An Exact All-Width Plateau for the Three-Summand Tu-Deng Count modulo $2^k-1$

Put $N=2^k-1$ and write $\mathrm{wt}$ for the binary Hamming weight. Let $F_k(t)$ count the ordered triples $(a,b,c)\in\{0,\dots,N-1\}^3$ with $a+b+c\equiv t \pmod N$ and $\mathrm{wt}(a)+\mathrm{wt}(b)+\mathrm{wt}(c)<k$, the three-summand analogue of the two-summand count of the Tu-Deng conjecture at the same modulus and weight budget. Evaluate $F_k(t)$ exactly on the targets $t$ whose $k$-bit cyclic word has exactly two zero digits, no two adjacent: is the value the same for every such $t$ at a given $k$, and what is it?

Answered in full: for every $k\ge4$, a target with exactly two nonadjacent zero digits has $F_k(t)=(k+23)3^{k-4}$, independently of the distance between the zeros. Exact at every width, no error term, no hypothesis on $k$ (Theorem 1.1). This is an evaluation, not an extremal result, and the paper is explicit about the difference: the plateau value is not maximal. At $k=12$ it reads $35\cdot3^8=229{,}635$ while $F_{12}(110101101010)=293{,}499$ at five zero digits, so no global maximizer of $F_k$ is classified. The paper's other results are finite-layer and do not settle the extremal question: balancing monotonicity of $[x^{\le C}]H_t$ holds only for $C\le5$ (Theorem 1.3), and the all-mass statement is Conjecture 8.1, which the paper states outright does not follow from Theorem 1.3. The chamber where zero digits are adjacent is not addressed.