User:Wythagoras/Rado's sigma function/Green's machines

Green's numbers and the \(B\) hierarchy
Define for odd \(n\) the following hierarchy:

\[B_n(m) =\begin{cases}  1 && \text{if  } m=0 \\  m+1 && \text{if  } n=1 \\ B_{n-2}[B_n(m-1) + 1] + 1 && \text{otherwise} \end{cases}\]

Then, Green's numbers \(\text{BB}_{\text{Green}}(n)\) are defined as:


 * \(\text{BB}_{\text{Green}}(n) = B_{n-2}[B_{n-2}(1)]\) for odd \(n\)


 * \(\text{BB}_{\text{Green}}(n) = B_{n-3}[B_{n-3}(3) + 1] + 1\) for even \(n\)

Definition form S. Ligocki.

The following things can be observed:


 * \(B_3(m)=3m+1\) and \(B_5(m)=\frac72\cdot 3^m-\frac52\). (As noted by Ligocki)
 * From \(B_5(m)=\frac72\cdot 3^m-\frac52\), we have \(B_5(m) = 3^{m+1} +\frac12\cdot 3^m-\frac52 >3^{m+1}\) for \(m\geq1\).
 * We prove now prove \(B_{2k+3}(m) > \{3,m+1,k\}\) for \(m\geq1\), \(k\geq1\) using induction to \(k\). The base case is given above. Now, suppose \(B_{2k+3}(m) > \{3,m+1,k\}\) for all \(m\geq1\) and some \(k\geq1\). Then \(B_{2k+5}(1)=B_{2k+3}(B_{2k+5}(0)+1)+1>B_{2k+3}(2)>\{3,3,k\}=\{3,2,k+1\}\). Now we use induction to \(m\). Suppose \(B_{2k+5}(m)>\{3,m+1,k+1\}\) for some \(m,k\geq1\). Then \(B_{2k+5}(m+1)=B_{2k+3}(B_{2k+5}(m)+1)+1>B_{2k+3}(B_{2k+5}(m))>\{3,\{3,m+1,k+1\}+1,k\}>\{3,\{3,m+1,k+1\},k\}=\{3,m+2,k+1\}\). This completes the induction to \(m\), and hence \(B_{2k+5}(m)>\{3,m+1,k+1\}\) for all \(m\). This also completes the induction to \(k\).

Showing the bound for \(\text{BB}_{\text{Green}}(9)\)
Using the claim above, we have the following inequalities:

\begin{align*} \text{BB}_{\text{Green}}(9) &= B_7(B_7(1)) \\ &=B_7(B_5(2)+1) \\ &=B_7(30) \\ &>\{3,31,2\}\end{align*}

Showing the bound for \(\text{BB}_{\text{Green}}(10)\)
Using the claim above, we have the following inequalities:

\begin{align*} \text{BB}_{\text{Green}}(10) &= B_7(B_7(3)+1)+1 \\ &> B_7(B_7(3)) \\ &>B_7(B_5(B_7(2)+1)) \\ &=B_7(B_5(B_5(B_7(1)+1)+2)) \\ &=B_7(B_5(B_5(31)+2)) \\ &=B_7(B_5(2161856886993814)) \\ &>\{3,\{3,2161856886993815\},2\} \end{align*}

Showing the bound for \(\text{BB}_{\text{Green}}(11)\)
Using the claim above, we have the following inequalities:

\begin{align*} \text{BB}_{\text{Green}}(11) &= B_9(B_9(1)) \\ &> B_9(B_7(2)+1) \\ &=B_9(2161856886993814) \\ &> \{3,2161856886993815,3\}\end{align*}

Showing the bound for \(\text{BB}_{\text{Green}}(12)\)
Using the claim above, we have the following inequalities:

\begin{align*} \text{BB}_{\text{Green}}(12) &> B_9(B_9(3)) \\ &> B_9(B_7(B_7(B_9(1)+1))) \\ &= B_9(B_7(B_7(B_7(2)+2))) \\ &= B_9(B_7(B_7(2161856886993815))) &> \{3,\{3,\{3,2161856886993816,3\},3\},4\} \end{align*}