An extremely long power tower

[golden3159]

An extremely long "power tower"

Hey, this isn't really a homework or classwork question. My professor always puts questions that he finds interesting on our notes. This really got my attention, I've been trying to see if it's possible to write a computer program to solve it.

However, I've come to the conclusion that this number is simply too large to be calculable in decimal notation. I was just wondering if you could tell me if I've done the right things.

First of all here is the information for the problem:
http://img62.imageshack.us/img62/4931/problemy.th.png

And the problem itself
[PLAIN]http://img687.imageshack.us/img687/1779/49195211.png

(Sorry for using so many pictures instead of typing this out, my browser is having some issues with using the toolbar items above). Anyways, after working this out, the conclusion that I come to is that I need to keep a power of 2's, in the following fashion:
[PLAIN]http://img687.imageshack.us/img687/5981/part2s.png

I have to have a height of more than 65,000 on the tower! As far as I am aware, this would produce a number FAR too massive to compute. Am I on the right track and is there any possible way to even obtain a decimal notation of the above problem? Thanks.

EDIT: Just a side thought, is there any relation to this and the Ackermann function at all?`

 

[Martin Rattigan]

The number you describe is $wow(4)$, i.e.$f_4(4)$ in the Ackermann hierarchy - the Ackermann function itself being $f_\omega$.

There is no chance of an exact computation in decimal digits in meaningful time, nor would there be any way of physically holding or displaying the digits.

 

[Mentallic]

is there any possible way to even obtain a decimal notation of the above problem?

I understand you probably realize this is too massive to be expressed in decimal notation (and even too massive to be expressed reasonably as it is in the power tower form!) but just watch at how fast the tower power grows in decimal notation so your hopes of expressing this can be destroyed into fragments of shattered dreams...

$$2^2=4$$

$$2^{2^2}=2^4=16$$

$$2^{2^{2^2}}=2^{16}=65,536$$

$$2^{65,536}\approx 10^{19,728}$$

Only 4 applications and the resulting number is 19728 digits long!

 

[g_edgar]

Presumably "A" stands for Ackerman.

 

[Count Iblis]

http://en.wikipedia.org/wiki/Ultrafinitism"

 

[golden3159]

Looking back at my original post, I am not sure why I even asked if it was possible to express in the first place. I should have known better. I guess I just wanted to make sure I had arrived at the proper notation at least. Thanks for the answers guys.