User blog:DrCeasium/Extended Hyperfactorial Array Notation

Thanks to everyone who commented on my previous blog, it helped me spot all the errors, typos and false claims in it, and these have all now been corrected. This post will contain an extension to my hyperfactorial array notation (as the title suggests), which (I think) will take it up to order type $$\omega^\omega$$ (please correct me if i'm wrong). This extension is also different to anything in Bowers' work (or at least more different than it was before). Anyway, this is the extension:

Before, the notation was in the form a![b,c,d,e...]. This worked in a way that meant its order type is $$\omega^2$$. To reach $$\omega^\omega$$, we need to add another level of expansion to the notation, a new way to recursively expand the notation. This is done by adding more exclaimation marks, so making it in the form a!!...!![b,c,d,e...] (shorthand $$a!^n[b,c,d,e\dots]$$ for n !'s). This could be evaluated by working out the value with one exclaimation mark and putting that back in the brackets, but that would not get the level of recursion we need. Instead, we define it with a new rule to the three usual ones, taking precedence over the others: This is a pretty sizeable extension to the previous notation, and it certainly ramps up the numbers generated by this notation. Here are two examples of how it works: Again, there is a list of named numbers coming soon, and there is another expansion i'm working on to get to $$\epsilon_0$$, called Hyperfactorial Array Avalanche Notation. Any and all constructive comments are appreciated. If there are any problems please point them out, and I will make changes where necessary.
 * If n > 1, $$a!^n[x_1,x_2,x_3,\dots,x_k] = a!^{n-1}[(a!^{n-1}[x_1,x_2,\dots,x_k]),$$$$(a!^{n-1}[x_1,x_2,\dots,x_k]-1),(a!^{n-1}[x_1,x_2,\dots,x_k]-2),\dots,3,2]$$
 * 3!![2,2]. First of all, we need to calculate 3![2,2] = 3![2![2,1],1] = 3![2![2]] = 3![2] = $$3\uarr\uarr2$$ = 27. Then feed this back in using the new rule to get 3![27,26,25,24,23,...,4,3,2]. You may remember what the 5![5,5,5,5] did from my last blog. It wasn't really extendable on a reasonably sized post. That had 4 entries. This has 26, and even though it is only 3 factorial to that base, $$3\uarr^y2$$ is still pretty huge for any reasonably sized y. I think that this y qualifies as reasonably sized. Another thing to note is that this is easily far larger than 3![3,64], which in the last post I worked out was greater than Graham's number.
 * 4!!![3,2]. This time, first of all we need work out 4!![3,2], and to do this, we need to calculate 4![3,2] = 4![3![3,1],1] = 4![3![3]] = $$4![3\uarr\uarr(7.6\times10^{12})]$$. That's pretty large by itself. Then just take that and feed it into the rule for 4!![3,2]... and then work out  that... and... feed it back in... and... thats going to be big.