Collatz Conjecture

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11 minutes ago, wtf said:

Ok good. But I'm still confused about G(139). We have 139 -> 140 -> 2^2 x 5 x 7. Now do you recurse each of 5 and 7 separately? Or recurse five copes of 7 separately? I found your earlier remark extremely confusing.

So here is where it gets a little tricky. You cant recurse 5,7 separately since they're multiplied together, but you can select which you want to recurse through.
Once you select which prime you want to recurse, you manipulate the product such that you have a copy of that prime you selected.

Suppose you select 7 then
5*7 = (5-1)7 + 7
and you can recurse through the first copy of 7 you have extracted.  Then recurse through all 5 copies of 7.    Unfortunately my gut is saying im missing something here or something isnt quite right.  I need to go find and look at all my notes again.    Something is a little off, but its also very late.    I'm gonna think about it and see if I can't remember whatever it is that's lurking in the shadows of my mind.

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20 hours ago, Zolar V said:

I'm american, though the problem with my grammar is likely the product of not talking very much to anyone else.

I didn't want to implicate anything, just curious. In fact I just finished a research paper joint with two japanese and one korean colleague. I had to play grammar police quite a bit, so it is getting more easy for me to predict where the pitfalls lie.

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I think as we work through this i'm having new ideas about it,  here is one of them:

maybe it would make sense to first state that any natural number can be written as a non distinct sum of primes.
E.G 3*7 = 3 +3+3... +3  or  7+7+7

The proof if we need to show it could probably be a straight forward one.
Suppose there exists a natural number  that is not a sum of primes.
by the fundamental theorem of arithmetic, n is either prime or a product of primes.
Case 1:  n is prime
if n  is prime then
n =  n  ;  lower index = 1, upper index = 1
Case 2: n is a product of primes
if n is a product of primes then
n = 2^r * p_1.. *p_i  =   ∑ p_i  ; lower index=1, upper index = 2^r *p_1 *p_2 *...p_(i-1)

Since n is either a sum of itself or a sum of primes given by the prime factorization then there does not exist a natural number that is not a sum of primes and all natural numbers greater than 2 are sums of primes.  I suppose this also says that any number is either a sum of itself or has a i number of equivalent sums of primes.

The purpose in stating that any natural number can be written as a sum of primes is to validate our G function on all natural numbers.

G is still G(p) = p + 1  ;  p is an element of Natural numbers and is prime.

Here i am saying you need 2^r *p_1 *p_2 *...p_(i-1) iterations of G on H(k) in order to add 1 to each prime p_i.
you would be left with each (p_i + 1) = 2^m * q  .  so you would have  2^r *p_1 *p_2 *...p_(i-1) many (2^m * q)
Note:   q does not divide (p_i + 1),  in fact q < p_i
I think the next step would be to rewrite (2^m *q) to a new sum of primes.  and then apply G to it again. and keep on rewriting and applying until you drill down to 2's

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10 hours ago, Zolar V said:

So here is where it gets a little tricky. You cant recurse 5,7 separately since they're multiplied together, but you can select which you want to recurse through.
Once you select which prime you want to recurse, you manipulate the product such that you have a copy of that prime you selected.

Suppose you select 7 then
5*7 = (5-1)7 + 7
and you can recurse through the first copy of 7 you have extracted.  Then recurse through all 5 copies of 7.    Unfortunately my gut is saying im missing something here or something isnt quite right.  I need to go find and look at all my notes again.    Something is a little off, but its also very late.    I'm gonna think about it and see if I can't remember whatever it is that's lurking in the shadows of my mind.

> Unfortunately my gut is saying im missing something here or something isnt quite right.

I'll stand by for clarification, because your exposition is pretty murky here. But what do you do if you're left with 3 primes or a million primes? What if you're left with 5x7x47x47x113 x 113 x 113? What now?

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It's good to see two members having such a cooperative discussion about a subject.

Here is a current discussion by someone else about much the same problems with latex that may help both of you.

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2 minutes ago, studiot said:

It's good to see two members having such a cooperative discussion about a subject.

Here is a current discussion by someone else about much the same problems with latex that may help both of you.

Evidently LaTeX does work on this site ... $e^{i \pi} + 1 = 0$

The delimiters are backslash-open-squarebracket and backslash-close-squarebracket.

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8 hours ago, wtf said:

> Unfortunately my gut is saying im missing something here or something isnt quite right.

I'll stand by for clarification, because your exposition is pretty murky here. But what do you do if you're left with 3 primes or a million primes? What if you're left with 5x7x47x47x113 x 113 x 113? What now?

I'm sorry WTF, I can't find all of my notebooks.  The best I can discern is this was one of the places I needed some good thought and collaboration on.  I think I naively applied G on none prime numbers.  Though later I did add the Remark at line 37.     It appears that what I did here was essentially rewrite a number as a sum of a prime number, and then that allowed G to be applied.

Let us take writing any natural number as a sum of primes.

$H(k) = \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i-1} } p_{i}$

I think here there exists a sum that has the least steps needed for G(p) to equal a power of 2

Suppose there are at least i number of sums for any natural number then these are our sums
$\sum_{i=1}^{2^r * p^1 *p^2 *... p^{i-1} } p_{i} = \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i} } p_{i-1} = ... =\sum_{i=1}^{ p^1 *p^2 *... p^{i} } 2^r$
and
$G_1^{2^r * p^1 *p^2 *... p^{i-1} } ( \sum_{i=1}^{2^r * p^1 *p^2 *... p^{i-1} } p_{i}) = 2^{m}$
$G_2^{2^r * p^1 *p^2 *... p^{i} } (\sum_{i=1}^{2^r * p^1 *p^2 *... p^{i} } p_{i-1}) = 2^{m}$
$G_i^ {p^1 *p^2 *... p^{i} } ( \sum_{i=1}^{ p^1 *p^2 *... p^{i} } 2^r ) = 2^{m}$

Where
$G_1 \leq G_2 \leq ... \leq G_i$

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Sorry before I dive into this ... Theorem 1 is retracted? We're done with the paper now? Please be clear.

Natural number as sum of primes? Wow you're way beyond anything in your paper. You need the argument to become more grounded and focussed, not more wild. You're halfway to the famous partition problem, how many ways can you express a given number as a sum of natural numbers. You will not get any kind of similar argument under control.

What is the above symbology supposed to be about? What's your claim, what are you proving, what is your conclusion? I get that you're unhappy you can't find your original argument and that your paper's cooked. But this latest isn't possible to read without a sensible explanation.

To clarify: You are giving up on explaining G(139) and you have a new argument now?

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I don't think Theorem 1 is wrong yet, but I think there is an important step missing between the application of Theorem 1 on primes to the application of Theorem 1 on all natural numbers.
I really wish I could remember, or find, why I deduced that if Theorem 1 was true for primes it was true for all natural numbers.

I believe my logic went something like this:
Step 1: Theorem 1 is true for Primes.
Step 2: Any number can be written as a product of primes.
Step 3: Any product of primes can be rewritten to be a number (a) + a prime (b)...   a +b.
Step 4: Since I have a prime (b) that stands alone,  I can apply Theorem 1,   Then that means Theorem 1 is true for all natural numbers.
Note:  The result on Theorem 1 is it reduces value of the primes in the output to be less than the input prime ( 2 < p_i < p) .

I think you are correct that the 3rd step isnt fully developed and it indeed cooks the goose.  If we can flesh out Step 3 then we can get back on track.
I also think at a minimum, this Prime Reducing function is ultimately why the collatz conjecture converges.  It's just a matter of proving step 3, and then showing how the collatz conjecture is simply an extension of the result.   Namely the odd function can be rewritten to be (2x) + (x+1).  and x+1 is our prime reducing function.

OK.  onward to fixing step 3!

" What is the above symbology supposed to be about? What's your claim, what are you proving, what is your conclusion? "
-
I believe a sum of primes is what i tried to state on line 40,  though I did not state it well.
- I claim any Natural number can be rewritten as at least a sum of a prime.  There are many sums for natural numbers and that number of sums is related to how many primes there are in the composite number, though we only need to look at just one sum at the moment.
- If we can prove that a sum of a prime is equivalent to a product of primes for any given natural number then we can apply our theorem 1 to that number.
- I conclude that the number of times we need to apply G to a natural number is equal to the product of primes of that natural number divided by the prime we are summing.
example:  5 * 7 * 47 * 47 * 113 * 113 * 113 = p

$\sum_{i=1}^{5 * 7 *47 *47*113} 113 = p = 5*7*47*47*113*113*113$
$n = \frac {5*7*47*47*113*113*113} {113}$
$G^n (p) = 2^m$

I believe my original assessment was applying G to the product of primes, which is not a valid move.   So I suggest we look at rewriting the product of primes as a sum of a prime. We will need to later look at which sum is the right sum for the least number of times we need to apply G.

Edit:  I hope you dont think im an idiot.

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Help me out here. The last thing I understood was theorem 1, right up until the recursion step. As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7.

Now you do SOMETHING with 5 and 7 but for several posts you've failed to clearly explain it. Your two most recent posts seem to be veering off into some new direction. You have to put what you're doing in context. Is this a new idea, are you trying to patch theorem 1, are you trying to explain G(139), or what?

What you want to do here is say LESS and not MORE. I got to 140 = 2^2 x 5 x 7 and asked what happens to 5 and 7, and you have gone off with two long posts that are not anchored to anything I understand of your argument. Try just answering me with one- or two-sentence posts till I have some idea what you're talking about.

I'm still at 2^2 x 5 x 7 and I've been there for two days now. Give me one clear sentence at a time as to what you are doing. Are you

* Explaining 5 and 7?

* Going off in a new direction with a new argument?

I just see no relation at all between what we agreed on a couple of days ago, and your most recent new direction.

I think you are correct that the 3rd step isnt fully developed and it indeed cooks the goose.

No, you have not even established theorem 1 yet. What happens with 5 and 7?

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17 hours ago, wtf said:

Help me out here. The last thing I understood was theorem 1, right up until the recursion step. As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7.

Now you do SOMETHING with 5 and 7 but for several posts you've failed to clearly explain it. Your two most recent posts seem to be veering off into some new direction. You have to put what you're doing in context. Is this a new idea, are you trying to patch theorem 1, are you trying to explain G(139), or what?

What you want to do here is say LESS and not MORE. I got to 140 = 2^2 x 5 x 7 and asked what happens to 5 and 7, and you have gone off with two long posts that are not anchored to anything I understand of your argument. Try just answering me with one- or two-sentence posts till I have some idea what you're talking about.

I'm still at 2^2 x 5 x 7 and I've been there for two days now. Give me one clear sentence at a time as to what you are doing. Are you

* Explaining 5 and 7?

* Going off in a new direction with a new argument?

I just see no relation at all between what we agreed on a couple of days ago, and your most recent new direction.

I think you are correct that the 3rd step isnt fully developed and it indeed cooks the goose.

No, you have not even established theorem 1 yet. What happens with 5 and 7?

"As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7. "
- the next step is to extract either 7 or 5.  if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.)

I'm sorry i desperately looked for my notebooks and work has been real busy.

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42 minutes ago, Zolar V said:

"As an example we have G(139) which is computed as 139 -> 140 -> 2^2 x 5 x 7. "
- the next step is to extract either 7 or 5.  if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.)

I'm sorry i desperately looked for my notebooks and work has been real busy.

No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure.

Still, just in case I want to try to read your most recent couple of posts, can you just tell me whether you are

* Explaining how to handle 5 and 7 in the computation of G(139); or

* Trying to patch theorem 1 in some other way; or

* Introducing an entire new argument.

Also, are we still following the paper, or has that been abandoned? I just need to understand the context of the last few posts. Once I asked you about 5 and 7, your posts went off in a completely different direction without context, so I'm confused.

> if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.)

Still makes no sense to me. The sums are not discussed in theorem 1. So you have not established theorem 1. In your paper you talk about sums AFTER theorem 1. And you haven't provided an explanation I can understand of how 5 and 7 are handled.

What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?

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35 minutes ago, wtf said:

No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure.

Still, just in case I want to try to read your most recent couple of posts, can you just tell me whether you are

* Explaining how to handle 5 and 7 in the computation of G(139); or

* Trying to patch theorem 1 in some other way; or

* Introducing an entire new argument.

Also, are we still following the paper, or has that been abandoned? I just need to understand the context of the last few posts. Once I asked you about 5 and 7, your posts went off in a completely different direction without context, so I'm confused.

> if we chose 7 then we need to extract it 2^2 *5 times, (hence why i started to talk about sums.)

Still makes no sense to me. The sums are not discussed in theorem 1. So you have not established theorem 1. In your paper you talk about sums AFTER theorem 1. And you haven't provided an explanation I can understand of how 5 and 7 are handled.

What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?

"No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure."
- I would hope so, Ive been sitting on this for almost a year.

I am attempting to explain how I handled 5 and 7 in the computation of G(139).  We are following the paper,  line 40 to be exact.

"What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"

extract, to take out one piece of a whole. Think about a number as an entire collection of collections of objects that are equivalent.   If we only consider the additive operation then the number of equivalent objects within a number is how many different numbers can add up to equal that number.   E.G. 7.   1+6 = 2+5 = 3+4 ... = 2+2+2+1 = ... 7
A whole is 7, a piece would be take out a 2 or a 4 or some other part of 7.

In our case our object is 2^2*5*7 which means you can extract a 2,5, or 7 and there are  2*5*7 , 2^2 *7, or 2^2*5 copies respectively.
If we extract 7, then we are extracting all 2^2*5 copies of 7.  So 7 +7+7+7+7+7+7+7+7+7+7+7+7+7+....+7
And we iterate through that sum of primes.

"As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"

-
Currently we select a prime and reconfigure the product of primes into a sum of that prime.  Then iterating through that on each prime.
The core concept of adding 1 to a prime and reducing the primes within it is the primary mechanism we want to chase.   On a very rudimentary level, the mathematics that I know, only allow you to do something like add to a particular value only if you can group it together.    Obviously you cannot just select random primes within a product to add 1, but you can do just that over a sum of primes.

Edited by Zolar V
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12 minutes ago, Zolar V said:

"No prob, take your time. I doubt anyone will beat you to Collatz and claim credit. You have a few more days at least, I'm sure."
- I would hope so, Ive been sitting on this for almost a year.

I am attempting to explain how I handled 5 and 7 in the computation of G(139).  We are following the paper,  line 40 to be exact.

"What do you mean "extracting" 7 20 times? This really is very unmotivated, it's not possible for me to follow your logic. As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"

extract, to take out one piece of a whole. Think about a number as an entire collection of collections of objects that are equivalent.   If we only consider the additive operation then the number of equivalent objects within a number is how many different numbers can add up to equal that number.   E.G. 7.   1+6 = 2+5 = 3+4 ... = 2+2+2+1 = ... 7
A whole is 7, a piece would be take out a 2 or a 4 or some other part of 7.

In our case our object is 2^2*5*7 which means you can extract a 2,5, or 7 and there are  2*5*7 , 2^2 *7, or 2^2*5 copies respectively.
If we extract 7, then we are extracting all 2^2*5 copies of 7.  So 7 +7+7+7+7+7+7+7+7+7+7+7+7+7+....+7
And we iterate through that sum of primes.

"As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"

-
Currently we select a prime and reconfigure the product of primes into a sum of that prime.  Then iterating through that on each prime.
The core concept of adding 1 to a prime and reducing the primes within it is the primary mechanism we want to chase.   On a very rudimentary level, the mathematics that I know, only allow you to do something like add to a particular value only if you can group it together.    Obviously you cannot just select random primes within a product to add 1, but you can do just that over a sum of primes.

> - I would hope so, Ive been sitting on this for almost a year.

Numerous scatalogical jokes omitted.

> extract, to take out one piece of a whole. Think about a number as an entire collection of collections of objects that are equivalent.

I am certain I have no conceivable idea what you are talking about, even after imagining various set-theoretic constructions of various classes of numbers.

For the third or fourth time in a row you are pointedly ignoring my request for context.

Currently we select a prime and reconfigure the product of primes into a sum of that prime.  Then iterating through that on each prime.

"reconfigure." What am I supposed to make of that? You still won't tell me how to iterate through 5 x 7.

I don't want to sound negative, but your last half dozen posts haven't made any sense and haven't addressed my concerns, nor to you seem to be making any progress in clarity. I am near the end of my ability to be helpful. I did some useful work in helping you to define G, but once you do the initial factorization of p+1, the rest of your idea is simply unexplained. I think we're at around three days with no progress at all.

The core concept of adding 1 to a prime and reducing the primes within it is the primary mechanism we want to chase.   On a very rudimentary level, the mathematics that I know, only allow you to do something like add to a particular value only if you can group it together.    Obviously you cannot just select random primes within a product to add 1, but you can do just that over a sum of primes.

Simply doesn't parse. Maybe I should stop posting. I have nothing to add.

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"As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?"
Lets select 103.  then we have 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 copies of 103 added together.
G(103+103+103+103+103+103+103+103+.......+103) = 103 +103+103 ....   +    (103+1)

$2^n * 7 * 7 * 7 * 101 * 101 * 103 * 103 * 103 * 103 = \sum_1^{ 2^n * 7 * 7 * 7 * 101 * 101 * 103 * 103} 103 /] Edited by Zolar V Link to comment Share on other sites 2 minutes ago, Zolar V said: "As I asked earlier, what do you do with 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103? How do you recurse down this configuration of primes?" Lets select 103. then we have 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 copies of 103 added together. G(103+103+103+103+103+103+103+103+.......+103) = 103 +103+103 .... + (103+1) I can't make heads or tails. I can't be helpful anymore. I can see you're multiplying and adding a bunch of numbers, but there doesn't seem to be anything for me to work with. I need to let this go. On the left hand side you have G of a composite number. We're already in perfect agreement that this is not defined. You can input a prime to G and recurse down a bit till your exposition falls apart, but you can NOT input a composite number to G. I have to let this go, I have nowhere to go with this. Edited by wtf Link to comment Share on other sites Our function G(p) = p+1. p = 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103 Which also equals 103 added together 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 times. if we use the sum, we can apply G to it. Resulting in G(p) = .... + (103+1) Link to comment Share on other sites 6 minutes ago, Zolar V said: Our function G(p) = p+1. So by definition, G(103 + ... + 103) isn't defined because the argument to the function is not prime. > p = 2^n x 7 x 7 x 7 x 101 x 101 x 103 x 103 x 103 x 103 No, that number is not prime. 'p' has denoted a prime since the beginning. If that's no longer true, you sure didn't mention it. I think instead of posting anymore, you should go back to the drawing board and try to work through your idea. Maybe someone else can offer suggestions. I'm out of ammo. Edited by wtf Link to comment Share on other sites 7 minutes ago, wtf said: So by definition, G(103 + ... + 103) isn't defined because the argument to the function is not prime. I'm sorry, I can't add anything. I have to let this go. ok then, I see the error. I will renew my efforts to clearly define a set of functions and the rules by which to use them. Cheers and thanks for all the help! Probably just be G( x+ h) = x + (h +1) ; where x is a natural number and h is prime. That would define the function well enough without losing the original idea behind a prime +1. Theorem 1 needs a complete rewrite. Edited by Zolar V Link to comment Share on other sites I will renew my efforts to clearly define a set of functions No, please, let's go back to the point where we were in agreement and work from there. I have a plan. We were in perfect agreement at a certain point then (I feel) you spun off in a million different directions. The expression that came to my mind was "thrashing," which is what a computer does when it's so busy managing virtual memory that it can't get any actual work done. It gets busier and busier but accomplishes nothing. Please don't take that as pejorative or "condescending." God it's not the first time I've been called that online. It's just the way I write. Or as Jessica Rabbit said, "I'm not bad, it's just the way I'm drawn." Believe it or not I'm trying to be helpful with those remarks. Because we're having TWO separate conversations. One is about the math. The other is about the exposition of the math. So I am not attacking your idea. I perfectly well stipulate that you might be Ramanujan and I'm damn well not Hardy. Maybe you're right that a smarter person than me could figure out what you're talking about. But on exposition, I am simply playing the part of a typical, somewhat dim, reader of mathematics. I read a line, try to figure out it, write down some simple examples. So when you write some exposition and I say you're thrashing, I am simply telling you that you have lost your reader. You have to make it much simpler to get through to the likes of me. Now here is what I posted before. Speaking of mathematical communication, we ALL have a lot to learn. I thought I was making a clear point, a meta-point, a big-picture point. But I forgot to make my point! So here it is. There isn't a convenient "function" for G. Just express it as an algorithm. For that matter, this is exactly how Collatz works! I can't think of a convenient way to express Collatz as a function, even though there is a function in theory. But in practice it's far better to work with it as a procedure or recipe. So if you would forget about calling G a function, and notating it as if it were a function, your exposition would improve dramatically and I wouldn't have to spend so much time yelling at you that G's not properly defined. I'm sure we're both tired of it by now. G is not a function. It's a procedure. Think about it that way and talk about it that way. Else we're doomed. Having said that, we are in agreement through step 3. You did at one point say that you "keep the power of 2 and only divide it out at the end." That is the first moment of murky divergence from our point of agreement. Not only that, in your recent posts you are clearly using the 2^n in some way along with the other primes. So this is important to you and I don't quite understand it. It seems to complicate things. Now ignoring the 2^n, whether you divide it out or include it in something later, we have the remaining primes. For example in the case of G(139), we have 139 + 1 = 140 = 2^2 * 5 * 7. Note that G(139) is our notation for "Input the prime 139 into the G recipe." It is not a function. But now that we've defined the notation, it's ok. As long as we only input primes!! Now my idea, which is evidently not what you have in mind, is then to recurse through the procedure on each of 5 and 7. So 5 + 1 = 6 = 2 x 3 and 3 + 1 = 2^2. And 7 + 1 = 8 = 2^3. So we claim (although I have not yet formally proved) that each subtree terminates at a power of 2. But that is my idea. (*) At that point, you tried to explain to me that you do something with five 7's, and that's where I lost the thread. And in an attempt to explain this to me, you're (FROM MY VIEWPOINT, this is not meant to be pejorative) spiraling out of control. So if you see any hope of going back to our most recent point of perfect agreement, that might be helpful. (*) I'm nervous about this. I can see it's a strong induction ... having proved it for all primes below a certain point, it's true for larger ones because p+1 has factors that are smaller than p. On the other hand, it's a funny induction. The induction step never depends on any of the values that have gone before. I am starting to wonder if it's true. For example ... suppose we know that 3, 5, 7, 11 all go to 2^n. What if we then put in a p such that p+1 only has prime factors of 13 and above? There's no actual induction step, you have to look at every prime by hand. I am starting to wonder if there might in fact be a counterexample. Your argument depends on p_i < p so I believe this impacts your argument. Edited by wtf Link to comment Share on other sites 2 hours ago, wtf said: Now my idea, which is evidently not what you have in mind, is then to recurse through the procedure on each of 5 and 7. So 5 + 1 = 6 = 2 x 3 and 3 + 1 = 2^2. And 7 + 1 = 8 = 2^3. So we claim (although I have not yet formally proved) that each subtree terminates at a power of 2. But that is my idea. (*) Alright, I can work with that. I didn't know that is what you wanted. You are exactly right then for step 3. all you are doing is adding 1 to each prime, one prime at a time per iteration. in programming it would look something like this: P = prime While( P =/= 2^r ){ P +1 = 2^r P_1 *P_2 ... foreach (Prime in (P+1)) { Call (while P =/= 2^r) } } it's a nested loop. Quote Your argument depends on p_i < p so I believe this impacts your argument. -very true that it does. but can you think of any number that is even, prime, and greater than 2? the largest prime in an even number must be half the size of the even number, Example: P is our first prime, if P+1 = 2^r * k then \[ \frac {P+1} {k} = 2^r$

Now suppose there is a prime factor in k that is larger than P,  then if we were divide k we would not get a positive even number.   I believe we also know that r >= 0

Quote

keep the power of 2

yeah, the powers of 2 we collect at the end were a very good indication of the number steps we needed to take.

Edited by Zolar V
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10 minutes ago, Zolar V said:

Alright,  I can work with that.  I didn't know that is what you wanted.   You are exactly right then for step 3.  all you are doing is adding 1 to each prime, one prime at a time per iteration.

-very true that it does.  but can you think of any number that is even, prime, and greater than 2?
the largest prime in an even number must be half the size of the even number,
Example:  P is our first prime,
if  P+1  = 2^r * k
then  /[ \frac {P+1} {k} = 2^r /]

Now suppose there is a prime factor in k that is larger than P,  then if we were divide k we would not get a positive even number.   I believe we also know that r >= 0

The claim that every branch terminates in a power of 2 is obviously true but a formal proof eludes me. I haven't spent much time on it. Something about the induction bothers me but I can't think of how a counterexample would work. I've decided to put the issue aside for now.

So let's just say I believe every subtree terminates in a power of 2. But this is different than your "adding" idea. Is it sufficient to get your argument off the ground? Either you have to accept that my "tree" method serves your purposes; or else you have to explain to me your adding idea, which I confess makes my eyes glaze.

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3 minutes ago, wtf said:

The claim that every branch terminates in a power of 2 is obviously true but a formal proof eludes me. I haven't spent much time on it. Something about the induction bothers me but I can't think of how a counterexample would work. I've decided to put the issue aside for now.

So let's just say I believe every subtree terminates in a power of 2. But this is different than your "adding" idea. Is it sufficient to get your argument off the ground? Either you have to accept that my "tree" method serves your purposes; or else you have to explain to me your adding idea, which I confess makes my eyes glaze.

I added a little extra to my last post that i thought of.

Lets accept and work with the tree method first.  If this gets nowhere then we can explore the adding idea.  Though even if the tree method does work, I think it's also worth exploring the adding idea.

Just now, Zolar V said:

Something about the induction bothers me but I can't think of how a counterexample would work. I've decided to put the issue aside for now.

I'm not surprised,  being inductive would indicate you can systematically build primes.   I mean technically you can,  but this would be for all primes not special primes.

Here is an odd thought:
Suppose this idea was true. Some inverse equation/algorithm would build only primes.  Each equation we have now that creates some subset of primes would somehow be a part of or equivalent to that equation/algorithm

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21 minutes ago, Zolar V said:

I added a little extra to my last post that i thought of.

Lets accept and work with the tree method first.  If this gets nowhere then we can explore the adding idea.  Though even if the tree method does work, I think it's also worth exploring the adding idea.

I'm not surprised,  being inductive would indicate you can systematically build primes.   I mean technically you can,  but this would be for all primes not special primes.

Here is an odd thought:
Suppose this idea was true. Some inverse equation/algorithm would build only primes.  Each equation we have now that creates some subset of primes would somehow be a part of or equivalent to that equation/algorithm

There IS a function that produces exactly the primes, in order. It's the function f that maps the natural number n to the n-th prime p_n.

This function exists. There's an algorithm for it:

For each natural number n:

If n is prime, print it, otherwise don't.

It's slow but we don't care about that, all we care is that the function exists. It can be coded as an algorithm, that is, as a Turing machine. As such, it inputs a natural number and outputs the corresponding prime. Note that we'll need a subroutine that determines if a number is prime. For that we just use trial divisors.

Also note that if we just want, say, f(45545434343), we just run it that far and print out only the value we want.

There is no question that such a function exists, that it's computable, and that a brand new programming student would be able to write within a few weeks.

The problem is, we don't have a nice simple closed-form expression for it. We have to brute-force it. So the question isn't the existence of such a function. It's that we don't have an expression for it.

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2 minutes ago, wtf said:

There IS a function that produces exactly the primes, in order. It's the function f that maps the natural number n to the n-th prime p_n.

This function exists. There's an algorithm for it, the well-known sieve of Erotosthenes. It's slow but we don't care about that, all we care is that the function exists. It can be coded as an algorithm, that is, as a Turing machine. As such, it inputs a natural number and outputs the corresponding prime.

There is no question that such a function exists, that it's computable, and that a brand new programming student would be able to write within a few weeks.

The problem is, we don't have a nice simple closed-form expression for it. We have to brute-force it. So the question isn't the existence of such a function. It's that we don't have an expression for it.

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

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