1RB2LC1RC_2LC---2RB_2LA0LB0RA

https://bbchallenge.org/1RB2LC1RC_2LC---2RB_2LA0LB0RA

This is a BB(3, 3) holdout under active exploration. It simulates a complex set of Collatz-like rules with two decreasing parameters and seems as if it may be a new Cryptid (and perhaps even one that "probviously" halts! But this is really speculation at this point.)

NOTE: These rules are under active development and may have mistakes or typos.

Shawn's Rules

https://discord.com/channels/960643023006490684/1084047886494470185/1254307091863048264

We can reduce the set of rules from savask's list a bit by noticing that we can evaluate so that all rules end with c even:

  (0, b, 2c)    -> (2b+2, 1, 2c)

  (1, b, 0) -> Halt
  (1, 2b,   2c)  -> (0, 1, 2(b+c+1))
  (1, 2b+1, 2c)  -> (2, 1, 2(b+c+3))

  (a, 2b,   0)  -> (a-2, 2b+3, 0)
  (2, 2b+1, 0)  -> (0, 1, 2b+6)
  (a, 2b+1, 0)  -> (a-3, 1, 2b+4)

  (a, b, c) -> (a-2, b+3, c-2)

Phases

We can think of this going through two different phases. "Even Phase" (where a is even) and "Odd Phase" (where a is odd).

Even Phase: a,c even:
  (0, b, 2c) -> (2b+2, 1, 2c)
  (2a+2, 2b, 0) -> (2a, 2b+3, 0)
  (2, 2b+1, 0) -> (0, 1, 2(b+3))

  To Odd Phase:
    (2a+4, 2b+1, 0) -> (2a+1, 1, 2b+4)
 
Odd Phase: a odd, c even
  To Halt:
    (1, b, 0) -> Halt
    (3, 2b, 0) -> (1, 2b+3, 0) -> Halt

  To Even Phase:
    (1, 2b, 2c+2) -> (0, 1, 2(b+c+2))
    (1, 2b+1, 2c+2) -> (0, 1, 2b+2c+5) -> (2, 1, 2(b+c+4))
    
    (2a+5, 2b, 0) -> (2a+3, 2b+3, 0) -> (2a, 1, 2b+6)
    (2a+3, 2b+1, 0)  -> (2a, 1, 2b+4)

So the only way for this to halt is if it is in "Even Phase" and hits (2k+8, 2k+1, 0) or (4k+12, 4k+3, 0) (which will lead to (1, b, 0) or (3, 2b, 0) eventually). If a is bigger or smaller, then "Odd Phase" will end going back to "Even Phase" again.

Repeated (0, b, 2c)

Let $f(n)=3n+4$, then $$(0,b,2c)\to (0,f(b),2(c-b-1))$$.

Let

$$h(n)=f^n(1)+1=3^{n+1}-1$$

$$g(n)=\sum _{k=0}^{n-1}h(k)=\frac{3}{2}(3^n-1)-n$$

Then if $c>g(n)$:

$$(0,1,2c)\to (0,f^n(1),2(c-g(n)))\to (2h(n),1,2(c-g(n)))$$

Repeated (0, 1, 2c)

https://discord.com/channels/960643023006490684/1084047886494470185/1254635277020954705

Let $C(n)=(0,1,2n)$ = 0^inf 1 <A2 22 (20)^2n 0^inf

$$C(g(n)+8k+1)\to C(g(n)+8k+1+n+9)$$

$$\mathrm{\forall }k:\frac{h(n)-45}{65}<k<\frac{h(n)-22}{38}$$

Notably, when 8 divides (n+1) then this rule can potentially be applied repeatedly.

Ex: if n = 7, then we get:

$$\mathrm{\forall }k\in [101,172]:C(3273+8k)\to C(3273+8(k+2))$$

And we see this starting with $C(4137)=C(3273+8\cdot 108)$ which repeats this rule until we get to $C(4665)=C(3273+8\cdot 174)$.

And as n gets way bigger, these ranges of repeat will increase exponentially.