# Is the following heuristic probabilistic argument correct?

Probability theory and statistics

### Is the following heuristic probabilistic argument correct?

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

function T until another odd integer $$n_{1}$$ occurs. Then 1/2 of the time, $$n_{1} = \frac{3(n_{0}+1)}{2}$$, 1/4 of the time $$n_{1} = \frac{3(n_{0}+1)}{4}$$, 1/8 of the time $$n_{1} = \frac{3(n_{0}+1)}{8}$$ , 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 $$2^{k}$$) from the set of odd integers (mod $$2^{k}$$) for all k, then the expected growth in size between two consecutive odd integers in such a trajectory is the multiplicative factor,

$$( \frac{3}{2} )^{\frac{1}{2}} * ( \frac{3}{4} )^{\frac{1}{4}} * ( \frac{3}{8} ) ^{\frac{1}{8}} *$$ ... = $$\frac{3}{4} < 1$$.

Consequently, ... "

'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?
Attachments Algorithm for the Collatz Conjecture.jpg (7.11 KiB) Viewed 525 times
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### Re: Is the following heuristic probabilistic argument correc

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 $$2^{k}$$ divides 3($$n_{0 }$$ + 1) ...

And therefore,

$$( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *$$ ... = ? The answer depends on t and k.
Guest

### Re: Is the following heuristic probabilistic argument correc

'Proof of Collatz Conjecture',

https://www.math10.com/forum/viewtopic.php?f=63&t=1485.
Attachments Collatz Conjecture is true!
collatz_conjecture.png (23.48 KiB) Viewed 514 times
Guest

### Re: Is the following heuristic probabilistic argument correc

David Cole wrote: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 $$2^{k}$$ divides 3($$n_{0 }$$ + 1) ...

And therefore,

$$( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *$$ ... = ?

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!
Attachments The Collatz Conjecture is true!
collatz_conjecture.png (23.48 KiB) Viewed 417 times
Guest

### Re: Is the following heuristic probabilistic argument correc

Guest wrote: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 $$2^{k}$$ divides 3($$n_{0 }$$ + 1) ...

And therefore,

$$( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *$$ ... = ? The answer depends on t and k.
Guest

### Re: Is the following heuristic probabilistic argument correc

Guest wrote:
Guest wrote: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 $$2^{k}$$ divides 3($$n_{0 }$$ + 1) ...

And therefore,

$$( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *$$ ... = ? 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

### Re: Is the following heuristic probabilistic argument correc

Guest wrote: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 $$2^{k}$$ divides 3($$n_{0 }$$ + 1) ...

And therefore,

$$( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *$$ ... = ? 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, $$e_{max }$$, in the Collatz sequence.

$$k = \frac{log(e_{max })}{log(2)}$$.

Therefore,

$$( \frac{3}{2} )^{\frac{t}{2}} * ( \frac{3}{4} )^{\frac{t}{4}} * ( \frac{3}{8} ) ^{\frac{t}{8}} *...*( \frac{3}{k} ) ^{\frac{t}{k}} \rightarrow 0$$. Attachments The Collatz conjecture is true!
A Collatz Sequence.jpg (9.6 KiB) Viewed 91 times
Guest

### Re: Is the following heuristic probabilistic argument correc

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

$$k = \left \lfloor{\frac{log(e_{max })}{log(2)}}\right \rfloor$$.
Guest

### Re: Is the following heuristic probabilistic argument correc

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

### Re: Is the following heuristic probabilistic argument correc

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

### Re: Is the following heuristic probabilistic argument correc

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

Thank you! Guest

### Re: Is the following heuristic probabilistic argument correc

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

### Re: Is the following heuristic probabilistic argument correc

Guest wrote:"The following heuristic probabilistic argument support the 3x + 1 (Collatz) Conjecture ... Pick an odd integer $$n_{0 }$$ at random and iterate the

function T until another odd integer $$n_{1}$$ occurs. Then 1/2 of the time, $$n_{1} = \frac{3(n_{0}+1)}{2}$$, 1/4 of the time $$n_{1} = \frac{3(n_{0}+1)}{4}$$, 1/8 of the time $$n_{1} = \frac{3(n_{0}+1)}{8}$$ , 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 $$2^{k}$$) from the set of odd integers (mod $$2^{k}$$) for all k, then the expected growth in size between two consecutive odd integers in such a trajectory is the multiplicative factor,

$$( \frac{3}{2} )^{\frac{1}{2}} * ( \frac{3}{4} )^{\frac{1}{4}} * ( \frac{3}{8} ) ^{\frac{1}{8}} *$$ ... = $$\frac{3}{4} < 1$$.

Consequently, ... "

'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

### Re: Is the following heuristic probabilistic argument correc

Dave wrote: ...

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

### Re: Is the following heuristic probabilistic argument correc

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

### Re: Is the following heuristic probabilistic argument correc

Update:

Final Remark: The heuristic probabilistic argument is true!

Dave.
Guest

### Re: Is the following heuristic probabilistic argument...

Dave wrote:Update:

Final Remark: The heuristic probabilistic argument is true!

...

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

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

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

Therefore, $$r = r(n_{0 }, 1) \le \frac{4}{3}$$.

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

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

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

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