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4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^ 4^4^4^4^4^4^4^4^4^4^4^4^4^4
Thread has been throwed....
I believe this is bigger than any of your numbers.
Since pics are disabled, the following in an explanation on how to count better.
Ever tired of counting on your fingers and being stuck at 10? Here's the solution: Binary finger counting. Instead of counting how many fingers are up, you count which fingers are up. With only 8 fingers, you can count up to 255.
1. Ignore thumbs, they make this process harder.
2. Start with your rightmost finger. This finger represents 2^0, or 1. Your next finger represents 2^1, or 2. Try all the different combinations with these two fingers, you have a range from 0-3. Your next finger, 2^2 = 4, followed by 2^3 = 8 and so on.
3. Now you can count high! the finger combination 00110101 = 1+4+16+32 = 53.
Is the number bigger, or is its representation?Originally Posted by Fetyukov
By rule 1, define explicitly your number(not just its representation); and by rule 2, prove that it's larger than the person's above you.
Last edited by MichaelBurge; 05-30-2011 at 12:59 AM.
A(g64,g64) is not admissible, you haven't defined it in your post. That would leave my answer as the biggest since one centillion is SMALLER THAN MY NUMBER.
Credit to Devious`, with thanks to AvunaOs for my last signature
Also, to up the ante:
Y(1000,1000,1000)
Where Y is the YawningAngel function in which the first argument is raised to the power of the second, the second to the power of the third, the third to the power of the first, all the resulting numbers are multiplied together, the factorial of the product is taken, and this number is cubed to yield the final result. Thus Y(1000,1000,1000)=
(((1000^1000)^3)!)^3
Which can be simplified to (1000^3000)!^3
Credit to Devious`, with thanks to AvunaOs for my last signature
At this point, your number needs to accompanied by an argument that it's larger than the previous answer of
4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^4^ 4^4^4^4^4^4^4^4^4^4^4^4^4^4
Glee:jera:
Burge, that was never an answer.
Credit to Devious`, with thanks to AvunaOs for my last signature
An interesting thing, if you've heard of busy beaver numbers, is there's a specific N so that BB[n] for any n>N is independent of ZFC. (The reason is that the logic of ZFC can be recognized by a Turing machine with N states.)
All answers should have lime text from now on.
I have to ignore alot of answers; his was the last that seemed legitimate('^' referring to a power in his post).
Another interesting thing is that all ZFC-definable ordinals are allowed, which nobody's taken advantage of yet.An interesting thing, if you've heard of busy beaver numbers, is there's a specific N so that BB[n] for any n>N is independent of ZFC. (The reason is that the logic of ZFC can be recognized by a Turing machine with N states.)
Last edited by MichaelBurge; 05-30-2011 at 02:57 PM.
1 gogoplexian^5 gogoplex
gg wp
1 gogoplexian^6 gogoplex
O.O
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gogoplex^(gogoleplex^gogoleplex)
100^1000
come at me bro
G(6)
Where the function G takes the number n, and computes a power tower (n^n...) with height of n centillion. i.e., G(n) computes n to the power of n (to the power of n) x n*10^303.
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Bonuses include sober girls around you they be ackin' like they drunk.
Ackin', ackin' like they drunk.
They be ackin' like they drunk,
all these sober girls around me they be ackin' like they dru-uh-unk.
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G(7)
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