User blog:Kolmogrov142 Oracle Tuto X/ my array notation

Let [a,0] = 1 and [a] = a.

This implies that [a,1] = [a] = a.

There [a,b] = a^b.

If [a,b,0] = [a,1] = [a] = a then [a,b,c] = a^(b^c).

Thus if  we have [a,b,1] then [a,b,1] = a^(b^1) = a^b.

[a,b,c,d] = [a,b,c] recursed d times.

For example [a,b,c,2] = [a,b,c] = a^(a^(b^c))

Thus [a,b,c,d] = a^(a^(a^(a^....(a^(b^c))))) where there are d copies of a in the stack.

[a,b,c,d,e] = [a,b,c,d] recursed [a,b,c,d] number of times.

Therefore [a,b,c,d,e,f] must equal [a,b,c,d,e] recursed [a,b,c,d,e] number of times.

Furthermore for an array of length n i.e. [a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p...] the generalization is

[n] = [|n-1|] recursed [|n-1|] number of times.

Whereby |n| denotes an array of length n.

This can be written as [a,b,c,d,(n)].

For example [a,b,c,d,(1)] = [a] and [a,b,c,d (5) = [a,b,c,d,e].

Let [a,b,c,d,(n),(x)] = [a,b,c,d,(n)] recursed [a,b,c,d,(n)] number of times.

Please note that (x) is simply an operation value rather than a numerical value, the same applies for anything from e onwards.

Following this we get [a,b,c,d,(n),(x),(y)] = [a,b,c,d,(n),(x)] recursed [a,b,c,d,(n),(x)] number of times.

If we call this level of the array hiearchy the operational component class of arrays then the generalization is,

[a,b,c,d, ()] whereby  denotes the nth operational component.

For example [a,b,c,d,(n)] = [a,b,c,(<1>)].

Following this we have [a,b,c,d,(),()] = [a,b,c,d,()] recursed [a,b,c,d,()] number of times.

The generalization of this is almost identical to [a,b,c,d,()] only this time the notation uses 2 pairs of <> brackets.

I call this superclass of arrays nth degree arrays where n refers to the number of pairs of <> brackets.

This can be written down as [a,b,c,d,(<(n)>)].

Next we come to naming some numbers.

[3] = three.

[3,3] = twenty seven.

[3,3,3] = seven trillion six hundred and twenty five billion five hundred and ninety seven million four hundred and eighty four thousand nine hundred and eighty seven.

Skipping forward to [a,b,c,d,(<(n)>)] we get [3,3,3,3,(<(3)>)] = Megatrois (couldn't think of any good names).