# Proof of Collatz Conjecture

### Proof of Collatz Conjecture

n1 = 3n0 + 1 where n0 is any odd integer > 1;
l1 = Log( n1 / greatest odd factor > 1 of numerator) / Log(2);

n2 = 3((3n0 + 1) / 2^ (l1) ) + 1 = 3n1 / 2^ (l1) + 1;

l2 = Log( n2 / greatest odd factor > 1 of numerator) / Log(2);

...
nm = 3*nsub(m-1) / 2^ (lsub(m-1)) + 1;

lm = Log( nm / greatest odd factor > 1 of numerator) / Log(2);

Note: nm = 2^lm * (greatest odd factor > 1 of nm);

Probability( n1 is a multiple of 2) = 1;

Probability( n1 is a multiple of 2^2 | n1 is a multiple of 2) = 1/2;

Probability( n1 is a multiple of 2^3 | n1 is a multiple of 2^2) = 1/2;
...

Probability( n1 is a multiple of 2^2) = Probability( n1 is a multiple of 2^2 | n1 is a multiple of 2)
* Probability( n1 is a multiple of 2) = (1/2) * 1 = 1/2;

Probability( n1 is a multiple of 2^3) = Probability( n1 is a multiple of 2^3 | n1 is a multiple of 2^2) * Probability( n1 is a multiple of 2^2) = (1/2) * (1/2) = 1/4;...

So, Probability(n1 is at least a multiple of 2^2) = Sum[i = 0 to l1 - 1 of (1/2)^(i+1) ]
...

Probability(nm is at least a multiple of 2^2) = Sum[i = 0 to lm - 1 of (1/2)^(i+1) ]

Hence,

Probability(Collatz sequence does not converge to 1)

= Product[ m=1 to infinity of Sum[ i=0 to lm - 1 of (1/2)^(i+1) ] ] --> 0.

So, the Collatz conjecture is true!

Note: 'sub' indicates a long subscript of a variable.

Please Be A Good Sport! Take any detection of mistakes as learning exercises...
And please let me know too. Thank you! P.S. There's more analysis that I did include here for brevity sake...

FYI: The Last Calculation In My Proof Of The Collatz Conjecture Revisited

Note: I assumed there exists a Collatz sequence which does not converge to 1.

Recall Final Result: Probability(Collatz sequence does not converge to 1)

= Product[ j=1 to infinity of Sum[ i=0 to lj - 1 of (1/2)^(i+1) ] ] --> 0. Why?

Let N0 be the event that 3*n0 + 1 is a multiple of 2 for any odd integer > 1;

Let N1 be the event that n1 = 3*n0 + 1 is at least a multiple of 2^2;

Let N2 be the event that n2 = 3*n1/(2^(l1)) + 1 is at least a multiple of 2^2;

...

Let Nm be the event that nm = 3*nsub(m-1)/(2^(lsub(m-1)) ) +1 is at least a multiple of 2^2.

Note: 'sub' indicates a long subscript of a variable.

Note: lj = Log( nj / greatest odd factor > 1 of numerator) / Log(2).

Note: Probability(N0) = Prob(N0) = 1.

Prob(N1) = Prob(N1 | N0) * Prob(N0) = Prob(N1 | N0) = Sum[ i=0 to l1 - 1 of (1/2)^(i+1) ];

Prob(N2) = Prob(N2 | N1) * Prob(N1) = Prob(N2 | N1) * Prob(N1 | N0)
= Sum[ i=0 to l2 - 1 of (1/2)^(i+1) ] * Sum[ i=0 to l1 - 1 of (1/2)^(i+1) ] ;
...
...

Prob(Nm) = Product[ j = 1 to m of Sum[ i=0 to lj - 1 of (1/2)^(i+1) ];

Therefore, I have the following result:

Probability(Collatz sequence does not converge to 1)

= Prob(Nm as m --> infinity)

= Product[ j =1 to infinity of Sum[ i=0 to lj - 1 of (1/2)^(i+1) ] ] --> 0. Q.E.D.

P.S. Please share your comments. Thank you! Recall: Basics of Conditional Probability

Prob(N2|N1)P(N1) = Prob(N2 ∩ N1) = Prob(N2).

In context of my proof of the Collatz conjecture, N2 ∩ N1 means that the greatest odd factor of events, N2 and N1 is greater one, and the events, N2 and N1 are at least multiples of 4.

And since the prior existence of event N1 in the Collatz sequence indicates the existence of the succeeding event N2 ... all according to my prior assumption that there exists a Collatz sequence that does not converge one. And thus, N2 ∩ N1 = N2.

I hope this explanation helps you to follow the previous post about my proof of the Collatz conjecture.

Please share your comments too. Thanks! David Cole
aka primework123
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!
primework123

Posts: 22
Joined: Sat Oct 31, 2015 2:07 pm
Reputation: 0

### Re: Proof of Collatz Conjecture--Update via Latex

An update with latex script for math symbols for sake of clarity is coming soon... Stay tuned.
Guest

### Re: Proof of Collatz Conjecture--Update via Latex

primework123 wrote:Please see https://en.wikipedia.org/wiki/Collatz_conjecture for explanation.

We let $$n_{1 }$$ = $$3n_{0 }$$+ 1 where $$n_{0}$$ is any odd integer > 1;

$$l_{1 }$$ = $$Log( n_{1 }$$ / (greatest odd factor > 1 of $$n_{1 })) / Log(2)$$;

$$n_{2}$$ = $$3((3n_{0 } + 1$$) / $$2^ {l_{1 }} ) + 1$$ = $$3n_{1 }$$ /$$2^ {l_{1 }} + 1$$;

$$l_{2}$$ = $$Log( n_{2 }$$ / (greatest odd factor > 1 of $$n_{2 })) / Log(2)$$;

...
$$n_{m}$$ = $$3n_{m-1 }$$ / $$2^ {l_{m-1 }} + 1$$;

$$l_{m}$$ = $$Log( n_{m }$$ / (greatest odd factor > 1 of $$n_{m})) / Log(2)$$;

Note: $$n_{m}$$ = $$2^{l_{m }}$$ * (greatest odd factor > 1 of $$n_{m }$$);

Probability( $$n_{1}$$ is a multiple of 2) = 1;

Probability( $$n_{1}$$ is a multiple of $$2^{2}$$ | $$n_{1}$$ is a multiple of 2) = 1/2;

Probability( $$n_{1}$$ is a multiple of $$2^{3}$$ | $$n_{1}$$ is a multiple of $$2^{2}$$) = 1/2;
...

Probability( $$n_{1}$$ is a multiple of $$2^{2}$$)
= Probability( $$n_{1}$$ is a multiple of $$2^{2}$$ | $$n_{1}$$ is a multiple of 2)
* Probability( $$n_{1}$$ is a multiple of 2) = (1/2) * 1 = 1/2;

Probability( n1 is a multiple of $$2^{3}$$) =
Probability( n1 is a multiple of $$2^{3}$$ | n1 is a multiple of $$2^{2}$$)
* Probability( n1 is a multiple of $$2^{2}$$) = (1/2) * (1/2) = 1/4; ...

So, Probability($$n_{1 }$$ is at least a multiple of $$2^{2}$$) = $$\sum_{i=0}^{l_{1 } - 1 }(1/2)^{i+1}$$
...

Probability($$n_{m }$$ is at least a multiple of $$2^{2}$$) = $$\sum_{i=0}^{l_{m-1 } - 1 }(1/2)^{i+1}$$

Hence,

Probability(Collatz sequence does not converge to 1)

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

So, the Collatz conjecture is true!

Furthermore,

Let $$N_{0 }$$ be the event that $$3n_{0 }$$+ 1 is a multiple of 2 for any odd integer, $$n_{0 }$$ > 1;

Let $$N_{1 }$$ be the event that $$n_{1 }$$ = $$3n_{0 }$$+ 1 is at least a multiple of $$2^{2}$$;

Let $$N_{2 }$$ be the event that $$n_{2 }$$ = $$3n_{1 }/2^{l_{1 }}$$+ 1 is at least a multiple of $$2^{2}$$;

...

Let $$N_{m }$$ be the event that $$n_{m }$$ = $$3n_{m-1 }/2^{l_{m-1 }}$$+ 1 is at least a multiple of $$2^{2}$$;

Note: $$l_{j }$$ = $$Log( n_{j }$$ / (greatest odd factor > 1 of $$n_{j })) / Log(2)$$;

Note: Probability($$N_{0 }$$) = 1.

Prob($$N_{1 }$$) = Prob($$N_{1}$$ | $$N_{0 }$$) * Prob($$N_{0 }$$) = Prob($$N_{1}$$ | $$N_{0 }$$)

$$= \sum_{i=0}^{l_{1 } - 1 }(1/2)^{i+1}$$;

Prob($$N_{2 }$$) = Prob($$N_{2 }$$ | $$N_{1 }$$) * Prob($$N_{1 }$$)
= Prob($$N_{2 }$$ | $$N_{1 }$$) * Prob($$N_{1 }$$ | $$N_{0 }$$)

= $$\sum_{i=0}^{l_{2 } - 1 }(1/2)^{i+1}$$ * $$\sum_{i=0}^{l_{1 } - 1 }(1/2)^{i+1}$$;
...
...

Prob($$N_{m }$$)

$$=\prod_{j=1}^{m} \sum_{i=0}^{l_{j } - 1 }(1/2)^{i+1}$$;

Again, we have the following result:

Probability(Collatz sequence does not converge to 1)

= Prob($$N_{m }$$ as $$m \rightarrow \infty$$)

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

Recall: Basics of Conditional Probability

Prob($$N_{2 }$$|$$N_{1 }$$)Prob($$N_{1 }$$) = Prob($$N_{2 }$$ ∩ $$N_{1 }$$) = Prob($$N_{2 }$$).

In context of our proof of the Collatz conjecture, $$N_{2 }$$ ∩ $$N_{1 }$$ means that the greatest odd factor of events, N2 and N1 is greater one, and the events, $$N_{2 }$$ and $$N_{1 }$$ are at least multiples of 4.

And since the prior existence of event N1 in the Collatz sequence indicates the existence of the succeeding event N2 ... all according to our prior assumption that there exists a Collatz sequence that does not converge one.
And thus, $$N_{2 }$$ ∩ $$N_{1 }$$= $$N_{2 }$$.

Please share your comments too. Thanks! David Cole
aka primework123
Please support my research work at: https://www.gofundme.com/david_cole
Thank you! Thank Lord GOD!
Guest

### Re: Proof of Collatz Conjecture

Thanks for sharing  leesajohnson

Posts: 208
Joined: Thu Dec 31, 2015 7:11 am
Location: London
Reputation: -33

### Re: Proof of Collatz Conjecture

In his argument,https://www.google.com/search?q=proof+of+collatz+by+david+cole&rlz=1C1CAFA_enUS632US632&oq=proof+of+collatz+by+david+cole&aqs=chrome..69i57.6395j0j4&sourceid=chrome&ie=UTF-8, Dr. G. Lassmann claims to refute David Cole's proof of Collatz Conjecture.

But, he does not understand that David's proof of the Collatz Conjecture is conditional:

Important Notes:
I. 3$$n_{m }$$ + 1 will always be a multiple of 2 for all positive odd integers, $$n_{m }$$ or that the
Probability[3$$n_{m }$$ + 1 is a multiple of 2] = 1.

II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some $$n_{m }$$ or that the Collatz sequence never converges to one.

III. $$N_{m }$$ be the event that $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1 = is at least a multiple of $$2^{2}$$
= 3 * f($$n_{0 }$$) + 1 where f($$n_{0 }$$) is a function of $$n_{0 }$$ according to the Collatz Process and according to the definition of $$n_{0 }$$ in David Cole's proof of the Collatz Conjecture.

Consider for example, if 3$$n_{0 }$$ + 1 = 3 * 2 = 6, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.

So, for the above example, the Probability[3$$n_{0 }$$ + 1 is at least a multiple of 4] is 1/2.

Consider for example, if 3$$n_{0 }$$ + 1 = 5 * 2 = 10, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is multiple of 4] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.

Consider for example, if 3$$n_{0 }$$ + 1 = 11 * 2 = 22, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is multiple of 4] = 1/2.
And furthermore, Probability [3$$n_{0 }$$ + 1 is a multiple of 16 | 3$$n_{0 }$$ + 1 is multiple of 8] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.

So, $$2^{l_{m }}$$ depends on the size of the largest odd factor of 3$$n_{m }$$ + 1, and they, $$2^{l_{m }}$$ and $$l_{m }$$ , will vary generally too.

So, the Probability [ 4 divides $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1] varies:

Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 1] = 1/2;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 2] = 3/4;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 3] = 7/8;
...
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = k] = $$\sum_{i=1}^{k }(1/2)^i$$;
Therefore, 1/2 $$\le$$ Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ < $$\infty$$] < 1.

Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest

### Re: Proof of Collatz Conjecture

But, he does not understand that David's proof of the Collatz Conjecture is conditional:

Important Notes:
I. 3$$n_{m }$$ + 1 will always be a multiple of 2 for all positive odd integers, $$n_{m }$$ or that the
Probability[3$$n_{m }$$ + 1 is a multiple of 2] = 1.

II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some $$n_{m }$$ or that the Collatz sequence never converges to one.

III. $$N_{m }$$ be the event that $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1 = is at least a multiple of $$2^{2}$$
= 3 * f($$n_{0 }$$) + 1 where f($$n_{0 }$$) is a function of $$n_{0 }$$ according to the Collatz Process and according to the definition of $$n_{0 }$$ in David Cole's proof of the Collatz Conjecture.

Consider for example, if 3$$n_{0 }$$ + 1 = 3 * 2 = 6, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.

So, for the above example, the Probability[3$$n_{0 }$$ + 1 is at least a multiple of 4] is 1/2.

Consider for example, if 3$$n_{0 }$$ + 1 = 5 * 2 = 10, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is multiple of 4] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.

Consider for example, if 3$$n_{0 }$$ + 1 = 11 * 2 = 22, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is multiple of 4] = 1/2.
And furthermore, Probability [3$$n_{0 }$$ + 1 is a multiple of 16 | 3$$n_{0 }$$ + 1 is multiple of 8] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.

So, $$2^{l_{m }}$$ depends on the size of the largest odd factor of $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1, and they, $$2^{l_{m }}$$ and $$l_{m }$$ , will vary generally too.

So, the Probability [ 4 divides $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1] varies:

Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 1] = 1/2;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 2] = 3/4;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 3] = 7/8;
...
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = k] = $$\sum_{i=1}^{k }(1/2)^i$$;
Therefore, 1/2 $$\le$$ Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ < $$\infty$$] < 1.

Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:

But, he does not understand that David's proof of the Collatz Conjecture is conditional:

Important Notes:
I. 3$$n_{m }$$ + 1 will always be a multiple of 2 for all positive odd integers, $$n_{m }$$ or that the
Probability[3$$n_{m }$$ + 1 is a multiple of 2] = 1.

II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some $$n_{0 }$$ or that the Collatz sequence never converges to one.

III. $$N_{m }$$ be the event that $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1 = is at least a multiple of $$2^{2}$$
= 3 * f($$n_{0 }$$) + 1 where f($$n_{0 }$$) is a function of $$n_{0 }$$ according to the Collatz Process and according to the definition of $$n_{0 }$$ in David Cole's proof of the Collatz Conjecture.

Consider for example, if 3$$n_{0 }$$ + 1 = 3 * 2 = 6, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.

So, for the above example, the Probability[3$$n_{0 }$$ + 1 is at least a multiple of 4] is 1/2.

Consider for example, if 3$$n_{0 }$$ + 1 = 5 * 2 = 10, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is multiple of 4] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.

Consider for example, if 3$$n_{0 }$$ + 1 = 11 * 2 = 22, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is multiple of 4] = 1/2.
And furthermore, Probability [3$$n_{0 }$$ + 1 is a multiple of 16 | 3$$n_{0 }$$ + 1 is multiple of 8] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.

So, $$2^{l_{m }}$$ depends on the size of the largest odd factor of $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1, and they, $$2^{l_{m }}$$ and $$l_{m }$$ , will vary generally too.

So, the Probability [ 4 divides $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1] varies:

Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 1] = 1/2;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 2] = 3/4;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 3] = 7/8;
...
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = k] = $$\sum_{i=1}^{k }(1/2)^i$$;
Therefore, 1/2 $$\le$$ Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ < $$\infty$$] < 1.

Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:
Guest wrote:

But, he does not understand that David's proof of the Collatz Conjecture is conditional:

Important Notes:
I. 3$$n_{m }$$ + 1 will always be a multiple of 2 for all positive odd integers, $$n_{m }$$ or that the
Probability[3$$n_{m }$$ + 1 is a multiple of 2] = 1.

II. David Cole's proof assumes there exists the possibility the Collatz Conjecture is false for some $$n_{0 }$$ or that the Collatz sequence never converges to one.

III. $$N_{m }$$ be the event that $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1 = 3 * f($$n_{0 }$$) + 1 is at least a multiple of $$2^{2}$$ where f($$n_{0 }$$) is a function of $$n_{0 }$$ according to the Collatz Process and according to the definition of $$n_{0 }$$ in David Cole's proof of the Collatz Conjecture.

Consider for example, if 3$$n_{0 }$$ + 1 = 3 * 2 = 6, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is a multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 6 = 2 * 3, the largest odd factor is 3 > 2 therefore, we could have 2 instead of 3. The choice between choosing an even and odd integer is the same or 1/2.

So, for the above example, the Probability[3$$n_{0 }$$ + 1 is at least a multiple of 4] is 1/2.

Consider for example, if 3$$n_{0 }$$ + 1 = 5 * 2 = 10, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is a multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 10 = 2 * 5, the largest odd factor is 5 > 2 therefore, we could have 4 instead of 5. And because of the possibility of 4 instead of 5, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is a multiple of 4] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 * 1/2 = 1/2 + 1/4 = 3/4.

Consider for example, if 3$$n_{0 }$$ + 1 = 11 * 2 = 22, then the Probability [3$$n_{0 }$$ + 1 is a multiple of 4 | 3$$n_{0 }$$ + 1 is a multiple of 2] = 1/2. Why? Because 3$$n_{0 }$$ + 1 = 22 = 2 * 11, the largest odd factor is 11 > 8 therefore, we could have 8 instead of 11. And because of the possibility of 8 instead of 11, we have Probability [3$$n_{0 }$$ + 1 is a multiple of 8 | 3$$n_{0 }$$ + 1 is a multiple of 4] = 1/2.
And furthermore, Probability [3$$n_{0 }$$ + 1 is a multiple of 16 | 3$$n_{0 }$$ + 1 is multiple of 8] = 1/2.

So, for this latest example, the Probability[3$$n_{0 }$$ + 1 is at a least multiple of 4] is 1/2 + 1/2 */2 + 1/2 * 1/2 *1/2 = 7/8.

So, $$2^{l_{m }}$$ depends on the size of the largest odd factor of $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1, and they, $$2^{l_{m }}$$ and $$l_{m }$$ , will vary generally too.

So, the Probability [ 4 divides $$n_{m }$$ = 3$$n_{m-1 }/2^{l_{m-1 }}$$+ 1] varies:

Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 1] = 1/2;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 2] = 3/4;
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = 3] = 7/8;
...
Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ = k] = $$\sum_{i=1}^{k }(1/2)^i$$;
Therefore, 1/2 $$\le$$ Probability [ 4 divides $$n_{m }$$ | $$l_{m }$$ < $$\infty$$] < 1.

Therefore, Dr. G. Lassmann's claim is refuted as nonsense!
Guest

### Re: Proof of Collatz Conjecture

Correction:

Delete 'Note I' in the above argument since is is not correct! And it is not needed.
Guest

Guest

Guest

### Re: Proof of Collatz Conjecture

Interesting! You have used the probabilistic method to prove the Collatz Conjecture... How sound is that?
Guest

### Re: Proof of Collatz Conjecture

Probability(d = $$2^{l_{m}}$$) = Probability($$l_{m }$$ = m) = $$2^{-m}$$ over the set, M, of all possible random variables, m, or any positive integer where d is any divisor in the Collatz processing.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:Probability(d = $$2^{l_{m}}$$) = Probability($$l_{m }$$ = m) = $$2^{-m}$$ over the set, M, of all possible random variables, m, or any positive integer where d is any divisor in the Collatz processing.

Probability(d = $$2^{l_{m}}$$) = Probability($${l_{m}}$$ = m) = $$2^{-m}$$ over the set, M, of all possible values of the random variable, m, or any positive integer where d is any divisor in the Collatz processing. Guest

### Re: Proof of Collatz Conjecture

$$l_{m}$$ = m ??

$$l_{m}$$ is any positive integer, and therefore,
Probability(d = $$2^{l_{m}}$$) = Probability($$l_{m}$$) = $$2^{-l_{m }}$$
...
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:Interesting! You have used the probabilistic method to prove the Collatz Conjecture... How sound is that?

FYI: Please read the relevant article on the mathematical concept of, 'Almost Surely', at
https://en.m.wikipedia.org/wiki/Almost_surely.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:
Guest wrote:Interesting! You have used the probabilistic method to prove the Collatz Conjecture... How sound is that?

FYI: Please read the relevant article on the mathematical concept of, 'Almost Surely', at
https://en.m.wikipedia.org/wiki/Almost_surely.

Does one accept the clear fact,

l = $$\sum_{l_{m }=1}^{l_{m }=\infty}2^{-l_{m }}$$ = 1?

With the acceptance of that fact, the mathematical concept of, 'Almost Surely', is surely irrelevant here. And the proof of the Collatz Conjecture via the probabilistic method is sound!
Guest

### Re: Proof of Collatz Conjecture

P.S.
There are no exceptions (counterexamples or cycles) that disprove the Collatz Conjectue. And we challenge anyone to find one. Otherwise, it is an act of pure fiction to suggest such things without proof.
Guest

### Re: Proof of Collatz Conjecture

Guest wrote:
P. S.
There are no exceptions (counterexamples or cycles) that disprove the Collatz Conjecture. And we challenge anyone to find one. Otherwise, it is an act of pure fiction to suggest such things without proof.

A Keen and Proven Observation:

The divisor, 2, occurs 50% of all possible divisions by 2, 4, 8, ..., $$2^{n}$$ of the Collatz processing/operations according to our probabilistic model/proof and according to the Law of Large Numbers. Therefore, that fact explains why there's growth in the Collatz sequence, and it also explains why the Collatz sequence will always converge to one. Go Blue!

Dave,

https://www.researchgate.net/profile/David_Cole29.
Guest

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