Dominius, a competition was what I was aiming for.
You should read the Googology wiki to know more http://googology.wikia.com/wiki/Googology_Wiki . Your concern is valid however the goal is always to get a function that transcends all the other ones. Fast growing hierarchy is a kind of "speedometer" for large number notations and functions http://googology.wikia.com/wiki/Fast-growing_hierarchy , f0 (n) is the growth rate for addition, f1 for multiplication, f2 for exponentiation, f3 for repeated exponentiation (tetration), f4 for pentation etc. to infinity.
There are 2 ways to go further from here. One is the naive way, you might say the "lame" way, to just keep simply increasing the hyper operators, so you could say for example f(googol) (100).
There is however another way, and this is to make a function that will always grow further than all the functions below it. In this case it is f_w (n), where w is omega, the first countable ordinal.
f_w(n) = f_n(n) - this is basically the growth rate of Ackermann numbers http://googology.wikia.com/wiki/Ackermann_number . It makes the number have the same number of up-arrows/hyper-operators as the starting number. So A(1000) = 1000^^^^....(1000 arrows)^^^^1000.
Above omega, there is really no shame in using the "lame" way a bit. The famous Graham's number is basically f_w+1(64), so the growth rate is just 1 step above w. In human terms this is still an enormous leap. Instead of using the number n with the nth hyper operator, Graham's number starts with 3^^^^3 and is defined so that every further member of this sequence has the same numbers of hyperoperators as the previous number.
So:
G1 - 3^^^^3
G2 - 3^...^3 with G2 arrows
etc.
Until G64.
Now this is still growth rate w+1. To get just to w+2, you must repeat the G function on itself G times, so something like G(G(G...(G64))) but with G64 brackets. We are just 2 simple steps above w and our number has more brackets than elementary particles in the observable universe.
f_w*2 (n) is basically f_w+n (n). So the iteration of this function is the same as our n. For example f_w*2 (G64) will be f_w+G64 (G64). Are you dizzy yet?
But we haven't even trascended the first countable ordinal yet. To get to the second one, epsilon-0, we'd have to go w*3, w*4.... then to w^w and finally to w^w^w^w.... with w ws (basically infinite ws). w^w^w^w.... = epsilon-0. There are TEN more ordinals that can describe the growth of finite numbers (ordinals are infinite, the functions using them are not).
It has been popular wisdom that TREE(3) is a larger number than Graham's in professional math. This is very true. But saying it is just "larger" does not do it justice. You can describe Graham's number in 1 paragraph of text using just simple functions and Knuth up-arrows. The TREE functions grows at the rate of fϑ(Ωω^ω)(n). This is 8 ordinals above omega. So the difference between 1 and G64 is far lower than between G64 and TREE(3).
That being said, tincup, a valiant attempt but 30 billion light years is just about 1.756 * 10^61 Planck lengths. So it's not even a googol which is 10^100. Anyways, you don't really need functions below the f_w (n) level so don't feel burdened by this mathematical stuff and COMPETE 😈 😀 . To give people without interest in ungodly numbers a chance, I revise my entry to just 10^100, it's one of the most well known big numbers and fairly easy to surpass significantly.
To not be to easy on people, it is NOT permitted to just state "my number is f_e0 (100) in the fast growing hierarchy". The FGH is just a benchmark, you have to construct a number on your own. And from now on, yes, this is a competition, so your number has to be larger than the previous guy's number.
Also, before anyone thinks I am some math crazy guy tormenting innocent Vogons residents, I actually almost failed high school math. I just happen to find large numbers fascinating. They really help a person to appreciate what infinity really means. It is eye opening to know that no matter how high we go, infinity is still the same distance ahead.