This was my approach and where it has taken me, as more of a programmer of sorts rather than a math guy (though i’m a huge fan of the stuff). At any rate this is possibly already widely known, but I hope seeing it maybe in a different way might spark some insights:
Let N0 be the set of non-negative integers {0, 1, 2, …}.
I = { 4n + 1 | n in N0 } = { x in N+ | x mod 4 = 1 }
L = { 4n + 2 | n in N0 } = { x in N+ | x mod 4 = 2 }
H = { 4n + 3 | n in N0 } = { x in N+ | x mod 4 = 3 }
K = { 4n + 4 | n in N0 } = { x in N+ | x mod 4 = 0 }
H to L: (H → 3x+1 → L)
HL = 4n + 3 = { n in N+ | n mod 4 = 3 } ends in L because 3x+1-ing an H always results in an L
Let’s show this:
An H (x_H) has the property x_H mod 4 = 3.
We apply the 3x+1 operation: 3x_H + 1.
Now, let’s see what this is modulo 4:
3x_H + 1 === 3(3) + 1 (mod 4)
=== 9 + 1 (mod 4)
=== 10 (mod 4)
=== 2 (mod 4).
Since the result is 2 (mod 4), it’s an L
I to K: (I → 3x+1 → K)
IK = 4n + 1 = { n in N+ | n mod 4 = 1 } ends in K because 3x+1-ing an I always results in a K
Let’s show this:
An I (x_I) has the property x_I mod 4 = 1.
We apply the 3x+1 operation: 3x_I + 1.
Now, let’s see what this is modulo 4:
3x_I + 1 === 3(1) + 1 (mod 4)
=== 3 + 1 (mod 4)
=== 4 (mod 4)
=== 0 (mod 4).
Since the result is 0 (mod 4), it’s a K
Now if we try that with K or L we have some problems. A K can stay a K or become an L under C(K).
L’s are an interesting topic as well because they can become an H or an I. Lets take a look at that.
LI = 8n + 2 = { n in N0 | n mod 8 = 2 }
Let’s show this for LI:
Consider an element x from the set {n in N0 | n mod 8 = 2}.
This means x can be written as x = 8k + 2 (for some integer k >= 0).
The operation is division by 2 (x / 2).
x / 2 = (8k + 2) / 2 = 4k + 1.
This result, 4k + 1, is precisely of the form 4n+1.
Thus, dividing these L’s by 2 results in an I
Now for the other L’s.
LH = 8n + 6 = { n in N+ | n mod 8 = 6 } ends in H because div 2 of L’s of this form become numbers of the form 4n+3 HL
Let’s show this for LH:
Consider an element x from the set {m in N+ | m mod 8 = 6}.
This means x can be written as x = 8k + 6 (for some integer k >= 0).
The operation is division by 2 (x / 2).
x / 2 = (8k + 6) / 2
= 4k + 3.
This result, 4k + 3
Thus, dividing these specific L by 2 results in an H.
One find’s the longest chains of these by starting from any n that is a power of 2-1 try 63 for instance.
So a few more points of interest before we get into some tables:
KLI = 16n + 4 = { n in N+ | n mod 16 = 4 } ends in I because 2 div 2’s of K’s of this form become numbers of the form 4n+1 IK
KLH = 16n + 12 = { n in N+ | n mod 16 = 12 } ends in H because 2 div 2’s of the K’s of this form become numbers of the form 4n+3 HL
KKLI = 32n + 8 = { n in N+ | n mod 32 = 8 } ends in I because
3 div 2’s of K’s of this form become numbers of the form 4n+1 IK
KKLH = 32n + 24 = { n in N+ | n mod 32 = 24 } ends in H because 3 div 2’s of K’s of this form become numbers of the form 4n+3 HL
With all of this in mind I went ahead and changed my set definitions for the 4n+x:
IK = { 4n + 1 | n in N0 } = { x in N+ | x mod 4 = 1 }
L = { 4n + 2 | n in N0 } = { x in N+ | x mod 4 = 2 }
HL = { 4n + 3 | n in N0 } = { x in N+ | x mod 4 = 3 }
K = { 4n + 4 | n in N0 } = { x in N+ | x mod 4 = 0 }
so lets look at the 8n bois. I am going to put a | between what we knew from 4n and what we ‘learned’ from 8n.
8n + 1 = IK | L
8n + 2 = L | IK
8n + 3 = HL | IK
8n + 4 = K | L
8n + 5 = IK | K
8n + 6 = L | HL
8n + 7 = HL | HL
8n + 8 = K | K
Some things to notice are 1-4 are our original set. but we have added an L. IK, IK, L respectively.
5-8 is again our starting set but there K’s went K and these L’s went H giving us K, HL, HL, K respectively.
Another quick fun thing to note is if you are at say 4n+1 and you want to go to 8n land. you can ‘stay’ on your current chain, by staying +1 or find your other branch by adding your current power of two (in this case 4) to your 1 = 5 and your buddy will be at 8n + 5.
So what does the land of 16n look like? you guessed it, two copies of 8n trajectories, with one more div 2 tacked on to the end:
16n + 1 = IKLIK
16n + 2 = LIKL
16n + 3 = HLIKK
16n + 4 = KLIK
16n + 5 = IKKK
16n + 6 = LHLIK
16n + 7 = HLHLIK
16n + 8 = KKL
16n + 9 = IKLHL
16n + 10 = LIKK
16n + 11 = HLIKL
16n + 12 = KLHL
16n + 13 = IKKL
16n + 14 = LHLHL
16n + 15 = HLHLHL
16n + 16 = KKK
Neat huh? Anyway here is wonderwall, I mean 32n:
32n + 1 = IKLIKL
32n + 2 = LIKLIK
32n + 3 = HLIKKK
32n + 4 = KLIKL
32n + 5 = IKKKL
32n + 6 = LHLIKK
32n + 7 = HLHLIKL
32n + 8 = KKLIK
32n + 9 = IKLHLHL
32n + 10 = LIKKK
32n + 11 = HLIKLIK
32n + 12 = KLHLIK
32n + 13 = IKKLIK
32n + 14 = LHLHLIK
32n + 15 = HLHLHLIK
32n + 16 = KKKL
32n + 17 = IKLIKK
32n + 18 = LIKLHL
32n + 19 = HLIKKL
32n + 20 = KLIKK
32n + 21 = IKKKK
32n + 22 = LHLIKL
32n + 23 = HLHLIKK
32n + 24 = KKLHL
32n + 25 = IKLHLIK
32n + 26 = LIKKL
32n + 27 = HLIKLHL
32n + 28 = KLHLHL
32n + 29 = IKKLHL
32n + 30 = LHLHLHL
32n + 31 = HLHLHLHL
32n + 32 = KKKK
So one neat thing you can do is notice our case of 32n + 1, we know 1 4 2 goes 1 4 2 for forever. and we know then that as we grow our mod 32 / 64 / 128 and so on, we can find other numbers that follow that, or any of these other trajectories.
For instance 33 will go IKLIKL before veering off somewhere as will 65 and 97…
as will any of these sequence match any particular trajectory you might be looking for.
So at one point I was looking into this in terms of a proper grammar system, but have recently been digging on the other angles from the videos. Big thanks to @Perryman for pointing me towards those.
More stuff in this thread soon. Hopefully this helps someone.
