scott wrote:

> Next task, give us some indication of how big n is :-)

I can see I won't get much work done today...



OK, well we have

   d = λ n → n 2
   c = λ n → n 2^n
   b = λ n → c@n n
   a = λ n → b@n n

where "b@n" means "b applied n times".

If n = 100, then 2^n ≈ 10^30, while is a mere 10^2 times larger. As n 
becomes larger, the difference becomes more and more insignificant. So, 
we can approximate with

   c = λ n → 2^n

On that bases, the final result is guaranteed to be an exact power of 2. 
We just need to figure out which one.

Using Knuth's "up-arrow notation", we have

   c = λ n → 2↑n
   b = λ n → 2↑↑n
   a = λ n → 2↑↑↑n

So the answer we seek is (approximately) 2↑↑↑2. In other words, the 
pentation of 2 with 2. Unfortunately, I don't know of any efficient way 
to compute the size of this ludicrously massive quantity.

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