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.

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)

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 $C(4137)=C(3273+8*108)$ which repeats this rule until we get to $C(4665)=C(3273+8*174)$.

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