User blog comment:Wythagoras/SuperJedi's X function/@comment-27173506-20151220171603

Proof that X(9) = 9^9^9^9^9:

To define a one argument function we need at least 5 symbols, and with that we can define only Ax=Sx. For exponentiation (the highest we can reach) we need 6 symbols: Ax=x^x. Then we need at least 2 more symbols to define a number: A9. If we try to define X(8) with the 2 rules:

Ax=x^x

A9

We get 9^9, which is a mere X(3)!

There are two options to increase with one symbol: The first is to add a successor symbol to the definition, as in:

Ax=x^Sx

A9

But that only gets us to 9^10. If we could use the predecessor function (Pn = n-1) then we could reach 10^10, but even that is far less than X(4). The better option would to add another A to the number definition, as in AA9. That brings us up to (9^9)^(9^9) = 9^(9^9*9) = 9^9^10, in between X(5) and X(6), much less than 9^9^9^9^9. Therefore, X(9) = 9^9^9^9^9.