# Proof of Collatz Conjecture

### Re: Proof of Collatz Conjecture

An Important Note Update:

The value, $$l_{m} = 1$$, which implies division by two in the Collatz process will occur as often as all other values combined, i.e., $$l_{m} > 1$$, which imply division either by four or by eight or by sixteen or by thirty-two, … in the Collatz process according to the our probability calculations and according to the Law of Large Numbers. Thus, we expect the division by two will occur 50% of all possible divisions in the Collatz processing. That fact explains any growth in the Collatz sequence, and it also explains why the Collatz sequence will always converge to one.

'Is the Collatz Conjecture solvable?'

https://www.math10.com/forum/viewtopic.php?f=63&t=1708.
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### Re: Proof of Collatz Conjecture

'Solution to Collatz's Conjecture' by Prof. J. W. Porras,

https://www.researchgate.net/publication/325389892_Solution_to_Collatz's_Conjecture.
Guest

### Re: Proof of Collatz Conjecture

FYI: 'Mathematician Proves Huge Result on ‘Dangerous’ Problem
By
KEVIN HARTNETT
December 11, 2019

Mathematicians regard the Collatz conjecture as a quagmire and warn each other to stay away. But now Terence Tao has made more progress than anyone in decades.'

https://www.quantamagazine.org/mathematician-terence-tao-and-the-collatz-conjecture-20191211/.

Hah! Any 'almost proof' of Collatz Conjecture by T. Tao (a so-called math expert on Collatz Conjecture, etc.) et al. is not a true proof! And any 'almost proofs' of the Riemann Hypothesis or the Goldbach Conjecture are not true proofs!

The Collatz Conjecture, Riemann Hypothesis, and the Goldbach Conjecture are true!!!

-- David Cole,

https://wwww.researchgate.net/profile/David_Cole29/amp.

"Don't pay attention to "authorities" (so-called experts)!
Think for yourself!" -- Richard Feynman.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:FYI: 'Mathematician Proves Huge Result on ‘Dangerous’ Problem
By
KEVIN HARTNETT
December 11, 2019

Mathematicians regard the Collatz conjecture as a quagmire and warn each other to stay away. But now Terence Tao has made more progress than anyone in decades.'

https://www.quantamagazine.org/mathematician-terence-tao-and-the-collatz-conjecture-20191211/.

Hah! Any 'almost proof' of Collatz Conjecture by T. Tao (a so-called math expert on Collatz Conjecture, etc.) et al. is not a true proof! And any 'almost proofs' of the Riemann Hypothesis or the Goldbach Conjecture are not true proofs!

The Collatz Conjecture, Riemann Hypothesis, and the Goldbach Conjecture are true!!!

-- David Cole,

https://wwww.researchgate.net/profile/David_Cole29/amp.

"Don't pay attention to "authorities" (so-called experts)!
Think for yourself!" -- Richard Feynman.

Moreover, the difficulty of proving/understanding the Collatz Conjecture has been exaggerated by Erdos, Lagarias, et al. That indicates their lack of understanding the conjecture/problem. And therefore, they are so-called experts on the Collatz Conjecture. How can anyone solve a problem if one does not understand the problem?

Remarks: An 'almost proof' here indicates an incomplete/flawed proof. Math proofs of important math conjectures should always be complete/true...

Dave.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:An Important Note Update:

The value, $$l_{m} = 1$$, which implies division by two in the Collatz process will occur as often as all other values combined, i.e., $$l_{m} > 1$$, which imply division either by four or by eight or by sixteen or by thirty-two, … in the Collatz process according to the our probability calculations and according to the Law of Large Numbers. Thus, we expect the division by two will occur 50% of all possible divisions in the Collatz processing. That fact explains any growth in the Collatz sequence, and it also explains why the Collatz sequence will always converge to one.

'Is the Collatz Conjecture solvable?'

https://www.math10.com/forum/viewtopic.php?f=63&t=1708.

'Is the Collatz Conjecture solvable?'

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collatz_conjecture.png (23.48 KiB) Viewed 814 times
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### Re: Proof of Collatz Conjecture

FYI: We recall that the divisors, d, are powers of 2, and we had computed their probabilistic distributions in accordance with the algorithm of the Collatz Conjecture and in accordance with all possibile multiples of d:

Probability (d = $$2^{l_{m}}$$) = Probability ($$l_{m}$$) = $$2^{-l_{m }}$$

where the random variable, $$l_{m}$$, is any appropriate nonzero positive integer and where the index, m, is any appropriate nonzero positive integer.

Therefore,

Probability (d = 2) = Probability ($$l_{m} = 1$$) = $$\frac{1}{2}$$;

Probability (d = 4) = Probability ($$l_{m} = 2$$) = $$\frac{1}{4}$$;

Probability ($$d = 8$$) = Probability ($$l_{m} = 3$$) = $$\frac{1}{8}$$;
...

Probability (d = $$2^{l_{m}}$$) = Probability ($$l_{m}$$) = $$\frac{1}{2^{l_{m}}}$$ = $$2^{-l_{m}}$$.

Thus,

$$\sum_{l_{m }=1}^{l_{m }=\infty}2^{-l_{m }}$$ = 1.
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### Re: Proof of Collatz Conjecture

Probability(Collatz sequence does not converge to 1 for any initial positive odd integer, $$n_{0}$$ )

$$=\prod_{m=1}^{m < \infty} \sum_{i=1}^{l_{m} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence in accordance with the algorithm of the Collatz Conjecture).

So, the Collatz conjecture is true!
Guest

### Re: Proof of Collatz Conjecture

An Update:

Probability(Collatz sequence does not converge to 1 or $$n_{m < \infty} \ne 1$$ for any initial positive odd integer, $$n_{0}$$ )

$$=\prod_{m=1}^{m < \infty} \sum_{i=1}^{l_{m} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).

So, the Collatz conjecture is true!

Remarks: In our Collatz sequence from
$$n_{0}$$ to $$n_{m < \infty}$$, we have,

$$1 \le {l_m} \le log ( max (3 * n_{k} + 1 )) / log( 2)$$

for $$0 \le k \le m - 1$$ where $$1 \le m < \infty$$.
Attachments
collatz_conjecture.png (23.48 KiB) Viewed 807 times
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### Re: Proof of Collatz Conjecture

An Update:

Probability(Collatz sequence does not converge to 1 or $$n_{m < \infty} \ne 1$$ for any initial positive odd integer, $$n_{0}$$ )

$$=\prod_{j = 1}^{m < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).

So, the Collatz conjecture is true!

Remarks: In our Collatz sequence from
$$n_{0}$$ to $$n_{m < \infty}$$, we have,

$$1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)$$

for $$0 \le k \le m - 1$$ where $$1 \le j \le m < \infty$$.

Remarks: Our previous indicies were wrong or not well-defined for our sum and product symbols and elsewhere. Math is hard work! Dave.
Attachments
collatz_conjecture.png (23.48 KiB) Viewed 805 times
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### Re: Proof of Collatz Conjecture

Oops! The plural of index are indexes and indices and not indicies. Dave.
Guest

### Re: Proof of Collatz Conjecture

An Update: ( $$n_{m }$$ )

Probability(Collatz sequence does not converge to 1 or $$n_{m < \infty} \ne 1$$ for any initial positive odd integer, $$n_{0}$$ )

$$=\prod_{j = 1}^{n_{m } < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).

So, the Collatz conjecture is true!

Remarks: In our Collatz sequence from
$$n_{0}$$ to $$n_{m < \infty}$$, we have,

$$1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)$$

for $$0 \le k \le m - 1$$ where $$1 \le j \le m < \infty$$.

Dave.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:An Update: ( Oops! index $$n_{m }$$ is wrong! We revert back to m instead.)

Probability(Collatz sequence does not converge to 1 or $$n_{m < \infty} \ne 1$$ for any initial positive odd integer, $$n_{0}$$ )

$$=\prod_{j = 1}^{m } < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).

So, the Collatz conjecture is true!

Remarks: In our Collatz sequence from
$$n_{0}$$ to $$n_{m < \infty}$$, we have,

$$1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)$$

for $$0 \le k \le m - 1$$ where $$1 \le j \le m < \infty$$.

Dave.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:
Guest wrote:An Update: ( Oops! index $$n_{m }$$ is wrong! We revert back to m instead.)

Probability(Collatz sequence does not converge to 1 or $$n_{m < \infty} \ne 1$$ for any initial positive odd integer, $$n_{0}$$ )

$$=\prod_{j = 1}^{m < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence according to the algorithm of the Collatz Conjecture).

So, the Collatz conjecture is true!

Remarks: In our Collatz sequence from
$$n_{0}$$ to $$n_{m < \infty}$$, we have,

$$1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)$$

for $$0 \le k \le m - 1$$ where $$1 \le j \le m < \infty$$.

Dave.
Guest

### Re: Proof of Collatz Conjecture

A Minor Update:

Probability[ Collatz sequence does not converge to 1 ($$n_{m < \infty} \ne 1$$) for any initial positive odd integer, $$n_{0}$$ ]

$$=\prod_{j = 1}^{m < \infty} \sum_{i=1}^{l_{j} }(1/2)^{i} \rightarrow 0.$$

where $$n_{1} = 3 * n_{0} + 1$$ ... (the generation of the complete Collatz sequence according to the algorithm for the Collatz Conjecture).

So, the Collatz conjecture is true!

Remarks: In our Collatz sequence from $$n_{0}$$ to $$n_{m < \infty}$$, we have,

$$1 \le {l_j} \le log ( max (3 * n_{k} + 1 )) / log( 2)$$

for $$0 \le k \le m - 1$$ where $$1 \le j \le m < \infty$$.

Dave.

'Is the Collatz Conjecture solvable?'

Attachments
"The Collatz Conjecture is true!" -- David Cole.
The Algorithm for the Collatz Conjecture.jpg (7.06 KiB) Viewed 478 times
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### Re: Proof of Collatz Conjecture

Remarks: Probability [ $$l_{j } = 1$$ ] = .5 for $$1 \le j \le m < \infty$$ in accordance with the Law of Large Numbers.

We had assumed the Probability [ $$l_{m } = 0$$ ] = 0. However, it becomes clear that our assumption is wrong!

'Law of Large Numbers',

https://en.wikipedia.org/wiki/Law_of_large_numbers.
Guest

### Re: Proof of Collatz Conjecture

Remark: The Probability [ $$l_{m } = 0$$ ] = 0 indicates $$n_{m } \ne 2^{0} = 1$$.
Guest

### Re: Proof of Collatz Conjecture

Dave wrote:Remarks: Probability [ $$l_{j } = 1$$ ] = .5 for $$1 \le j \le m < \infty$$ in accordance with the Law of Large Numbers.

We had assumed the Probability [ $$l_{m } = 0$$ ] = 0. However, it becomes clear that our assumption is wrong!

'Law of Large Numbers',

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

Remarks: Probability [ $$l_{j } = 1$$ ] = .5 for $$1 \le j \le m < \infty$$ is also consistent with data (the distribution of maximum divisors, $$2^{x}$$, for all positive even integers). The positive integer, x, is the maximum value such $$2^{x}$$ divides any positive even integer.
Guest

### Re: Proof of Collatz Conjecture

Remark: For our previous remark, we assume and we refer to the ordered set of all positive even integers, E = {2, 4, 6, 8, 10, 12, ..., $$\infty$$}.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:Remark: For our previous remarks, we assume and we refer to the ordered set of all positive even integers, E = {2, 4, 6, 8, 10, 12, ..., $$\infty$$}.

An Example: We compute the distributions of maximum divisors, $$2^{x}$$, of even integers, between 2 and 48, inclusively.

Maximum divisor 2 occurs 12 times;

Maximum divisor 4 occurs 6 times;

Maximum divisor 8 occurs 3 times;

Maximum divisor 16 occurs 2 times;

Maximum divisor 32 occurs 1 time.

Thus, the maximum divisor 2 occurs 50% of the time (24) between even integers, 2 and 48, inclusively.
Guest

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