User blog:Mh314159/Alpha numbers (and beyond)

Several months ago I posted a series of natural number recursions with the goal of working toward TREE3. I was not successful but did learn a lot, especially from Pbot. I put it all aside, needing a break from the disappointment and not being interested in getting a degree in set theory to understand all the talk about ordinals that seems to be the main focus here. I do have some basic idea of how the FGH works, but have no understanding of OCFs.

I have worked on and off in my spare minutes here and there on various notations and I have a system that I think is much more powerful than any of my previous ideas. I'll post it here in parts and I would appreciate any comments, especially any definitive evaluations of growth rate.

The first part defines the alphabetic sequence, which I then generalize with the alpha numbers. It seems odd that an alpha number of one argument is larger than an alpha number of two arguments, but that's because I am using a system that recurses the function to subscript or to the second argument when there's isn't already one present. Hopefully it makes sense when you see the definitions, and if not, please ask.

Rule set 1:

f0(x) = x + 1

for a > 0,

fa(0) = fa-1(a)

fa(b) = fa-1m(m), m = fa(b-1)

The unsubscripted version of a function recurses the function into the subscript:

f(x) = ff(x-1)(x); f(0) = 1

Examples:

f1(1) = f0m(m), m = f1(0) = f0(1) =2, so f1(1) = f02(2) = 4

f1(2) = f04(4) because f1(1) = 4, so f1(2) = 8

f1(3) = f08(8) = 16

f1(4) = f016(16) = 32

so f1(x) = (2)(x+1)

fn(x) is somewhat greater than x^ n x in Knuth up-arrow notation

f(x) = ff(x-1)(x) is similar to the (x-1)th iteration of the Graham-Gardner sequence.

We now define a series of similar functions such that h is to g as g is to f and i is to h as h is to g, etc., up to the nth symbol in the sequence with g being the n=1 function. A theoretical alphabetic sequence can be used, or the notation α(a,x) can be used for the ath function beyond f in the sequence; α(0,x) = f(x), α(1,x) = g(x), etc.

The single argument alpha function copies the argument to the alpha subscript

αm(x) = αm-1(α(x))

α1(x) = α(x) = αm(x) m = α(x-1)

α(0) = 1

Recursing the α subscript leads to a two argument α expression

αb(x) = αb-1m(x) m = αb(x-1)

αb(0) = αb-1(b)

α0(x) = α(m,x) m = α0(x-1)

α0(0) = 1

The double argument α function with unsubscripted first argument generates the subscript of the first argument

α(a,x) = α(am,x) m = α(a,x-1)

α(a,0) = α(a-1,a)

α(0,0) = 1

Recursing a subscripted two argument α expression leads to the base function

α(ab,x) = α(ab-1m,x) m = α(ab,x-1)

α(ab,0) = α(ab-1,b)

for x > 1, α(a0,x) = α(a-1m,x) m = α(a0,x-1)

for a > 0, α(a0,0) = α(a-1,a)

α(00,x) = x+1 (the f function)

Examples:

g0(1) = α(10,1) = α(0m,1) m = α(10,0) = α(0,1) = α(01,1) = 4. So we have g0(1) = α(0,α(0,α(0,α(0,1)))) which in f notation is f(f(f(f(1)))) = f(f(f(4))) where f(4) is similar to G3 in Graham-Gardner sequence so f(f(4)) is similar to iterating the Graham-Gardner sequence more than G3 times and f(f(f(4))) = g0(1) is therefore much larger than Conway's 3->3->3->3.

g0(2) = α(10,2) = α(0m,2) m = α(10,1) so α(10,2) = α(0,α(0,... (α(0,2))) with α(10,1) iterations. Now the number of iterations of f(x) is equal to the large number described above.

g0(3) = α(10,3) = α(10,3) = α(0m,3) m = α(10,2) so α(10,2) = α(0,α(0,... (α(0,3))) with α(10,2) iterations.

α(11,1) = α(10m,1), m = α(11,0) = α(10,1). Therefore, we have α(10,α(10,...α(10,1))) with α(10,1) (see above) iterations of the α(10) function.

g(1) = α(1,1) = α(1m,1) m = α(1,0) = α(0,1) = α(0n,1) n = α(0,0) = 1 so α(0n,1) = 4 and α(1,1) = α(14,1) = α(13p,1) p = α(14,0) = α(13,4) What would be its comparable expression in Conway chain notation?

g(2) = α(1,2) = α(1m,2) m = α(1,1) (see above)

h(1) = α(2,1) = α(2m,1) m = α(2,0) = α(1,2)

h(2) = α(2,2) = α(2m,2) m = α(2,1)

α(1) = αα(0)(1) = α1(1) = α0m(1) m = α1(0) = α0(1) = α(n,1) n = α0(0)= 1 so m = α(1,1) and therefore α(1) = α0(α0(...α0(1)) with α(1,1) repetitions. Since α0(1) = α(1,1) after one of the many iterations we have α0(α(1,1)) which = α( α(α(1,1)-1)), α(1,1) ) so this goes beyond α(y,y) where y represents a large number in the range of Conway chains, after only α0(α0(1)) and there are in fact α(1,1) repetitions of α0in the full expression of α(1).

How large is α(2)? or α(100)? Or α3(3)?

Looking beyond, I will next define three term arguments and then extend the alpha numbers up the greek alphabet, and then generalize them.