Large number game

[martin3141]

I’ll submit g_gg where g = g₆₄ is Grahams number.

[shinuito]

@jlt

Let f0(n)=n+1. Define by induction fk+1(n)=fkn(n) (the power means n-fold composition).

So f₁(n)= f₀ⁿ(n)= 2n?

f₂(n)= f₁ⁿ(n)=2ⁿn ?

f₃(n)= f₂ⁿ(n)=… tetration?

I was trying to figure this out and stopped. Maybe I’ll think tomorrow.

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@richyfourtytwo

A(A(G,G),A(G,G))

@martin3141

I’ll submit ggg

And whose job is it to figure out which number is bigger?

[Feijoa]

@Vsotvep

1000

Not seeing a lot of proofs here. How about before jumping straight to something ridiculous, I submit this:

• 10¹⁰⁰

Together with a proof that it’s bigger than yours.

Proof

1000 = 10³.

Now who can actually prove theirs is bigger than mine?

[martin3141]

@Feijoa

Now who can actually prove theirs is bigger than mine?

[jlt]

@jlt

Let f₀(n)=n+1. Define by induction f_k+1(n)=f_kⁿ(n) (the power means n-fold composition).
Then f₉(9) is already quite big.

@Feijoa

who can actually prove theirs is bigger than mine?

Let’s prove first that f_k is (strictly) increasing for all k. For k=0 this is obvious. Suppose that’s the case for f_k then
f_k+1(n+1)=f_kⁿ⁺¹(n+1)=f_k(f_kⁿ(n+1))
≥ f_kⁿ(n+1) (because if g is any increasing function from N to N then g(p)≥ p for all p)
> f_kⁿ(n) since f_k is increasing
so f_k+1(n+1)>f_k+1(n).

Next we observe that for all n ≥1, f_k+1(n)≥f_k(n) since f_k+1(n)=g(f_k(n)) for some nondecreasing function.

Therefore, if k’≥k and n’≥n then f_k’(n’) ≥ f_k(n).

Next we compute f_k(n) for small values of k and n. As noted by shinuito, f₂(n)=2ⁿn. Then
f₃(3)=f₂(f₂(f₂(3)))=f₂(f₂(24))=f₂(2²⁴.24)=2^{2^24.24}.2²⁴.24>2^{2^24.24} > 10^{2^24} >10¹⁰⁰.

Finally f₉(9) ≥ f₃(3) > 10¹⁰⁰.

[john.tromp]

Rayo’s number [1] is pretty big, dwarfing everything mentioned here so far.

On the other hand, if you want to limit yourself to computable numbers (numbers computeble by human scale programs), then Loader’s number [2] is my favorite.

[1] Rayo's number - Wikipedia

[2] math - Golf a number bigger than Loader's number - Code Golf Stack Exchange

[shinuito]

I wonder how efficient first order set theory is for describing numbers. Even if I have a googol symbols, would I be better off using them with arithmetic operations or function definitions without the fluff of set theory and get an even bigger number?

Or is it more a case of formally defining things in such a way that makes comparing the complexity of description/definition of two numbers viable maybe

[Atorrante]

Some background info

Names of large numbers

https://kids.tpl.ca/wonders/what-is-the-biggest-number#

[jlt]

@shinuito

Even if I have a googol symbols, would I be better off using them with arithmetic operations or function definitions without the fluff of set theory and get an even bigger number?

Suppose you describe a number using “ordinary” arithmetic operations or function definitions, and your description uses a string of length10¹⁰⁰. Let f(n) be the smallest number that is larger than any number which can be described with a string of length n or less, using “ordinary” arithmetic operations or function definitions. Then f(10¹⁰⁰) is obviously bigger than your number. And f(f(f(…f(10¹⁰⁰)…))) obviously much bigger (where f is composed with itself 10¹⁰⁰ times).

[shinuito]

Right so it is about being about to formally compare numbers in a unified way.

[Vsotvep]

Ah, that’s a very nice entry, and one I was not familiar with either.

My entry is / was a number associated to the Paris-Harrington theorem. This theorem is provable in second-order arithmetic, but not in Peano arithmetic. A definition can be found here (timestamp after 13 minutes):

Infinite Necessity (Hugh Woodin) - Distinguished Lecture Series at UIC

Yesterday evening, I was thinking that PH(9) would’ve been a pretty good entry.

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I hope I have some time next week to make some comparisons between the various submissions so far. I always wanted to understand Paris-Harrington better, and Rayo’s number seems like a reasonable number to compare it to.

The others, I suspect, may very well be too small. Specifically anything computable necessarily has to be eventually dwarfed by the likes of Rayo or Paris-Harrington.

[Vsotvep]

Thinking about it, I believe PH(9) is larger than all other numbers suggested, except for Rayo’s number, which I believe is larger.

[john.tromp]

Specifically anything computable necessarily has to be eventually dwarfed by the likes of Rayo or Paris-Harrington.

Correct about Rayo, but incorrect about PH.

PH(9) is only at level ε0 of the Fast Growing Hierarchy (FGH), and is thus dwarfed by TREE(3). That in turn is dwarfed by lim(BMS), at level PTO(Z2) in the FGH. That in turn is dwarfed by Loader’s number, which is at PTO(Zω) in the FGH. The latter, being computable by a tweet sized program, is in turn dwarfed by BB(10^100), which is in turn dwarfed by Rayo’s number.

[calebjdastrup]

I submit the smallest odd perfect number, or 7 if no such number exists. This is smaller than Rayo’s number, but good luck proving that any of the others beat it. You can do a similar thing with the Riemann hypothesis.

[jlt]

You win the contest if someone can prove that your number is bigger, not if no one can prove that their number is bigger.

[shinuito]

@jlt

You win the contest if someone can prove that your number is bigger

I suppose it’s not in anyone’s interest to prove your number is bigger if they’ve submitted something

Of course the whole format of a “contest” is a little silly. As a setting and a way to entice people to contribute it makes sense, but ultimately it turns into who has studied X, watched video Y, or read Wikipedia Z.

@Vsotvep

and that you’ll have to give a reference to an amateur or professional mathematician having mentioned the number you propose

but I suppose that was already in the premise.

We could have a “contest” of who can submit the hardest Tsumego, but then it’s just a case of who has access to books or other collections of hard Tsumego and who’s going to pick the Igo Hatsuyouron 120.

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It would be mildly interesting to me to approximate and beat @jlt’s

@jlt

Let f0(n)=n+1. Define by induction fk+1(n)=fkn(n) (the power means n-fold composition).
Then f9(9) is already quite big.

without appealing to something like “it’s a number bigger than anything you can write down” = rayo’s number. Of course that is interesting in its own right in its own way.

Except maybe the more I look at it, it’s just another formation of the Ackermann function is it?

@richyfourtytwo

A(A(G,G),A(G,G))

where rather than writing A(m,n) we’re writing f_k(n), and looking at for example

Ackermann function - Wikipedia

the 1-ary function part.

[shinuito]

If you want an “interesting” extension of the problem, maybe you have to both provide a large number and be able to compute a few of its digits, say decimal digits, last digit, last two digits (remainders mod 10^n) etc.

Of course it doesn’t wildly change the problem, you can go and google which big numbers we know the most digits of, so you’d need a weighting system on how big the number is vs how many digits you can provide.

Ultimately I guess it’s just “share what you know about big numbers” in a competitive disguise (I guess I’ve learned a few things from reading the responses so far )

[yebellz]

So, maybe Rayo(10¹⁰⁰) is leading right now?

We could do something like Rayo(Rayo(10¹⁰⁰)) to make something even absurdly larger. While, we’re at that, we could repurpose Knuth’s up-arrow notation to define something like:

2 ↑ 3 := Rayo(Rayo(Rayo(2)))
2 ↑↑ 3 := ((2 ↑ 3) ↑ 3) ↑ 3
2 ↑↑↑ 3 := ((2 ↑↑ 3) ↑↑ 3) ↑↑ 3
2 ↑⁴ 3 := 2 ↑↑↑↑ 3

and then propose Rayo(n) ↑^Rayo(n) Rayo(n), where n = 10¹⁰⁰

[shinuito]

@yebellz

Rayo(n) ↑Rayo(n) Rayo(n)

Maybe it depends on the interpretation of

@Vsotvep

you’ll have to give a reference to an amateur or professional mathematician having mentioned the number you propose (or a provably larger one, including proof) in some capacity involving a general public, like a publication, or a talk, or whatever. This is to avoid the “your number +1” kind of trick.

since using another entries function or description feels similar in spirit to “my number is your number +1” except with different arithmetic or algebraic operations.

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Unrelated,

since how do you go about defining a new number “bigger than one you can write down with N symbols…”,

but related to the idea of the thread:

How big would the following number be?

Take the diverging harmonic series

It’s diverging like ln(n).

Every time the partial sum crosses an integer, take the partial summand to N (=number of terms used) decimal places and multiply by 10^N making it an integer.

Then concatenate this integer onto all smaller integers produced before it.

Now do this for the first G= 10^{10^100} integers crossed.

I guess you expect N~e^n terms before crossing an integer n, and the way I’ve worded it you get a number of the order 10^N at each stage, maybe a little larger.

Then you concatenate those numbers 10^10^100 times.

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Thinking of say if I had three 3 digits numbers, each less than 10^3, and then totally less than 10^9 = (10^3)^3.

Not exactly sure how to bound concatenation.

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Of course you could replace concatenation with faster growing functions, but I was just thinking of a way to generate exponentially large numbers in order to combine them with each other, in a way that seemed different to the other numbers mentioned.

[Vsotvep]

@shinuito

We could have a “contest” of who can submit the hardest Tsumego, but then it’s just a case of who has access to books or other collections of hard Tsumego and who’s going to pick the Igo Hatsuyouron 120.

I mean, yeah, it’s not really a contest, but rather an opportunity to learn something.

The problem with a consecutive game, is that people will go the uninspiring route of “I’ll just use your trick, but with more applications of whatever it is that makes your number big. Now my number is bigger”. It defeats the purpose of such a game, specifically, to find a clever way to get to large numbers, or to find a surprising way to get to large numbers.

I think Rayo’s number is very clever, in that it is very easy to formulate, yet grows unfathomably fast. Sure, a Knuth-arrow version of Rayo grows faster, but it does not really bring new insight.

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As soon as you come up with a larger quantity that is clever enough to appear in a mathematical article, as well as being very large, the very act of sharing it in this forum could constitute as giving …

@Vsotvep

a reference to an amateur or professional mathematician having mentioned the number you propose (…) in some capacity involving a general public