User blog comment:Leighton90chamberlain/108xen/@comment-27516045-20160911105303

To get you started I'm going to convert or approximate your numbers in what are considered more standard notations.

108(9)(9)(9)108 = (10^54)^^(10^54)^^3. The ^^ operation is called tetration, and it is defined as a^^b = a^a^a^... for b a's.

108(9)(9)(9)(9)108 ~ (10^54)^^^(10^54)^^3. This is using pentation, which is iterated tetration.

Every (9) goes another recursion higher. 108(9)...(9)108 with n 9's is about (10^54){x-1}(10^54)^^3, where {n} means n ^'s, and each arrow repeats the last one. See arrow notation for more.

Then X2 is about 10{108(9)(9)(9)(9)108}(10^54)^^3. X3 is about 10{X2}10^54, etc. And your 108xen is, using BEAF, equivalent to around {10,(10^54)^^^(10^54)^^3,1,2}. {a,b,1,2} means a{a{...{a{a}a}...}a}a with b a's from center out. Alternatively it is about the 108(9)(9)(9)(9)108-th term of the sequence used to define Graham's number.

Not bad for a first attempt at large numbers - you've already passed up Graham's number (though not by terribly much - it is less than G(G(1)) (G(n) is n-th member of Graham's sequence).

I hope my comments have been helpful.