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(9*(10^23)^X)^2, where X is the number of posts in this thread

Time's up.

I'll still count all the solutions that were posted in the thread, even though I asked for them to be sent in private messages. It doesn't help you to post it in the thread (can only hurt you, because you give anyone else the ability to simply copy/paste your answer and add some arbitrary amount of nesting of functions or simply add 1)

That said, to be clear with everyone, when someone posts something like "graham's number" and then someone else posts something like "graham's number ^ graham's number", I don't count the second one as larger.

Simply because it doesn't take any creativity to generate that second number. And Graham's Number is already dealing with arithmetic operations of an order that itself couldn't be expressed in any other way than something akin to arbitrary degrees of re-iterated nestings of knuth's up arrows (or conway's chained arrows etc.)

In terms of computational complexity.. there really is no difference between G and G^G. You have to find a new way of thinking to generate a number in a different computational class.

Many people used the busy beaver. And one of them even made it into the top five. The top four all used infinite ordinals subscripting a fast growing hierarchy via an "nth element of the fundamental sequence" mapping. I know there will be tears regarding the busy beaver being beaten, but what you may not have understood is that, though the busy beaver outgrows any possible computable function, that doesn't mean it outgrows it just anywhere. To overcome the infinite ordinal subscripted fast growing hierarchy (for some of the hilariously huge ordinals some of the answers used), you have to take the busy beaver VERY VERY far out.

I received about 40 messages. At least a dozen will take me some serious math work to determine which is larger than which. So it might be a day or two before I can post the winners. Sorry about that. (to give you an example, think about something like: graham's number vs. f_epsilon_nought(2) from the fast growing hierarchy. In orer to prove the second is larger, you have to observe that f_(omega+1)(64) = f_omega^64(64)>f_omega^64(6)>graham's number, that last inequality depending on a very subtle observation of the relationship between successive diagonalizations of the hierarchy and successive terms in the "g" sequence that ultimately leads to g_64 or "graham's number." so some of these answers will be very difficult to rank. especially given that it is effectively impossible to compare various utilizations of the busy beaver since it's hardly computable or even estimatable in any sense. but i have a good sense for this type of math so i should be able to get it all figured out. if not, i'll call in some big gun friends from the math stack exchange)

Btw I have four keys though at one point in the thread I accidentally said I had five. Don't know if anyone noticed that. So to avoid any confusion, I'm going to buy a fifth key for the first place winner. They can choose what game. Just $60 or less. I personally would suggest Skyrim Legendary Edition :p But whatever they want.

Thanks for all the submissions. In case I didn't make it clear there were a ton of really awesome answers and I even learned about Rayo's number and for anyone who knows me, me getting to learn something new in the vein of esoteric/candy-ish math/compsci is rare and much appreciated :P

At 8/22/13 12:07 PM, Manly-Chicken wrote: Do you think I have a chance?

Afraid not man.

Your answer was basically googolplex^googolplex^...^googolplx, googolplex times, which is (googolplex)↑↑(googolplex) which is nothing compared to something like 3↑↑↑↑3.

It's (computationally speaking) trivially bigger than 3↑↑↑3 and unimaginably smaller than 3↑↑↑↑3.

This 3↑↑↑↑3 is g_1 in a sequence defined: 3(↑^(g_n))3 = g_(n+1)

And Graham's number = g_64.

And to make matters worse (for you) Graham's number will have nothing to do with the top five winning entries :p. Some of them used Graham's number as arguments to functions, but I would have weighed their answer equally if they'd used 2 instead of G, since the function to which they were feeding said argument was so computationally far beyond the ideas that generate Graham's number, that Graham's number is no longer of any interest and is unnecessary icing on the cake.

Locking this thread until sharpnova is ready to announce the winners.