1RB0LF_1RC1RA_0RD0RB_1LE---_0LA1LE_1RD1LA

1RB0LF_1RC1RA_0RD0RB_1LE---_0LA1LE_1RD1LA (bbch) was a BB(6) potential Cryptid first analyzed by discord user Big_numbers, and then independently found by Marco Dos Santos using Gemini and reported in his paper At the Limits of Computability: Steps Towards Autonomous AI Discovery on the Busy Beaver Frontier. The probvious behavior of these TMs has not yet been analyzed. In November 2026, Yeyito, with assistance from Claude Opus 5.5, provided a Rocq proof that this machine halts with a runtime of about $6.4\times 10^{78}$.

Analysis by Big_numbers

A(a,b) = 0^inf 1^a 1A 1^b 01 0^inf
A(a,b)->A(2a+6, b-a-3)
If (b-a-3)<0:
    If b is 3:
        If a mod 4 is 0:
            A(a, 3)->A(a+8, 2)
        If a mod 4 is 2:
            A(a, 3)->A(6, a+13)
    If b is 1:
        If a mod 4 is 0:
            A(a, 1)->A(22, a-2)
        If a mod 4 is 2:
            A(a, 1)->A(6, a+7)
    If b mod 2 is 1:
        If (a-b+1)/2 mod 2 is 1:
        A(a,b)->A(a-b+11, 2b-4)
        If (a-b+1)/2 mod 2 is 0:
            If (a-b)>0:
            A(a,b)->A(a-b+19, 2b-8)
            If (a-b)<0:
            A(a,b)-A(6, a+b+2)
    If b mod 2 is 0:
        If (a-b)/2 mod 2 is 1:
        A(a,b)->A(6, a+b+7)
        If (a-b)/2 mod 2 is 0:
        A(a,b)->A(22, a+b-2)

A(a, a+3)->halt
A(4n+2, 8n+22)->nonhalt for n>1 (gets stuck in a loop)
Starts at A(6, 1)

Analysis by Marco Dos Santos

Consider:
J(u, v, w) = 0^inf 1^5 10^u 1 A> 11 01^v 1^w 01 0^inf
S_E(b, m, k) = 0^inf 1^5 10^b E> 1 1101^m 1^6 11^k 01 0^inf
S_{E,0}(b, m) = 0^inf 1^5 10^b E> 1 1101^m 1101 0^inf
S_{E,1}(b, m, n) = 0^inf 1^5 10^b E> 1 1101^m 111 01^n 01001 0^inf

J(0, v, 0) -> S_{E,0}(v+3, 0)
J(0, v, 1) -> S_{E,1}(v+2, 0, 0)
J(0, v, 2) -> S_E(v+2, 0, 0)
J(0, v, w) -> J(v+3, 0, w-3) if w >= 3
J(1, v, 0) -> S_E(0, 0, v+3)
J(u, v, 0) -> S_{E,1}(u-2, 0, v+2) if u >= 2
J(1, v, 1) -> Halt
J(u, v, 1) -> S_E(u-2, 0, v+2) if u >= 2
J(u, v, w) -> J(u-1, v+2, w-2) if u >= 1 and w >= 2

S_E(0, 0, k) -> J(0, 0, 2k+6)
S_E(0, m, k) -> J(2, 2m+2, 2k+1) if m >= 1
S_E(1, m, k) -> J(0, 2m+3, 2k+1)
S_E(b, m, k) -> S_E(b-2, m+1, k) if b >= 2

S_{E,0}(0, m) -> S_E(2, 0, 2m)
S_{E,0}(1, m) -> S_E(0, 0, 2m+1)
S_{E,0}(b, m) -> S_{E,0}(b-2, m+1) if b >= 2

S_{E,1}(0, m, n) -> S_E(2, 0, 2m+n+2)
S_{E,1}(1, m, n) -> S_E(0, 0, 2m+n+3)
S_{E,1}(b, m, n) -> S_{E,1}(b-2, m+1, n) if b >= 2

Start: J(0, 0, 0), reached after 35 raw TM steps

This can be simplified to:

Consider:
J(u, v, w) = 0^inf 1^5 10^u 1 A> 11 01^v 1^w 01 0^inf
J(0, v, 0) -> J(0, 0, 2v+12) if v is even
J(0, v, 0) -> J(2, 4, 2v+7) if v is odd
J(0, v, 1) -> J(2, 4, 2v+9) if v is even
J(0, v, 1) -> J(0, 0, 2v+14) if v is odd
J(0, v, 2) -> J(2, v+4, 1) if v is even
J(0, v, 2) -> J(0, v+4, 1) if v is odd
J(0, v, w) -> J(v+3, 0, w-3) if w >= 3
J(1, v, 0) -> J(0, 0, 2v+12)
J(u, v, 0) -> J(2, 4, 2u+2v+5) if u >= 2 and u is even
J(u, v, 0) -> J(0, 0, 2u+2v+10) if u >= 3 and u is odd
J(1, v, 1) -> Halt
J(2, v, 1) -> J(0, 0, 2v+10)
J(u, v, 1) -> J(2, u, 2v+5) if u >= 4 and u is even
J(u, v, 1) -> J(0, u, 2v+5) if u >= 3 and u is odd
J(u, v, w) -> J(u-1, v+2, w-2) if u >= 1 and w >= 2
Start: J(0, 0, 0), reached after 35 raw TM steps

Analysis by hipparcos

G(c, b) = A(2c + 6, b)
G(c, b) = 0^inf 1^6 11^c A> 1 1^b 01 0^inf
G(c, b) ->
(a) Halt                       b = 0
(b) G(0, 2c+13)                b = 1, c ≡ 0 mod 2
(c) G(8, 2c+4)                 b = 1, c ≡ 1 mod 2
(d) G(0, 2c+19)                b = 3, c ≡ 0 mod 2
(e) G(c+4, 2)                  b = 3, c ≡ 1 mod 2
(f) G(0, 2c+b+13)              2 <= b < 2c+9,  b ≡ 0, c + b/2 ≡ 0 mod 2
(g) G(8, 2c+b+4)               2 <= b < 2c+9,  b ≡ 0, c + b/2 ≡ 1 mod 2
(h) G(c - (b-1)/2 + 5, 2b-4)   5 <= b < 2c+9,  b ≡ 1, c + (b-1)/2 ≡ 0 mod 2
(i) G(c - (b-1)/2 + 9, 2b-8)   5 <= b <= 2c+5, b ≡ 1, c + (b-1)/2 ≡ 1 mod 2
(j) G(0, 2c+b+8)               b = 2c+7,       b ≡ 1, c + (b-1)/2 ≡ 1 mod 2
(k) G(2c+6, b-2c-9)            b >= 2c+9
Starts from G(0, 1)
Loop is at G(2n, 8n+30)