Is the following heuristic probabilistic argument correct?

[Guest]

"The following heuristic probabilistic argument support the 3x + 1 (Collatz) Conjecture ... Pick an odd integer [tex]n_{0 }[/tex] at random and iterate the

function T until another odd integer [tex]n_{1}[/tex] occurs. Then 1/2 of the time, [tex]n_{1} = \frac{3(n_{0}+1)}{2}[/tex], 1/4 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{4}[/tex], 1/8 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{8}[/tex] , and so on. If one supposes that the function T is sufficiently 'mixing'' that successive odd integers in the trajectory of n behave as though they were drawn at random (mod [tex]2^{k}[/tex]) from the set of odd integers (mod [tex]2^{k}[/tex]) for all k, then the expected growth in size between two consecutive odd integers in such a trajectory is the multiplicative factor,

[tex]( \frac{3}{2} )^{\frac{1}{2}} * ( \frac{3}{4} )^{\frac{1}{4}} * ( \frac{3}{8} ) ^{\frac{1}{8}} *[/tex] ... = [tex]\frac{3}{4} < 1[/tex].

Consequently, ... "

Source Links: http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/node3.html;

'Collatz (3x+ 1) Con jecture',

https://en.wikipedia.org/wiki/Collatz_conjecture.

Does the above argument makes sense? Is it correct?

If that argument is correct, then it suggests a counterexample to Collatz Conjecture may exist. Do you agree?

[Guest]

Hmm. We suspect the argument is flawed. We must first quantify time with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.

[Guest]

Relevant Reference Link:

'Proof of Collatz Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1485.

[Guest]

Hmm. We suspect the argument is flawed. We must first quantify time with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ?

The answer depends on t and k.

"I am the author of the above post." -- David Cole, https://www.researchgate.net/profile/David_Cole29.

P.S. Please try different values of t and k to see if the original heuristic probabilistic argument is correct. I am convinced the argument is wrong!

[Guest]

Hmm. We suspect the argument is flawed. We must first quantify time (or the number of trials) with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.

[Guest]

Hmm. We suspect the argument is flawed. We must first quantify time (or the number of trials) with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.

Remarks: The number of times (t) implies the number of appropriate trials (t) in any Collatz sequence as described by the stated argument... Time alone is a bit incoherent and inappropriate here. -- Dave.

[Guest]

Hmm. We suspect the argument is flawed. We must first quantify time (or the number of trials) with a discrete (positive integer) random variable, t.

Secondly, we must determine k so that [tex]2^{k}[/tex] divides 3([tex]n_{0 }[/tex] + 1) ...

And therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *[/tex] ... = ? The answer depends on t and k.

Remarks: The number of times (t) implies the number of appropriate trials (t) in any Collatz sequence as described by the stated argument... Time alone is a bit incoherent and inappropriate here. -- Dave.

We let t be the number of trials it takes for the Collatz sequence to converge.

We can determine k by the largest positive even integer, [tex]e_{max }[/tex], in the Collatz sequence.

[tex]k = \frac{log(e_{max })}{log(2)}[/tex].

Therefore,

[tex]( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *...*( \frac{3}{k} ) ^{\frac{t}{k}} \rightarrow 0[/tex].

[Guest]

Oops! We corrected the k value in the previous post.

[tex]k = \left \lfloor{\frac{log(e_{max })}{log(2)}}\right \rfloor[/tex].

[Guest]

Added Update: We let t be the number of trials it takes for the Collatz sequence of odd integers to converge.

[Guest]

Added Update: That convergent value, 0, represents the probability that the Collatz sequence of odd integers does not converge to one.

[Guest]

Please vote on the truth of the Collatz conjecture at the link below.

Link: https://theory-of-energy.org/2020/09/15/please-vote-on-the-truth-of-the-collatz-conjecture-thank-you/.

Thank you!

[Guest]

Relevant Reference Link:

Our Response to 4.2 A probabilistic heuristic:

https://en.wikipedia.org/wiki/Talk:Collatz_conjecture#Our_Response_to_4.2_A_probabilistic_heuristic:.

Dave.

[Guest]

"The following heuristic probabilistic argument support the 3x + 1 (Collatz) Conjecture ... Pick an odd integer [tex]n_{0 }[/tex] at random and iterate the

function T until another odd integer [tex]n_{1}[/tex] occurs. Then 1/2 of the time, [tex]n_{1} = \frac{3(n_{0}+1)}{2}[/tex], 1/4 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{4}[/tex], 1/8 of the time [tex]n_{1} = \frac{3(n_{0}+1)}{8}[/tex] , and so on. If one supposes that the function T is sufficiently 'mixing'' that successive odd integers in the trajectory of n behave as though they were drawn at random (mod [tex]2^{k}[/tex]) from the set of odd integers (mod [tex]2^{k}[/tex]) for all k, then the expected growth in size between two consecutive odd integers in such a trajectory is the multiplicative factor,

[tex]( \frac{3}{2} )^{\frac{1}{2}} * ( \frac{3}{4} )^{\frac{1}{4}} * ( \frac{3}{8} ) ^{\frac{1}{8}} *[/tex] ... = [tex]\frac{3}{4} < 1[/tex].

Consequently, ... "

Source Links: ...;

'Collatz (3x+ 1) Conjecture',

https://en.wikipedia.org/wiki/Collatz_conjecture.

Does the above argument makes sense? Is it correct?

If that argument is correct, then it suggests a counterexample to Collatz Conjecture may exist. Do you agree?

Oops! Thus far, the heuristic probabilistic argument is apparently correct! More analysis may be needed.

Dave,
https://theory-of-energy.org/2020/09/17/a-brief-analysis-of-the-collatz-conjecture/.

[Guest]

...

Oops! Thus far, the heuristic probabilistic argument is apparently correct! More analysis may be needed.

Dave,
https://theory-of-energy.org/2020/09/17/a-brief-analysis-of-the-collatz-conjecture/.

More analysis is required to determined if the heuristic probabilistic argument is true for all positive odd integers greater than one.

Dave.

[Guest]

Final Remark: The heuristic probabilistic argument is true for infinitely many positive odd integers greater than one, but it is also false for infinitely many positive odd integers greater than one.

Dave.

[Guest]

Update:

Final Remark: The heuristic probabilistic argument is true!

Dave.

[Guest]

Update:

Final Remark: The heuristic probabilistic argument is true!

...

Dave's Conjecture: [tex]r = r(n_{0 }, t) =[/tex] O[tex](t)[/tex] or [tex]r = c_{t } * t[/tex] for some real number, [tex]c_{t }[/tex], such that either [tex]c_{t } >1[/tex] or [tex]0< c_{t } < 1[/tex]. (Update)

Example: If we have [tex]t = 1[/tex] for some [tex]n_{0 }[/tex], then as [tex]k \rightarrow \infty[/tex],

[tex]r = \frac{1}{ \prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}}} \rightarrow \frac{4}{3 } = c_{1 }[/tex].

Therefore, [tex]r = r(n_{0 }, 1) \le \frac{4}{3}[/tex].

[tex]r = r(n_{0 }, 1) \approx \frac{4}{3}[/tex] for infinitely many positive odd integers, [tex]n_{0} > 1[/tex].

Remark: [tex]\prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}} \rightarrow \frac{3}{4 }[/tex] as [tex]k \rightarrow \infty[/tex].

Relevant Reference Link:

"The Collatz equation that supports the truth of the Collatz conjecture",

https://www.math10.com/forum/viewtopic.php?f=63&t=9166.