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Frostburn (talk | contribs)
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Formulate the musical equivalent of the Collatz conjecture.
 
Frostburn (talk | contribs)
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Give (seemingly) unconditional form of the Collatz iteration.
 
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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 ==
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2/1
2/1
</pre>
</pre>
== Additional results ==
Not all variations of the recursion 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>.
=== Additional conjectures ===
The recursion:
<math>
s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 3 p_n}{4 q_n}\ \mathrm{red}\ 5
</math>
produces finite scales:
<math>\left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 5 \right\}</math>
for all starting values <math>s_0</math>.
Evidence
<pre>
! varcollatz1_3_4_5_seed_4_3.scl
!
Collatz scale 1, 3, 4, 5 variation of 7/5
13
!
17/16
11/10
11/8
7/5
7/4
2/1
37/16
5/2
49/16
13/4
65/16
17/4
5/1
</pre>
The recursion:
<math>
s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 5 p_n}{4 q_n}\ \mathrm{red}\ 3
</math>
produces finite scales:
<math>\left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 3 \right\}</math>
for all starting values <math>s_0</math>.
Evidence
<pre>
! varcollatz1_5_4_3_seed_12_7.scl
!
Collatz scale 1, 5, 4, 3 variation of 12/7
13
!
383/336
7/6
479/336
3/2
12/7
599/336
2/1
61/28
107/48
153/56
11/4
67/24
3/1
</pre>
There are many (too many to list) other variations of the recursion that seem to work too, but these stood out.
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