(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)(1,0)(0,0)[10] = 176

Can't postpone it any longer, can I?

Alright. Here are the rules for ANY expression that ends with either (0,0)[n] or (1,0)[n]:

(1) If it ends with (0,0)[n] then we delete the (0,0) and add 1 to n.

(2) if it ends with (1,0)[n] then we:

(2-i) delete the final (1,0)[n] (but remember the value of n)

(2-ii) Copy everything from the last (0,0) in the resulting expression till the end, an additional n-1 times.

(2-iii) reappend the [n] at the end (without changing its value)

For example, to expand (0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)(1,0)[3]:

(a) We see that it ends with a (1,0) so we'll follow rule 2.

(b) We delete the (1,0)[3] and remember "n=3"

(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)

(c) We look for the last (0,0) in there:

(0,0)(1,0)(2,0)(3,0)

(0,0)(1,0)(2,0)

(d) For clarity, let's bold everything starting from that (0,0)

(0,0)(1,0)(2,0)(3,0)

(0,0)(1,0)(2,0)(e) Copy the bolded part two more times (since n is 3):

(0,0)(1,0)(2,0)(3,0)

(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)(f) Append the [3] at the end:

(0,0)(1,0)(2,0)(3,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)(0,0)(1,0)(2,0)[3]

Of-course, we don't yet know how to expand

that (it ends in a (2,0)) , but I've chosen this example to illustrate that the above rules will continue to be valid even once we add things like (2,0) and beyond.

We also need a new rule for the forbidden substrings:

Let (a,b)

_{n} be shorthand for (a,b)...(a,b) (with n (a,b)'s)

Then the following substrings are forbidden:

(a) (0,0)(0,0)(1,0) (we've already mentioned that one)

(b) (0,0)(1,0)

_{n}(0,0)(1,0)

_{m} for m>n.

And that's it for now. The limit of

this is already going to be a very big number (more info on exactly how big it is, will be given as we progress).

At any rate, you now have all the information needed to calculate the actual value of the next expression. For those who are just interested in the end result, here is the general formula for (0,0)(1,0)(1,0)[n] (if you want to find it on your own, don't peek!):