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First, Let me try to express the notation so far in Terms of the FGH.

Tb = 10^b  about f_2
aTb is similar to tetration f_3
aTb@c@d about f_4
aTb@..@c with x terms - f_4
aTb@@c - f_5
 ? aTb@@c@f - f_{5}
aTb@@c@@f - f_{5}
aTb@@@f - f_{6}
aTb(@^@)c - f_{\omega }
aTb(@^^@)c - f_{\omega^\omega}

Going even further...

aTb(@^^@)10@5 =
aTb(@^^@)5T10 = >
aTb(@^^5T10)5T10 = >
The height of the power tower of @ is 10^^6

T7(@^^@)10(@^^@)6
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 @.

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