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Last time I talked briefly about the fact that symbols produced from each subsequent vocabulary of symbols is 1 less the letter value than the letter preceding it. Now I'm going to describe a generalization of this vocabulary rule.

For any set of length x>=3 and of the form (x,y,z,w,v,u...) , there exists a unique division of the set into the form.

{a,(x,y,z,w,v,u...)b}R{a,(x,y,z,w,v,u...)b} such that everything in the ()'s has 1 less symbol than the combined set.

For example: {2,(4,4,4),8} = {2,(4,4),8}R{2,(4,4)8}.

But suppose that we want to extend beyond simple linear expressions?

We need to define a new symbol and rule. [] [{a,(x,y,z,w,v,u...)b}] = {a,(x,y,z,w,v,u...)b} recursed into () {a,(x,y,z,w,v,u...)b} number of times.

To give an example of this [{2,(4,4,4),8}] = {2,(4,4,4),8}R{2,(4,4,4),8}.

However we could in theory use iterated brackets and simple recursion to generate a faster growing notation.

I'll start by defining what 2 []'s does to the notation then build up to defining what n []'s does.

[[]]

[[{a,(x,y,z,w,v,u...)b}]] = [{a,(x,y,z,w,v,u...)b}]R[{a,(x,y,z,w,v,u...)b}].

Now for n []'s of which we will label []#n.

For any array of the form [{a,(x,y,z,w,v,u...)b}]#n the array can be partitioned into the form [{a,(x,y,z,w,v,u...)b}]#(n-1)R[{a,(x,y,z,w,v,u...)b}]#(n-1).

Now would be a good place to stop or at least just for now , in the meantime here are some examples of numbers defined in this way.

[{10,(10,10,10,10,10,10,10,10,10)10}]#10 = Titusillion.

For those of you who don't know Titus was the 10th roman emperor.

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