User blog:Wythagoras/The nineteenth Busy Beaver number is greater than Graham's Number!

I found some improvements to Deedlit's original Ackermannian growth and expandal Turing Machines.

Using this, I can prove that $\Sigma(19) > f_{\omega+1}(163) > G$.

10 state, 2 color machine with Ackermannian growth
Somehow states 3, 4 and 5 are completely obsolete in the original machine.

In fact, removing them even increases the first string by three ones instead of two.

0 1 1 l 0 0 _ 1 r 2 2 1 1 r 2 2 _ _ r 6 6 _ _ l halt 6 1 1 r 7 7 _ _ r 6 7 1 1 l 8 8 1 _ l 9 9 _ _ l 10 9 1 1 r 11 10 1 1 l 9 10 _ 1 r 0 11 1 _ r 12 11 _ _ l 13 12 _ 1 r 11 12 1 1 l 13 13 1 1 l 13 13 _ 1 l 10

You can simulate the machine on Anthony Morphett's site.

14 state, 2 color machine with expandal growth
This also works in the expandal growth machine, as that is just an extension of the Ackermannian machine. 0 1 1 l 0 0 _ 1 r 2 2 1 1 r 2 2 _ _ r 6 6 _ _ r 14 6 1 1 r 7 7 _ _ r 6 7 1 1 l 8 8 1 _ l 9 8 _ _ l 15 9 _ _ l 10 9 1 1 r 11 10 1 1 l 9 10 _ 1 r 0 11 1 _ r 12 11 _ _ l 13 12 _ 1 r 11 12 1 1 l 13 13 1 1 l 13 13 _ 1 l 10 14 _ _ l halt 14 1 _ l 8 15 _ 1 l 16 16 1 1 l 17 16 _ 1 l 0 17 _ _ l 16 17 1 _ l 16

14 state, 2 color machine with expandal growth, variant
This also works in the expandal growth machine, as that is just an extension of the Ackermannian machine. 0 1 1 l 0 0 _ 1 r 2 2 1 1 r 2 2 _ _ r 6 6 _ _ r 14 6 1 1 r 7 7 _ _ r 6 7 1 1 l 8 8 1 _ l 9 8 _ 1 l 16  9 _ _ l 10 9 1 1 r 11 10 1 1 l 9 10 _ 1 r 0 11 1 _ r 12 11 _ _ l 13 12 _ 1 r 11 12 1 1 l 13 13 1 1 l 13 13 _ 1 l 10 14 _ _ l halt 14 1 _ l 15 15 _ _ l 8 16 1 1 l 17 16 _ 1 l 0 17 _ _ l 16 17 1 _ l 16 I switched state 15 and a part of state 8. Both of these parts are added in the expandal machine and are only accessed using state 14. They can therefore be switched.

This machine can also be simulated on Anthony Morphett's site.

5 state, 2 color machine which outputs 165 consecutive symbols 1
The following machine:

0_ 1 l b 0 1 1 r 0 b _ 1 l c b 1 1 r e c _ 1 l d c 1 1 l e d _ _ r 0 d 1 1 l c e 1 _ r b e _ 1 l halt

halts in 15,589 steps, outputing the following tape:

_11111111...[150 ones omitted]...1111111 ^ Source: Mathematica Journal 2009.

19 state, 2 color machine which outputs over $f_{\omega+1}(163)$ symbols
0 _ 1 l 18 0 1 1 r 0 1 1 1 l 1 1 _ 1 r 2 2 1 1 r 2 2 _ _ r 6 6 _ _ r 14 6 1 1 r 7 7 _ _ r 6 7 1 1 l 8 8 1 _ l 9 8 _ _ l 15 9 _ _ l 10 9 1 1 r 11 10 1 1 l 9 10 _ 1 r 1 11 1 _ r 12 11 _ _ l 13 12 _ 1 r 11 12 1 1 l 13 13 1 1 l 13 13 _ 1 l 10 14 _ _ l halt 14 1 _ l 8 15 _ 1 l 16 16 1 1 l 17 16 _ 1 l 1 17 _ _ l 16 17 1 _ l 16 18 _ 1 l 19 18 1 1 r 21 19 _ 1 l 20 19 1 1 l 21 20 _ _ r 0 20 1 1 l 19 21 1 _ r 18 21 _ _ l 8

This machine can be simulated online on Anthony Morphett's site.

Sketch of the proof that this machine outputs more than $f_{\omega+1}(163)$ ones
First, we have to prove that the expandal machine indeed exhibits expandal growth.

For this, we can prove by induction that the Ackermannian growth machine indeed has Ackermannian growth. Then, another induction shows that the expandal machine indeed exhibits expandal growth. (Details might be added later, but it is a lot of writing. If anyone else wants to do this, please do so!).

When the 14 state machines are given the input $1\_\_11$, after approximately 35,000 steps, the following tape is created: 11111...[110 ones omitted]...11111_1_11111...[107 ones omitted]...11111_1_1

Every one in the third group more than triples the number of ones in the first group. So there are over $120\cdot 3^{117} > 7 \cdot 10^{57}$ ones on the tape when this halts.

Then, for every extra one after the double space, the machine takes $m$ ones to more than $g(n)=f_{\omega}(m/2)$ ones. Note that $g^k(n+1)> f_{\omega}^k(n/2-1)$. (This requires a proof. It would be easier to show that $g^k(n) > f_{\omega}^{k/2}(n/2), but it is weaker. It would still be enough to beat G, though).

This means that the total number of ones will be over $f^{163}_{\omega}(3 \cdot 10^{57}} > f_{\omega+1}(163)$.