User blog:B1mb0w/Strong D Function Calculations

The Strong D Function requires a certain number of iterations to match the growth rate of \(f_{\epsilon_0}\). This blog will calculate a formula for the number of iterations required.

To get from from omega to epsilon nought the number of iterations required is \(n^{n\uparrow\uparrow(n-1)-1}\) for any n. The number of iterations required from 1 to epsilon nought is larger by one power of n.  Therefore the number of iterations required is \(n\uparrow\uparrow n\).

Strong D Function Overview
In the Strong D Function blog, the relevant iteration occurs as per this example:

\(D(1,0,1) = D(0,D(1,0,0),D(1,0,0))\) where \(D(1,0,0) = D(0,D(0,1,1),D(0,1,1))\) and \(D(1,1) = 4\)

also \(D(m,n) >> f_{m-1}(n+2)\) so

\(D(1,0,0) = D(4,4) >> f_3(6) >> f_{\omega}(3)\)

\(D(1,0,1) = D(D(4,4),D(4,4)) >> D(f_{\omega}(3),f_{\omega}(3)) = f_{f_{\omega}(3)}(f_{\omega}(3)) = f_{f_{\omega}(3)}(f_{\omega}(3)) = f_{\omega}(f_{\omega}(3)) = f_{\omega}^2(3)\)

This iteration needs to be repeated 3 times (or n times in the general case) to generate \(f_{\omega+1}(3)\)

Number of iterations Strong D Function uses:

\(n\) from  \(\omega\)  to  \(\omega+1\)

Number of iterations required to get to \(\epsilon_0\) is:

\(n\uparrow\uparrow n\) from  \(1\)  to  \(\epsilon_0\)

and \(n\uparrow\uparrow n - n\) from  \(\omega\)  to  \(\epsilon_0\)

and \(n\uparrow\uparrow n - n - 1\) from  \(\omega+1\)  to  \(\epsilon_0\)

Number of iterations Strong D Function needs to get to \(\epsilon_0\) is:

\(n + n.(n\uparrow\uparrow n - n - 1)\)

Formula and Calculations
The formula to compare \(D(l,m,n)\) to \(f_{epsilon_0}(n)\) is:

\(f_{epsilon_0}(n) =  f_{\omega}^{n + n.(n\uparrow\uparrow n - n - 1}(n)\)

\(D(1,0,1) = f_{\omega}^2(3)\)

\(D(1,0,n) = f_{\omega}^{n+1}(3)\)

\(D(1,0,(n + n.(n\uparrow\uparrow n - n - 1)-1) = f_{\omega}^{n + n.(n\uparrow\uparrow n - n - 1)}(n)\)