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Or you could use a operation of my own design it’s kinda like a hyperoperation but I haven’t seen it officially listed anywhere.
Basically F(n) whatever n is equal to tells you the number of operations used and the level of the last operation you’ll use that step. Ex
F(0)=0
F(1)=1+1=2
F(2)=2•(2+2)
F(3)=3^(3•(3+3))
F(4)=4^^(4^(4•(4+4)))
F(5)= 5^^}(5^^(5^(5•(5+5)))){^^5
basically in that last one there are approximately 5^^85decillion arrows with 5s on either end it’s kinda hard to notate on the phone but it’s already starting to stack in the same way Graham’s Number (G64) does by increasing the actual ammount of arrows. However, it should surpass Graham’s number a good while before F(64) I would imagine and so F(G64) would be absolutely humongous. I built this operation trying to create a fun pattern that would at some point outpace the TREE(n) function but I’ll never be certain if it would work or just how long it would take to grow faster cause TREE(3) is already so massively huge the gap is a VERY difficult one to breach. Like imagine how small 100 is next to G64 and TREE(3) makes G64 look beyond tiny I don’t even fully understand how.
also fun fact I tried calculating this pattern in reverse as well and kinda kept breaking things giving me the following answers before things really stop making any sense at all but it’s still fun to break the rules sometimes just to see what would happen if you could lol
F(1)=0
F(2)~ Infinity?
F(3)~ (i)
F(4)= DEFINITELY UNDEFINED