User blog comment:MachineGunSuper/A HTN extension/@comment-30754445-20181109143444

This one finally reaches ω2 on the FGH. That final number you've given is equivalent to about NE100 ≈ fω2(10100)

A seasoned googologist can see immediately that this function cannot be any weaker or any stronger, because:

(1) With the exception of the final line, everything in this notation is either a function that does a straightforward iteration of a previous function (contributing +1 to your FGH power) or a single-indexed list of such functions (contributing +ω to your FGH power). There's nothing with more power in there.

(2) The one definition that breaks this rut is the final one (the Tr@ one). It indexes your Tr###...#### with an arbitrary number of #'s, with each # giving you another +ω to your FGH.

Amazingly enough, this also means that everything you've done before the "Tr#1" level has a negligible effect on the final result. Your final function wouldn't be substantially weaker, had you started with:

Tr#1(n)=n+1 (or any other reasonable function of n).

And then proceed as you already did.

Sounds unbelievable, I know. But it's a true. Your #'s are simply so powerful, that everything you've done before can be accomplished with by adding about half a dozen of them. And since Tr@ actually counts the #'s, this translates to adding about half a dozen to the argument we're putting in Tr@. So:

Tr@SIMPLIFIED(100) ≈ Tr@(100+half a dozen) = Tr@(106).

So yeah. Your #'s are DA BOMB.