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00:00
@SimplyBeautifulArt the largest square fibonacci number
aka 144
the smallest number that can be expressed as the sum of cubes in two ways
so fast lol
:P
I'm just being safe, since I don't remember what you're number was
But it can't be terribly big
00:05
1729... Ramanujan's number
:P
Oh right
I should've remembered that one
infinity
No infinities, cuz that'll end very badly for you
@SimplyBeautifulArt not that infinity
Oh haha
Let's call it the 'finite-infinity'
00:07
:|
Hey @Zacharý
Hello.
Oh, it's my turn
$\psi(\psi(16))$
i.e.?
@SimplyBeautifulArt So, how does one play this "game"
00:10
@Zacharý We try and make a bigger number than the previous
No copying other numbers +1
or anything silly like that
If someone makes their own function, don't use it.
$\psi^2(16)\approx 2^{2^{\rm googol}}$
I will not do well in this game, for sure.
I'll just observe
:P okay
@LeakyNun ur turn
2^2^10^100 < 2^2^2^400 < 2^2^2^2^9 < 2^2^2^2^2^4
2^^7 is my number
$$a[b]c=\begin{cases}a+1,& b\cdot c=0\\t[b-1]t,&t=a[b](c-1)\end{cases}$$
$2[3]10$
^ That's my number
It is a bit larger than 2^^10
@SimplyBeautifulArt how did you analyze so quickly
00:16
Induction
It's pretty similar to Knuth's up-arrow
a[0]b = a+1
a[1]b = a+b
a[2]b = a(b+1)
a[3]b ~ a^2^b
our analysis don't match
:|
a[2]b = a(b+1) #=> a(b+2) = (a(b+1))+(a(b+1)) = 2a(b+1) = false
Ur analysis is wrong
a[2](b+1) = (a[2]b)[1](a[2]b)
aha
a[2]b = (a+1)*2^b
:P
@Antonios-AlexandrosRobotis Hello and welcome to my realm
@SimplyBeautifulArt how did you think of a function in like a minute
let's see if I can come up with any novel functions
I'm gonna go eat
00:35
@LeakyNun back
so fast
:P
I had a big+late lunch
$n^! = \begin{cases} 0 & n = 0 \\ n^{k^!} & n = k+1 \end{cases}$
10^! is my number
@SimplyBeautifulArt
00:48
2[3]10 is bigger though
is it?
Hm
It is, by a small bit
(The top of the exponential tower matters more than the bottom)
(start by using 3^2^1^0 < 2^2^2)
so 11^! would be ok?
@SimplyBeautifulArt you see, I can come up with nothing new
01:04
@LeakyNun yes
your turn then
10[4]1
= 11[3]11
Knock knock
who's there
10[4]1
@JeanAraujo Hello and welcome to my realm
Don't forget to knock
@LeakyNun How far are you in antimatter dimensions?
@LastIronStar Hello and welcome to my realm
Don't forget to knock
> You have made a total of 177.85ai antimatter.
You have done 7 soft resets.
You have 0 Antimatter Galaxies.
You have played for 1 days, 7 hours, 58 minutes and 5 seconds.
If every antimatter were a planck volume, you would have enough to fill 17.5 teaspoons.
01:14
@SimplyBeautifulArt Is it okay to be here?
> My dog ate too much antimatter, now he says 'meow!'
@LastIronStar Of course it is!
@LeakyNun Haha
Coming close to an antimatter galaxy?
@SimplyBeautifulArt what's the good news?
Once you get two antimatter galaxies, you'll be zooming
@LastIronStar ivark.github.io
I only get addicted to the weirdest of games
(-1/a^2)^!, where 1+2+3+...=a
01:15
And secondly, I study extremely large finite numbers
@LeakyNun NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
"N"+"O"*10^100+"!"
@SimplyBeautifulArt Is this based on the predicativist's hierarchy of set construction concept?
:(
@LastIronStar No, I usually do recursion
But very non-trivial recursion
ok reading how to play
01:18
Made this notation
I can never build anything new
And my number is 10[3]10
@LeakyNun :P Mess around a bit with the recursion
And try ur best to maximise every function argument
@SimplyBeautifulArt seems interesting, care to explain to the laystacker?
You have made a total of Infinite antimatter.
You have done 5 soft resets.
You have 0 Antimatter Galaxies.
You have infinitied 1 time.
Your fastest Infinity is in 1 days, 11 hours, 23 minutes and 41 seconds.
You have spent 2 hours, 22 minutes and 8 seconds in this Infinity.
You have played for 1 days, 15 hours, 48 minutes and 45 seconds.
If every antimatter were a planck volume, you would have enough to make 2.10e7 Hydrogen atoms.
@LastIronStar Uh, so I've got 10[3]10 as my current number
10[3]10 = t[2]t, where t=10[3]9
10[3]9 = u[2]u, where u=10[3]8
10[3]8 = ...
10[3]0 = 10+1 = 11
10[3]1 = 11[2]11
> To understand dimensional sacrifice, you do actually need a PhD in theoretical physics. Sorry!
meta-meme!
01:21
11[2]11 = v[1]v, where v=11[2]10
11[2]10 = ...
@LeakyNun :P
etc.
11[2]0 = 12
> If the 9th dimension is all evil, then is 3 the root of all evil?
11[2]1 = 12[1]12
12[1]12 = w[0]w = w+1, where w=12[1]11
etc.
@LastIronStar In general, a[0]c = a[b]0 = a+1
a[1]b = a+b+1
a[2]b > a*2^b
a[3]b > a^a^a^...b powers...^a
etc.
Nah
It's considered relatively small
I'm new to this business of numerical fertility.
@LastIronStar it's the most elegant thing I've ever seen
why thank you
106
Q: Largest Number Printable

VereosYour goal is to write a program that prints a number. The bigger the number, the more points you'll get. But be careful! Code length is both limited and heavily weighted in the scoring function. Your printed number will be divided by the cube of the number of bytes you used for your solution. So...

how does one get initiated in this dark art?
@LastIronStar do you know ordinals?
01:27
One starts playing
or knows ordinals...
@LeakyNun I do not know them.
@LastIronStar You program?
@SimplyBeautifulArt C++, python, Fortran. Now learning Erlang
Oh yeah, and my current number is 10[3]1
If we had to rank this number...
@SimplyBeautifulArt I did another number
01:29
I'd be in 14th place
13 mins ago, by Leaky Nun
(-1/a^2)^!, where 1+2+3+...=a
Oh right
Then I activate my face-down card!
"10[3]2"
Which would put me in 13th place
6
Q: BigNum Bakeoff Reboot

Simply Beautiful ArtSome of you may be familiar with the BigNum Bakeoff, which ended up quite interestingly. The goal can more or less be summarized as writing a C program who's output would be the largest, under some constraints and theoretical conditions e.g. a computer that could run the program. In the same spi...

@WavesWashSands Hello and welcome to my realm.
Don't forget to knock
@LastIronStar If you have a discord, you can join me: discord.gg/5v6ucfN
@BenjiAltman Hello and welcome to my realm.
Don't forget to knock
@SimplyBeautifulArt registering
Any who
@LastIronStar What's the largest number you can write really quickly?
You may use programmy stuff if you want
@SimplyBeautifulArt log*-100^100(100^100)
01:36
@LastIronStar why would you log* it?
x'D
i'm inverse log*ing it
Oh
Heh, that's actually half decent of an idea
And believe it or not
why thank you
You beat Leaky's latest number today
But it still loses to mine
Unless...
log*-100^100 means take the inverse log*, 100^100 times?
01:38
Ah
You beat me!
congrats
what!? HOW
But you lose to 10[4]10
I was just going for something quick and dirty
Becuz my number isn't that crazy yet
10[4]10 is like the output of the following pseudo program:
n=10
n.times{ n = log*-n(n) }
print n
To be a quick and dirty approximation
@LastIronStar log* grows very very slow, so log*- grows pretty fast.
@SimplyBeautifulArt yes, what i didn't realise is that your a[b]c numbers are related to log*-1 numbers!
01:41
They aren't
I'm just good with relating things :P
So yeah, that's my current number
Try and beat it.
clearly I cannot continue with hardcoded numerics, i need a notation
(I suppose you could just use my pseudo-program above...)
01:53
@SimplyBeautifulArt ok i'm there.
@LastIronStar shrugs
Not much is happening there right now
too bad
Well
Ordinals?
Shall we?
Well
Ez function:
H(0,n) = n
H(x+1,n) = H(x,n+1)
i am not familiar with ordinals
Calculate H(10,10)
01:57
ok
20?
Ez
Next:
H(0,n) = n
H(x,n) = H(f(x,n),n+1), x≠0
f(x+1,n) = x
Still the same, right?
no! it reduces to H(x,n) = n
wait what's the boundary on f again?
Oh whoops
My bad
There we go
Boundaries? There are none
yea it is same now
Okay
Now:
f(ω,n) = n
Compute H(ω,10)
(should still be easy)
02:02
21
H(ω+1,10)
(Read the rules literally!)
oops 23
yup
Easy, right?
02:05
seems like a setup but i'll bite, yes.
f(x+y,n) = x+f(y,n)
Now compute H(ω+ω+ω,10)
doesn't f have two arguments always?
ty
Sry, I'm occasionally very bad at this.
can i use pen&paper or is this supposed to be handsfree?
You can totally use pen&paper, or even work it out in here
02:12
ok, so H(ω+ω+ω,10) = H(f(ω+ω+ω,10),11) = H(ω+ω+10,11) = H(f(ω+ω+10,11),12)
7 mins ago, by Simply Beautiful Art
f(x+y,n) = x+f(y,n)
Hint hint wink wink
yes so we need f(ω+ω+ω,10) = ω+ω+10 by the hint
I meant for the next step
oh ok
I was in teh middle of writing it lol
02:17
= H(ω+10+11,12) = H(f(ω+21,12),13) = H(10+11+12,13)
haha let me write it again
f(ω+ω+10,11) = ω+f(ω+10,11) = ω+ω+f(10,11) = ω+ω+9
(You can think of it as the f tends to move to the right until it can no more)
but how did you get that f(10,11) = 9?
also, this is the normal addition right? meaning it's commutative?
Not normal addition
And not commutative, but literal
And all natural numbers n > 0 have some natural number k such that n = k+1
02:21
oh my, how silly of me
H(f(ω+ω+10,11),12) = H(ω+ω+9,12) right?
Haha
And yes
But "The orange is orange." is fun to put into google translate
=H(f(ω+ω+9,12),13) = H(ω+ω,22)?
Uh....
No
Close though
02:27
ok, let me think
H(ω+ω,21)?
What's H(ω,10)?
Yup
And then keep expanding
Hopefully you've figured out that each "+ω" is equivalent to removing it, doubling n, then adding 1 to n.
The end result of H(ω+ω+ω,10) is 87
yeah, that's about right
phew
btw this is fun
And then some new rules:

a*(b+1) = a*b+a
otherwise,
f(a*b,n) = a*f(b,n)

a^(b+1) = a^b*a (PEMDAS: (a^b)*a)
otherwise,
f(a^b,n) = a^f(b,n)
Yeah, ikr?
PEMDAS?
Order of operations:
Parenthesis,
then exponents,
then multiplication/division,
then addition/subtraction
02:33
oh ok
we call it BEDMAS
B = Bracket
You can tell if it is of the form a*(b+1) or of the form a^(b+1) if the right term literally ends with a "+1", or a "+n", n being a natural number.
:P I'm a weird American
Okay
So the previous expression you did was equal to H(ω*3,10)
Try doing H(ω^3,10)
Good luck!
it's about as large as log*-1(10)
So you won't be able to actually finish
:P
i wasn't gonna try writing it down this time!
more of reason and approximation
Still good to try a few steps
Oh yeah, I always forget
a*1 = a^1 = a
ω^3 = ω^(2+1) = ω^2*ω = ω^(1+1)*ω = ω*ω*ω
H(ω^3,10) = H(f(ω*ω*ω,10),11)
f(ω*ω*ω,10) = ω*ω*f(ω,10) = ω*ω*10
02:40
= H(ω^f(ω*ω,10),11)
Shouldn't be an exponent
oops you're right
ω*ω*10 = ω*ω*9 + ω*ω
H(ω*ω*10,11) = H(f(ω*ω*9 + ω*ω,11),12)
f(ω*ω*9 + ω*ω,11) = ω*ω*9 + ω*f(ω,11) = ω*ω*9 + ω*11
As you can probably see, this is going to take a while
Notice ω*11 = ω+ω+ω+ω+ω+ω+ω+ω+ω+ω+ω
And each ω makes n double, then add 1
After all that, we still have ω*ω*9 to deal with...
And this is only H(ω^3,10)
Imagine H(ω^ω^ω^ω^ω,10)
@SimplyBeautifulArt YUGE!
what is ω btw?
does it have more than placeholder significance?
@LastIronStar The first infinite ordinal
You can imagine H(0,n) < H(1,n) < H(2,n) < ... because 0 < 1 < 2 < ...
Well, we also have
0 < 1 < 2 < 3 < ... < ω < ω+1 < ω+2 < ... < ω*2< ω*2+1 < ... < ω*3 < ... < ω^2 < ω^2+1 < ...
In any case, I've got to head to bed.
H(ω^ω^ω^ω^ω,10) is large enough to get in 5th place on Largest Printable Number
And H(ω^x,n) ~ f_x(n)
In the fast growing hierarchy used for the top 5 answers on that link
More special symbols generally = bigger
Good night!
02:52
Gn! ttyl
 
19 hours later…
21:39
@AnantSaxena Hello and welcome to my realm!
Don't forget to knock.
Hey!
Can I ask you about trying to make an argument of mine rigorous (Im not sure how to find the errors)?
*error bars
sure
Though I may be slow to respond
thats fine :)
any pointers would be gr8
0
Q: Making a rigorous argument on the series of mobius function?

Anant SaxenaI was recently fiddling with the mertens function and realized I could do some manipulations on it. Consider the series: $$ \sum_{r=1}^\infty \mu(r) x^r = f(x)$$ This satisfies: $$ \sum_{r=1}^\infty f(x^r) = f(x) + f(x^2) + \dots = x $$ We will use $x = 1 - \epsilon = 1- \frac{1}{n}$ wh...

this is a result of me fiddling with number theory
Hm, I don't usually do number theory unfortunately
you can think of it as functional analysis .... theres not that much number theory used
:P
21:56
any thoughts or pointers?
@SimplyBeautifulArt
ohk ... thanks for looking at it though :)

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