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First, Let me try to express the notation so far in Terms of the FGH.
Going even further...
aTb(@^^@)10@5 =
aTb(@^^@)5T10 = >
aTb(@^^5T10)5T10 = >
The height of the power tower of @ is 10^^6T7(@^^@)10(@^^@)6
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First, the 10(@^^@)6 turns into 10(@^^6)6
T7(@^^@)10(@^@^@^@^@^@)6
T7(@^^@)10(@^@^@^@^@^6)6
T7(@^^@)10(@^@^@^@^@^6)6 T7(@^^@)10(@^@^@^@^@^5)6(@^@^@^@^@^5)6(@^@^@^@^@^5)6(@^@^@^@^@^5)6(@^@^@^@^@^5)6(@^@^@^@^@^5)6 Then the last 5 turns into several copies of (@^@^@^@^@^4), which turns into several copies of (@^@^@^@^@^3), (@^@^@^@^@^2), then finally (@^@^@^@^@). Then you have to keep going down the power tower until you get 10@a@b....@z with a GIANT number of @. 
Rules from previous blogpost: 1.Ta = 10^a, a \in Z+ The expression Ta means simply, 10 to the power of A. 2.aTb = 10^10^10^...10^b , where there are a 10s. 3.aTb@c = (c*a)Tb. 4.aTb@c...@y@z = aTb@c...@zTy 5.aTb@@c = aTb@c@c@c@c (c times) 6.aTb(@^c)d = aTb(@^c1)d(@^c1)d(@^c1)d... (d times) 7.aTb(@^@)d = aTb(@^d)d (diagonalized)
To go further:
T1(@^@)2@3 = T1(@^@)3T2 = T1(@^3T2)3T2 This has a 10^10^10^2(googolplex) entries in the next line!
T1(@^@)10@@2 = T1(@^@)10@2@2 = T1(@^@)10@2T2 T1(@^@)(2T2)T10
Eventually, we can define aTb(@^@@)c as aTb(@^@)c(@^@)...c
c terms in expression
aTb(@^@@@)c = aTb(@^@@)c(@^@@)...c
... Eventually, we reach things like aTb(@^@^5)c aTb(@^@^10)c aTb(@^@^@)c = aTb(@^@^c)c We can generalized this to tetration:
aTb(@â€¦
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Right now my site has exploding T and my expanded illions This is my numbers site: https://sites.google.com/site/kelsnumberssite/home
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First, let's define an Illion function I(x) = 10^(3x+3) (short scale)
Million = I(1) = 10^(3+3) = 10^6 Centillion = I(100) = 10^(3*100+3) = 10^303
This system uses the standard latin based names up to a Millillion.
Then we can do things like:
Millionillion = I(I(1) = I(10^6) = 10^3,000,003
In this naming system, we will have:
Latin Prefix + Greek Prefix + "illion"
The Latin Prefix tells us the number that goes in the innermost I function. The Greek Prefix tells us how many times to plug I into itself
In other words, ABillion = I(I(I(...I(A)))..) B times
Examples: undiillion A = 1 B = 2 I(I(1) = 10^3,000,003
centitriillion A = 100 B = 3 I(I(I(100))) = I(I(10^303) I(10^(3*10^303 + 3)) ~ I(10^10^303) ~ 10^10^10^303
quadriheptillion A = 4 B = 7 I(I(â€¦
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This is a new notation I came up with that's somewhat similar to Hyper E. It is called Kel's Exploding T Notation.
The rules are:
1. 10^^(10^^(10^^4^2)^2) > 10^^^4
T10@10@10@100 Googolplexkel
T10@10@10@10@10@10 Sexdenalinakel
T10@10@10@10@10@10@10 Septendenalinakel
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