Lucy's Moonlight

1RB0RD_0RC1RE_1RD0LA_1LE1LC_1RF0LD_---0RA (bbch) is a probviously halting tetrational BB(6) Cryptid found by Racheline on 1 Mar 2025 (Discord link).

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)
c_0 = 14(?)
c_(n+1) is somewhere around the smallest g^k(a_i) that is (something) mod 3, where i is maximal such that b_i <= c_n
where g is similar to f, just with probably different constants and it only has two defined cases 
so c_(n+1) is around (8/3)^(3/5*c_n)
but it actually depends
if b_i = c_n then c_(n+1) should have exactly that form (*some constant ±some constant) always 
if b_i = c_n-1 (which is the only other option) then it depends on a_i mod 3
for one value of a_i mod 3 it halts, for another you directly get c_(n+1) without iterating g, and for the remaining one you do iterate g but the input is probably slightly different
so
the probability of b_i = c_n-1 is 2/5 since the average b_(m+1)-b_m is 5/3
and if that happens, the probability of halting is 1/3
so probability of halting is 2/15 at each step
which means probability of beating the champion is (13/15)^(something around 15)
wait thats actually a pretty good chance
we should be able to compute c_2 (and check if it exists or if the TM halts before that)
if it doesnt halt, i think c_2 should be 10^(something in the hundreds/thousands), so with the information that it doesnt halt there, we will know the TM runs for >10^^3 steps, which means (depending on whether the top of the tower is smaller or larger than that of the current champion) we need either 12 or 13 more elements of the c sequence
in the case of 13, thats a ~15.56% chance of beating the champion, and in the case of 12, its ~17.95%