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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 exampâ€¦
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Since I have been away from this wiki for what seems like years and forgotten almost completely about my old notation I decided to start anew with a brand new notation which I call alpha array notation.
Since I've "literally" just thought about this notation a few minutes before writing these words I won't develop a rule set until later on since I don't want to rush into things.
First of all we need to start at the bottom and work up.
I have decided to call this regiment the unary regiment and with it comes the unary vocabulary.
The unary vocabulary is listed as so:
 { } are used to hold the array in place.
 Variables {a,b,c...z ,a,c,b,....z , b,a,c,....z, b,c,a,....z and so on) are used to represent values in which to input into the array.
 numbersâ€¦

Can someone explain explain to me how the SKI combinator calculus works because when I try to understand it is goes over my head ,I would also like to know how it can lead to functions such as Îž function which grow uncomputably fast?
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The name of this number is Sindadrinx. Its value in my notation is {10:â†‘((Ï‰^(Ï‰^2))(10)}.
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This is part 2: I will be starting at {K:â†‘(Ï‰+1)(X)} then moving on from there.
{K:â†‘(Ï‰+1)(X)} = {K:â†‘(Ï‰)(X)} recursed ({K:â†‘(Ï‰)(X)}1) times.
{K:â†‘(Ï‰2)(X)} = {K:â†‘(Ï‰+Ï‰)(X)}= {K:â†‘(Ï‰+X)(X)} = {K:â†‘(Ï‰+X1)(X)} recursed ({K:â†‘(Ï‰+X)(X)}1) times.
{K:â†‘(Ï‰3)(X)}= {K:â†‘(Ï‰2+x)(X)} = {K:â†‘(Ï‰2+X1)(X)} recursed ({K:â†‘(Ï‰2+X1)(X)}1)times.
{K:â†‘(Ï‰^2)(X)}= {K:â†‘(Ï‰*Ï‰)(X)}= {K:â†‘(Ï‰*X)(X)} = {K:â†‘(Ï‰*X1)(X)} recursed ({K:â†‘(Ï‰*X1)(X)}1) times.
{K:â†‘(Ï‰^Ï‰)(X)} = {K:â†‘(Ï‰^X)(X)}= {K:â†‘(Ï‰^X1)(X)} recursed ({K:â†‘(Ï‰^X1)(X)}1)times.
{K:â†‘(Ï‰^Ï‰+(1))(X)} = {K:â†‘(Ï‰^Ï‰)(X)} recursed ({K:â†‘(Ï‰^Ï‰)(X)}1) times.
{K:â†‘(Ï‰^Ï‰+(Ï‰))(X)} = {K:â†‘(Ï‰^Ï‰+(X))(X)} = {K:â†‘(Ï‰^Ï‰+(X1))(X)} recursed ({K:â†‘(Ï‰^Ï‰+(Ï‰))(X1)}1) times.
{K:â†‘(Ï‰^Ï‰+(Ï‰+1))(X)} = {K:â†‘(Ï‰^Ï‰+(Ï‰))(X)} recursed ({K:â†‘(Ï‰^Ï‰+(Ï‰))(X)}1)times.
{K:â†‘(Ï‰^Ï‰+(Ï‰2)â€¦
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