User:Chembudwikipedian/bignum

Here is my big number.

Lets start with g1 which is 3^^^^3 click on this link for mor info

g2 is 3(g1 number of ^ s)3

g3 is 3(g2 number of ^ s)3

etc.

Balgonkah

Inferior Balgongkah

what about g(g(1))? that is (g(3^^^^3)) = 3(g(3^^^^3-1) number of arrows)3 which is very big.

i could say g(g(g(g(g(g(2)))))) is my number but i'm not done here yet.

lets define the number of gs.

remember g(g(1)) from earlier? well in the new notation it is r(2,1).

i will define the inferior balgongkah as being r(A(g64,g64),A(g64,g64)) let's name A(g64,g64) xkcd for reasons.

let's layer up the functions like so.

Regular Balgongkah

r1 = our r function

r2 = r(r(1)) = r2(2,1)

r3 = r2(r2(1)) = r3(2,1)

balgongkah is r(xkcd)(xkcd,xkcd)

these numbers are getting big.

Balgongkah Magnus

let's get some ackermann recursion into the mix. (note from 2023 Chembudwikipedian: this is not actually Ackerman recursion)

b for the balgongkah function.

b(m,0) = m

b(0,m) = m+1

b(m+1,n+1) = b(b(r(m,2),n-1),b(m-1,r(n,2))

ok, now balgonkah magnus is b(xkcd,xkcd).

The Lumbonkers series.

lets "bunker" the function like so.

bb(m+1,n+1) = bb(bb(r(n)(m,2),n-1),bb(m-1,r(m)(n,2))

i feel like this number is big enough that it desrves a new name,

lumbonkers is bb(xkcd,xkcd)

lumbonkers is mind-bogglingly large, but we can go further.

Mega Lumbonkers

lets layer up bb's the same way we made balgongkah.

so,

bb1 = our r function

bb2 = bb(bb(1,1)) = r2(2,1)

bb3 = bb2(bb2(1,1)) = r3(2,1)

etc.

mega lumbonkers is bb(xkcd)(xkcd,xkcd).

Giga lumbonkers

let's recurse what we did.

more ackermann recursion,

bb1 > bbⁱ¹ is congruent to r1 > bb1. Now layer it up.

bbⁱ¹ > bb¹ is congruent to bb1 > bb2

etc.

Giga Lumbonkers is bb^balgonkah (balgonkah,balgonkah)

brief interlude

Lets define a function for functions, because we're gonna need it.

bb¹ > bb² is congruent to fff.

or, in other words,

fff(bb¹) = bb².

Tera lumbonkers

Let's apply fff on fff.

So,

fff(fff) = fff2(2,1).

Now, lets apply fff^balgonkah(balgongkah,balgongkah) to the bb function, giving the mega(m,n) fuction.

Let's define tera lumbonkers as mega(balgongkah,balgongkah)

Bluzzalimbo series

Hyperfactorial

Everybody knows the factorial, right?

5! = 5*4*3*2*1

Not enough. How about this?

5!2 = 5^4^3^2^1

Screw it, let's use the mega function,

5? = mega(5,mega(4,mega(3,mega(2,mega(1,1)))))

Bluzzalimbo

Here's the definition,

Lumbonkers? = Bluzzalimbo.

Yes,that's it.

Mega Bluzzalimbo

Let's define giga as a function.

giga(m,0) = m

giga(0,m) = m+1

giga (m+1,n+1) = giga(m?,giga(m?,giga...(n layers)...m?,n? ))

Mega bluzzalimbo = fff^lumbonkers(giga)(lumbonkers,lumbonkers)

Super Bluzzalimbo

Giga and the balgonkah function both use m+1 for (0,m).

The IGiga function is simply the giga function but IGiga(0,m) is BB(m).

fff^lumbonkersIGiga(lumbomkers,lumbonkers) is Super Bluzzalimbo.

Super-duper Bluzzalimbo

Let's shorthand fff^lumbonkers as bfff.

Here is Big arrow notation.

m→n is bfff(IGiga(m,n))

Any ones at the ends become deleted

Any ones in the middle become subtracted i.e. x→1 is x-1

...→x_n-1→x_n becomes ...→x_n-1-1→(...→x_n-1→x_n-1)

ok now super-duper bluzzalimbo is IGiga(lumbonkers→lumbonkers→(repeated lumbonkers times),(the same number again))

Ultra Bluzzalimbo

Let's do this.

m→→n = m→m→m→(with n arrows)→m→m.

Ultra bluzzalimbo is lumbonkers→→(with lumbonkers arrows)→→lumbonkers.

Sumbapliboop series

Let us now appreciate the great bigness of sumbapliboop

IGiga(lumbonkers→→→(with lumbonkers arrows)→→→lumbonkers(with lumbonkers lumbonkers')lumbonkers,(the same number again))

As you can see this is much bigger than lumbonkers or even giga lumbonkers. We are racing higher and higher!

Grand sumbapliboop

Let's make a shorthand function.

IGiga(x→→→(with x arrows)→→→x(with x x's)x,(the same number again))

Is all tidied up into tera(x)

Now we do our fancy layering...

tera(tera((With y layers)X))) = tera2(x,y)

And so on...

Grand sumbapliboop is defined as tera(bluzzalimbo)(bluzzalimbo,bluzzalimbo)

Mollosogrifgh and tremendogrifgh coming soon!

note from future 2023 Chembudwikipedian: no mollosogrigh is never ever coming and also I used if u use that sort of recursion on the Ackerman function then it literally just adds ordinal w to the growth rate.