User:Wythagoras/Rado's sigma function/BB(14)

\(\Sigma(14)>\{3,3,1.7 \times 10^{18267}\}\) (Wythagoras 2016)

Ackermannian growth using 8 states
0 1 1 r 0 0 _ 1 r 5 1 1 1 l 1 1 _ _ r 2 2 1 1 r 3 2 _ 1 l 1 3 1 _ r 0 3 _ 1 r 0 5 _ _ r halt 5 1 _ r 6 6 1 1 l 8 6 _ _ l 2 8 1 1 l 8 8 _ _ l 9 9 1 1 l 8 9 _ 1 r 3 This machine executes a function \(f(a_1,a_2,a_3,\cdots,a_n) \) of \(n\) variables, defined as We have \(f(3,a_2)=f(3a_2,1)\), \(f(3,a_2,2)=f(3a_2,1,2)=f(3,3a_2+1,1)>f(3,3a_2,1)=f(9a_2,1,1)\) and in general \(f(3,a_2,a_3)>f(3^{a_3}a_2,1,1)\). \(f(3,1,a_3)>f(3^{a_3},1,1)\). This gives \(f(3,1,1,a)>f(3\uparrow\uparrow a,1,1,1)\), and in general, \[f(3,\underbrace{1,1,\cdots,1,1}_{b},a)>f(3\uparrow^b a,\underbrace{1,1,\cdots,1,1}_{b+1})>3\uparrow^b a\]
 * \(f(a_1,a_2,a_3,\cdots,a_n)=f(a_1+3,a_2-1,a_3,\cdots,a_n)\) if \(a_2>1\).
 * \(f(a_1,\underbrace{1,1,\cdots,1,1}_{k},1,a_k,\cdots,a_n)=f(3,\underbrace{1,1,\cdots,1,1}_{k},a_1+1,\cdots,a_n)\) if \(a_k>1\).
 * \(f(a_1,\underbrace{1,1,\cdots,1,1}_{k})=a_1+k+1\).

Input using 6 states
There exists a 6 state TM with the following output :

(110)K111 with \(K>1.7 \times 10^{18267}\)

It halts with the head on the second one. Instead of halting, we let it go to state 0 of the Ackermannian growth machine. This amounts to simulating \[ f(2,\underbrace{2,2,\cdots,2,2}_{1.7 \times 10^{18267}},2) > f(3,\underbrace{1,1,\cdots,1,1}_{1.7 \times 10^{18267}},3) = 3 \uparrow^{1.7 \times 10^{18267}} 3 \]