User:Frostburn/Collatz scales - Revision history

Frostburn: Give (seemingly) unconditional form of the Collatz iteration.

2024-02-27T12:05:18Z

Give (seemingly) unconditional form of the Collatz iteration.

← Older revision		Revision as of 12:05, 27 February 2024
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	The [[Wikipedia:Collatz conjecture\|Collatz conjecture]] states that a certain sequence always reaches one regardless of the starting value.		The [[Wikipedia:Collatz conjecture\|Collatz conjecture]] states that a certain sequence always reaches one regardless of the starting value.

			== Definition ==
	We can formulate the Collatz iteration musically as follows:		We can formulate the Collatz iteration musically as follows:

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	We call the set <math>\left\{ s_n \| n \in \mathbb{N} \right\} \cup \left\{ 2 \right\}</math> the '''Collatz scale''' of <math>s_0</math>.		We call the set <math>\left\{ s_n \| n \in \mathbb{N} \right\} \cup \left\{ 2 \right\}</math> the '''Collatz scale''' of <math>s_0</math>.

			It might seem that we need to somehow compute and access the components of the reduced form, but we can make the iteration unconditional by generalizing gcd and lcm to fractions.

			Now the recursion is simply.
			<math>
			s_{n+1} = \bigl ( (1 + 3\ \mathrm{lcm}(1, s_n)) \cdot \mathrm{gcd}(1, s_n) \bigr ) \mathrm{red}\ 2
			</math>

	== Conjecture ==		== Conjecture ==

Frostburn: /* Additional results */ Make additional conjectures.

2024-01-05T18:19:29Z

Additional results: Make additional conjectures.

@@ Line 68: / Line 68: @@
 == Additional results ==
-Variations of the recursion do not seem work for all seeds e.g.
+Not all variations of the recursion work for all seeds e.g.
 <math>
@@ Line 90: / Line 90: @@
 </pre>
 But produces an infinite scale for <math>\frac{19}{17}</math>.

Frostburn: Add notes about the 3/1 iteration being special and give a counter example for 5/3 iteration.

2024-01-05T16:18:37Z

Add notes about the 3/1 iteration being special and give a counter example for 5/3 iteration.

← Older revision		Revision as of 16:18, 5 January 2024
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	2/1		2/1
	</pre>		</pre>

			== Additional results ==
			Variations of the recursion do not seem work for all seeds e.g.

			<math>
			s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 5 p_n}{3 q_n}\ \mathrm{red}\ 2
			</math>

			Works for <math>\frac{14}{11}</math>
			<pre>
			! varcollatz14_11.scl
			!
			Collatz scale 5/3 variation of 14/11
			7
			!
			71/66
			7/6
			14/11
			4/3
			3/2
			178/99
			2/1
			</pre>
			But produces an infinite scale for <math>\frac{19}{17}</math>.

Frostburn: Formulate the musical equivalent of the Collatz conjecture.

2024-01-05T15:41:59Z

Formulate the musical equivalent of the Collatz conjecture.

New page

The [[Wikipedia:Collatz conjecture|Collatz conjecture]] states that a certain sequence always reaches one regardless of the starting value.

We can formulate the Collatz iteration musically as follows:

Start with a seed interval <math>s_0 = \frac{p_0}{q_0}</math> such that <math>1 \le s_0 < 2</math> and apply the following recursion:

<math>
s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 3 p_n}{q_n}\ \mathrm{red}\ 2
</math>

where <math>\frac{p_n}{q_n}</math> is in reduced form (no common factors) and <math>x\ \mathrm{red}\ 2</math> denotes octave reduction i.e. repeated division or multiplication by 2 until <math>1 \le x < 2</math>.

We call the set <math>\left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 2 \right\}</math> the '''Collatz scale''' of <math>s_0</math>.

== Conjecture ==
All Collatz scales are finite.

== Evidence ==
Seed <math>\frac{7}{4}</math>:
<pre>
! collatz7.scl
!
Collatz scale of 7/4
6
!
17/16
5/4
11/8
13/8
7/4
2/1
</pre>

Seed <math>\frac{10}{7}</math>:
<pre>
! collatz10_7.scl
!
Collatz scale of 10/7
10
!
35/32
31/28
5/4
71/56
10/7
23/16
53/32
47/28
107/56
2/1
</pre>

Seed <math>\frac{13}{11}</math>:
<pre>
! collatz13_11.scl
!
Collatz scale of 13/11
7
!
23/22
13/11
53/44
61/44
35/22
20/11
2/1
</pre>