Count up with recursive prime factorization

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Elmach
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Count up with recursive prime factorization

Postby Elmach » Sun Nov 02, 2014 8:43 am UTC

So we have decimal. We have hex. We have base factorial. Let's do something different.

EDIT: Because of what happened in the first page, this is slightly different.
As you may know, every number can be prime-factorized. For instance,

21 = 3*7,
24 = 23*3 = 2*2*2*3.

Now, 3 is the second prime -- we can represent 3 as p<2>, or p2. Similarly, 7 is the fourth, so we note it p4. Thus, 21 is p2p4; we use concatenation to multiply.

I called this recursive prime factorization: 2 is the first prime, so we can represent 3 as p<p<1>>. By doing it this way, we reduce all numbers into a series of brackets, down to the "root" 1. We don't actually need that; we could just as well represent that as nothing.

This is the basic idea; the goal in this thread is to post the next number using this format in some way. Some common formats are:

Omit the "p": 3 becomes <<>>. This is bracket notation/b-notation
Leave the "p", but omit unnecessary brackets (say, for composite-numbered primes.) 6 is p1pp1, 7 is p<p1p1>. We call this p-notation. This was the original one used for this thread.
Represent it as a tree: There are many ways to do this. Examples are in the spoilers below.
Lineal-Spatial (L-S) Notation
Spoiler:
This was introduced on the 5th page by Reecer.
84 = 2*2*3*7 can be represented as

Code: Select all

|  ||
||||

(3(2)*2*2*7(2*2))
This ended up being called lineo-spatial notation (L-S). This grows trees from the bottom up. The 84th prime is

Code: Select all

|  ||
||||
|

and the 84th prime squared is

Code: Select all

|  |||  ||
||||||||
|   |

From this, you should be able to figure out how L-S works.

Depth Notation
Spoiler:
This was based off of L-S Notation, and introduced by faubi on page 8.
84 (2*2*3*7) is

Code: Select all

*** * * *
* * *
.
The first line separates it into four units (four prime factors); the last two don't go deeper, so they are 2. (Alternatively, beneath them is nothing, which represents 1 because 1 is the multiplicative identity.) The one on the left is 7 because it is the fourth prime; beneath the unit is the representation for 4. 3 is the second prime, so we put 2 beneath it. Thus, the 84th prime would be

Code: Select all

*********
*** * * *
* * *

and the 84th prime squared would be

Code: Select all

********* *********
*** * * * *** * * *
* * *     * * *

There are, of course, tons of possible notations; people keep making more. Again, as long as you show this recursive structure somehow, it is fine.

Alternatively, you could use one of the many common shorthands for some of the common numbers, as seen below.

As examples, I will give the numbers from 1 to 10 using bracket notation, p-notation, and decimal. Consensus has built to use bracket notation, with some random other notations, although I strongly suggest including decimal as well. I also include some common shorthand (namely: circle-notation, abc-notation, and dot-notation. These are often mixed.).

, 1, (1)
<>, p1 (2) (Shorthand: . (dot), o (circle), a (abc/alphabetical), ۝ (fancy circle))
(<>), pp1 (3) (Shorthand: ⊙ (circle), b (abc/alphabetical), ۞ (fancy circle))
<><>, p1p1 (4) (Shorthand: : (dot))
<[<>]>, ppp1 (5) (Shorthand: c (abc/alphabetical)
<><[]>, p1pp1 (6)
<[][]>, p<p1p1> (7) (Shorthand: (:) (dot))
<>[]<>, p1p1p1 (8)
<[]><[]>, pp1pp1 (9)
<><[<>]>, p1ppp1 (10).

Let us continue.
<<<<>>>>, pppp1 (11). (Shorthand: d (abc/alphabetical))
Last edited by Elmach on Wed Nov 12, 2014 11:10 am UTC, edited 8 times in total.

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 9:57 am UTC

(wait, I got it, we can't write a 2, because 2 is written as p1)

p1p1pp1 (12)

spoiler with original post
Spoiler:
Elmach wrote:Now, 3 is the second prime, so we will note it as p2.

<snip>

But 2 is the first prime, so we instead note 3 as pp1.


These statements seem contradictory, why 2 being the first prime negates 3 being noted as p2?

(will edit here correct count once this is clarified)

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 10:16 am UTC

pp1pp1 (13)

On ambiguity:
Spoiler:
One way to resolve ambiguity would be to require brackets around the argument of 'p' if it is not prime (i.e some number of 'p's then a '1'). Then both 'p1pp1' and 'pp1p1' would unambiguously refer to 6, while 'p{p1p1}' would refer to 7.

In that case 13 is p{p1pp1}

The original grammar that generates these numbers (written as BNF) is

Code: Select all

<number> ::= <number> <number>
<number> ::= '1'
<number> ::= 'p' <number>

which is an ambiguous grammar.

The grammar for the alternative form using brackets is

Code: Select all

<number> ::= <prime>
<number> ::= <composite>
<composite> ::= <number> <number>
<prime> ::= '1'
<prime> ::= 'p' '{' <composite> '}'
<prime> ::= 'p' <prime>

which as far as I can tell is not ambiguous.

(Excepting the chance that I have made a mistake here)

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 10:26 am UTC

p1p<p1p1> (14)

On ambiguity:
Spoiler:
Other thank liking how the <prime> stuff looks on faubiguy's code more than curly brackets, I agree with his idea.

Here's latest code:

Code: Select all

<number> ::= <prime>
<number> ::= <composite>
<composite> ::= <number> <number>
<prime> ::= '1'
<prime> ::= 'p' '<' <composite> '>'
<prime> ::= 'p' <prime>

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 10:30 am UTC

pp1ppp1 (15)

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phillip1882
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Re: Count up with recursive prime factorization

Postby phillip1882 » Sun Nov 02, 2014 1:29 pm UTC

p1p1p1p1
on the abiguity problem, another option is to use parenthesis.
6 = p(p1)p1, 7 = p(p1p1)
good luck have fun

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 7:32 pm UTC

pp<p1p1>

phillip1882 wrote:p1p1p1p1
on the abiguity problem, another option is to use parenthesis.
6 = p(p1)p1, 7 = p(p1p1)


Yeah, what would work too. Though parenthesis give the illusion of multiplication (as in p*(p1p1)) while I think <>'s do a good job of showing recursion.
Last edited by Vytron on Sun Nov 02, 2014 7:40 pm UTC, edited 1 time in total.

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 7:34 pm UTC

p1pp1pp1

Technically for 17, the outer pair of <>s is unnecessary because there's only one number inside of them.

Elmach
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Re: Count up with recursive prime factorization

Postby Elmach » Sun Nov 02, 2014 7:41 pm UTC

p<p1p1p1> (19)

Regarding ambiguity:

I like the bracket idea.

(In my sleep, I was going to suggest having muliplication be a prefix operator, so **p1p1p1 for 8, but I can see people hating this.)
(Also, I guess <> means 1 2, because as Vytron stated later 1s only come from p1s now, so we can get rid of 1s... and <...> imply a p, so we can technically remove everything but the brackets?)

(so like this: Should we try it like this?)
<<<>><<>><<>>> (19) (bracket notation, <> = 1)
<<><><>>> (19) (bracket notation, = 1)
Last edited by Elmach on Sun Nov 02, 2014 7:53 pm UTC, edited 2 times in total.

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 7:41 pm UTC

Fixed

p1p1ppp1(20)

Oh, intriguing suggestion, I do agree the if the 1s are unnecessary we should drop them.

I think nothing can be a 1 (because we have everything be multiplied infinitely many times by 1) it'd be:

(1)
<> (2)
<<>> (3)
<><> (4)
<<<>>> (5)
<><<>> (6)
<<><>> (7)
<><><> (8)
<<>><> (9)
<><<<>>> (10).
Last edited by Vytron on Sun Nov 02, 2014 7:49 pm UTC, edited 1 time in total.

Elmach
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Re: Count up with recursive prime factorization

Postby Elmach » Sun Nov 02, 2014 7:45 pm UTC

pp1p<p1p1>
<<<>>><<<>><<>>> (<> = 1, <<>> = 2)
<<>><<><>> ( = 1, <> = 2)
21
Last edited by Elmach on Mon Nov 03, 2014 12:25 am UTC, edited 2 times in total.

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 7:50 pm UTC

Check previous message, if the 1 is the empty set the outer pair of <>s is unnecessary.

<><<<<>>>> (22)

List of numbers in this system
(1)
<> (2)
<<>> (3)
<><> (4)
<<<>>> (5)
<><<>> (6)
<<><>> (7)
<><><> (8)
<<>><> (9)
<><<<>>> (10)
<<<<>>>> (11)
<><><<>> (12)
<<><<>>> (13)
<><<><>> (14)
<<>><<<>>> (15)
<><><> (16)
<<<><>>> (17)
<><<>><> (18)
<<><><>> (19)
<><><<<>>> (20)
<<>><<><>> (21)
<><<<<>>>> (22)
Last edited by Vytron on Sun Nov 02, 2014 7:58 pm UTC, edited 1 time in total.

Elmach
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Re: Count up with recursive prime factorization

Postby Elmach » Sun Nov 02, 2014 7:55 pm UTC

Inner pair, but yes.
23
p<pp1pp1>
<<<>><<>>>

The reason I kind of don't want to do it like this is because it suddenly becomes a lot less readable.
That's also why I kind of want to do it.
(Also, maybe use parens or other grouping symbols... brackets are kind of hard to read.)

((())(()))?
([<>][<>])?
<[<>][<>]>?

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 8:07 pm UTC

Elmach wrote:The reason I kind of don't want to do it like this is because it suddenly becomes a lot less readable.


I see (I have edited all numbers so far into my previous post). Though I think stuff like <><><>=8 <<><><>>=8th prime looks recursively good.

As a last "improvement" (if it can be called that, since I keep making things looking more unreadable) we can drop all the closing brackets that were at the very right, because they can be implied (add closing brackets until all opening brackets are closed, with your mind)

The result:
(1)
< (2)
<< (3)
<>< (4)
<<< (5)
<><< (6)
<<>< (7)
<><>< (8)
<<>>< (9)
<><<< (10)
<<<< (11)
<><><< (12)
<<><< (13)
<><<>< (14)
<<>><<< (15)
<><>< (16)
<<<>< (17)
<><<>>< (18)
<<><>< (19)
<><><<< (20)
<<>><<>< (21)
<><<<< (22)
<<<>>< (23)
<><<>><< (2^3 *3=24)

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 9:21 pm UTC

p1p1p1pp1
<><><><<>>

I think exponentiation is unnecessary, since the operations of multiplication and taking the nth prime number are sufficient to construct all positive integers.

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 9:45 pm UTC

<><<><<>>> (26)

<>s represented by os version: o<o<o>>

The thing with exponentiation is that is uses less characters, and I'm crazy about optimization. But you're right that that, and omitting all closing brackets at the very right seem unnecessary.

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 10:06 pm UTC

pp1pp1pp1
<<>><<>><<>>

It reduces length, but makes the representation more complex by adding another operation and 2 more characters (< and >)

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 10:10 pm UTC

Yeah.

Elmach wrote:(Also, maybe use parens or other grouping symbols... brackets are kind of hard to read.)

((())(()))?
([<>][<>])?
<[<>][<>]>?


I suggest using ◇ as a shorthand for <>, I think that'd make them easier to read.

◇◇<◇◇> (28)

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 10:16 pm UTC

p<p1ppp1>
<<><<<>>>>

Allowing exponentiation also removes the property that each number has a unique representation (which can otherwise be enforced by requiring that lower factors precede greater factors).

I think that using ◇ in place of <> might sometimes easier to read, but it makes it harder to type, and I also think that it looks a bit worse than <> (probably because the lines slope differently).

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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 10:19 pm UTC

Oh, what about an o? That was my idea initially.

o<o><<o>> (30)

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Re: Count up with recursive prime factorization

Postby faubiguy » Sun Nov 02, 2014 10:33 pm UTC

<<<<<>>>>>

I still prefer plain <>, but I think using 'o's would also be fine.

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Re: Count up with recursive prime factorization

Postby Vytron » Sun Nov 02, 2014 10:56 pm UTC

If there's a standard then I'll follow the standard.

<><><><><>

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 12:01 am UTC

<<>><<<<>>>>

Elmach
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Re: Count up with recursive prime factorization

Postby Elmach » Mon Nov 03, 2014 12:12 am UTC

<><[<><>]> (34 = 2 * 17 = 2 * 5th prime)

Please include the decimal form, so at least one person can understand what is going on.

EDIT: This kind of looks like a tree (with 5 nodes in this case) now. You know,

Code: Select all

   (?; 34)
  /       \
 (2; 1)  (17; 4)
         /     \
        (2;1)   (2;1)

Maybe we could do something with this.
Last edited by Elmach on Mon Nov 03, 2014 12:21 am UTC, edited 1 time in total.

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 12:19 am UTC

<<<>>><<><>> (35)

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Re: Count up with recursive prime factorization

Postby Elmach » Mon Nov 03, 2014 12:22 am UTC

<o>oo<o> (36 = 62)

(brackets are not in order, but that is fine because now there is no ambiguity!)

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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 12:27 am UTC

<<><><<>>> (37)

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Mon Nov 03, 2014 1:53 am UTC

<><o<>o> (38)

Combining both methods, lol!
Last edited by Vytron on Mon Nov 03, 2014 2:36 am UTC, edited 1 time in total.

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 2:16 am UTC

That's 67, not 38.

<<>><<><<>>> (39)

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Mon Nov 03, 2014 2:38 am UTC

Fixed. Curiously I'd have made that mistake no matter what notation we used...

o<>o<<o>> (40)

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 2:45 am UTC

<<<><<>>>> (41)

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Mon Nov 03, 2014 2:54 am UTC

o<o><oo> (answer to the ultimate question about life, the universe, and everything)

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 3:12 am UTC

<<><<><>>>

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Re: Count up with recursive prime factorization

Postby Elmach » Mon Nov 03, 2014 3:16 am UTC

o<<<<>>>>o (44)

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Mon Nov 03, 2014 3:17 am UTC

Now with arbitrary colors!

<<>><<<>>><<>> (45)

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Re: Count up with recursive prime factorization

Postby Elmach » Mon Nov 03, 2014 3:28 am UTC

Code: Select all

"<o> <o>"
    o

(46)
Where I am using quotes as another pair of angle brackets
and ignoring all whitespace.

Alternatively,
(⊙⊙)o, where ⊙ is <<>>.
Last edited by Elmach on Mon Nov 03, 2014 3:32 am UTC, edited 1 time in total.

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 3:32 am UTC

000110001111 (47)

In which I use '0' and '1' in place of '<' and '>'

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Re: Count up with recursive prime factorization

Postby Elmach » Mon Nov 03, 2014 3:34 am UTC

oo⊙oo (48)
Whereby the symbol ⊙ means 3 because why not.

<><><<>><><> using traditional bracket-notation.

Also, <><><><><<, using the Vytronian trimmed bracket-notation.

Also, <>^<>^<> <<>>, using flat exponentiation bracket-notation.

Also, 010100110101 using the faubian cypher for bracket-notation.
Last edited by Elmach on Mon Nov 03, 2014 3:39 am UTC, edited 2 times in total.

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Vytron
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Re: Count up with recursive prime factorization

Postby Vytron » Mon Nov 03, 2014 3:34 am UTC

That looks... naughty, but it's probably my dirty mind at fault.

(..)(..) (49)

Where (..) = <<><>>

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faubiguy
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Re: Count up with recursive prime factorization

Postby faubiguy » Mon Nov 03, 2014 3:39 am UTC

<><<<>>><<<>>> (50)

Alternatively, 'acc', where 'a' is '<>', b is '<<>>', c is '<<<>>>', etc.

Or p1ppp1ppp1 in original notation.


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