Antihydra nonhalting conjecture

Prove that, for the Collatz-like map H: N → N defined by H(x) = 3x/2 when x is even and H(x) = 3(x−1)/2 when x is odd, iterating H starting at x = 8 never yields strictly more than twice as many odd values as even values; equivalently, show that the 6-state, 2-symbol Turing machine with transition table 1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA (Antihydra) does not halt from the all-zero tape.

Background

The paper introduces Antihydra, a specific 6-state 2-symbol Turing machine, as a prototypical “Cryptid,” i.e., a machine whose halting behavior from the all-zero tape is believed to be mathematically hard. The authors relate its halting to a Collatz-like process on natural numbers, providing a number-theoretic formulation equivalent to the machine’s nonhalting.

A probabilistic analysis suggests an exceedingly small chance of halting, yet no formal proof is known. The authors present this as one of the smallest open problems in mathematics on the Busy Beaver scale and highlight its close analogy to the notoriously difficult Collatz conjecture.

References

Antihydra (https://bbchallenge.org/#1{1RB1RA_0LC1LE_1LD1LC_1LA0LB_1LF1RE_---0RA}; see Turing machine notation, Section~\ref{sec:TMs}) is a machine that does not halt from the all-zero tape if and only if the following Collatz-reminiscent conjecture holds: Consider the Collatz-like map $H: \mathbb{N} \to \mathbb{N}$ defined by $H(x) = 3\frac{x}{2}$ if $x$ is even and $H(x) = 3\frac{x-1}{2}$ if $x$ is odd. Iterating $H$ from $x=8$, there are never (strictly) more than twice as many odd numbers as even numbers.

Determination of the fifth Busy Beaver value  (2509.12337 - Collaboration et al., 15 Sep 2025) in Section 1.2 Discussion — Cryptids (Conjecture)