User blog comment:Mh314159/Natural number recursion - first 4 rule sets/@comment-35470197-20191019143654/@comment-39585023-20191024223041

Thanks everyone but I have a hugely busy life and don't have the time, background, or motivation to study ordinals and fundamental sequences. Thanks for your confidence in me, but I'm just going to take one more intuitive shot and if it falls short, so be it. Let's ignore the alpha superscripts which I thought was adding a lot because it recursed the argument so many times to put the entire previous structure into the subscript. But I guess I was mistaken. So if Am(x) is the strongest thing I have, then I will iterate the entire A function by defining a new series of functions starting with B and continuing with [n] with n is the nth function:

B‹0›(x) = Am(x) with m = B‹0›(x-1)

B‹0›(0) = A2(2)

Other than the ‹0›(x) and ‹0›(0) terminations, the same recursion rules apply as to B as to A

Continuing, square brackets indicate the ath function in the sequence starting with A

[1](x) = A(x)

[2](x) = B(x) etc.

[a](x) = [a]‹m,m,...m›(x) with m instances where m = [a](x-1) and [a](0) = [a-1](a) and [0](0) = 1

[a]n(x) = [a]n-1m(x) m = [a]n(x-1)

[a]n(0) = [a]n-1(n)

[a]0(x) = [a](x)

[a]‹0›(x) = [a-1]m(x) with m = [a]‹0›(x-1)

[a]‹0›(0) = [a-1]a(a)

[1]‹0›(x) = x + 1

The next plan would be to define bracket strings that recurse the entire structure back into the brackets.