[InclinedPlane]

I think that breaks the rules.

Here's something simple I came up with:

P(1) = 10

P(n+1) = 10^P(n)

Q(n) = P(n)^P(n)

Z(1) = Q(1)

Z(n+1) = Q(Q(n))

And then: Z(10^100)

[btilly]

In a somewhat restricted version of this contest some time ago the best entry was my:

B(b, c, 0) = 0

i < b => B(b, c, i + j * b) = i + B(b, c, j) * c

T(b, 0) = b * b

T(b, n+1) = T(T(b, n), B(b, T(b, n), n))

Z = T(2, T(2, T(2, 9)))

It is much, much larger than the number you just threw out. See http://bentilly.blogspot.com/2010/03/large-numbers.html for an explanation.