What? W takes an array and returns an integer.

W([1, 1]) = W([W([1, 0]), 0]) + 1.

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
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- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
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Yes, but in your explanation it looked like it has the same array as you started with.testitemqlstudop wrote: ↑November 14th, 2019, 8:18 pmWhat? W takes an array and returns an integer.

W([1, 1]) = W([W([1, 0]), 0]) + 1.

Well in that case it's fine.

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

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

{}

([)]

...

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

{}

([)]

...

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

(![]+[])[+!![]]+(![]+[])[!![]+!![]]+(!![]+[])[!![]+!![]+!![]]+(!![]+[])[+!![]]+(!![]+[])[+[]]

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

That's

So

Very

Irrelevant

Since size stacking no longer involves interleaved tags

I am a prolific creator of many rather pathetic googological functions

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

But wait! there's more naïve extensions to be done!Moosey wrote: ↑November 26th, 2019, 7:40 pmwhat 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

(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?

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

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?

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?

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

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

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

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.

¤(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.

(![]+[])[+!![]]+(![]+[])[!![]+!![]]+(!![]+[])[!![]+!![]+!![]]+(!![]+[])[+!![]]+(!![]+[])[+[]]

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

Define ×(n) -- is it n×n = n^2?PkmnQ wrote: ↑January 13th, 2020, 1:08 pmHmm, 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.

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

dcf1 has tetrational or maybe barely rate. Probably > f_3(n), but definitely < f_4(n)nolovoto wrote: ↑January 13th, 2020, 11:21 amdon'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

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

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Hdjensofjfnen
**Posts:**1473**Joined:**March 15th, 2016, 6:41 pm**Location:**r cis θ

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

"A man said to the universe:

'Sir, I exist!'

'However,' replied the universe,

'The fact has not created in me

A sense of obligation.'" -Stephen Crane

'Sir, I exist!'

'However,' replied the universe,

'The fact has not created in me

A sense of obligation.'" -Stephen Crane

Code: Select all

```
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o$6bo$o3b3o$2o$bo!
```

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

Well, first, it's not much stronger than the strongestHdjensofjfnen wrote: ↑January 25th, 2020, 3:10 amKind of cheating, but you could define a function P(n) which is the product of all functions in this thread.

And second, it's illdefined since what if someone adds another function to this thread

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- testitemqlstudop
**Posts:**1323**Joined:**July 21st, 2016, 11:45 am**Location:**in catagolue-
**Contact:**

f(x) = 0

[Laughs evilly]

[Laughs evilly]

Are you talking about the ×4 inMoosey wrote: ↑January 13th, 2020, 4:45 pmDefine ×(n) -- is it n×n = n^2?PkmnQ wrote: ↑January 13th, 2020, 1:08 pmHmm, 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.

¤(×,2,2,2) = ¤(×4,1,8,4) = 128?

If so, then ×4(n) = n×4

(![]+[])[+!![]]+(![]+[])[!![]+!![]]+(!![]+[])[!![]+!![]+!![]]+(!![]+[])[+!![]]+(!![]+[])[+[]]

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

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

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

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

I'm stuck at my ,,, stuff

(See v3 here)

I'm stuck at my ,,, stuff

(See v3 here)

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

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!

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!

(![]+[])[+!![]]+(![]+[])[!![]+!![]]+(!![]+[])[!![]+!![]+!![]]+(!![]+[])[+!![]]+(!![]+[])[+[]]

This post was brought to you by the Element of Magic.

Plz correct my grammar mistakes. I'm still studying English.

Working on:

Nothing.

Favorite gun ever:

Plz correct my grammar mistakes. I'm still studying English.

Working on:

Nothing.

Favorite gun ever:

Code: Select all

```
#C Favorite Gun. Found by me.
x = 4, y = 6, rule = B2e3i4at/S1c23cijn4a
o2bo$4o3$4o$o2bo!
```

- Moosey
**Posts:**3038**Joined:**January 27th, 2019, 5:54 pm**Location:**A house, or perhaps the OCA board. Or [click to not expand]-
**Contact:**

This is a modification of the Ackermann function and probably doesn't get much farther than the Ackermann functionPkmnQ wrote: ↑February 6th, 2020, 7:55 amA*(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!

My CA rules can be found here

Also, the tree game

Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"

It's actually a modification of a modification that you did before, when somebody posted about the Ackermann function.Moosey wrote: ↑February 8th, 2020, 10:15 amThis is a modification of the Ackermann function and probably doesn't get much farther than the Ackermann functionPkmnQ wrote: ↑February 6th, 2020, 7:55 amA*(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!

(![]+[])[+!![]]+(![]+[])[!![]+!![]]+(!![]+[])[!![]+!![]+!![]]+(!![]+[])[+!![]]+(!![]+[])[+[]]