Okay, so that's (G(G(1, G(1)+2)^100^100^100^100))*((G(1, G(1)+2)^100^100^100^100)^2). 100^100^100^100 is 100^100^10^200, which is equal to 100^10^(2*(10^200)), which is less than 100^10^10^201, which is 10^(2*(10^10^201)), which is less than 10^10^10^202. And (10^10^10^202)^2 is 10^(2*(10^10^202)), which is less than 10^10^10^203, so your number is less than (G(G(1, G(1)+2)^10^10^10^203))*(G(1, G(1)+2)^10^10^10^203). Since G(1, G(1)+3) is massively greater than G(1, G(1)+2), your number is less than G(G(1, G(1)+3))*G(1, G(1)+3). Since G(G(x, y)) is really just G(x, y+1), your number is less than G(1, G(1)+4)*G(1, G(1)+3), which is less than G(1, G(1)+5).
Let us, for the sake of convenience, call G(1, G(1)) A. Your number, obviously, is less than (A) ^A^A^A^ (A * (A ^A^A^A^A) ). That number is less than A^A^A^A^A^A^A^A^A^A, which is less than Ack(A, 11, 4).
Now let us take G(1, G(1)+1), as opposed to Ack(G(1, G(1)), 11, 4). G(1, G(1)+1) is equal to G(G(1), G(1)), so we are comparing G(A) and Ack(A, 11, 4). I shouldn't have to explain why G(x) grows faster than Ack(x, 11, 4).
0
I use a Heal-inator on myself.
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You, sir, have no idea how this game works, do you?
Well, I'm onto you. You think you can just post whatever you want in a game as long as it's tangentially related, but I'm watching you.
0
(
Ack(a, b, 1) = a+b
Ack(a, 1, c) = a
Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)
G(1) = Ack(10, 10, 10)
G(a) = Ack(10, 10, G(a-1))
G(a, 1) = G(a)
G(a, b ) = G(G(a), b-1)
)
Dude. If you don't understand the current number, don't embarrass yourself. Your number was beaten last page.
G(1, G(3)+2)[/b]
0
B
0
F
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T is for Treading closer to victory
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Not really, it takes me less time than you might think.
(
Ack(a, b, 1) = a+b
Ack(a, 1, c) = a
Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)
G(1) = Ack(10, 10, 10)
G(a) = Ack(10, 10, G(a-1))
G(a, 1) = G(a)
G(a, b ) = G(G(a), b-1)
)
G(1, G(3))
Word of advice: Raising numbers to some teeny tiny power is pretty much useless if you just slap it on.
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Fabastic!
(
Ack(a, b, 1) = a+b
Ack(a, 1, c) = a
Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)
G(1) = Ack(10, 10, 10)
G(a) = Ack(10, 10, G(a-1))
G(a, 1) = G(a)
G(a, b ) = G(G(a), b-1)
)
Okay, so that's (G(G(1, G(1)+2)^100^100^100^100))*((G(1, G(1)+2)^100^100^100^100)^2). 100^100^100^100 is 100^100^10^200, which is equal to 100^10^(2*(10^200)), which is less than 100^10^10^201, which is 10^(2*(10^10^201)), which is less than 10^10^10^202. And (10^10^10^202)^2 is 10^(2*(10^10^202)), which is less than 10^10^10^203, so your number is less than (G(G(1, G(1)+2)^10^10^10^203))*(G(1, G(1)+2)^10^10^10^203). Since G(1, G(1)+3) is massively greater than G(1, G(1)+2), your number is less than G(G(1, G(1)+3))*G(1, G(1)+3). Since G(G(x, y)) is really just G(x, y+1), your number is less than G(1, G(1)+4)*G(1, G(1)+3), which is less than G(1, G(1)+5).
G(1, G(2))
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[to] placate artichokes
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A is for At least allow us a glimmer of hope.
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practicing googolology
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A
0
(
Ack(a, b, 1) = a+b
Ack(a, 1, c) = a
Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)
G(1) = Ack(10, 10, 10)
G(a) = Ack(10, 10, G(a-1))
G(a, 1) = G(a)
G(a, b ) = G(G(a), b-1)
)
G(1, G(1))^100^100^100^100 * G(1, G(1))^100^100^100^100
Alright... let's put in some more parentheses. (This is being favorable with the grouping, by the way)
(G(1, G(1))) ^100^100^100^ (100 * (G(1, G(1)) ^100^100^100^100) )
Let us, for the sake of convenience, call G(1, G(1)) A. Your number, obviously, is less than (A) ^A^A^A^ (A * (A ^A^A^A^A) ). That number is less than A^A^A^A^A^A^A^A^A^A, which is less than Ack(A, 11, 4).
Now let us take G(1, G(1)+1), as opposed to Ack(G(1, G(1)), 11, 4). G(1, G(1)+1) is equal to G(G(1), G(1)), so we are comparing G(A) and Ack(A, 11, 4). I shouldn't have to explain why G(x) grows faster than Ack(x, 11, 4).
G(1, G(1)+2)
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artichokes and
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artichokes because