User blog comment:Wythagoras/SuperJedi's X function/@comment-27173506-20151220171603/@comment-27173506-20151221195611

Using this, it's quite simple to prove that X(10) equals 9^9^9^9^99: 8 symbols to define:

Ax=x^x

A9

Then two more symbols. Here we have the option of doing:

Ax=x^x^x

A9

But that brings us up to X(5)

We can do:

Ax=Sx^Sx

A9

Which is equal to 10^10: much weaker!

We could do:

Ax=x^Sx (stronger than Sx^x)

AA9

Which is equal to (9^10)^(9^10+1) ~ 9^9^11. We're getting closer, but still very far. Our best option is merely iterating the function:

Ax=x^x

AAA9

Which is equal to (9^9^10)^(9^9^10) ~ 9^9^9^11 < 9^9^9^9^99. Therefore, X(10) = 9^9^9^9^99.

X(11) is still 9^9^9^9^9^9.

Ax=x^x||A9999 = 9999^9999 < X(5) = 9^9^9 < Ax=x^x^x||A99 = 99^99^99    < Ax=x^x||AA999 = (999^999)^(999^999)  < Ax=x^x||AAA99 = ((99^99)^(99^99))^(99^99)^(99^99) < X(7) = 9^9^9^9 < Ax=x^x^x||AA9 = (9^^3)^^3 ~ 9^9^9^9^11 <  Ax=x||AAAA9 = (((9^9)^(9^9))^((9^9)^(9^9)))^(((9^9)^(9^9))^((9^9)^(9^9))) ~ 9^9^9^9^10 < X(10) = 9^9^9^9^99 <  X(11) = 9^9^9^9^9^9

X(12) is the first value that is a defined function: 9^9^9^9^99 < Ax=x^x||AAAAA9 = (((9^9)^(9^9))^((9^9)^(9^9)))^(((9^9)^(9^9))^((9^9)^(9^9)))^(((9^9)^(9^9))^((9^9)^(9^9)))^(((9^9)^(9^9))^((9^9)^(9^9))) ~  9^9^9^9^9^9^10

As you can see, this passes not only what we can define by exponentiation in 12 symbols, but also in 13! Quite a jump.

Furthermore, my better bound for X(13) was underestimated. It is actually around 9^9^9^9^9^9^9^9, much higher!

We can build on that and try to get a better bound for X(14) as well: Ax=x^x||AAAAAAA9. At this point however the power of x^x^x kicks in and is ever so slightly more powerful. The Ax=x^x may seem promising, but it actually holds for only one value.