A Simple Linear Diophantine Equation That Encodes All Cycles

General
1

We set M = 3, meaning we classify each value in the trajectory by its remainder modulo 3. This gives three residue classes: [0], [1], and [2].

There are six possible transitions, but we are only concerned with the following (we ignore seed values that are multiples of 3 in this analysis):

  • From [1] to [2] via (3n + 1)/2
  • From [1] to [2] via n/2
  • From [2] to [1] via n/2
  • From [2] to [2] via (3n + 1)/2

For example, we observe that any value in [1] will either increase by 3k + 1 or decrease by 3k + 2 (for some integer k) when moving to [2].

The full transition table is as follows:

[1] → [2] [1] → [2] [2] → [1] [2] → [2]
-2 +1 -1 +0
-5 +4 -4 +3
-8 +7 -7 +6
… … … …

For any value of d in the rule (3n + d)/2 such that d is congruent to 1 modulo 6. the transition table below remains the same (shared across all cases with some offsets).

To construct a Diophantine equation that encodes all trajectories we use a simple trick.

For instance, summing values from the first column [1] → [2] can be expressed as:

S = -2x - 3y

For example, (-2) + (-8) + (-20) = -2(3) - 3(8) = -30.

The last column does not require a second variable, so we only need seven variables in total.

Thus, the full linear Diophantine equation (for d congruent to 1 modulo 6) becomes:

-2 x_1 - 3 x_2 + x_3 + 3 x_4 - x_5 - 3 x_6 + 3 x_7 = 0

With the following constraints:

  • When x1 = 0, x2 also equals 0
  • When x3 = 0, x4 also equals 0
  • When x5 = 0, x6 also equals 0

Since we end up with the exact same equation for each value of d that is congruent to 1 mod 6, each solution V can represent a trajectory in one of the systems (or worlds). However, there are some solutions that are phantoms and do not represent any trajectory.

2

Interesting … what solution corresponds to the trajectory starting at 31 and ending at 1?

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5

Yes (and I think it is in one of mathkook’s paper for the case d=1).
It’s really simple to show: In a cycle, \sum\limits_t(\frac{3o+d}{2})+\sum\limits_{k-t}(\frac{e}{2})=\sum\limits_t o+\sum\limits_{k-t} e
with o the odd elements of the cycle, and e the even elements
You do the math and find out that d\cdot t+\sum o=\sum e

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10

That’s easy to show: every even number between A and B is B times a power of 2, so the sum can be written as 2B(2^{v2(3A+1)-2}+2^{v2(3A+1)-3}+....+1)=2B(2^{v2(3A+1)-1}-1).
Just replace 2B2^{v2(3A+1)-1} by 3A+1

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13

Let’s say that you have …→A → B_1 → B_2 →… → B_i →C→… where A and C are even numbers and B_1, B_2,…, B_i are i consecutive odds, then the sum of these odds is 2(C-B_1)-i

Can be found the same way as for even, knowing that if you write B_1=b2^i-1 then B_2=b2^{i-1}3-1, B_3=b2^{i-2}3^2-1, … C=b3^i-1 (with b odd)

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16

It is always worth, but O>t^2 gives -O<-t^2 and I don’t see how you got the inequality (inverted?)

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18

Why not E + O\geqslant E_{min}+O_{min} \geqslant 2T(t) + 2T(h) - t ? Bounds…we rarely see how useful they can be until it become obvious that they are…

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