# Collatz Conjecture

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2 minutes ago, Zolar V said:

Oh thats really neat, i didnt know that.  I thought we only had expressions for special primes.

I edited it a little, give it another glance if you would. The whole subject of what functions are computable by algorithms and which aren't is very interesting.

So please be clear. Is my tree concept sufficient for your needs? Because it seems damn simple to me, which means that I understand it. And I like ideas I understand better than ones I don't!

But if it's not sufficient for your purposes, do let me know.

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1 minute ago, wtf said:

I edited it a little, give it another glance if you would. The whole subject of what functions are computable by algorithms and which aren't is very interesting.

So please be clear. Is my tree concept sufficient for your needs? Because it seems damn simple to me, which means that I understand it. And I like ideas I understand better than ones I don't!

But if it's not sufficient for your purposes, do let me know.

I think it is sufficient to show how the idea works and I think it gives a good indication of what mechanisms are in play that cause the collatz to converge.    I am unsure if it is going to be sufficiently rigorous or if it will give us the right number of steps.   For our purposes here, it may be enough to continue.  If we reach a terminal branch and are stuck again, then it might be worth going back and looking at a different way to go about applying the Prime + 1 concept.  E.G the addition idea

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Edit:

I think it is sufficient. *

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11 hours ago, Zolar V said:

Edit:

I think it is sufficient. *

Ok so what happens at the end? You fill out the tree and you have a bunch of terminal nodes that are powers of 2. Do you add them up? Multiply them? Does it matter? Once we nail this down we have a complete description of G and we can move on.

Edited by wtf

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53 minutes ago, wtf said:

Ok so what happens at the end? You fill out the tree and you have a bunch of terminal nodes that are powers of 2. Do you add them up? Multiply them? Does it matter? Once we nail this down we have a complete description of G and we can move on.

Possibly both operations, but the final result is going to be a 2^n number. (line 34 in my poorly written pdf)
If all we need to do is multiply then that's all we need to do. Once we have that result,  n will equal the number of steps/iterations we needed to get to a 2^n number.
if we also divide by 2^n, then 2*n will equal the number of steps needed to get to 1.

Edited by Zolar V

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42 minutes ago, Zolar V said:

Possibly both operations, but the final result is going to be a 2^n number. (line 34 in my poorly written pdf)
If all we need to do is multiply then that's all we need to do. Once we have that result,  n will equal the number of steps/iterations we needed to get to a 2^n number.
if we also divide by 2^n, then 2*n will equal the number of steps needed to get to 1.

Ok we can table that for later. So we build the tree, all the terminal nodes are powers of 2, and we do "something" to those powers of 2.

Now what. Please go slowly, one step at a time. Short and clear please.

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31 minutes ago, wtf said:

Ok we can table that for later. So we build the tree, all the terminal nodes are powers of 2, and we do "something" to those powers of 2.

Now what. Please go slowly, one step at a time. Short and clear please.

all the terminal nodes are multiplied together to give a final result of 2^n and youre done

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8 minutes ago, Zolar V said:

all the terminal nodes are multiplied together to give a final result of 2^n and youre done

Ok. Is it important to you that the mapping from p to the final 2^n is injective? Meaning that two different p's give different n's? I don't think that's true but I haven't looked for a counterexample and won't bother if it's not important.

So, onward. How does this prove Collatz?

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9 minutes ago, wtf said:

Ok. Is it important to you that the mapping from p to the final 2^n is injective? Meaning that two different p's give different n's? I don't think that's true but I haven't looked for a counterexample and won't bother if it's not important.

So, onward. How does this prove Collatz?

Hmm,  thats a tough question.  I want to say it isnt going to matter for the proof, but I think it will be injective,  i doubt it would be surjective.

"So, onward. How does this prove Collatz?"
-
I believe our function is embedded in the odd function of the collatz conjecture.  That is to say we can manipulate 3x+1  to be  2x + x+1  where x is our p.
then 2x + (x+1)   and (x+1) is our function.  (yes i know,  its not defined here)
since G^n(p)= 2^n  then
2x^n + 2^n    ->   2^n(x+1)  ->   2^n(G(x+1))  ->  2^n (2^m)

then apply the even collatz function n*m times.
2^{n*m}  / 2^{n*m} = 1

So I know there is a lot missing/wrong.  But the underlaying mechanism to the collatz function is the prime reduction done by adding 1 to a prime.

the odd collatz function will always converge at some point to a 2^n number,  and as soon as it does the even function will reduce it to 1

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1 hour ago, Zolar V said:

Hmm,  thats a tough question.  I want to say it isnt going to matter for the proof, but I think it will be injective,  i doubt it would be surjective.

"So, onward. How does this prove Collatz?"
-
I believe our function is embedded in the odd function of the collatz conjecture.  That is to say we can manipulate 3x+1  to be  2x + x+1  where x is our p.
then 2x + (x+1)   and (x+1) is our function.  (yes i know,  its not defined here)
since G^n(p)= 2^n  then
2x^n + 2^n    ->   2^n(x+1)  ->   2^n(G(x+1))  ->  2^n (2^m)

then apply the even collatz function n*m times.
2^{n*m}  / 2^{n*m} = 1

So I know there is a lot missing/wrong.  But the underlaying mechanism to the collatz function is the prime reduction done by adding 1 to a prime.

the odd collatz function will always converge at some point to a 2^n number,  and as soon as it does the even function will reduce it to 1

> I think it will be injective,  i doubt it would be surjective.

Of course it's not surjective, its output is restricted to powers of 2.

> So I know there is a lot missing

Correct. Burden's on you.

the odd collatz function will always converge at some point to a 2^n number,

Of course. If you can prove that you become famous.

Edited by wtf

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9 minutes ago, wtf said:

> I think it will be injective,  i doubt it would be surjective.

Of course it's not surjective, its output is restricted to powers of 2.

> So I know there is a lot missing

Correct. Burden's on you.

Quote

"Of course it's not surjective"

yea i know, i just didnt want to say definitively...
i didnt want to look like an idiot

Quote

"> So I know there is a lot missing  Correct. Burden's on you. "

yeah so here is how its supposed to work

1: For P prime, show P+1 = 2^r *p_1 *p_2 *p_3... p_i    .    show 2 < p_i <P
2:  Show how any odd composite number can be rewritten such that a function from step 1 can be applied to it
3:  Since any odd composite number can have the function from step 1 applied to it, it also has the same prime reduction mechanism happening to it

4: Show how the odd collatz conjecture can be rewritten such that it looks like our function.
5: Since our function converges to 2^n , so does the odd collatz function
6: since the even collatz function divides 2^n, then after n times  it converges to 1
7:  and thats it.

Edited by Zolar V

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I didn't think about your idea in detail, but if that's your research program, you should definitely get to it. Take your time, work your program, report back.

5: Since our function converges to 2^n , so does the odd collatz function

Now that I understand the prime reduction function, I can't see at all how it relates to Collatz. It's a totally different procedure. You take a prime, add 1, drill each of its prime factors (to multiplicity) down to a power of 2. Well ok you have me believing that.

If you have a reduction from 3n+1 to that, perhaps you should work on a very clear exposition of how this works. Because Collatz doesn't have anything to do with taking the prime factors of anything. I think there must be something wrong with your idea right here.

Edited by wtf

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52 minutes ago, wtf said:

I didn't think about your idea in detail, but if that's your research program, you should definitely get to it. Take your time, work your program, report back.

5: Since our function converges to 2^n , so does the odd collatz function

Now that I understand the prime reduction function, I can't see at all how it relates to Collatz. It's a totally different procedure. You take a prime, add 1, drill each of its prime factors (to multiplicity) down to a power of 2. Well ok you have me believing that.

If you have a reduction from 3n+1 to that, perhaps you should work on a very clear exposition of how this works. Because Collatz doesn't have anything to do with taking the prime factors of anything. I think there must be something wrong with your idea right here.

Suppose the prime reduction function is true
Take any odd natural number,   It is either prime or composite.

If it is prime, then adding 1 would result in all its primes being less than itself.  (we've already agreed on this)
if its a composite,  then adding 1 would still result in at least 1 of its primes being less than the primes in the composite.  (this needs to be shown.  This will be the result of step 2)

-----

if you split 3n+1 into 2n + (n+1),  then you have an odd integer being added to 1.  which fits our function.  the drill down part of our function is the same iteration in the collatz function.   Collatz has everything to do with the adding 1 mechanism.    I believe the (3) part in 3n+1 is irrelevant.  Since if this works it works for An+1 where A is any odd natural number.

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1 hour ago, Zolar V said:

Suppose the prime reduction function is true
Take any odd natural number,   It is either prime or composite.

If it is prime, then adding 1 would result in all its primes being less than itself.  (we've already agreed on this)
if its a composite,  then adding 1 would still result in at least 1 of its primes being less than the primes in the composite.  (this needs to be shown.  This will be the result of step 2)

-----

if you split 3n+1 into 2n + (n+1),  then you have an odd integer being added to 1.  which fits our function.  the drill down part of our function is the same iteration in the collatz function.   Collatz has everything to do with the adding 1 mechanism.    I believe the (3) part in 3n+1 is irrelevant.  Since if this works it works for An+1 where A is any odd natural number.

Yes but the collapsing to 1 is completely different in the two cases. You're convincing me you haven't a proof. In any event this convo is now meaningless unless you supply a proof of your assertions. You can't keep tossing out vague plausibility arguments, especially ones I don't consider plausible. You claimed a proof. You don't actually have a proof. I'd say that's a problem for your claim.

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5 hours ago, wtf said:

Yes but the collapsing to 1 is completely different in the two cases. You're convincing me you haven't a proof. In any event this convo is now meaningless unless you supply a proof of your assertions. You can't keep tossing out vague plausibility arguments, especially ones I don't consider plausible. You claimed a proof. You don't actually have a proof. I'd say that's a problem for your claim.

I understand you're frustration. It is my hope to show you I am right, though it may take some time in making it a proof given your commentary here.

So lets get some terminology right:

Convergence:  I want to say a function converges to 2, if for any input the output is 2 after some amount of iterations.
Definition: " property (exhibited by certain infinite series and functions) of approaching a limit more and more closely as an argument (variable) of the function increases or decreases or as the number of terms of the series increases"

Would it be better to define convergence like this: For M iterations a function converges to 2 ?

Theorem: A function that we are proving
Lemma: A supporting argument for the theorem
Corollary:  An unintended result

Edited by Zolar V

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1 hour ago, Zolar V said:

I understand you're frustration. It is my hope to show you I am right, though it may take some time in making it a proof given your commentary here.

So lets get some terminology right:

Convergence:  I want to say a function converges to 2, if for any input the output is 2 after some amount of iterations.
Definition: " property (exhibited by certain infinite series and functions) of approaching a limit more and more closely as an argument (variable) of the function increases or decreases or as the number of terms of the series increases"

Would it be better to define convergence like this: For M iterations a function converges to 2 ?

Theorem: A function that we are proving
Lemma: A supporting argument for the theorem
Corollary:  An unintended result

None of that is important. I understood what you meant by convergence very early on. You don't have to define what a theorem is. You have to figure out what you're trying to say about Collatz.

Edited by wtf

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These are proof of concept right now:

Lemma 1:  2 < p_i < P
For a natural number P that is prime and P >2 , P+1 =  2^r * p_1 *p_2 * p_3 ... p_i
for every p_n, n = 1,2,3,.. ,i
2 < p_n <P

Lemma 2: Any natural composite number can be rewritten as a sum of prime(s)   {might need to add about odd}
Let k be a composite number
$k = 2^r *p^{1}_1*p^{1}_2*p^{1}_3*p^{1}_14...*p^{1}_n = \sum_{i=1}^ { 2^r *p^{1}_1*p^{1}_2*p^{1}_3*p^{1}_4*...*p^{1}_(n-1) } p_n$

Lemma 3:  Add 1 for each prime in a sum of primes.     {might need to say only for odd, not sure yet}
adding 1 to each prime in a sum is equivalent to adding the upper index to the prime.

Lemma 4:  do the adding part until a power of 2 is reached

example:  3*7 =   7+7+7                             ( potentially 21 may be expressed as   2^3 + 5,    then applying lemma 3 to (5) :  5+ 1 = 6  :  6+2 = 8 : 16+8 = 24 :  24+8 = 32 )
applying lemma 3:   7+7+7(+3)= 24
24 = 2*2*2 *3= 8*3
Applying lemma 3:   24+8 =32 = 2^5

Example: 5*3*7 = 7+7+7+7+7+7+7+7+7...
Applying Lemma 3:  105+15 =120
Prime factorization:    12 = 2*2*2*3*5   : 2*2*2*3 = 24
Applying lemma 3:      120+24=144
Prime factorization:    2*2*2*2*3*3   :  2*2*2*2*3 = 48
Applying lemma 3:     144+48 = 192
Prime factorization:    2^6 * 3
Applying lemma 3:     192+2^6 = 256  = 2^8
Done

NOTE:  All I am saying with " (potentially 21.... )"  is that there may be more than 1 way to solve the problem,  and each way may be more or less efficient.   If we were to measure efficiency by how many steps it takes to solve.

Also, suppose we were to divide out the 2^r numbers each step, that would also shrink the number before we add 1 to it.
Again the purpose of keeping the 2^r numbers is that inside them they hold the key to how many steps are taken.

Edited by Zolar V

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11 hours ago, Zolar V said:

potentially 21 may be expressed as   2^3 + 5

Jeez man.

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2^4  sorry,  i was up for about 25hrs or by that point

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18 minutes ago, Zolar V said:

2^4  sorry,  i was up for about 25hrs or by that point

I don't think I've served you by encouraging you in these last few pages. You haven't got a proof and your exposition is incoherent. I'm sorry.

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I know I dont,  i was outlining what I am going to show.  I suppose I posted it to see if it made sense

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11 minutes ago, Zolar V said:

I know I dont,  i was outlining what I am going to show.  I suppose I posted it to see if it made sense

It makes no sense at all to me. You've reverted to the "spinning out of control" narrative that I thought I talked you down from the past couple of days. I was mistaken. I did no good at all. I can't do anything else here.

Edited by wtf

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6 minutes ago, wtf said:

It makes no sense at all to me. You've reverted to the "spinning out of control" narrative that I thought I talked you down from the past couple of days. I was mistaken. I did no good at all. I can't do anything else here.

I don't see your perspective.   Maybe it is a difference in thinking styles.  I think in big picture and then I work on the details.  So when I see something,  I see all these connected dots, though I might not yet fully understand or know what the dots are. Sometimes my big picture is off, so when I work on the dots they are off too.  But usually if my dots are off too then I get a good indication of what is wrong and fix my big picture.

Maybe your idea of "spinning out of control" is my expression of the different ideas that I get while working through something.  I let my brain connect the dots and explore new dots, even if it has little to do with what I am doing.  I'm essentially letting my brain be creative and intuitive, supposedly these flashes of creativity are how our brain works on problems in the background.

What I did was show a set of lemmas that begin to outline how I am going to prove steps 1-3 from:

On 8/23/2018 at 5:21 PM, Zolar V said:

1: For P prime, show P+1 = 2^r *p_1 *p_2 *p_3... p_i    .    show 2 < p_i <P
2:  Show how any odd composite number can be rewritten such that a function from step 1 can be applied to it
3:  Since any odd composite number can have the function from step 1 applied to it, it also has the same prime reduction mechanism happening to it

I am going to design a theorem supported by the lemmas.  The Theorem will state: For any natural number N, there exists a function G such that any N converges to a power of 2 over M iterations.
Right now I have a bunch of hypotheses and no proof.  The set of hypotheses (lemmas) are the framework in which we will show Theorem 1 is true.   The proof of the lemmas are some of the dots.
Imagine it like a couple nested big pictures with nested dots.
Big picture 1:  The collatz conjecture converges to 1 via this prime reduction method
Dots_1:  Show the collatz looks like the prime reduction function
Dots_2:  Show the prime reduction method works
Big Picture 2:   Show the prime reduction method works
Dots_2_1:  Show lemma 1
Dots_2_2:  Show lemma 2
Dots_2_3:  Show lemma 3
Dots_2_4:  Show lemma 4
Dots_2_5:  Show Theorem 1

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4 minutes ago, Zolar V said:

I simply can't help anymore. If I ever did.

For any natural number N, there exists a function G such that any N converges to a power of 2 over M iterations.

So now there's a DIFFERENT G for each n?

Edited by wtf

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2 hours ago, wtf said:

For any natural number N, there exists a function G such that any N converges to a power of 2 over M iterations.

So now there's a DIFFERENT G for each n?

Whether or not there is does not even matter. The function G that maps every N to 2 does the trick in 1 iteration.