We derive the coding capacity for duplication-correcting codes capable of correcting any number of duplications. We do so both for reverse-complement duplications, as well as palindromic (reverse) duplications. We show that except for duplication-length $1$, the coding capacity is $0$. When the duplication length is $1$, the coding capacity depends on the alphabet size, and we construct optimal codes.

翻译：我们推导了能够纠正任意数量重复的重复纠正码的编码容量。我们同时针对反向互补重复以及回文（反向）重复进行了研究。结果表明，除了重复长度为$1$的情况外，编码容量为$0$。当重复长度为$1$时，编码容量取决于字母表大小，并且我们构造了最优码。