Three-summand Tu–Deng counts with binary-weight constraints

Modulo 2ᵏ−1, let Fₖ(t) count ordered triples summing to t whose total binary weight is below k. What are its exact values, especially when the k-bit target has exactly two nonadjacent zero digits?

Why it matters and what counts as progress

Why it matters: An all-width formula gives a three-summand analogue of structures studied in the two-summand Tu–Deng problem.

What counts as progress: A proof of the plateau (k+23)3^(k−4) for k≥4 on that target family, or broader exact values; partial carry balancing does not classify global maximizers.

How the problem developed

2011 · Conjecture [1]
Ziran Tu and Yingpu Deng Introduced the two-summand binary-string conjecture that motivates the later cyclic-carry counting problem.
2019 · Partial result [2]
Lukas Spiegelhofer and Michael Wallner Proved the two-summand Tu–Deng conjecture for an asymptotic-density-one set of targets; the result counts pairs, not triples.
18 Oct 2026 · Reported claim [3]
GPT-5.6 Sol and the manuscript author Reported an exact three-summand count for targets with two nonadjacent zero digits, plus balancing results only through carry mass five.

AI contribution

An exact plateau for three binary summands modulo a Mersenne number, with finite-layer carry balancing

Anonymous

For k≥4, the ordered three-summand count Fₖ(t) modulo 2ᵏ−1, with total binary weight below k, equals (k+23)3^(k−4) when the k-bit target has exactly two nonadjacent zero digits. The paper proves balancing only through carry mass five and does not classify global maximizers.

AI involvementAI-autonomous

ImportanceModest

Technical verificationPartial

Human understandingFew