1RB1RZ 0RC0RE 1LD1LA 1LC0LG 0RF1LF 0RD1LF 1LB0LE

1RB1RZ_0RC0RE_1LD1LA_1LC0LG_0RF1LF_0RD1LF_1LB0LE (bbch) is a tetrational halting BB(7) TM with sigma score over $10\uparrow \uparrow 519$. It was found by Andrew Ducharme on 6 Jun 2025 (Discord link).

Analysis by Shawn Ligocki

This TM goes through 2 phases: Phase A and Phase B.

A(a, b) = 0^inf <F 10 1^a 00 1^b 0^inf
B(a, b, c, d, e) = 0^inf <F 1^2a+1 00 1^2b 0^2c+1 1^d 00 1^e 0^inf

Shorthand:
  A(a) = A(a, 0)
  B(b, d, e) = B(0, b, 0, d, e)

f1(x) = 2x+2
f2(x) = f1^x(4) = 6 2^b - 2

Phase A:
  Start --(13)--> A(1)
  A(3k) -> B(4, 4k-3, 4) 
  A(3k+1) -> A(4k+4)
  A(3k+2) -> A(4k+6)

Phase B:
  B(b, 3k,   e) -> B(f1(f2^k(b)), 4k+e,   0)
  B(b, 3k+1, e) -> B(f2^{k+1}(b), 4k+e+4, 0)
  B(b, 3k+2, e) -> Halt(2 f2^k(b) + 4k + e + 12)

The trajectory is:

Phase A:
  1 4 8 14 22 32 46 64 88 120
Phase B:
  B(4, 157, 4)
  B(a, 216, 0)    a = f2^{53}(4)
  B(b, 288, 0)    b = f1(f2^{72}(a)) = f1( f2^{125}(4) )
  B(c, 384, 0)    c = f1(f2^{96}(b))
  B(d, 512, 0)    d = f1(f2^{128}(c))
  Halt(2e + 692)  e = f2^{170}(d)

If we ignore the f1 calls and note that $f_{2}(x)>2^{x}$ then the sigma score at halt is larger than $e>f_{2}^{519}(4)>(2\uparrow )^{519}4>10\uparrow \uparrow 519$.

It gets very lucky in Phase A: resetting 9 times before hitting a multiple of 3 and a little bit lucky in Phase B: resetting 4 times before hitting a remainder 2 factor.