Equality Cases in the Tu-Deng Conjecture Bound

The Tu-Deng Conjecture, made in 2011, asserts that for $k\ge 2$ and $1\le t<2^k-1$, the set $$S_{t,k}=\{(a,b)\in \{0,1,\dots ,2^k-2{\}}^2:a+b\equiv t(\mathrm{m}\mathrm{o}\mathrm{d}{2}^k-1),\text{wt}(a)+\text{wt}(b)<k\}$$ satisfies $$\mathrm{\mid }{S}_{t,k}\mathrm{\mid }\le 2^{k-1}\mathrm{.}$$ Proofs of the conjecture were given independently by the author and by Liu, Luo and Xie in arXiv in 2026. Let $r=\text{wt}(t)$, and let $z=k-r$ be the number of zeros in the $k$-bit expansion of $t$. We prove that equality holds in the Tu-Deng Conjecture if and only if every cyclic distance between consecutive zeros is at least $z$. We give several other simple necessary and sufficient conditions, including a forbidden-pattern condition, a packing condition for the zero positions, a composition condition for the gaps, and equivalent conditions involving the coefficients of the polynomial $H_W(u,v)$ in the proof of the Tu-Deng conjecture. We also count the equality words having a prescribed number of zeros.