My Number is WAYYY larger [game]

[rutabaga]

g_3024(g_32(1024)). g_3024 iterations, starting with g_32(1024).

g_10000(g_10000(10000))

I can see where this is going, and I can immediately think of two ways to increase it. One of them is fun, and the other is the Fast Growing Hierarchy.
I'll choose the one that's at least somewhat fun.

I'll define a function cracker(x) (cause you know, graham crackers are tasty) that has graham's function with x as the argument and iterated x times. Here's the thing - it isn't a single-argument function. Cracker(1,x) is cracker(cracker(...cracker(x)...)) with cracker(x) crackers. Cracker(2,x) is cracker(1,cracker(...1,cracker(1,x)...)) with cracker(1,x) crackers. In general, cracker(n,x) is cracker(n-1,cracker(...n-1,cracker(n-1,x)...)) with cracker(n-1,x) crackers. Feel free to steal my two argument function (I already have plans for a larger function though).

My number is cracker(2,2).

[unname4798]

g_3024(g_32(1024)). g_3024 iterations, starting with g_32(1024).

g_10000(g_10000(10000))

I can see where this is going, and I can immediately think of two ways to increase it. One of them is fun, and the other is the Fast Growing Hierarchy.
I'll choose the one that's at least somewhat fun.

I'll define a function cracker(x) (cause you know, graham crackers are tasty) that has graham's function with x as the argument and iterated x times. Here's the thing - it isn't a single-argument function. Cracker(1,x) is cracker(cracker(...cracker(x)...)) with cracker(x) crackers. Cracker(2,x) is cracker(1,cracker(...1,cracker(1,x)...)) with cracker(1,x) crackers. In general, cracker(n,x) is cracker(n-1,cracker(...n-1,cracker(n-1,x)...)) with cracker(n-1,x) crackers. Feel free to steal my two argument function (I already have plans for a larger function though).

My number is cracker(2,2).

crackerr(x) is cracker(x,x) repeated x times.
crackerr(2)

[rutabaga]

cracker(cracker(...cracker(x,x),x...),x) with cracker(cracker(x,x),x) crackers is cracker(1,1,x).
that's a lot of crackers, though, so i'll shorten it to just c(x).
calculate c(1,2,x) from c(1,1,x) the same way we went from c(1,x) to c(2,x). c(1,c(1,c(...1,c(1,c(x,x),x),x...),x),x) is c(2,1,x).

my number is c(2,22,222). if this isn't well-defined enough, i can explain it more.

[unname4798]

cracker(cracker(...cracker(x,x),x...),x) with cracker(cracker(x,x),x) crackers is cracker(1,1,x).
that's a lot of crackers, though, so i'll shorten it to just c(x).
calculate c(1,2,x) from c(1,1,x) the same way we went from c(1,x) to c(2,x). c(1,c(1,c(...1,c(1,c(x,x),x),x...),x),x) is c(2,1,x).

my number is c(2,22,222). if this isn't well-defined enough, i can explain it more.

cplus(x) is c(x,x,x) repeated x times.
cplus(1000)

[rutabaga]

cracker(cracker(...cracker(x,x),x...),x) with cracker(cracker(x,x),x) crackers is cracker(1,1,x).
that's a lot of crackers, though, so i'll shorten it to just c(x).
calculate c(1,2,x) from c(1,1,x) the same way we went from c(1,x) to c(2,x). c(1,c(1,c(...1,c(1,c(x,x),x),x...),x),x) is c(2,1,x).

my number is c(2,22,222). if this isn't well-defined enough, i can explain it more.

cplus(x) is c(x,x,x) repeated x times.
cplus(1000)

cplus_[cplus(64)](64). I have 4 ideas left-- an idea that will have to come up soon before it's too small, a stronger version of that idea, an idea that has a small chance of killing the thread (you and ababa11e are both smart though), and an idea that has to do with FGH and probably isn't allowed lol

[ababa11e]

cracker(cracker(...cracker(x,x),x...),x) with cracker(cracker(x,x),x) crackers is cracker(1,1,x).
that's a lot of crackers, though, so i'll shorten it to just c(x).
calculate c(1,2,x) from c(1,1,x) the same way we went from c(1,x) to c(2,x). c(1,c(1,c(...1,c(1,c(x,x),x),x...),x),x) is c(2,1,x).

my number is c(2,22,222). if this isn't well-defined enough, i can explain it more.

cplus(x) is c(x,x,x) repeated x times.
cplus(1000)

cplus_[cplus(64)](64). I have 4 ideas left-- an idea that will have to come up soon before it's too small, a stronger version of that idea, an idea that has a small chance of killing the thread (you and ababa11e are both smart though), and an idea that has to do with FGH and probably isn't allowed lol

im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)

edit: rutabaga do your worst

[unname4798]

cplus(x) is c(x,x,x) repeated x times.
cplus(1000)

cplus_[cplus(64)](64). I have 4 ideas left-- an idea that will have to come up soon before it's too small, a stronger version of that idea, an idea that has a small chance of killing the thread (you and ababa11e are both smart though), and an idea that has to do with FGH and probably isn't allowed lol

im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)

edit: rutabaga do your worst

Second fool (or level):
cpluses(x,y) is c[x pluses](y,...,y)
cpluses(2024,2024)

[rutabaga]

im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)

edit: rutabaga do your worst

lmao cracker hierarchy is awesome, but you just gave me another new idea
the generalized version of cplus is capital C(x), but it can take any amount of arguments.
first, though, let's define cplus+++(x1,x2,x3,x4) to make sure everyone understands the pattern.
cplus+++(x1,x2,x3,x4)=cplus++(x1,x2,cplus++(x1,x2,cplus++(x1,x2...x4 times...cplus++(x1,x2,x3)...)

now we can tell that in general:
C(x1,x2,...x(n-1),x(n)) = C(x1,x2...x(n-2),C(x1,x2...x(n-2),C(x1,x2...x(n) times...C(x1,x2...x(n-1))...)

...but i just had ANOTHER idea, so i don't want to stop there. first, let's make C((x)) be C(x,x,...C(x) times...x).
now let's run through the whole entire hierarchy again. let's pretend that the old C((x)) is the new C(x), and we have to do the whole hierarchy thing again to get back to C((x)). doing this gives us C((1,x)). doing the C(x) to C((1,x)) procedure, C((1,x)) times, gets us to C((2,x)).
you can probably see where this is headed. doing the C(x) to C((n,x)) procedure, repeated C((n,x)) times, leads us to C((n+1,x)).

naturally, we'll eventually get more arguments in the function. first, C((C((...C((x,x)) times...C((x,x)),x...)),x)) is C((1,1,x)).
we'll grow the second argument in the same way as normal-- C(x) to C((1,1,x)) repeated C((1,1,x)) times is C((1,2,x)).
increasing the new first argument goes like this:
C((a,C((a,C((...C((a,C((x,x)),x)) times...a,C((x,x)),x...)),x)),x)) = C((a+1,1,x)).
when the "a" reaches C((C((C((...C((C((x,x,x)),x,x)) times...C((x,x,x)),x,x...)),x,x)),x,x)), we get a fourth argument.

we increase the 2nd-to-last argument the same way we did before. increasing the 3rd-to-last argument is mostly the same:
C((n1,a,C((n1,a,C((...C((n1,a,C((x,x,x)),x)) times... n1,a,C((x,x,x)),x...)),x)),x)) = C((n1,a+1,1,x)).
n1 is just placeholder for numbers that come before the a-argument.
when a = C((n1,C((n1,C((...C((n1,C((x,x,x)),x,x)) times...n1,C((x,x,x)),x,x...)),x,x)),x,x)), increase the 4th-to-last number by 1.

a pattern is faintly visible here, and while it'd be easier to see with more arguments, it'd take a lot of time to explain each step of adding each argument, so if this isn't well-defined enough, i can explain it more in another post because this one is already kinda long.

adding a 5th argument is just like adding a 4th was, and it behaves like the other arguments. you can put any amount of arguments in C((...)). to top this all off, C(((x))) is C((x,x,...C((x)) times...x)).

my number is C(((418))).

not sure if i did my worst or not

[unname4798]

cplus_[cplus(64)](64). I have 4 ideas left-- an idea that will have to come up soon before it's too small, a stronger version of that idea, an idea that has a small chance of killing the thread (you and ababa11e are both smart though), and an idea that has to do with FGH and probably isn't allowed lol

im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)

edit: rutabaga do your worst

Second fool (or level):
cpluses(x,y) is c[x pluses](y,...,y)
cpluses(2024,2024)

[rutabaga]

im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)

edit: rutabaga do your worst

Second fool (or level):
cpluses(x,y) is c[x pluses](y,...,y)
cpluses(2024,2024)

mine is bigger. C((2)) would exceed your number, i believe-- that would be cpluses(C(2),2) in your function. that's a LOT of 2's.

[ababa11e]

im doin this first fool: cracker heirarchy: cplus+(x1,x2) = cplus(cplus(...x2 times...cplus(x1)...)))
cplus++(x1,x2,x3) = cplus+(x1,cplus+(x1,cplus+(x1...x3 times...cplus+(x1,x2)...)
my number is cplus++(100,100,100)

edit: rutabaga do your worst

lmao cracker hierarchy is awesome, but you just gave me another new idea
the generalized version of cplus is capital C(x), but it can take any amount of arguments.
first, though, let's define cplus+++(x1,x2,x3,x4) to make sure everyone understands the pattern.
cplus+++(x1,x2,x3,x4)=cplus++(x1,x2,cplus++(x1,x2,cplus++(x1,x2...x4 times...cplus++(x1,x2,x3)...)

now we can tell that in general:
C(x1,x2,...x(n-1),x(n)) = C(x1,x2...x(n-2),C(x1,x2...x(n-2),C(x1,x2...x(n) times...C(x1,x2...x(n-1))...)

...but i just had ANOTHER idea, so i don't want to stop there. first, let's make C((x)) be C(x,x,...C(x) times...x).
now let's run through the whole entire hierarchy again. let's pretend that the old C((x)) is the new C(x), and we have to do the whole hierarchy thing again to get back to C((x)). doing this gives us C((1,x)). doing the C(x) to C((1,x)) procedure, C((1,x)) times, gets us to C((2,x)).
you can probably see where this is headed. doing the C(x) to C((n,x)) procedure, repeated C((n,x)) times, leads us to C((n+1,x)).

naturally, we'll eventually get more arguments in the function. first, C((C((...C((x,x)) times...C((x,x)),x...)),x)) is C((1,1,x)).
we'll grow the second argument in the same way as normal-- C(x) to C((1,1,x)) repeated C((1,1,x)) times is C((1,2,x)).
increasing the new first argument goes like this:
C((a,C((a,C((...C((a,C((x,x)),x)) times...a,C((x,x)),x...)),x)),x)) = C((a+1,1,x)).
when the "a" reaches C((C((C((...C((C((x,x,x)),x,x)) times...C((x,x,x)),x,x...)),x,x)),x,x)), we get a fourth argument.

we increase the 2nd-to-last argument the same way we did before. increasing the 3rd-to-last argument is mostly the same:
C((n1,a,C((n1,a,C((...C((n1,a,C((x,x,x)),x)) times... n1,a,C((x,x,x)),x...)),x)),x)) = C((n1,a+1,1,x)).
n1 is just placeholder for numbers that come before the a-argument.
when a = C((n1,C((n1,C((...C((n1,C((x,x,x)),x,x)) times...n1,C((x,x,x)),x,x...)),x,x)),x,x)), increase the 4th-to-last number by 1.

a pattern is faintly visible here, and while it'd be easier to see with more arguments, it'd take a lot of time to explain each step of adding each argument, so if this isn't well-defined enough, i can explain it more in another post because this one is already kinda long.

adding a 5th argument is just like adding a 4th was, and it behaves like the other arguments. you can put any amount of arguments in C((...)). to top this all off, C(((x))) is C((x,x,...C((x)) times...x)).

my number is C(((418))).

not sure if i did my worst or not

C(((...C(((418))) amount of brackets...(((999)))...)))

[rutabaga]

alright then.
C[1](x) is C(((...x brackets...(((x)))...)))
similarly to what i did earlier-- you can increase the number in the square brackets by "restarting" and pretending that C[1](x) is C(x) and you have to run through the whole hierarchy again, repeating the process an amount of times equal to the current number (C[1](x) in this case).
my number is C[10^100](10^^^^10) + 1000. don't ask me why i put the thousand there.

[JoshM]

W(x) reports an object's weight in Planck masses (units of ~21.76 µg) and spits it out as a number.

W(Yo mama)

[ababa11e]

alright then.
C[1](x) is C(((...x brackets...(((x)))...)))
similarly to what i did earlier-- you can increase the number in the square brackets by "restarting" and pretending that C[1](x) is C(x) and you have to run through the whole hierarchy again, repeating the process an amount of times equal to the current number (C[1](x) in this case).
my number is C[10^100](10^^^^10) + 1000. don't ask me why i put the thousand there.

C[C[10^100](10^^^^10)](C[10^100](10^^^^10))

[rutabaga]

Let C[n,x] be equivalent to C[n](x).

Assume g(x) is a stronger function that was created from f(x). Let g↪n(x) denote the function that would result if the process applied to f(x) that resulted in g(x), was applied to g(x) itself n times. (This is a slightly more formal definition of my "restart" thing-- C↪1((x)), for example, is C((1,x)).)

C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10)))))))))] is my number. That's 10 10's right there-- and there are 10 pairs of parentheses. There's also a combined total of 10 of all other types of brackets, counting individually. I would not give this number a 10 though

[ababa11e]

Let C[n,x] be equivalent to C[n](x).

Assume g(x) is a stronger function that was created from f(x). Let g↪n(x) denote the function that would result if the process applied to f(x) that resulted in g(x), was applied to g(x) itself n times. (This is a slightly more formal definition of my "restart" thing-- C↪1((x)), for example, is C((1,x)).)

C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10)))))))))] is my number. That's 10 10's right there-- and there are 10 pairs of parentheses. There's also a combined total of 10 of all other types of brackets, counting individually. I would not give this number a 10 though

What if i assign C[n1,n2,x] = = C[C[n1,n2]+x,C[n1,n2]+x]], c[n1,n2,n3,x] = C[C[n1,n2,n3]+x,C[n1,n2,n3]+x]]...

My number is C[64,64,64...g[64] times...](C[64,64,64...g[64-1] times...]( C[64,64,64...g[64-2] times...]...C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10))))))))) times... C[64](g(64)...), which is greater than C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10)))))))))

[rutabaga]

What if i assign C[n1,n2,x] = = C[C[n1,n2]+x,C[n1,n2]+x]], c[n1,n2,n3,x] = C[C[n1,n2,n3]+x,C[n1,n2,n3]+x]]...

My number is C[64,64,64...g[64] times...](C[64,64,64...g[64-1] times...]( C[64,64,64...g[64-2] times...]...C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10))))))))) times... C[64](g(64)...), which is greater than C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10)))))))))

Define m(x) (M being short for marshmallow-- the next big function I make will be d(x) for Dark Chocolate so we can have smores ) as
C↪m(x-1)[m(x-1),m(x-1),...m(x-1) times...m(x-1)] where m(1) = g(64). c(x) is to C[x] as m(x) is to M[x].
M↪(M↪(...M[x] times...M[x]...)[x])[x] = M↠[x]. My number is M↠[3301].

[ababa11e]

What if i assign C[n1,n2,x] = = C[C[n1,n2]+x,C[n1,n2]+x]], c[n1,n2,n3,x] = C[C[n1,n2,n3]+x,C[n1,n2,n3]+x]]...

My number is C[64,64,64...g[64] times...](C[64,64,64...g[64-1] times...]( C[64,64,64...g[64-2] times...]...C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10))))))))) times... C[64](g(64)...), which is greater than C↪(C[10{10}10,10{10}10])[10{10}10,C(((((((((10)))))))))

Define m(x) (M being short for marshmallow-- the next big function I make will be d(x) for Dark Chocolate so we can have smores ) as
C↪m(x-1)[m(x-1),m(x-1),...m(x-1) times...m(x-1)] where m(1) = g(64). c(x) is to C[x] as m(x) is to M[x].
M↪(M↪(...M[x] times...M[x]...)[x])[x] = M↠[x]. My number is M↠[3301].

M↠[M↠[M↠[...M↠[3301] TIMES...M↠[3301]]]
EDIT: what is this in fgh, so i can estimate where we are in googology rn

[eRroR_6o6]

number of families of 14-glider collisions

[unname4798]

New round!
2024

[ababa11e]

New round!
2024

okay fine, since the 'amount of families' one is infinite.
3^^3

[unname4798]

New round!
2024

okay fine, since the 'amount of families' one is infinite.
3^^3

10^10

[eRroR_6o6]

20!

[ababa11e]

20!

10^30

[unname4798]

20!

10^30

2^102 (number of INT rules)