1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE

1RB0RF_1LC1RB_0RD0LB_1RZ0LE_1RE0RA_1RD1RE (bbch) halts with roughly $2^{2^{97/3}}$ steps.

Analysis by racheline

https://discord.com/channels/960643023006490684/1239205785913790465/1336829029948588113

It halts with roughly 2^(2^96/3)/3 1s on the tape and after roughly 2^(2^97/3) steps. it follows these rules:

A(n,m) := 111 E> 1^(3n) 0 1^m
A(n,6m+1) -> A(n+2m+1,2^(n+2m)+2^((2^(2m)-1)/3)+1)
A(n,3m+2) -> halt with n+m+6 1s
start from A(4,13)

A(4,13) -> A(9,289) -> A(106,2^105+2^((2^96-1)/3)+1)
 since both 105 and (2^96-1)/3 are odd, that is A(106,3m+2), which halts as soon as the machine gets to the right side of the tape

here are the details of the A(n,6m+1) case:

A(n,6m+1) =
111 E> 1^(3n) 0 1^(6m+1)                                                               --(translated cycles)-->
111 (001)^n 1 (001)^(2m) 0 A> 0                                                        --(counting with 2m bits)-->
111 (001)^n 1 (111)^(2m) 0 A> 0 1^(2^(2m)-1)                                           --(translated cycles)-->
111 (001)^(n-1) 111 (001)^(2m) 0 111 (001)^((2^(2m)-1)/3) 0 A> 0                       --(counting with (2^(2m)-1)/3 bits)-->
111 (001)^(n-1) 111 (001)^(2m) 0 (111)^((2^(2m)+2)/3) 0 A> 0 1^(2^((2^(2m)-1)/3)-1)    --(translated cycles)-->
111 (001)^(n-1) 111 (001)^(2m-2) 111 001 0 A> 0 1^(2^(2m)+2^((2^(2m)-1)/3)+3)          --(counting with n+2m bits)-->
111^(n+2m+1) 0 A> 0 1^(2^(n+2m)+2^((2^(2m)-1)/3))                                      --(translated cycles)-->
A(n+2m+1,2^(n+2m)+2^((2^(2m)-1)/3)+1)