User:Frostburn/Collatz scales

The 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]\displaystyle{ s_0 = \frac{p_0}{q_0} }[/math] such that [math]\displaystyle{ 1 \le s_0 < 2 }[/math] and apply the following recursion:

[math]\displaystyle{ s_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{1 + 3 p_n}{q_n}\ \mathrm{red}\ 2 }[/math]

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

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

Conjecture

All Collatz scales are finite.

Evidence

Seed [math]\displaystyle{ \frac{7}{4} }[/math]:

! collatz7.scl
!
Collatz scale of 7/4
6
!
17/16
5/4
11/8
13/8
7/4
2/1

Seed [math]\displaystyle{ \frac{10}{7} }[/math]:

! 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

Seed [math]\displaystyle{ \frac{13}{11} }[/math]:

! collatz13_11.scl
!
Collatz scale of 13/11
7
!
23/22
13/11
53/44
61/44
35/22
20/11
2/1

Additional results

Not all variations of the recursion work for all seeds e.g.

[math]\displaystyle{ 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]\displaystyle{ \frac{14}{11} }[/math]

! 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

But produces an infinite scale for [math]\displaystyle{ \frac{19}{17} }[/math].

Additional conjectures

The recursion:

[math]\displaystyle{ 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]\displaystyle{ \left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 5 \right\} }[/math]

for all starting values [math]\displaystyle{ s_0 }[/math].

Evidence

! 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

The recursion:

[math]\displaystyle{ 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]\displaystyle{ \left\{ s_n | n \in \mathbb{N} \right\} \cup \left\{ 3 \right\} }[/math]

for all starting values [math]\displaystyle{ s_0 }[/math].

Evidence

! 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

There are many (too many to list) other variations of the recursion that seem to work too, but these stood out.