Lucy's Moonlight

1RB0RD_0RC1RE_1RD0LA_1LE1LC_1RF0LD_---0RA (bbch)

Lucy's Moonlight is a probviously halting tetrational BB(6) Cryptid found by Racheline on 1 Mar 2025 (Discord link).

Analysis by Racheline

https://discord.com/channels/960643023006490684/1345810396136865822/1345820781363597312

A(x,y) := 0^inf (1011)^x 10 <A (01)^y 0^inf
B(x) := 0^inf 1^x B> 0^inf

A(x+1,3y) -> A(x,8y+3)
A(x+2,3y+1) -> A(x,8y+11)
A(x+2,3y+2) -> A(x,8y+12)
A(0,y) -> B(2y+1)
A(1,3y+1) -> A(4y+4,4)
A(1,3y+2) -> halt
B(3y) -> B(8y-2)
B(3y+1) -> A(2y,4)
B(3y+2) -> B(8y+6)

a is the sequence such that A(x,a_n) goes to A(x',a_(n+1)) in one step assuming x>=2
b is the sequence such that A(x,a_0) goes to A(x-b_n,a_n) in n rules (without using the A(0,y) or A(1,y) rules) assuming x>=b_n
c is the sequence such that A(c_n,a_0) goes to A(c_(n+1),a_0) after only one application of the A(0,y) or A(1,y) rules

f(3n) = 8n+3
f(3n+1) = 8n+11
f(3n+2) = 8n+12
a_0 = 4
a_(n+1) = f(a_n)
b_0 = 0
b_(n+1) = b_n+(1 if 3|a_n else 2)

g(n) = 8ceil(n/3)-2     (i.e. g(3n) = 8n-2 and g(3n+2) = 8n+6)
h(n) = largest i such that b_i <= n
c_0 = 14
if b_h(c_n) = c_n, then c_(n+1) = 2m where m is minimal such that 3m+1 = g^k(2a_h(c_n)+1) for some k
otherwise if a_h(c_n) = 3m+1, then c_(n+1) = 4m+4
otherwise halt after roughly (8/3)^(6c_n/5) steps