改进的任意k和字母表大小下不同k-序列数量上界：通过计数独立参数
Improved upper bound on the number of distinct k-decks for any k and alphabet size by counting the independent parameters
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中文总结 AI 辅助
通过计数Lyndon词确定k-序列的仿射子空间维数，改进了不同k-序列数量的上界，并在前两个非平凡情形证明了匹配下界。
中文摘要 AI 辅助
存储在合成DNA中的数据通过鸟枪测序进行检索，该测序返回短子序列而非存储的单词本身。这种读出的自然抽象是单词的$k$-序列：记录每个长度为$k$的单词作为子序列出现次数的向量。两个存储的单词当且仅当它们的$k$-序列不同时才能从它们的读出中区分，因此，长度为$n$、字母表大小为$q$的单词的不同$k$-序列的数量$D_{q,k}(n)$衡量了长度为$k$的读出所保留的信息。我们分析了在所有较短序列固定后，$k$-序列中剩余的自由度。在具有指定字母多重性的每类单词中，长度为$k$的条目被限制在一个仿射子空间中，该子空间的维数恰好是具有相同多重性的Lyndon词的数量，我们以Möbius和的形式给出了其闭式表达式。记$L_q(j)$为长度为$j$、字母表大小为$q$的Lyndon词的数量，我们推导出改进的上界\\[ D_{q,k}(n)=O\\!\left(n^{E_q(k)}\right),\qquad E_q(k)=\sum_{j=1}^{k}j\\,L_q(j)-1. \\] 在二进制字母表的情况下，该上界满足$D_{2,k}(n)=O\\!\left(n^{4\cdot 2^{k-1}}\right)$。然后我们在前两个非平凡情形中证明了匹配的下界：对于每个字母表大小$q$，$D_{q,2}(n)=\Theta\\!\left(n^{q^2-1}\right)$，以及对于二进制字母表，$D_{2,3}(n)=\Theta(n^{9})$。后者证实了对于$q=2$和$k=3$，我们的猜想：对于每个$q$和$k$，上界具有正确的次数。
英文摘要
Data stored in synthetic DNA is retrieved by shotgun sequencing, which returns short subsequences rather than the stored word itself. A natural abstraction of this readout is the $k$-deck of a word: the vector recording how often each word of length $k$ occurs as a subsequence. Two stored words are distinguishable from their readouts exactly when their $k$-decks differ, so the number $D_{q,k}(n)$ of distinct $k$-decks of words of length $n$ over an alphabet of size $q$ measures what a length-$k$ readout retains. We analyse the degrees of freedom remaining in a $k$-deck once all shorter decks are fixed. Within each class of words having prescribed letter multiplicities, the length-$k$ entries are confined to an affine subspace whose dimension is exactly the number of Lyndon words with the same multiplicities, which we give in closed form as a Möbius sum. Writing $L_q(j)$ for the number of Lyndon words of length $j$ over an alphabet of size $q$, we deduce the improved upper bound \[ D_{q,k}(n)=O\!\left(n^{E_q(k)}\right),\qquad E_q(k)=\sum_{j=1}^{k}j\,L_q(j)-1 . \] In the case of a binary alphabet this bound satisfies $D_{2,k}(n)=O\!\left(n^{4\cdot 2^{k-1}}\right)$. We then prove matching lower bounds in the first two nontrivial cases: $D_{q,2}(n)=Θ\!\left(n^{q^2-1}\right)$ for every alphabet size $q$, and $D_{2,3}(n)=Θ(n^{9})$ for the binary alphabet. The latter confirms, for $q=2$ and $k=3$, our conjecture that the upper bound has the correct degree for every $q$ and $k$.
