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Posted: **November 14th, 2019, 8:18 pm**

Moosey wrote: November 14th, 2019, 9:35 am I think that's still has a problem. You're making W(1,1) = W(W(1,1),0) = W(W(1,1)), right? You can't use the same array because otherwise you have issues like this.

What? W takes an array and returns an integer.
W([1, 1]) = W([W([1, 0]), 0]) + 1.

Posted: **November 15th, 2019, 7:39 am**

testitemqlstudop wrote: November 14th, 2019, 8:18 pm
Moosey wrote: November 14th, 2019, 9:35 am I think that's still has a problem. You're making W(1,1) = W(W(1,1),0) = W(W(1,1)), right? You can't use the same array because otherwise you have issues like this.
What? W takes an array and returns an integer.
W([1, 1]) = W([W([1, 0]), 0]) + 1.

Yes, but in your explanation it looked like it has the same array as you started with.
Well in that case it's fine.

Posted: **November 16th, 2019, 7:44 am**

What would happen if you defined TREEEE(n) similarly to TREE(n) but you allowed Aidan-mode-esque nesting of parentheses?
Note: homeomorphic embeddability-wise, it would be two nodes each connected to each other, neither as a parent or a child-- {([)]} would be a triangle where each node was connected to another. {} would be the root.

e.g.
TREEEE(3) = (maybe) the length of
{}
([)]
...

Posted: **November 19th, 2019, 4:35 am**

Moosey wrote: November 16th, 2019, 7:44 am Aidan-mode-esque

Size
Stacking
Says
Hello

Posted: **November 19th, 2019, 7:57 am**

PkmnQ wrote: November 19th, 2019, 4:35 am
Moosey wrote: November 16th, 2019, 7:44 am Aidan-mode-esque
Size
Stacking
Says
Hello

That's
So
Very
Irrelevant
Since size stacking no longer involves interleaved tags

Posted: **November 26th, 2019, 7:44 pm**

Moosey wrote: November 26th, 2019, 7:40 pm what if...

1 ((()))
2 (()())(()())
3 (())(())(())(()())
4 ()()()()(())(())(()())
5 ()()()(())(())(()())
6 ()()(())(())(()())
7 ()(())(())(()())
8 (())(())(()())
9 ()()()()()()()()()(())(()())
10 ()()()()()()()()(())(()())
11 ()()()()()()()(())(()())
12 ()()()()()()(())(()())
13 ()()()()()(())(()())
14 ()()()()(())(()())
15 ()()()(())(()())
16 ()()(())(()())
17 ()(())(()())
18 (())(()())
19 ()()()()()()()()()()()()()()()()()()()(()())
20 ()()()()()()()()()()()()()()()()()()(()())
21 ()()()()()()()()()()()()()()()()()(()())
...
38 (()())
39 (()) x39
40 ()x40 (()) x38
...
80 (()) x38
...
~80*2^38 ()
~80*2^38

kirby-paris but you append n copies of the grandparent too
you can do it where you append n copies of all x-parents too

1 (((())))
2 ((()())(()()))((()())(()()))
...

probably comparable to 2^^n for some smallish n

But wait! there's more naïve extensions to be done!
(1) = (((...))) with n nodes, where n is the turn # (1=e_0 for ordinal purposes)

1 (1())
2 (1)(1)
3 (())(1)
4 ()()(1)
5 ()(1)
6 (1)
7 (((((())))))
Oh boy, even without the prior extensions, this will be crazy

But wait! there's more!
(2) = (1(1(...))) with n nodes (2 = e_1 for ordinal purposes)
(m+1) = (m(m(...))) with n nodes
([hydra]) = ([next step in hydra]([next step in hydra](...))) with n nodes
([]) = (1)

Now, ([hydra]) = e_hydraOrdinal for ordinal purposes
That means that the next mess is...

(0,1) = ([([(...)])]) with n nodes (note: () is a node, [] is not). This is equivalent to z_0
1 (0,1())
2 (0,1)(0,1)
3 ([()])(0,1)
4 (1(1))(0,1)
...

@ordinal people: How good is this in the fgh? for instance, if I have f(0) = (), and f(n+1) = (0,1 f(n)), and g(n) = length of reduction of f(n), how good is g(n) in the fgh?

Posted: **January 9th, 2020, 8:48 pm**

Let's revive this thread too
TGR from the googology discord requested the following so he could analyze a version of ah with a limit.
https://mooseymath.wordpress.com/2020/01/09/ah-v-2-0/
Analysis, anyone?

Posted: **January 13th, 2020, 11:21 am**

don't actually know too much about programming functions

Here's my dumb functions I invented

dumbcantorfunction1(1)= 1 (binary) = 1
dumbcantorfunction1(2)= 101 (1 replaced with 101 0 replaced with 000 iterated dumbcantorfunction1(1) times) = 5
dumbcantorfunction1(3)= 101000101000000000101000101000000000000000000000000000101000101 (1 replaced with 101 0 replaced with 000 iterated dumbcantorfunction1(2) (5) times) = 1534774961612150361293125 1.53477*10^24
dumbcantorfunction1(4)= (1 digit replaced with 101 and 0 digit replaced with 000 iterated dumbcantorfunction1(3) (1534774961612150361293125) times) = idunno a really big number

dumbcantorfunction2(n) = (1 digit replaced with binary of dumbcantorfunction2(n-1) or 101 if n<=2 and 0 replaced with 0's of the same digit number as dumbcantorfunction2(n-1) or 000 if n<=2 iterated dumbcantorfunction2(n-1) times)

so basically the same number s until it gets bigger at n=4

dumbcantorfunction3(n) = dumbcantorfunction1(n) but instead of always 101 and 000 one zero is added symmetrically for each side of the center zero of the replacement for each (1 zero for 101 = 10001, 1 zero for 000 = 00000). The number of zeros to add for each replacement is (dumbcantorfunction1(n-1)) with (dumbcantorfunction2(n-1)) number of factorials

at this point, i dunno if somethings broken here

if any of this stuff actually works name the number dumbcantorfunction3(121412181214121) after me

dunno if this is computable. dunno if graham's notation laughs at this pathetically small number. Just wanted to use this to inspire myself to do gud in college

Posted: **January 13th, 2020, 1:08 pm**

Hmm, what if I were to plug a function into itself?

¤(f,a,n,k)
f is the function
a is the amount of arguments for the function
x is just a number

¤(f,1,n,k) = f(n) × k
otherwise, ¤(f(...,kn),a-1,f(n,n,n...)×k,kn)

2×2 = 4
¤(×,2,2,2) = ¤(×4,1,8,4) = 128
¤(¤×,2,2,2) = ¤(¤×k4,1,128,4) = ¤(×,2,128,4) × 4 = ¤(×512,1,16384,512) × 4 = 16384 × 512 × 512 × 4 = 17179869184 (please check for me)

0th layer is 2^2, 1st layer is 2^7, and 2nd layer is 2^34.
How much does this grow proportional to x? I don't know, I'm not a googologist.

Posted: **January 13th, 2020, 4:45 pm**

PkmnQ wrote: January 13th, 2020, 1:08 pm Hmm, what if I were to plug a function into itself?

¤(f,a,n,k)
f is the function
a is the amount of arguments for the function
x is just a number

¤(f,1,n,k) = f(n) × k
otherwise, ¤(f(...,kn),a-1,f(n,n,n...)×k,kn)

2×2 = 4
¤(×,2,2,2) = ¤(×4,1,8,4) = 128
¤(¤×,2,2,2) = ¤(¤×k4,1,128,4) = ¤(×,2,128,4) × 4 = ¤(×512,1,16384,512) × 4 = 16384 × 512 × 512 × 4 = 17179869184 (please check for me)

0th layer is 2^2, 1st layer is 2^7, and 2nd layer is 2^34.
How much does this grow proportional to x? I don't know, I'm not a googologist.

Define ×(n) -- is it n×n = n^2?

Posted: **January 13th, 2020, 4:50 pm**

nolovoto wrote: January 13th, 2020, 11:21 am don't actually know too much about programming functions

Here's my dumb functions I invented

dumbcantorfunction1(1)= 1 (binary) = 1
dumbcantorfunction1(2)= 101 (1 replaced with 101 0 replaced with 000 iterated dumbcantorfunction1(1) times) = 5
dumbcantorfunction1(3)= 101000101000000000101000101000000000000000000000000000101000101 (1 replaced with 101 0 replaced with 000 iterated dumbcantorfunction1(2) (5) times) = 1534774961612150361293125 1.53477*10^24
dumbcantorfunction1(4)= (1 digit replaced with 101 and 0 digit replaced with 000 iterated dumbcantorfunction1(3) (1534774961612150361293125) times) = idunno a really big number

dumbcantorfunction2(n) = (1 digit replaced with binary of dumbcantorfunction2(n-1) or 101 if n<=2 and 0 replaced with 0's of the same digit number as dumbcantorfunction2(n-1) or 000 if n<=2 iterated dumbcantorfunction2(n-1) times)

so basically the same number s until it gets bigger at n=4

dumbcantorfunction3(n) = dumbcantorfunction1(n) but instead of always 101 and 000 one zero is added symmetrically for each side of the center zero of the replacement for each (1 zero for 101 = 10001, 1 zero for 000 = 00000). The number of zeros to add for each replacement is (dumbcantorfunction1(n-1)) with (dumbcantorfunction2(n-1)) number of factorials

at this point, i dunno if somethings broken here

if any of this stuff actually works name the number dumbcantorfunction3(121412181214121) after me

dunno if this is computable. dunno if graham's notation laughs at this pathetically small number. Just wanted to use this to inspire myself to do gud in college

dcf1 has tetrational or maybe barely rate. Probably > f_3(n), but definitely < f_4(n)
The other functions are probably similar
And probably all upperbounded by f_5(n)
For reference, Graham's function is f_w+1(n). I can explain what w is if you don't already know

Posted: **January 25th, 2020, 3:10 am**

Kind of cheating, but you could define a function P(n) which is the product of all functions in this thread.

Posted: **January 25th, 2020, 7:59 am**

Hdjensofjfnen wrote: January 25th, 2020, 3:10 am Kind of cheating, but you could define a function P(n) which is the product of all functions in this thread.

Well, first, it's not much stronger than the strongest
And second, it's illdefined since what if someone adds another function to this thread

Posted: **January 25th, 2020, 8:59 pm**

f(x) = 0

[Laughs evilly]

Posted: **January 28th, 2020, 8:07 am**

Moosey wrote: January 13th, 2020, 4:45 pm
PkmnQ wrote: January 13th, 2020, 1:08 pm Hmm, what if I were to plug a function into itself?

¤(f,a,n,k)
f is the function
a is the amount of arguments for the function
x is just a number

¤(f,1,n,k) = f(n) × k
otherwise, ¤(f(...,kn),a-1,f(n,n,n...)×k,kn)

2×2 = 4
¤(×,2,2,2) = ¤(×4,1,8,4) = 128
¤(¤×,2,2,2) = ¤(¤×k4,1,128,4) = ¤(×,2,128,4) × 4 = ¤(×512,1,16384,512) × 4 = 16384 × 512 × 512 × 4 = 17179869184 (please check for me)

0th layer is 2^2, 1st layer is 2^7, and 2nd layer is 2^34.
How much does this grow proportional to x? I don't know, I'm not a googologist.
Define ×(n) -- is it n×n = n^2?

Are you talking about the ×4 in
¤(×,2,2,2) = ¤(×4,1,8,4) = 128?
If so, then ×4(n) = n×4

Posted: **January 28th, 2020, 8:18 am**

PkmnQ wrote: January 28th, 2020, 8:07 am Are you talking about the ×4 in
¤(×,2,2,2) = ¤(×4,1,8,4) = 128?
If so, then ×4(n) = n×4

No, I'm talking about ×(n)
As I said in the question

Oh now I see
x(n,m)

That function is severely limited by the no. Of arguments

Posted: **February 5th, 2020, 9:35 am**

Anyone wanna help me come up with new ideas for ah, or analyze it?

I'm stuck at my ,,, stuff
(See v3 here)

Posted: **February 6th, 2020, 7:55 am**

I decided to analyze functions using only +1, -1, and recursion.
Rules:
1. You cannot stack +1s.
2. You cannot stack -1s (although I don’t see a use for it).
3. You can only have one level of recursion.

So, here's the one argument function.
f(x) = x+1
Amazing, the best function I have ever seen, an absolute MASTERPIECE!
Ok, joke’s over. Let’s move on.

f(3)-3/f(2)-2 = 4-3/3-2 = 1/1 = 1

Here’s the two argument function, which was actually posted in the past, and slightly tiny teeny weeny beanie boney pony coney kobe adobe photoshop changed.
A*(0,n)=n+1, n>=0
A*(m,0)=A*(m-1,1)+1, m>0
A*(m,n)=A*(m-1,A*(m,n-1)+1)+1, m>0 & n>0 (my addition)
I’ll let the 1’s slide, because they’re 0+1.
Also, the first line was originally 2 lines, but they simplify to that.
Python doesn’t like that this recurses more than 1000 times, so it calls quits and says “DO IT YOURSELF!”
No, Python, I AM NOT CALCULATING THIS BY HAND!

Posted: **February 7th, 2020, 5:00 am**

PkmnQ wrote: February 6th, 2020, 7:55 am pony

-------

f(0) = 3
f(x) = f(x-1)!

Posted: **February 8th, 2020, 10:15 am**

PkmnQ wrote: February 6th, 2020, 7:55 am A*(0,n)=n+1, n>=0
A*(m,0)=A*(m-1,1)+1, m>0
A*(m,n)=A*(m-1,A*(m,n-1)+1)+1, m>0 & n>0 (my addition)
I’ll let the 1’s slide, because they’re 0+1.
Also, the first line was originally 2 lines, but they simplify to that.
Python doesn’t like that this recurses more than 1000 times, so it calls quits and says “DO IT YOURSELF!”
No, Python, I AM NOT CALCULATING THIS BY HAND!

This is a modification of the Ackermann function and probably doesn't get much farther than the Ackermann function

Posted: **February 9th, 2020, 1:50 am**

Moosey wrote: February 8th, 2020, 10:15 am
PkmnQ wrote: February 6th, 2020, 7:55 am A*(0,n)=n+1, n>=0
A*(m,0)=A*(m-1,1)+1, m>0
A*(m,n)=A*(m-1,A*(m,n-1)+1)+1, m>0 & n>0 (my addition)
I’ll let the 1’s slide, because they’re 0+1.
Also, the first line was originally 2 lines, but they simplify to that.
Python doesn’t like that this recurses more than 1000 times, so it calls quits and says “DO IT YOURSELF!”
No, Python, I AM NOT CALCULATING THIS BY HAND!
This is a modification of the Ackermann function and probably doesn't get much farther than the Ackermann function

It's actually a modification of a modification that you did before, when somebody posted about the Ackermann function.

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