1RB1LA_0LC0RC_1LE1RD_1RE1RC_1LF0LA_---1LE

1RB1LA_0LC0RC_1LE1RD_1RE1RC_1LF0LA_---1LE (bbch) is a BB(6) Cryptid similar to Space Needle. It was discovered by Sheep on 7 April 2026.

Analysis by Daniel Yuan

https://discord.com/channels/960643023006490684/1448375857046360094/1490792769483051228

So I would say either

(0, 0) --> halt
(1, n) --> halt
(0, n) --> (2, n-1) for n >= 1
(2m, n) --> (3m+n-1, 2) for m >= 1
(2m+1, n) --> (m-1, m+n+3) for m >= 1


or

(0, 0) --> halt
(1, n) --> halt
(0, n+1) --> (2, n)
(2m+2, n) --> (3m+n+2, 2)
(2m+3, n) --> (m, m+n+4)


I like converting this into these rules:

(3, 0) --> halt
(4, n) --> halt
(3, n+1) --> (5, n)
(2m-1, n) --> (3m+n-4, 2) for m >= 3
(2m, n) --> (m, m+n+1) for m >= 3

That way we can see the effect of m dividing by 2
We can even reduce this to one number
Let A(n) = (n, 2)


Start at A(5), (the equivalent of (2, 2) using the original rules), then follow these rules:

A(2^k) --> halt
A(3*2^k) = (3*2^k, 2) --> (3, 3*(2^k-1)+2+k) --> (3, 3*(2^k)-1+k) --> (5, 3*(2^k)-2+k) --> (3*(2^k)+3+k), 2) --> A(3*2^k+3+k)
A((2m-1)*2^k) = ((2m-1)*2^k, 2) --> (2m-1, (2m-1)*(2^k-1)+2+k) --> (m-1+(2m-1)*2^k+k, 2) --> A((2m-1)*2^k+k+m-1) for m >= 3

Analysis by hipparcos

https://discord.com/channels/960643023006490684/1448375857046360094/1490909482493149306

in practice I think we only ever hit the $A((2m-1)*2^k)\to A((2m-1)*2^k+k+m-1)$ for $m\ge 3$ rule

so I think we can let $a=m-1$ and write this as the problem:

Let $f(b)=b+k+a$, where $k$ and $a$ are non-negative integers satisfying $b=(2a+1)\cdot 2^k$.

Now consider the iterated application of the function $f^{n+1}(b)=f(f^n(b))$, $f^0(b)=b$. Does there exist a non-negative integer $n$ such that $f^n(5)$ equals a power of 2 or 3 times a power of 2?