Talk: Pound-Star Notation

#*((2))*(1,3)*^*2**#*7

I want to solve #*((2))*(1,3)*^*2**#*7:

n=1, #*((2))*(1,3)*^*2**#*7

n=7, #*((2))*(1,3)*^*2**#

n=7, #*((2))*(1,3)*^*2**7

n=7, #*((2))*(1,3)*^*2*7*7*7*7*7*7*7

n=49, #*((2))*(1,3)*^*2*7*7*7*7*7*7

n=343, #*((2))*(1,3)*^*2*7*7*7*7*7

n=2401, #*((2))*(1,3)*^*2*7*7*7*7

n=16807, #*((2))*(1,3)*^*2*7*7*7

n=117649, #*((2))*(1,3)*^*2*7*7

n=823543, #*((2))*(1,3)*^*2*7

n=5764801, #*((2))*(1,3)*^*2

n=5764801, #*((2))*(1,3)[*^2]_5764801 (over 17 million characters long)

n=5764801, #*((2))*(1,3)[*^2]_5764800*_57648022 (over 23 million characters long)

The next string would contain more than one trillion characters. --84.61.137.139 13:34, October 9, 2014 (UTC)

Wrong. The seperator rules take precedence over the other rules. So you'd get n=7, #*((2))*(1,3)*^*2*#*#*#*#*#*#*#. -SJ224 14:39, October 9, 2014 (UTC)

A second try:

n=1, #*((2))*(1,3)*^*2**#*7

n=7, #*((2))*(1,3)*^*2**#

n=7, #*((2))*(1,3)*^*2*#*#*#*#*#*#*#

n=7, #*((2))*(1,3)*^*2*#*#*#*#*#*#*7

n=49, #*((2))*(1,3)*^*2*#*#*#*#*#*#

n=49, #*((2))*(1,3)*^*2*#*#*#*#*#*49

n=2401, #*((2))*(1,3)*^*2*#*#*#*#*#

n=2401, #*((2))*(1,3)*^*2*#*#*#*#*2401

n=5764801, #*((2))*(1,3)*^*2*#*#*#*#

n=5764801, #*((2))*(1,3)*^*2*#*#*#*5764801

n=7¹⁶, #*((2))*(1,3)*^*2*#*#*#

n=7¹⁶, #*((2))*(1,3)*^*2*#*#*7¹⁶

n=7³², #*((2))*(1,3)*^*2*#*#

n=7³², #*((2))*(1,3)*^*2*#*7³²

n=7⁶⁴, #*((2))*(1,3)*^*2*#

n=7⁶⁴, #*((2))*(1,3)*^*2*7⁶⁴

n=7¹²⁸, #*((2))*(1,3)*^*2

n=7¹²⁸, #*((2))*(1,3)[*^2]_(7¹²⁸) (has over a googol characters)

n=7¹²⁸, #*((2))*(1,3)[*^2]_(7¹²⁸-1)*_(7¹²⁸+1)2 (has over a googol characters)

The next string would contain more than one septuagintillion characters. --84.61.137.139 16:46, October 9, 2014 (UTC)

HELP!

Rulesets are too complicated for me. Help? -- A Large Number Googologist -- 15:54, October 12, 2014 (UTC)

what do you want us to do for you it's vel time 16:03, October 12, 2014 (UTC)

I'm really surprised that you can understand pretty much all of the R function, yet you cannot grasp the rules of this notation. LittlePeng9 (talk) 16:10, October 12, 2014 (UTC)

@LP9 I just understand up to {0**} arrays.

@vell make an easier ruleset to understand. (i'm still not generalized with some rulesets, especially where there are too long or make uses of symbols)

Forget to add signature. -- A Large Number Googologist -- 17:06, October 12, 2014 (UTC)

what part specifically are you confused on it's vel time 17:11, October 12, 2014 (UTC)

EVERYTHING (not mad at you). -- A Large Number Googologist -- 17:21, October 12, 2014 (UTC)

Well, recently on chat you claimed that you understand all of R function up to Nested Array Notation, which is {S**...*} with any number of asterisks. LittlePeng9 (talk) 17:28, October 12, 2014 (UTC)

Then i was wrong -- A Large Number Googologist -- 17:30, October 12, 2014 (UTC)

I mean, i did not mean that i did understand up to {S***...***}, but up to {0**} -- A Large Number Googologist -- 17:31, October 12, 2014 (UTC)

Soo, where is the easier explanation? -- A Large Number Googologist -- 17:50, October 12, 2014 (UTCi)

I find this definition to be quite easy if-then-else algorithm. I have no idea how to make it simplier. Try looking at the examples, maybe this will clarify some things. LittlePeng9 (talk) 17:56, October 12, 2014 (UTC)

Still don't get Pound-Star Notation. -- A Large Number Googologist -- 18:07, October 12, 2014 (UTC)

I don't know how to help in that case. In my opinion, the definition cannot be made much more clean. Sorry. LittlePeng9 (talk) 18:09, October 12, 2014 (UTC)

( -- A Large Number Googologist -- 18:17, October 12, 2014 (UTC)

Sorry, formating error. :( --~~~~

Just, forget it. -- A Large Number Googologist -- 18:18, October 12, 2014 (UTC)

lmao this whole conversation it's vel time 19:01, October 12, 2014 (UTC)

Incorrect evaluation

SuperJedi, Hyper-Exploding Pound-Star notation is comparable to $f_{\omega^{\omega^\omega}}$, not $f_{\varepsilon_{0}}$. Deedlit11 (talk) 00:03, October 17, 2014 (UTC)

So I could have overestimated just about everything past Titanicol. That would be problematic. -SJ224 00:19, October 17, 2014 (UTC)

Let's see: ^->w

^*->w+1

^^->w*2

^^^->w*3

{1}->w^2

{2}->w^3

{3}->w^4

{0:1}->w^w

{1:1}->w^(w+1)

{2:1}->w^(w+2)

{0:2}->w^(w*2)

{0:3}->w^(w*3)

{0:0:1}->w^((w^2)+w)

{1:0:1}->w^((w^2)+w+1)

{0:1:1}->w^((w^2)+w*2)

{0:0:2}->w^((w^2)*2)

{0:0:3}->w^((w^2)*3)

Yeah, I think you're right. :(

-SJ224 00:28, October 17, 2014 (UTC)

Missing definition of the question mark

What does ? mean in your notation? [n]? -- A Large Number Googologist -- 21:59, October 17, 2014 (UTC)

It meant [1] in an early draft, but I removed it from the definition. -SJ224 22:27, October 17, 2014 (UTC)

Oh. -- A Large Number Googologist -- 22:40, October 17, 2014 (UTC)

Proving that #*<<>> has level w^(w^w+w^2) in fgh

@ = w^w^w

<@> = w^(w^w+1)

<<@>> = w^(w^w+2)

[1] = w^(w^w+w)

<[1]> = w^(w^w+w+1)

[2] = w^(w^w+w*2)

[x] = w^(w^w+w*x)

As a result, #*<<>> has level w^(w^w+w^2) in fgh

-- A Large Number Googologist -- 14:04, October 19, 2014 (UTC)

Growth Rate of H#*<<>>

The limit of previous notation was  [] --> w^(w^w+w^2)

[*] --> w^(w^w+w^2+w)

[**] --> w^(w^w+w^2+w*2)

[^] --> w^(w^w+w^2*2)

[{1}] --> w^(w^w+w^3)

[{1}{1}] --> w^(w^w+w^3*2)

[{2}] --> w^(w^w+w^4)

[{0:1}] --> w^(w^w*2)

[{1:1}] --> w^(w^(w+1))

[{0:2}] --> w^(w^(w*2))

[{0:0:1}] --> w^(w^(w^2))

[@] --> w^(w^(w^w))

[<@>] --> w^(w^(w^w+1))

[[1]] --> w^(w^(w^w+w))

etc.

As a result, H#*<<>> has limit e(0)

-- A Large Number Googologist -- 19:05, October 20, 2014 (UTC)

Oops!

[[1]] --> w^(w^(w^w+w)), not just 1 (forget to add nowiki) -- A Large Number Googologist -- 19:08, October 20, 2014 (UTC)

I knew it would almost certainly significantly exceed the lower bound I provided (w^w^(w*2)), but I didn't think it would exceed it by that much. -SJ224 10:15, October 21, 2014 (UTC)

Yeah :D -- A Large Number Googologist -- 18:06, October 21, 2014 (UTC)

Hello?

Are you going to work on this?

It has been like almost a month.

-- From the googol and beyond -- 15:50, December 11, 2014 (UTC)

Probably, eventually. -SJ224 11:45, December 15, 2014 (UTC)

HELP

Make Simpler Definition -- Notorious V.L.E. 17:45, December 15, 2014 (UTC)

I need Simpler Definition

Please Explain Definition to Me -- Notorious V.L.E. 17:46, December 15, 2014 (UTC)

HELP

Someone PLease Think for Me -- Notorious V.L.E. 17:46, December 15, 2014 (UTC)

#*{{{1}}}^{1} = #*{{..{{..{{..{{..1..}}..}}..1..}}..}}^{0}. Does That Help Cookiefonster (talk) 20:20, December 15, 2014 (UTC)

HERE IS IT VELL

you < your momma

Happy now?

OK IT WAS A JOKE

If you have problems understanding this notation, ask SuperJedi, it may help. Or try looking again. Or understand this notation. -- From the googol and beyond -- 20:40, January 1, 2015 (UTC)

A faster-growing revision

I do this because there're long runs of steps with n being constant:

1. If there's only 1 entry, instead of doing multiplication, we use "square final number after "*" separator n times, then decrement n" (This is a dumping of n as its triangle into the final number).
2. If there're >1 numbers or the remaining separator isn't a single *, then increment n.
3. If the final entry in final parenthesized array is a 0, then increment n another time.

That makes it faster. 80.98.179.160 15:29, February 27, 2018 (UTC)

diction and operator clash

fast -> swift: free:fast::swift:slow::quick:qualm::speedy:idle::hasty:laggy::fleet:laden.

# is already used for the primorial. Can you find a new name? Alysdexia (talk) 19:28, April 28, 2018 (UTC)