Same underlying problem?

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The Collatz conjecture, the 5n+1 on 7 problem, the Antihydra termination problem … are these all instances of the same problem? Meaning, do you think a solution to one will give a solution to all?

They all have a flavor of “show this deterministic process behaves as if it were random” (whatever that means). Even the Riemann Hypothesis has that flavor, since it would be true if the primes were distributed randomly (in a certain technical sense, I guess).

Or would a solution to the Collatz conejecture be a one-off thing, leaving all other math problems still unsettled?

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I think all require a conversion process, thats what they have in common, the instructions pose as a question, when they are infact simply instructions for behaviour, and so I look for equivalence, for instance it appears equivalent to these rules if you consider movements along a tape where each cell in the tape hold 3 bits of information, a sequential position low to high, and odd number (n)low to high and an even number derived from n×3+1

Condition A.If s is even:s×1.5

Condition B.If s is 1 mod4:(3s+1)÷4

Condition C.If s is 3 mod4:(s+1)÷4

Note.1 Condition C only represents a pass through the odd number itself if its a starting point, under all other conditions it represents a fall through the even number produced by the odd number on the tape.(this must be logged asif it returned to the odd number itself, )

Note2. The first Condition A or B resulting from a condition C will represent a fall through the even number produced by A or B’s odd number

so these rules allow a kind of superposition By logging odd to odd movements whilst including falls through even numbers that have an odd counterpart asif they fall to the odd themselves. Observation of the collatz sequences do not highlight these odd numbers being passed through clearly.

Now assume the equivalents is proven (which can be done)

Note that all (1 mod4)×3 results in a 3 mod4, and as the rest of the process is the same for B and C we can change rule B to s×3 as rule C will complete the process, as all condition A. Converge on odd multiples of 3. And all condition B converge on condition C. We now know that under the new rules all movements not returning to 1 converge on 3 mod4….more specifically 3 mod12. As in the collatz conjecture this could be a fall through the even number or to the odd number you wouldnt necessarily see the convergence.

note that 2^y×1.5^y=3^y

condition A. Converges on multiples of 3

Condition B. Converges on multiples of 3

Condition C. Falls through multiples of 3 every 3rd repeat uninterrupted by condition A or B

Now im working to use all of this to see if I can rephrase…..or expose the real question to one that can be answered.

point being, i think were all trying to answer a question, before the question has even been formulated.

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That would be an amazing thing to prove…

In the bag we can also add Erdős’ conjectures on powers of 2, which has been (loosely) related to the weak Collatz conjecture by Terrence Tao:

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Collatz is related to the situation of a car trying to hit a series of green lights. If there is a 10% chance of hitting a red light we know that eventually one will be reached. This solves the problem that there is no infinite Collatz run.

The harder problem is whether there is a non-trivial circular run. In the car analogy we’d also need to know when it will stop. In that simple framework that is not knowable. To solve circular problem we need to incorporate other aspects of Collatz. Eliahou’s paper comes closest (that I have seen) and shows that in practice loops are not going to happen, but we don’t exactly know why.

If primes are involved that could put it in the realm of Riemann. I’d like to think it is not that hard. If it is then it could wind up in the set of proofs that start with , ‘if Rieman is true then …’

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