1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD

Unsolved problem:

Does this TM run forever?

1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD (bbch) is a Probviously Nonhalting BB(6) Cryptid first discovered and analyzed by @mxdys on 9 Jan 2025. Mxdys derived the lower level rules, and Racheline and Katelyn Doucette independently derived the higher level rules later on.

Determining whether this machine halts requires proving that a highly chaotically growing sequence never intersects with a value equal to $2^{n}-3$. It is believed to be a Cryptid, because proving this seems to require a deep understanding of how addition of two integers affects the number of factors of 2 in the sum, and/or much stronger ways of characterizing when two sequences can share terms in common. Both of which are believed to be out of reach of current mathematics.

Analysis

Low Level Rules

The original reported low level rules derived by @mxdys:

1RB1LA_1LC0RE_1LF1LD_0RB0LA_1RC1RE_---0LD

start: (3,1)
(0,2+c) --> (4+c,1)
(1,c) --> halt
(2+2b,c) --> (7+5b+c,1)
(3+2b,c) --> (b,4+b+c)

(b,c) := 0^inf <A 1^b 00 1^c 0^inf

Or the following rewritten form by Andrew Ducharme:

start: (3,1)
(1,c) --> halt
(2b,c) --> (2+5b+c,1)
(2b+1,c) --> (b-1,3+b+c)

(b,c) := 0^inf <A 1^b 00 1^c 0^inf

Higher Level Rules

Further derivation and analysis of the higher level rules took place on July 31 2025
The low level rules can be abstracted up a level by noticing that the c parameter always gets set to a consistent starting value of 1 when the rule triggered by b being even is encountered. It's possible to, through induction, generate a mapping from each $(b_{n},1)$ to the next $(b_{n+1},1)$. This allows one to completely eliminate the c parameter and have a set of rules entirely dependent on the value of b.

Rules derived by Katelyn Doucette:

Let n denote the exponent of the largest power of 2 that divides (b + 3)
START: b = 3

b --> b + n + 3/2 * ((b+3)/2^n - 1)

Halt if b = 2^n - 3

Alternative formulations can be derived by shifting b to b + 3.

Trajectory

At a high level, the machine's halting problem is governed by a highly chaotically growing sequence. The first few terms of the sequence are:

$${\begin{array}{|c|}\hline 3\rightarrow 7\rightarrow 14\rightarrow 38\rightarrow 98\rightarrow 248\rightarrow 623\rightarrow 1092\rightarrow 2733\rightarrow 2992\rightarrow \cdots \\\hline \end{array}}$$

This sequence has been calculated out to over 17 million terms, with the final value of b calculated before the program was terminated reaching $>10^{4,800,000}$. Not a single term was of the form $2^{n}-3$ that would cause halting. The largest value of n encountered was 24.

The growth rate of this sequence was characterized by @LegionMammal978 as below:
Assuming that the lower bits of b are random and uniform (which is well-supported empirically), it grows by an average log-factor of $0.652355$, which corresponds to $O(1.92006^{b})$, plus or minus a random walk on the log scale.

Heuristic Argument For Nonhalting

If you pick an arbitrary integer, the chance that it has n factors of 2 is ${\frac {1}{2^{n+1}}}$. Since the sequence grows roughly as fast as $2^{n}-3$, the chance of it hitting a large enough n decreases exponentially with the number of terms you go into the sequence. Thus, the chance of term k in the sequence halting is roughly ${\frac {1}{1.92^{k}}}$. By the time you get out to $k>17,000,000$, the probability of the sequences colliding is tiny.