My number is bigger than yours! 3

[cantab]

G↑^GG

Where G is of course Graham's number, and ↑ⁿ is n repetitions of the arrow in up-arrow notation

[Guest alemagno12]

Let stage 0 = 64

Let stage n = G(stage n-1)

Let G2(n) represent stage n

Let G3(0) = 64 and G3(n) = G2(G3(n-1))

Let G1(n) = G(n)

Let Gm(0) = 64 and Gm(n) = Gm-1(Gm(n-1))

My number is Gg(64), where g is graham's number

[Souper]

G^G↑^GG​^G

God forgive me.

[Guest alemagno12]

G^G↑^GG​^G

God forgive me.

Not even close.

[Guest alemagno12]

Let a{b}c = a↑^bc

Let a{{1}}b = a{a{a{...{a{a{a{a}a}a}a}...}a}a}a with b levels

Let a{{n}}b = a{n-1}a{n-1}a{n-1}...{n-1}a{n-1}a{n-1}a with b {n-1}'s

Let a{n,m}b = a{{{...{{{m}}}..}}}b with n levels

Let a{1,m}b = a{m}b

Let a{n,1}b = a{n-1,a{n-1,a{n-1,a{...{n-1,a{n-1,a{n-1,1}a}a}...}a}a}a}a with b levels

Let a{n,m}b = a{n,m-1}a{n,m-1}a{n,m-1}...{n,m-1}a{n,m-1}a{n,m-1}a with b {n,m-1}'s

Number is 3{G,G}3, where G is graham's number

[Souper]

-G

I win automatically.

*NEW ROUND*

1.

[Guest alemagno12]

-G

I win automatically.

*NEW ROUND*

1.

NEIN NEIN NEIN NEIN NEIN NEIN NEIN NEIN NEIN!!!!!!!!!

-G is A LOT smaller than my entry.

Edited October 10, 2014 by Vanamonde
Enthusiasm tempered.

[Guest alemagno12]

S^S, where S = Shrek.

I win, as no number can possibly be higher.

NEW.

ROUND.

1.

NEIN NEIN NEIN NEIN NEIN NEIN NEIN NEIN NEIN!!!!!!!!!

What the heck is Shrek?

Edited October 10, 2014 by Vanamonde

[Tery215]

Four.

i win this time

[Souper]

-G.

Going into negative numbers isn't moving along the scale of positive numbers, but rather going out the far end of positive numbers, so it is made infinity.

Because i did not use infinity in my number itself, but my number defines infinity, it's a sneaky way of going infinity, thus making it so that i can't be beat.

Edited October 10, 2014 by Souper

[cantab]

That would only be valid if we were talking about temperatures, in which case -0 K, game over.

But since we aren't, 4. Pretty sure that's the biggest number so far

[Guest alemagno12]

-G.

Going into negative numbers isn't moving along the scale of positive numbers, but rather going out the far end of positive numbers, so it is made infinity.

Because i did not use infinity in my number itself, but my number defines infinity, it's a sneaky way of going infinity, thus making it so that i can't be beat.

NEIN NEIN NEIN NEIN NEIN NEIN NEIN NEIN NEIN!!!!!!!!!

Only positive numbers allowed.

Edited October 11, 2014 by vexx32
Just stop.

[Tery215]

4. Pretty sure that's the biggest number so far

But I said 4.. :c

8

[Sun]

^{263463635252463636}(10)^{312000000005463460}

[Guest alemagno12]

^{263463635252463636}(10)^{312000000005463460}

Too long. Not valid.

[Sun]

999{g,g}99

[Guest alemagno12]

999{g,g}99

Valid.

Let define S-growing hierarchy:

S_0(n) = n{n,n}n

S_a+1(n) = S_a(S_a(...(S_a(n))...)) with n levels, where a is any ordinal

S_a(n) = S_a[n](n) where a is a limit ordinal and a[n] is the nth member of the fundamental sequence of a.

Fundamental sequences:

w = 0,1,2,...

w*2 = w,w+1,w+2,...

w^2 = 0,w,w*2,w*3,...

w^w = 0,w,w^2,w^3,...

e(0) = w^^w = 0,w,w^w,w^w^w,...

My number is S_e(0)(g) where g is graham's number

[cantab]

Let a{b}c = a↑^bc

Let a{{1}}b = a{a{a{...{a{a{a{a}a}a}a}...}a}a}a with b levels

Let a{{n}}b = a{n-1}a{n-1}a{n-1}...{n-1}a{n-1}a{n-1}a with b {n-1}'s

Let a{n,m}b = a{{{...{{{m}}}..}}}b with n levels

Let a{1,m}b = a{m}b

Let a{n,1}b = a{n-1,a{n-1,a{n-1,a{...{n-1,a{n-1,a{n-1,1}a}a}...}a}a}a}a with b levels

Let a{n,m}b = a{n,m-1}a{n,m-1}a{n,m-1}...{n,m-1}a{n,m-1}a{n,m-1}a with b {n,m-1}'s

Number is 3{G,G}3, where G is graham's number

I must confess confusion here. Why does a{n,m}b expand to something with n levels while a{n,1}b expands to something with b levels, for example?

Edit: And what's the order of evaluation in something like a{n-1}a{n-1}a{n-1}...{n-1}a{n-1}a{n-1}a , or can you show it's not important?

Edited October 10, 2014 by cantab

[Guest alemagno12]

I must confess confusion here. Why does a{n,m}b expand to something with n levels while a{n,1}b expands to something with b levels, for example?

Edit: And what's the order of evaluation in something like a{n-1}a{n-1}a{n-1}...{n-1}a{n-1}a{n-1}a , or can you show it's not important?

I don't see anything wrong.

[Sun]

10^10^10^10^10

[Guest alemagno12]

10^10^10^10^10

infinity times.

[Sun]

1019277009.

[Guest alemagno12]

1019277009.

infinity times.

[Sun]

Yes.

10char

[O Nerd]

(15x10³²)!