同的子序列，则称为洗牌平方。一个定义在字母表$\{1,2\}$上的有限词，若反复进行有限游程长度差分，直到达到空词为止，始终保持二元性，则称该词为光滑词。本文证明：每个光滑二元词，若其中两个字母的出现次数均为偶数，则当词长至少为$18$时，该词必为洗牌平方。恰好有$34$个例外，其长度均不超过$16$。证明的无限部分采用一种奇偶返回原子归纳法：一个光滑的偶Parikh词可唯一分解为光滑奇偶原子，且每个这样的原子长度至多为$8$。一个有限的、可独立验证的证书覆盖了长度从$18$到$40$的情况。作为推论，经典Kolakoski序列的非空前缀是洗牌平方，当且仅当两个符号均出现偶数次，且其长度不为$4$或$8$。这样的前缀有无穷多个。事实上，整个无限Kolakoski序列可被划分为两个相同的无限子序列，且每个子序列中的单色块长度一致有界。
英文摘要
A binary word is a shuffle square if its positions can be partitioned into two identical subsequences. A finite word over $\{1,2\}$ is smooth if repeated finite run-length differentiation remains binary until the empty word is reached. This paper proves that every smooth binary word in which both letters have even multiplicity is a shuffle square once its length is at least $18$. There are exactly $34$ exceptions, all of length at most $16$. The infinite part of the proof is a parity-return atom induction: a smooth even-Parikh word factors uniquely into smooth parity atoms, and every such atom has length at most $8$. A finite, independently checkable certificate handles lengths $18$ through $40$. As a consequence, a nonempty prefix of the classical Kolakoski sequence is a shuffle square if and only if both symbols occur an even number of times and its length is not $4$ or $8$. There are infinitely many such prefixes. In fact, the entire infinite Kolakoski sequence can be partitioned into two identical infinite subsequences with monochromatic blocks of uniformly bounded length.
CommentsIncludes an exact finite certificate and standalone Python verifier. Research, computation, and writing performed by GPT-5.6 Sol and GPT-6-Astra under human supervision; see contribution statement
