# 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?
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Algorithm for the Collatz Conjecture.jpg (7.11 KiB) Viewed 479 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 468 times
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### 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 371 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 45 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