Sorry, but that number wasn't actually bigger. Proof:

I forget what a quadrillion is, so let's call it 10^100 to be on the safe side. Eyeballing it, there were 10,000 repetitions of quadrillion at most in both the base and the exponent. So, we have ((10^100)^10,000)^^2. (10^100)^10,000 is 10^1,000,000, so the number is (10^1,000,000)^(10^1,000,000). Some more algebra leaves us with 10^10^1,000,006. That's less than 10^10^10^10, which is 10^^4, which is less than Ack(10, 10, 10). My number was Ack(10, 10, 100).

I forget what a quadrillion is, so let's call it 10^100 to be on the safe side. Eyeballing it, there were 10,000 repetitions of quadrillion at most in both the base and the exponent. So, we have ((10^100)^10,000)^^2. (10^100)^10,000 is 10^1,000,000, so the number is (10^1,000,000)^(10^1,000,000). Some more algebra leaves us with 10^10^1,000,006. That's the absolute most that A can be. Your number is, at most, (10^100)*(10^10^1,000,006), which is equal to 10^(10^1,000,006 + 100). That's less than 10^10^1,000,007, which is less than 10^10^10^10, which is 10^^4, which is less than Ack(10, 10, 10). My number was Ack(10, 10, 900).

And now it's Ack(10, 10, 9001).

Oh, and it's worth noting that A was beaten on post 25, with 100^^100.

(Note: If you're going to post in this, please post a number that is actually bigger.)

So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:
(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)

So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:
(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)

So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:
(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)

So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:
(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:
(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

My avatar is a ushanka wearing bird carrying a slice of pizza. Keep the assumptions within reason and I wont bother you. Also, the bird carries a red paint bucket, but that's cropped off.

Uh... Okay. Apparently, SSCG(3) > TREE(3) > A^(A(187196))(1), where A(x) = Ack(2, x+1, x) and A^n(x) = A^n-1(A(x)). G(x) is approximately equal to A^x(1), so an extremely weak lower bound for TREE(3) (And hence SSCG(3)) is G(G(187196)), which handily beats the previous record of G(66).

*gets super confused as to exactly what number that actually is*

Okay. I know some REALLY large numbers. I'm going to name some numbers instead of using weird notation. IDK how high your number is but I'm going to name the number Googolplex (1 followed by a googol zeroes[1 followed by 100 zeroes]). I do believe there are higher numbers than that but I'm not sure.

Rollback Post to RevisionRollBack

I am just another one of the Minecraft players here. Avatar: Tallest extreme hills mountain reaching level 150

Googolplex - One followed by a Googol zeroes. There isn’t enough space in the universe in to write this number.

Rollback Post to RevisionRollBack

Say something silly, Laugh 'til it hurts, Take a risk, Sing out loud, Rock the boat, Shake things up, Flirt with disaster, Buy something frivolous, Color outside the lines, Cause a scene, Order dessert, Make waves, Get carried away, Have a great day!

GDog_0, ConorGarland: 10^10^100 was beaten on post 25, with 100^^100.

(Note: x^^y is tetration, where:

1^^y = 1

x^^1 = x

x^^y = x^(x^^(y-1))

)

Proof that this is bigger: It is self-evident that when x, y > 0: y^x >= x. Let x be 100^100^100 (which is bigger than a googolplex) and let y be 100^^97.

Also, you kind of have to use notations instead of just throwing googolplexes out there, or you can't even beat G^66(1).

(

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^1(a) = G(a)

G^n(a) = G^n-1(G(a))

)

G^67(1)[/b]

There isn't enough space in the universe to write G^1(1). Not even if in a universe, every Planck length cube had a googolplex googolplex googolplex g-bits, each of which could store a number between 0 and a googolplex-1. Not even if every Planck length cube had a googolplex googolplex googolplex universes each packed to the brim with universes packed to the brim (repeat a googolplex times) with g-bits. That's only googolplex^(googolplex*5), which is (10^10^100)^(10^10^100 * 5), which is 10^((10^100)*(10^10^100 * 5)), which is less than 10^((10^100)*(10^10^101)), which is 10^10^(100 + 10^101), which is less than 10^10^10^102. G^1(1) is 10^^^^^^^^10.

"Also, you kind of have to use notations instead of just throwing googolplexes out there, or you can't even beat G^66(1)."

Did I say that in the post before you? Yes I did! Your number was beaten on post 25. 10^googolplex is 10^10^10^100, which is less than 100^100^100^100, which is 100^^4, which is less than 100^^100.

(

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^1(a) = G(a)

G^n(a) = G^n-1(G(a))

)

G^100(1)

Rollback Post to RevisionRollBack

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Kid's stuff! Time to break out the slightly less miniscule guns...

(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

)

Ack(10, 10, 100)

one quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion 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quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion to the power of one quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion q

uadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion quadrillion

Text.LiveTyping.com

This is my high quality stuff

I'm calling the above post variable A.

So how about 2A?

Sorry, but that number wasn't actually bigger. Proof:

I forget what a quadrillion is, so let's call it 10^100 to be on the safe side. Eyeballing it, there were 10,000 repetitions of quadrillion at most in both the base and the exponent. So, we have ((10^100)^10,000)^^2. (10^100)^10,000 is 10^1,000,000, so the number is (10^1,000,000)^(10^1,000,000). Some more algebra leaves us with 10^10^1,000,006. That's less than 10^10^10^10, which is 10^^4, which is less than Ack(10, 10, 10). My number was Ack(10, 10, 100).

And now it's Ack(10, 10, 900).

400,000,000,000,000,000,000,000,000,000,000A

Text.LiveTyping.com

This is my high quality stuff

THAT NUMBER IS NOT BIGGER! Proof:

I forget what a quadrillion is, so let's call it 10^100 to be on the safe side. Eyeballing it, there were 10,000 repetitions of quadrillion at most in both the base and the exponent. So, we have ((10^100)^10,000)^^2. (10^100)^10,000 is 10^1,000,000, so the number is (10^1,000,000)^(10^1,000,000). Some more algebra leaves us with 10^10^1,000,006. That's the absolute most that A can be. Your number is, at most, (10^100)*(10^10^1,000,006), which is equal to 10^(10^1,000,006 + 100). That's less than 10^10^1,000,007, which is less than 10^10^10^10, which is 10^^4, which is less than Ack(10, 10, 10). My number was Ack(10, 10, 900).

And now it's Ack(10, 10, 9001).

Oh, and it's worth noting that A was beaten on post 25, with 100^^100.

(Note: If you're going to post in this, please post a number that is actually bigger.)

Ack(10, 10, 9002).

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(2)

Go to 3 minutes 27 seconds

So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:

(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:

(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)So that's, being generous with the grouping, ((10^10^100)!)^(10^10^100)^(((10^10^100)^2)*Graham's number). X factorial, obviously, is less than X^X. so that number is smaller than ((10^10^100)^(10^10^100))^(10^10^100)^(((10^10^100)^2)*Graham's number). (10^10^100)^(10^10^100) is equal to 10^10^10^102, and (10^10^100)^2 is less than 10^10^101. Therefore, that number is less than (10^10^10^102)^(10^10^100)^((10^10^101)*Graham's number). I could go further with the simplification, but I'll move on from this.

Graham's number has multiple versions, but the largest one is defined like so:

(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)Graham's number has multiple versions, but the largest one is defined like so:

(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)Graham's number has multiple versions, but the largest one is defined like so:

(

Ack(a, b, 1) = a+b

Ack(a, 1, c) = a

Ack(a, b, c) = Ack(a, Ack(a, b-1, c), c-1)

Graham(1) = Ack(4, 4, 4)

Graham(a) = Ack(4, 4, Graham(a-1))

Graham's number = Graham(64)

)

For obvious reasons, that power tower doesn't even bump your number up to Graham(65).

(

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))

)

Again, it is obvious that G(a) > Graham(a) for all a. Therefore, G(65) is bigger than your number.

G(66)Since we're getting crazy at this point...

SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(G(121))))))))))

Have fun getting bigger.

My avatar is a ushanka wearing bird carrying a slice of pizza. Keep the assumptions within reason and I wont bother you. Also, the bird carries a red paint bucket, but that's cropped off.

i dont really know what to put here

(Since I can't grasp the concept of these ridiculously high numbers, I'll just take the number above me and try to one-up it)

SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(SSCG(G(363363363))))))))))[/b]

[Avatar] My unshaded MC avatar (Shaded version later)

[Status]

"...Thank you.""A magic attack right at their face!"~DavidUh... Okay. Apparently, SSCG(3) > TREE(3) > A^(A(187196))(1), where A(x) = Ack(2, x+1, x) and A^n(x) = A^n-1(A(x)). G(x) is approximately equal to A^x(1), so an extremely weak lower bound for TREE(3) (And hence SSCG(3)) is G(G(187196)), which handily beats the previous record of G(66).

(

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^1(a) = G(a)

G^n(a) = G^n-1(G(a))

)

G^66(1)Good luck proving that your number beats that.

*gets super confused as to exactly what number that actually is*

Okay. I know some REALLY large numbers. I'm going to name some numbers instead of using weird notation. IDK how high your number is but I'm going to name the number Googolplex (1 followed by a googol zeroes[1 followed by 100 zeroes]). I do believe there are higher numbers than that but I'm not sure.

I am just another one of the Minecraft players here. Avatar: Tallest extreme hills mountain reaching level 150

∞

(did i just break this?)

GASPYes you broke a rule: Cannot be infinity.

Googolplex - One followed by a Googol zeroes. There isn’t enough space in the universe in to write this number.

GDog_0, ConorGarland: 10^10^100 was beaten on post 25, with 100^^100.

(Note: x^^y is tetration, where:

1^^y = 1

x^^1 = x

x^^y = x^(x^^(y-1))

)

Proof that this is bigger: It is self-evident that when x, y > 0: y^x >= x. Let x be 100^100^100 (which is bigger than a googolplex) and let y be 100^^97.

Also, you kind of have to use notations instead of just throwing googolplexes out there, or you can't even beat G^66(1).

(

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^1(a) = G(a)

G^n(a) = G^n-1(G(a))

)

G^67(1)[/b]

There isn't enough space in the universe to write G^1(1). Not even if in a universe, every Planck length cube had a googolplex googolplex googolplex g-bits, each of which could store a number between 0 and a googolplex-1. Not even if every Planck length cube had a googolplex googolplex googolplex universes each packed to the brim with universes packed to the brim (repeat a googolplex times) with g-bits. That's only googolplex^(googolplex*5), which is (10^10^100)^(10^10^100 * 5), which is 10^((10^100)*(10^10^100 * 5)), which is less than 10^((10^100)*(10^10^101)), which is 10^10^(100 + 10^101), which is less than 10^10^10^102. G^1(1) is 10^^^^^^^^10.

1.000x10

^{googolplex}(I think that is scientific notation...not sure though)

I am just another one of the Minecraft players here. Avatar: Tallest extreme hills mountain reaching level 150

"Also, you kind of have to use notations instead of just throwing googolplexes out there, or you can't even beat G^66(1)."

Did I say that in the post before you? Yes I did! Your number was beaten on post 25. 10^googolplex is 10^10^10^100, which is less than 100^100^100^100, which is 100^^4, which is less than 100^^100.

(

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^1(a) = G(a)

G^n(a) = G^n-1(G(a))

)

G^100(1)