AI 中文总结
本文给出Tu-Deng猜想的完整证明，通过将其计数转化为循环进位解的数量，结合计数函数分解与严格负半平面质量估计，证实了该猜想的结论。
AI 中文摘要
设$N=2^k-1$，$\text{wt}(n)$表示二进制汉明重量。Tu-Deng猜想指出，对每个$1\le t\le N-1$，满足$a+b\equiv t\pmod N$且$\text{wt}(a)+\text{wt}(b)<k$的对$(a,b)\in\{0,\ldots,N-1\}^2$的数量至多为$2^{k-1}$，目前已有部分结果。本文给出该猜想的完整证明：首先证明Tu-Deng计数等于满足$\text{wt}(B)-\text{wt}(A)<0$且$A+t\equiv B\pmod N$的循环进位解的数量；循环进位解的计数函数可分解为$C_v = 1+(X+Y-1)J_v+X^{\text{z}(v)+1}Y^{\text{o}(v)+1}$，其中$t=10v$是$t$的二进制展开（最低有效位在前），$J_v$枚举语言$\text{Sub}\mathbin{\dot\cup}\{u\in\partial_1\text{Sub}(v):u<_{\rm lex}v\}$；通过估计$C_v$的严格负半平面质量得到所需界。
英文摘要
Let $N=2^k-1$ and let $\operatorname{wt}(n)$ denote the binary Hamming weight. The Tu-Deng conjecture asserts that, for every $1\le t\le N-1$, at most $2^{k-1}$ pairs $(a,b)\in\{0,\ldots,N-1\}^2$ satisfy $a+b\equiv t\pmod N$ and $\operatorname{wt}(a)+\operatorname{wt}(b)<k$. Partial results are known. We give a complete proof of this conjecture. We first show that the Tu-Deng counts equals the number of cyclic carry solutions for which $\operatorname{wt}(B)-\operatorname{wt}(A)<0$ and $A+t\equiv B\pmod N$. The enumerator of the cyclic carry solutions factors as $$C_v = 1+(X+Y-1)J_v+X^{\operatorname{z}(v)+1}Y^{\operatorname{o}(v)+1},$$ where $t=10v$ is the binary expansion of $t$(least significant bits first) and $J_v$ enumerates the language $$\operatorname{Sub}(v)\mathbin{\dot\cup}\{u\in\partial_1\operatorname{Sub}(v):u<_{\rm lex}v\}.$$ Estimating the strict negative half-plane mass of $C_v$ gives the desired bound.
