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Collatz Conjecture

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3 minutes ago, Zolar V said:

That would be all from the first equation.  the next part is to take the even equation from the collatz conjecture and apply it however many times it takes to reduce each 2 into a 1.
The purpose of G(p) = p + 1 is to reduce each prime to 2.

Please continue working one of your examples to completion. You had G(7) = 8, and 8 = 2^4. Ok what now? I hope there's more to this than noting that 2 is always a factor of 1 plus an odd prime.

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1 minute ago, wtf said:

Please continue working one of your examples to completion. You had G(7) = 8, and 8 = 2^4. Ok what now?

As stated awkwardly:  For any prime p > 2 the function G(p) = p + 1 (converges) to 2 after n iterations.
So G(7) = 8  and 8 = 2^3.  so each prime in 8 is 2.

A better example i think:
G(11) = 12 = 2*2*3  = (2*2 -1) *3 + 3
substep:  G(3)  =  4  = 2*2
So
G(G(11)) = (2*2 -1) *3 + G(3)
G(G(11)) = (2*2 -1) *3 +  2*2

The next step is to take out another prime and apply the function again
(2*2 -1) *3 = (2*2 -1) + (2*2 -1) + (2*2 -1)  or ((3-1)(2*2 -1)) + (2*2 -1)  = ((3-1)(2*2 -1)) + 3  = 2*3 + 3
G(G(G(11)) = 2*3 + 3 + 2*2   = 2*3 + G(3) + 2*2 = 2*3 + 2*2 +2*2  = 2*3 + 2*2*2
you can then rewrite 2*3 as (2-1)*3 + 3
G(G(G(G(11))) = (2-1)*3 + 3 + 2*2*2  = (2-1)*3 + G(3) + 2*2*2
and of course (2-1)*3 = (1)*3 = 3
G(G(G(G(G(11))))  = 3 + 2*2 + 2*2*2 = G(3) + 2*2 + 2*2*2   =   2*2 + 2*2 + 2*2*2
so, 5 iterations cause G(11) to be a series of primes where each prime is 2.

The purpose of G(p) = p + 1 is to reduce each prime within the composite number to 2.   once we have our 2's then the application of n iterations of the second function H(x) = x/2 , results in 1. after n iterations of course.

What this paper is trying to state is that G(p) = p + 1 is embedded in the odd function of the collatz conjecture.  Since G(p) over some m iterations results in a 2^k number, it's clear that dividing by 2^k equals 1.   thus the collatz conjecture always results in 1 after some m iterations.

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2 hours ago, Zolar V said:

As stated awkwardly:  For any prime p > 2 the function G^n(p) = p + 1 (converges) to 2 after n iterations.  (edited here to say G^n , instead of G)

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1 hour ago, Zolar V said:

On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof.

On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what.

Can you please supply the missing claim you're trying to prove?

Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general.

You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define.

Out of curiosity, are you aware that the function notation G(p)  = p + 1 means exactly what it says, and NOT some convoluted function you have in your mind but have never clearly described?

After all, @studiot was charitable enough to suggest that you are accustomed to the use of '=' as an assignment operator in some programming languages. That's a sensible interpretation in general, but it hardly explains how the notation G(p) = p + 1 is being used to denote some function G whose definition is not possible to discern from your exposition.

I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n.

ps -- Oh I see what the second "proof" is. It's the proof of the OUTER assertion. You nested the lemma inside the main thing you're trying to proof. Not a bad idea actually but really not how it's done. A little like an inner class in Java, is that where you got the idea? Way too confusing for readers. Next time do Lemma 1, Proof of Lemma 1. Main claim, Proof of main claim. Your readers will thank you. Take that as constructive criticism. The main issue is to untangle the definition of G.

pps -- Ok more. I'm slogging through your exposition trying to understand what G is, and you have a little problem on line 29 and a BIG BIG problem on line 30.

On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case.

On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k.

You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right?

You need to fix these errors and see if you still have an argument left.

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ppps -- No, you don't have an argument left. On line 32 you apply the "prime reducing function" to a list of POWERS of primes, VERY FEW OF WHICH are prime in the general case. Your proof is busted right there.

I'll also mention that line 33 doesn't follow from anything that came before EVEN IF we allow that prime powers are prime (which of course they are not)! And what is the 'x' on the right side of line 33?

So that's the error. Your Lemma 1 shows that any P_k < P (once you've handled the case where p + 1 is a power of 2 so that there are no P_k's); but the rest of your proof requires that all POWERS of p_k are less than P, and prime to boot. Both those assertions are false.

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20 hours ago, wtf said:

On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof.

On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what.

Can you please supply the missing claim you're trying to prove?

Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general.

You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define.

Out of curiosity, are you aware that the function notation G(p)  = p + 1 means exactly what it says, and NOT some convoluted function you have in your mind but have never clearly described?

After all, @studiot was charitable enough to suggest that you are accustomed to the use of '=' as an assignment operator in some programming languages. That's a sensible interpretation in general, but it hardly explains how the notation G(p) = p + 1 is being used to denote some function G whose definition is not possible to discern from your exposition.

I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n.

ps -- Oh I see what the second "proof" is. It's the proof of the OUTER assertion. You nested the lemma inside the main thing you're trying to proof. Not a bad idea actually but really not how it's done. A little like an inner class in Java, is that where you got the idea? Way too confusing for readers. Next time do Lemma 1, Proof of Lemma 1. Main claim, Proof of main claim. Your readers will thank you. Take that as constructive criticism. The main issue is to untangle the definition of G.

pps -- Ok more. I'm slogging through your exposition trying to understand what G is, and you have a little problem on line 29 and a BIG BIG problem on line 30.

On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case.

On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k.

You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right?

You need to fix these errors and see if you still have an argument left.

Great critique!  I didn't design the proof with any sort of programing language in mind.  However, I typically think in that sort of nested fashion.
So lets get started:

" On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof.  "
-Though the result is obvious and trivial, I think it is worth mentioning since it is the underlying concept to why adding 1 to a prime causes the resulting primes contained in the composite number to be less.

" On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what. "
- This is the proof of the G(p) = p +1,  but I believe you figured that out.

" Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general.  "
-I think a note here that explains why we can still apply the function to a composite number would work.    The reason why we can iterate G(p) on any integer is because every integer is either prime or a composite of primes.  The latter needs a little bit of work, but we can extract a prime out such that we have a number + a prime.  Like stated above, E.G. 7*3 = (3-1)*7+ 7

" You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define. "
-G(p) = p+1 is what I called the prime reducing function.

"I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n.
-I suppose a better way to write that would be G^n (p) = 2^m  , since after n iterations you get a 2^m number.  I mean you're really only adding 1 a number of times until you reach a 2^n.  its rather trivial if you think about it like that.

"On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case. "
-Fair enough,  but if p+1 happens to be a power of 2 then the problem is done.  We're only trying to create a string of 2's, so then we can divide by 2 however many times to get 1.  We only do this because that is part of the collatz cojecture.

" On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k. "
-You are both correct and wrong.  I shouldn't have written the notation with arbitrary powers of P_k,  I should have written it as a fully expanded string of powers.  3^3 should be 3*3*3...   We are attempting to handle individual primes, one at a time.  Not some grouping of primes in a shorthand notation.

" You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right? "
- I believe the prior answer to your prior question also solves this question.   by definition of the fundamental theorem of arithmetic each P_k is prime.

" ppps -- No, you don't have an argument left. On line 32 you apply the "prime reducing function" to a list of POWERS of primes, VERY FEW OF WHICH are prime in the general case. Your proof is busted right there. "
- I believe rewriting the composite number as a series of primes with a power of 1 solves your question here.  Very similar to the above two questions.

" I'll also mention that line 33 doesn't follow from anything that came before EVEN IF we allow that prime powers are prime (which of course they are not)! And what is the 'x' on the right side of line 33?  "
- Wrong notation here.  In my mind "x" was the first prime we punched into the problem.  so it should have been written as "p" not "x"
Line 33 is supposed to show that for each iteration of G(p), the resulting primes are less than p and also less than the last iteration of G(p).
E.G  for the first iteration, let our p = 5 .
G(5) = 6
rewrite 6 as 2*3, then rewrite it as (2-1)*3+3.  (again the purpose of the rewrite is to show that we can get a prime number to apply G(p) to.
So G(5) = 3 + 3
Then the next iteration is G(3) = 4 .
so G(5) = G(3) + G(3)   (there are two iterations here, since there are two 3's we're reducing)
So after 3 iterations we reach a 2^n number and we are done with G(p).
So   2 < 3 <5

Note:  I don't believe your questions yet disprove that this isn't the solution to the collatz conjecture.  They do help enormously to clarify the logical jumps I made.
In a simple sense, adding 1 to a number an arbitrary number of times such that you get a 2^m number is exceptionally trivial and intuitive.   of course if I add 1 enough times I will encounter an integer that is a power of 2.  But this result is why the collatz conjecture itself converges to 1.

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Ok this is way too long but what the hell. Hope you find a nugget of wisdom in here. If you reply, no point replying paragraph-by-paragraph since that will make it even longer. Bottom line let's figure out how to define G. Your exposition is seriously stuck there.

And at my end I'll make another run at your paper now that you've explained the prime products.

> Great critique!  I didn't design the proof with any sort of programing language in mind.  However, I typically think in that sort of nested fashion.

I have to say I was charmed by the idea. Almost like indenting a block of code.

Theorem 1: blah blah blah
Lemma 1.1: such and so
Proof: blah blah
Proof: blah blah blah blah.

It really makes perfect sense. Unless you want to use the same lemma more than once I guess. For what it's worth, math isn't written this way so it's better to do things the standard way.

By the way as I go I'm making stylistic remarks. One of the problems you're having is that you don't yet know how to write a clear mathematical argument. I'm doing my best to engage with your ideas and mentally filling in the blanks in your notation. But at some point the notation gets so garbled that it expresses no idea at all. You have something in your head, but I assure you that you have NOT expressed the idea in your paper.

Reading ahead, at the end of your post you write: "I don't believe your questions yet disprove that this isn't the solution to the collatz conjecture."

I quite agree. It's perfectly possible that you have a valid proof in your head. I can only tell you that you have not committed one to paper.

And that more importantly, it's not that your ideas are clearly wrong. It's that they are so unintelligible as to be, in the famous words of Wolfgang Pauli, "Not even wrong!"

I hope you will take these comments in the spirit of sharp but constructive criticism. You need to make an effort to write a much clearer and cleaner mathematical exposition.

Not on my account, I'm just some guy on the Internet. But your paper says it's been sent to the Journal of Number Theory. How are they expected to react when they skim down the page and see that (P_i)^(n_i) is "prime by definition?" Your case is lost at that moment.

So for your own interest in promoting your proof to the world, I urge you to rewrite your paper. Define every term and variable, use consistent notation throughout, and pretend you're some hapless journal editor who has a pile of papers from established mathematicans, and another pile from people he's never heard and who don't have Ph.D.'s in math. The papers in that latter file will get very short shrift. Bad exposition will get you roundfiled very quickly.

So. Onward. Looking forward, I did understand your clarififaction of the prime products, and I'll make another run at your paper with that understanding. But at some point I'm going to run out of steam. I really think you need to go to version two on this paper. What you have isn't enough to express an actual mathematical idea, no matter what is in your head. Which as far as anyone knows, is a valid proof of Collatz that will make you famous tomorrow morning. I can't disprove the possibility; for the simple reason that your argument as written is incomprehensible.

> " On line 21 you have Lemma 1, an extremely trivial result for which you supply a proof.  "
-Though the result is obvious and trivial, I think it is worth mentioning since it is the underlying concept to why adding 1 to a prime causes the resulting primes contained in the composite number to be less.

You're pointing out that if p is a prime, then p+1 is divisible only by primes strictly smaller than p. And that if you factor out all the 2's, then every remaining prime factor of p+1 is greater than 2.

This is perfectly true. It's so trivial as to not require proof. The fact that you wrote it out as a lemma and then laboriously proved it is a credibility point against you. Someone reads that and thinks, "This person believes this needs proof. There is no way they proved Collatz."

Just leave it out.

> " On line 28 immediately following your end-of-proof square, it says, "Proof" and then begins a long involved proof ... of something, we cannot tell what. "
- This is the proof of the G(p) = p +1,  but I believe you figured that out.

I truly thank you for reading my post in its entirety instead of snapping back line by line. I wish everyone would do that everywhere. I'm often guilty myself.

> " Also please clearly define what G is. You say G(p) = p + 1 and then you want to iterate it, but clearly you CAN'T ITERATE THIS because p + 1 is not prime in general.  "
-I think a note here that explains why we can still apply the function to a composite number would work.    The reason why we can iterate G(p) on any integer is because every integer is either prime or a composite of primes.  The latter needs a little bit of work, but we can extract a prime out such that we have a number + a prime.  Like stated above, E.G. 7*3 = (3-1)*7+ 7

Every time I ask you what G is, you respond with a convoluted paragraph and you never define G. You still have not defined G.

Regarding notation, I think I have the right to expect you to know and follow the standard notation for functions taught at the high school level everywhere in the world.

OUTPUT = f(INPUT).

If you say f inputs a prime, and p is prime, then f is not defined for p+1. Like I say I'm willing to cut you a lot of notational slack in order to understand your idea. But on this we really have a problem. G(p) = p + 1 has the same meaning for every math student and physical scientist in the world. G(7) = 8, G(8) = 9.

Someone earlier mentioned that you may be in need of the concept of a recurrence relation. This is a way of defining a function of the sort you have. For example here is how we can define the factorial function:

f(0) = 1

f(n) = n x f(n-1)

You can see that if you plug in 1 for n, then 2, and so forth, you always come down to the "base of the recursion" of n = 0, and the computation terminates. The idea is intimately related to how loops and recursion work in programming languages, and also the principle of mathematical induction in number theory.

You might have a go at defining G in this manner.

The paragraph you wrote is simply a massive red flag for anyone honestly trying to read your paper. You say G is defined on primes, and then you define G(Gp)) by waving your hands at the prime factors of p+1, but you never actually tell us how G is defined. I really hope you will take this to heart because you simply have not defined G and until you do, your paper can never get anywhere.

> " You've explained this by saying that G is NOT actually the function p+1, but is rather your "prime reduction function" or some such, which you never actually seem to define. "
-G(p) = p+1 is what I called the prime reducing function.

I truly do get that you call G the prime reducting function. In fact I've figured that much out. What I don't know is how G is defined. And in particular, why you repeatedly say G inputs only primes, then immediately claim to compute G(G(p)). That's a logical contradiction. Another HALT instruction for the poor reader of your paper.

> "I'll add that G^n(p) = p + 1 hardly solves the problem, since the right hand side is entirely independent of n. "
-I suppose a better way to write that would be G^n (p) = 2^m  , since after n iterations you get a 2^m number.  I mean you're really only adding 1 a number of times until you reach a 2^n.  its rather trivial if you think about it like that.

G^n (p) = 2^m? Well the right side is not only independed of n, it's now indepent of p as well. So it's a constant function. G^3(47) = 2^m, and G^154(p) = 2^m. G^anything(p) = 2^m. And you didn't say what m is!!

I simply have to draw a line. You either know how to use function notation, or else we need to talk about that before we talk about anything else. I can cut you a lot of slack on notation but not on this. What you wrote is meaningless and demonstrates a lack of familiarity with high school math past algebra I. That is not a good look for someone claiming to have solved a famous open problem.

Again I hope I'm not piling on. I'm being emphatic in the hopes of being helpful. Your notation is killing your argument because it makes no sense.

> "On line 29 you say that if p is prime, then p+1 can be factored as 2^q x product((P_i)^n_i). But if p + 1 happens to be a power of 2, such as 7 + 1 = 8 = 2^3, then none of the other primes P_1, P_2, etc., exist. I'm not sure if this is relevant to your argument but you should definitely handle this case. "
-Fair enough,  but if p+1 happens to be a power of 2 then the problem is done.  We're only trying to create a string of 2's, so then we can divide by 2 however many times to get 1.  We only do this because that is part of the collatz cojecture.

Yes perfectly well agreed, once we get to a power of 2 we're done. Yet what you originally wrote was wrong as stated. You need to explicitly call out this case so that we know you're paying attention to your own work!

> " On line 30 you have a flat out error though. You say that (P_k)^n_k < P. This is of course false. You have indeed proved that P_k < P, but that inequality does not hold for some arbitrary integer POWER of P_k. "
-You are both correct and wrong.  I shouldn't have written the notation with arbitrary powers of P_k,  I should have written it as a fully expanded string of powers.  3^3 should be 3*3*3...   We are attempting to handle individual primes, one at a time.  Not some grouping of primes in a shorthand notation.

" You compound the error on lines 31-32 by saying that (P_k)^n_k is prime. In fact you say it is "prime by definition." Of course for n_i > 1 it is not prime, it's a power of P_k. Right? "
- I believe the prior answer to your prior question also solves this question.   by definition of the fundamental theorem of arithmetic each P_k is prime.

Ok I accept your explanation. It's on me now to reinterpret your paper with this new understanding of what you are trying to say.

On your part, have you considered rewriting the paper from scratch in the light of some of these comments from me and others?

> " ppps -- No, you don't have an argument left. On line 32 you apply the "prime reducing function" to a list of POWERS of primes, VERY FEW OF WHICH are prime in the general case. Your proof is busted right there. "
- I believe rewriting the composite number as a series of primes with a power of 1 solves your question here.  Very similar to the above two questions.

Yes I agree. However, that's why I originally mentioned the following point ...

> " I'll also mention that line 33 doesn't follow from anything that came before EVEN IF we allow that prime powers are prime (which of course they are not)! And what is the 'x' on the right side of line 33?  "
- Wrong notation here.  In my mind "x" was the first prime we punched into the problem.  so it should have been written as "p" not "x"
Line 33 is supposed to show that for each iteration of G(p), the resulting primes are less than p and also less than the last iteration of G(p).

What I saw was that you established your facts about the non-2 prime factors of p+1, then waved your hands that since each factor is prime you can recurse your way down to your conclusion. But I did not see a proof, just a vague intution that there's a recursion to be had. Without a clear definintion of G, probably a recursive one, you won't have a proof.

That's why I mentioned that. I anticipated that even if I ALLOWED that p^n is a prime (!!!!) your argument STILL doesn't work. You glossed over every bit of detail you'd need to make your case.

> E.G  for the first iteration, let our p = 5 .
G(5) = 6
rewrite 6 as 2*3, then rewrite it as (2-1)*3+3.  (again the purpose of the rewrite is to show that we can get a prime number to apply G(p) to.
So G(5) = 3 + 3
Then the next iteration is G(3) = 4 .
so G(5) = G(3) + G(3)   (there are two iterations here, since there are two 3's we're reducing)
So after 3 iterations we reach a 2^n number and we are done with G(p).
So   2 < 3 <5

I still don't get it. Let's nail down what G is, maybe we can figure out the proper recursive definition needed to make the rest of this work.

"... we can get a prime number to apply G(p) to." Oh my. G(p) is the value of the function G at the input value p. It's not some other function. Or is G(p) some other function? That's legal too, functions can output other functions, for example the differntiation operator in calculus. High school function notation, please!

I don't mean to be impolite but I need to ask this. Are you misusing function notation because you don't know it? Or because you aren't familiar with the recursive definition of G you'd need to give and are hoping we'll understand what you mean? If the latter, let's fix that. If the former, let's talk about function notation before we do anything else.

> Note:  I don't believe your questions yet disprove that this isn't the solution to the collatz conjecture.

I have certainly proved that your exposition doesn't prove Collatz, since it's not a mathematical argument at all. On the other hand, I don't know what's in your head, and can't definitively say you haven't proved Collatz. If the latter, once you rewrite your paper and become world famous, maybe you'll give me credit for helping you fix your notation!

> They do help enormously to clarify the logical jumps I made.

Good to know. Now I don't feel so bad being so critical. I'm curious to understand your idea and your exposition is in the way so I'm trying to help you clarify the exposition.

> In a simple sense, adding 1 to a number an arbitrary number of times such that you get a 2^m number is exceptionally trivial and intuitive.

(Off topic) It's nothing of the sort. It's very deep. Suppose I give you the Peano axioms that say whenever you have p you can make p+1. I define addition as repeated "plus-one"-ing, multiplication as repeated addition, and exponentiation as repeted multiplication. I give you X = 2^2^2^2^2^2^2^2 where the operations bind right to left, so that this is a very very big number. If all you have is the plus-one operation, how do you know when you're reached X? It's a deep question.

>   of course if I add 1 enough times I will encounter an integer that is a power of 2.  But this result is why the collatz conjecture itself converges to 1.

Well that's a handwavy argument that doesn't convince me.

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16 hours ago, wtf said:

Ok this is way too long but what the hell. Hope you find a nugget of wisdom in here. If you reply, no point replying paragraph-by-paragraph since that will make it even longer. Bottom line let's figure out how to define G. Your exposition is seriously stuck there.

And at my end I'll make another run at your paper now that you've explained the prime products.

A couple notes:  I stumbled on this solution, that i hope we can both work through so you can see it too, while sitting at work.  I had a sudden flash of insight.
It might be noteworthy that my school career has been anything but standard.   I was home schooled till half way through 9th grade and then attended public school from 9th to 12th,  during my time being home schooled i had a smattering of public attendance as well.  4 months here, 2 months there.  After high school I joined the airforce and after that I attended college. In college I earned my B.S. in science with two supporting minors; mathematics and biology.   In mathematics I took a number of classes, but what really piqued my interest was abstract algebra.  Though I technically had the prerequisite (311W, which teaches you proofing and logic)  for it, the class was supposed to be the capstone math course.  I had to drop it since my proofing skills were subpar and I didnt want to risk failing.  Typically a math major would take that course after a series of 300 -400 level advanced math classes that focused on proofing.  My teacher noted that my proofs were basically written backwards.
Here is one math research project I did: https://drive.google.com/drive/folders/0B_B_IgMx0w_3cG5LZjBQaUZZYkU?usp=sharing

I developed a general formula (instead of stirlings approximation)  for a certain set of factorial divided by factorial problems.  There were some limitations that I hadn't quite explored by the end of the research project.

Anyways I digress,  my point is that I may not have a clear understanding of a lot of math concepts simply because my schooling was frequently interrupted.  I believe I understand a function, but I may not know how math would define a recurrence relation.  I do see what you're saying now about defining G(p),  I think the definition you're looking for is a higher order recurrence relation.   It seems I am defining one function on primes and then not defining the recurrence relation of that function on all integers.

" G^n (p) = 2^m? "
-  well you see, in my mind im saying that after n iterations of G(p) the final result will be some number that is a power of 2.  The number of iterations that cause that aren't related to the power of 2.

" Yes perfectly well agreed, once we get to a power of 2 we're done. Yet what you originally wrote was wrong as stated. You need to explicitly call out this case so that we know you're paying attention to your own work "
- Sure sure,  I can totally treat both cases explicitly.

" our part, have you considered rewriting the paper from scratch in the light of some of these comments from me and others?  ... ... submitted to number theory "
- Yes, I do entirely want to rewrite it to be bulletproof, if the proof actually stands up.  That is the purpose of wanting to work with a real mathematician.   I didn't submit it to number theory,  I added that because I was following a template and also needed to remind myself that I think number theory is my target audience.  Basically it was a mental note

" What I saw was that you established your facts about the non-2 prime factors of p+1, then waved your hands that since each factor is prime you can recurse your way down to your conclusion. But I did not see a proof, just a vague intution that there's a recursion to be had. Without a clear definintion of G, probably a recursive one, you won't have a proof. "
- Now that I think i understand what definition of G you're looking for we can make some headway and I would agree.

"...

So after 3 iterations we reach a 2^n number and we are done with G(p).
So   2 < 3 <5

I still don't get it. Let's nail down what G is, maybe we can figure out the proper recursive definition needed to make the rest of this work. .... "
- So I thought I was defining G as a function in just the same way that the original collatz conjecture had defined their odd function contained within the piecewise function.
So  G(p) = p + 1 ,  let p = 5.  then G(5) = 5 +1.   So iteration 1 is G(5) = 6.
Iteration 2 is taking the result of iteration 1, reworking the result such that you have a prime number and then adding 1 to that prime.
So iteration 2 is G(6) = 3 +(3 +1  )
Iteration 3 is adding 1 to the next prime number from iteration 2.   G(3+ 4) =(3 +1)  + 4
Iteration of G(p) is taking the prior result and inputting that as p.
A better defined recursive relation would be Hk = Hk-1  +1 .   As i said before,  I was trying to follow the original notation and thought I could define my recursive function in just the same way.

">   of course if I add 1 enough times I will encounter an integer that is a power of 2.  But this result is why the collatz conjecture itself converges to 1.

Well that's a handwavy argument that doesn't convince me."

Yes it is handwavy but true.  Just think about this.  Take the  largest number that is a power of 2 and add 1 however many times it takes to reach the next power of 2.  Since integers are infinite you would always reach another one.

think about it this way.  let 2k be the kth power of two.  then 2k - 2k-1 = n .  n is the number of times you need to add 1 to  2k-1  to reach  2

Example:  2^4  - 2 ^3 = 8    ,   2 ^3  + 8(1) is 2^4
multiplication is repeated addition of course. so 8 is the number of times you need to add 1 to go from the lower power of two, to the next power of two.

" Good to know. Now I don't feel so bad being so critical. I'm curious to understand your idea and your exposition is in the way so I'm trying to help you clarify the exposition. "
- be critical.  I'm trying to craft the best proof I can.  I did get irritated earlier since i read the first comments as being condescending.  They have since changed tone and thank you for that.

Edited by Zolar V
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Are you Asian, maybe Japanese or Korean? It might help to understand your explanations. I notice you are not using English grammar when you write  "integers are infinite", because integers and any objects that are bijective with integers are pretty much the central examples of objects that are not infinite. Whereas in Japanese you might say  多い人がいます。Ooi hito ga imasu (people are many) to mean that in some place there are numerous people present. And not to mean that every person is numerous. In usual mathematical English language you would have to say "the set of integers is infinite", to express what you probably intended to express. Not that it would help your reasoning, since the set of even integers is infinite as well, and what you said is true would be false in that case.

Edited by taeto
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9 minutes ago, taeto said:

Are you Asian, maybe Japanese or Korean? It might help to understand your explanations. I notice you are not using English grammar when you write  "integers are infinite", because integers and any objects that are bijective with integers are pretty much the central examples of objects that are not infinite. Whereas in Japanese you might say  多い人がいます。Ooi hito ga imasu (people are many) to mean that in some place there are numerous people present. And not to mean that every person is numerous. In usual mathematical English language you would have to say "the set of integers is infinite", to express what you probably intended to express. Not that it would help your reasoning, since the set of even integers is infinite as well, and what you said is true would be false in that case.

He wrote his full name and address in .pdf..

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23 minutes ago, taeto said:

Are you Asian, maybe Japanese or Korean? It might help to understand your explanations. I notice you are not using English grammar when you write  "integers are infinite", because integers and any objects that are bijective with integers are pretty much the central examples of objects that are not infinite. Whereas in Japanese you might say  多い人がいます。Ooi hito ga imasu (people are many) to mean that in some place there are numerous people present. And not to mean that every person is numerous. In usual mathematical English language you would have to say "the set of integers is infinite", to express what you probably intended to express. Not that it would help your reasoning, since the set of even integers is infinite as well, and what you said is true would be false in that case.

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

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6 hours ago, Zolar V said:

That is the purpose of wanting to work with a real mathematician.

This part I don't understand. No professional mathematician is going to read a paper claiming to have solved a famous open problem in which the author doesn't know basic function notation and can't write a proof. It's like applying for a job as a piano player and admitting that you can't read music or play the piano, but you have a lot of great tunes in your head. They'll tell you to come back when you can play. "I don't want to practice my scales, I just want to bang the groupies."

Anyway, this is what I'm getting about G. We have G(3) = 4, which is a power of 2 so we're done. G(11) = 12, and after factoring out the highest power of 2 we're left with 3, and G(3) = 4. So now what do you do with G(29)? G(29) = 30, factor out the highest power of 2 leaves 3 x 5. Then what? Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?

G^n (p) = 2^m

So what you mean is this:

Do you understand the notation? It says that for all p, there exist natural numbers n and m such that G^n(p) = 2^m. That's how you express that. Of course you still haven't defined G, but at least this expression tells us what n and m are. They're bound to the existential quantifier. You're not saying this is true for all n and m, but rather that it's true for SOME n and m.

I note that your paper is marked up nicely, which is good. Why did you choose to post your proof on a site that doesn't allow math markup? It makes mathematical conversation difficult. One can copy/paste most of the necessary symbols, but they're hard to read and they don't look good. I had the same problem here a couple of weeks ago.

Edited by wtf
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52 minutes ago, wtf said:

This part I don't understand. No professional mathematician is going to read a paper claiming to have solved a famous open problem in which the author doesn't know basic function notation and can't write a proof. It's like applying for a job as a piano player and admitting that you can't read music or play the piano, but you have a lot of great tunes in your head. They'll tell you to come back when you can play. "I don't want to practice my scales, I just want to bang the groupies."

Anyway, this is what I'm getting about G. We have G(3) = 4, which is a power of 2 so we're done. G(11) = 12, and after factoring out the highest power of 2 we're left with 3, and G(3) = 4. So now what do you do with G(29)? G(29) = 30, factor out the highest power of 2 leaves 3 x 5. Then what? Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?

G^n (p) = 2^m

So what you mean is this:

Do you understand the notation? It says that for all p, there exist natural numbers n and m such that G^n(p) = 2^m. That's how you express that. Of course you still haven't defined G, but at least this expression tells us what n and m are. They're bound to the existential quantifier. You're not saying this is true for all n and m, but rather that it's true for SOME n and m.

I note that your paper is marked up nicely, which is good. Why did you choose to post your proof on a site that doesn't allow math markup? It makes mathematical conversation difficult. One can copy/paste most of the necessary symbols, but they're hard to read and they don't look good. I had the same problem here a couple of weeks ago.

I do recall the notation; unfortunately, I only briefly used the notation in a few of my classes.  So I am familiar but probably not much past a familiarity.

"I note that your paper is marked up nicely, which is good."
- Thank you,  i wrote it over the course of a few weeks as I learned how to use LaTex.

I posted it here because I re-found the forum and remembered that this was a great place to have a good discussion.  I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor. I also don't really know where I would have posted this anyways.   I did show my paper to a few math major friends of mine and we did have some discussion about it.  They didn't seem to have the problems we are having here,  so I wonder if I did some explanation to them verbally that I just entirely missed.

"Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?"
- yes, every prime will eventually reduce to a power of 2.

I am going to try to really define our function, or functions here so that it makes sense.

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1 minute ago, Zolar V said:

I do recall the notation; unfortunately, I only briefly used the notation in a few of my classes.  So I am familiar but probably not much past a familiarity.

"I note that your paper is marked up nicely, which is good."
- Thank you,  i wrote it over the course of a few weeks as I learned how to use LaTex.

I posted it here because I re-found the forum and remembered that this was a great place to have a good discussion.  I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor. I also don't really know where I would have posted this anyways.   I did show my paper to a few math major friends of mine and we did have some discussion about it.  They didn't seem to have the problems we are having here,  so I wonder if I did some explanation to them verbally that I just entirely missed.

"Are you saying that each of 3 and 5 eventually reduce to a power of 2 this way? Is that the idea?"
- yes, every prime will eventually reduce to a power of 2.

I am going to try to really define our function, or functions here so that it makes sense.

>  I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor.

Yes they broke it. Also made it so you can't intersperse responses with quotes. No idea what the powers that be are thinking. Reason this is on my mind is that I'm formalizing your G process (not really a function, more of an algorithm) and the most expressive way to do that is with LaTeX. As before I'll have to write my response offline and post a screenshot. And nobody else can quote my post to leverage my markup. It's extremely annoying.

Anyway I just want to make sure I've got this. You input a prime and add 1. If it's a power of 2 you're done. Otherwise divide out the highest power of 2 from p+1 and factor it into a product of primes 2 < p_i < p, each prime written as many times as needed. Then you recursively apply G to each of the primes and eventually everything drills down to a power of 2. I have no idea what that proves or why it's important, but we'll get to that later. Do I have the basic algorithm right?

> They didn't seem to have the problems we are having here

I never realized it at the time, but a lot of people get degrees in math without learning any math.

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40 minutes ago, wtf said:

I never realized it at the time, but a lot of people get degrees in math without learning any math.

Getting the right answer and understanding  are not necessarily inclusive.

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1 hour ago, wtf said:

>  I believe a long time ago these forums were also integrated into LaTex or some other online math notation editor.

Yes they broke it. Also made it so you can't intersperse responses with quotes. No idea what the powers that be are thinking. Reason this is on my mind is that I'm formalizing your G process (not really a function, more of an algorithm) and the most expressive way to do that is with LaTeX. As before I'll have to write my response offline and post a screenshot. And nobody else can quote my post to leverage my markup. It's extremely annoying.

Anyway I just want to make sure I've got this. You input a prime and add 1. If it's a power of 2 you're done. Otherwise divide out the highest power of 2 from p+1 and factor it into a product of primes 2 < p_i < p, each prime written as many times as needed. Then you recursively apply G to each of the primes and eventually everything drills down to a power of 2. I have no idea what that proves or why it's important, but we'll get to that later. Do I have the basic algorithm right?

> They didn't seem to have the problems we are having here

I never realized it at the time, but a lot of people get degrees in math without learning any math.

"Anyway I just want to make sure I've got this. You input a prime and add 1. If it's a power of 2 you're done. Otherwise divide out the highest power of 2 from p+1 and factor it into a product of primes 2 < p_i < p, each prime written as many times as needed. Then you recursively apply G to each of the primes and eventually everything drills down to a power of 2. I have no idea what that proves or why it's important, but we'll get to that later. Do I have the basic algorithm right?"

-Mostly,  I don't divide out the 2's until the end.  That way I collect all of the division that I need to do all at once.
You input a prime and add 1,  if its a power of 2 you're done.  if not then you factor the composite number into a product of primes and manipulate it to give you a prime to work with.  Then recursively apply G again to each of the primes. once you have everything drilled down to a power of two for that piece of the proof.  Otherwise for the collatz part, you collect your two's and divide them all at once, giving you a single letter representation for how many steps it takes to go to 1 from whatever power of 2 number you drilled down to.

maybe the definition you're trying to get at looks like this

ignore H, I don't believe it gets us anywhere.

Maybe the definition you're looking for is G(j(X)).  So we have this prime factorized number that is manipulated to be a number + a prime.  We then plug that into G who takes the only prime number we have available and adds 1 to it.  We then go back and factorize, manipulate and replug into G.   we do this for as many steps as it takes until we get a power of 2.

the 2 < p_i < p inequality gives us a lot of information about the resulting primes.  Interesting enough, a few weeks after I had written my idea and worked with it on this problem I stumbled on this: https://en.wikipedia.org/wiki/Bertrand's_postulate      Which appears to be saying very similar things to me.

" I never realized it at the time, but a lot of people get degrees in math without learning any math.  "
I suppose, but they were the ones that spent the time studying, going to office hours, and getting high marks.

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29 minutes ago, Zolar V said:

maybe the definition you're trying to get at looks like this

I didn't look at what you wrote yet, I've been working on my version. Just tell me if I've got this right.

I must say that now TWO of us are screenshotting LaTeX and I think this is a very inefficient way to work. The problem is we can't quote each other's markup so everything has to be done from scratch. I'm not going to do this much more if at all. I hate giving the site owners the satisfaction of knowing they can delete LaTeX from the site and still have people discuss math. I'm just going to stick to ASCII from now on. This is dumb.

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4 minutes ago, wtf said:

I didn't look at what you wrote yet, I've been working on my version. Just tell me if I've got this right.

I must say that now TWO of us are screenshotting LaTeX and I think this is a very inefficient way to work. The problem is we can't quote each other's markup so everything has to be done from scratch. I'm not going to do this much more if at all. I hate giving the site owners the satisfaction of knowing they can delete LaTeX from the site and still have people discuss math. I'm just going to stick to ASCII from now on. This is dumb.

I think the omission is an Invision software thing, not to do with  the site owners.

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10 minutes ago, StringJunky said:

I think the omission is an Invision software thing, not to do with  the site owners.

I'd like to say up front that I'm mindful that I'm a guest here, and there's a Feedback section, so my comments regarding the site are off-topic in this forum, and a breach of etiquette. I've just got a bug up my butt about this because it's the second time in two weeks that someone asked a question I'm interested in, and I wanted to write out some mathematical exposition, and had to resort to posting screenshots. Having said my piece I'm going to stop talking about it here.

However, just to reply to what you said ... I was not around when the software change happened. But I'm perfectly well aware that the owner of a site chooses the software. So if the owners went from brand X forum software to brand Y, that's a choice. And if brand Y is a significant downgrade in capability, that's on the owners. Someone please straighten me out if I've got this wrong and there are circumstances that let the owners off the hook in this instance.

Thing is, implementing MathJax amounts to pasting a short block of Javascript into the HTML headers. So I am concluding that the site owners don't have access to the HTML headers to add CSS or Javascript. It's hard for me to understand why anyone would install software like that but I'm not privy to the details.

So as I say I'll just put a sock in it regarding this topic from now on.

Edited by wtf
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6 minutes ago, wtf said:

I didn't look at what you wrote yet, I've been working on my version. Just tell me if I've got this right.

I must say that now TWO of us are screenshotting LaTeX and I think this is a very inefficient way to work. The problem is we can't quote each other's markup so everything has to be done from scratch. I'm not going to do this much more if at all. I hate giving the site owners the satisfaction of knowing they can delete LaTeX from the site and still have people discuss math. I'm just going to stick to ASCII from now on. This is dumb.

I actually was about to use LaTex, but decided to use MS Word's insert equation function instead.  Its more clunky but a bit faster for a mockup.

I would say that you're generally right.  I don't divide mine by 2for the sole purpose of saving them all for the end.
I think the tricky detail you're missing is converting the product of primes into a summation of the primes. That conversion lets you work with one prime at a time and also gives you some detail about how many times you need to do step 3
Maybe a bit better of a way to think about it would be to convert all products back into basic addition.
5*7 = 7 + 7 + 7 + 7 + 7

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33 minutes ago, Zolar V said:

I actually was about to use LaTex, but decided to use MS Word's insert equation function instead.  Its more clunky but a bit faster for a mockup.

I would say that you're generally right.  I don't divide mine by 2for the sole purpose of saving them all for the end.
I think the tricky detail you're missing is converting the product of primes into a summation of the primes. That conversion lets you work with one prime at a time and also gives you some detail about how many times you need to do step 3
Maybe a bit better of a way to think about it would be to convert all products back into basic addition.
5*7 = 7 + 7 + 7 + 7 + 7

Well then now I'm perfectly well dispirited. Not factoring out 2^n till the end I can certainly accommodate. But the way you described it sounds like I've got something substantively wrong. Wow, 5*7 written out as five 7's? I confess, I gave it my best shot and cannot make sense of this. I'll read what you wrote but I'm afraid I'm at the end of my tether on this problem.

ps -- Let me get this straight. You get, what's a prime that works with 5 and 7? 2 x 35 is 70 and 69's not prime. 4 x 35 is 140 and 139 is prime, isn't it? Yes 139 is prime. So 139 -> 140 -> 4 x 5 x 7.

Now I thought you are going to recurse down both subnodes 5 and 7 separately, and note that they both end up at a power of 2. I believe that.

But now you're saying you write that as 7 + 7 + 7 + 7 + 7 and ... then what? Please be clear and take it slow. I see that the 7's go to 8 = 2^3 and done. So you want to add them? Why? What is the point of all this?

Edited by wtf
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38 minutes ago, wtf said:

Well then now I'm perfectly well dispirited. Not factoring out 2^n till the end I can certainly accommodate. But the way you described it sounds like I've got something substantively wrong. Wow, 5*7 written out as five 7's? I confess, I gave it my best shot and cannot make sense of this. I'll read what you wrote but I'm afraid I'm at the end of my tether on this problem.

Maybe I'm reading yours wrong, but it appears you're doing a modified form of the collatz conjecture.
if it is odd then add 1
if it is even divide by 2

What I am trying to show is this:
given any prime number ,P > 2, and adding 1 to it results in a new number who's primes are all less than the original prime.
(placeholder detail step)
Then doing that again to each subsequent prime number until you're left with 2's   2 <p_i < p

* The placeholder detail step is a very important step.  You decompose the product of primes so that you can add 1 to a particular prime without affecting the rest of the number(s).   That way can add 1 to only that prime and not some odd number that we know little about.
I first toyed with the idea that we didn't need to constrain the function to primes, but it was by primes that everything falls into place. Also given every number written as a product of primes, that number can also be written as a summation of primes.  The summation part is the part we need in order to apply our p+1 to particular primes.  It seemed to me that I only needed to show that primes worked and that any nonprime  odd number can be rewritten as a summation of primes.

E.G
Let p,q,r be prime
p*q*r + 1    doesn't seem to yield much information at all, other than being even.
but
(p*q-1)*r + (r +1)  seems to yield more information since we can add 1 to our particular prime r.  we know that (r+1) < 2r ,  which means the composite number (r+1) does not contain r in its prime factorization.

maybe a summation would be a better visual tool:

p*q*r = r     lower index = 1  upper index = p*q

Here the summation tells us that we need to add 1 to each r a minimum of p*q times.   p*q here is the number of iterations we add p+1
the subsequent (r+1) still yields the  (r+1) < 2r information as well. So we know whatever number (r+1) is,  that number doesnt contain r in its prime factorization, and each prime factor is also strictly less than r

" Well then now I'm perfectly well dispirited "
- My apologizes my friend; I'm as terrible as I am at relating my thoughts and ideas in person , I am even worse online.

Let P be prime
If P+1  then the prime factorization of P+1 = 2^m * K_1 *K_2*K_3... and each K_i  < P

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

Maybe I'm reading yours wrong, but it appears you're doing a modified form of the collatz conjecture.
if it is odd then add 1
if it is even divide by 2

Nothing of the sort. I'm not even at the second half of your paper. All I'm doing is trying to define G as in your first theorem. I'm not even at the part where you introduce the additive term. I"m just trying to make sense of theorem 1.

ps -- I'm just trying to understand section 2, the prime reducing function. Lines 16 through 34. That's the sole focus of my attention. Are you saying I'm not understanding that correctly? Can you clarify then? The business with adding a term to a product of primes isn't discussed in section 2 so I'm not up to that yet.

Edited by wtf
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2 hours ago, wtf said:

Ohhh! im sorry I read it wrong. I see now what you're doing.
You have defined 2^n as the largest power of 2 within p+1.  You then move it over and leave yourself the non-distinct product of prime factors.
I got mixed up on your step 4 and assumed that you were further along, forgetting entirely that we are only trying to define G.
In that case, yes I think you are on the right path.

4 hours ago, StringJunky said:

Hey man!  Glad to see you're still around here too!

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30 minutes ago, Zolar V said:

Ohhh! im sorry I read it wrong. I see now what you're doing.
You have defined 2^n as the largest power of 2 within p+1.  You then move it over and leave yourself the non-distinct product of prime factors.
I got mixed up on your step 4 and assumed that you were further along, forgetting entirely that we are only trying to define G.
In that case, yes I think you are on the right path.

Hey man!  Glad to see you're still around here too!

> You have defined 2^n as the largest power of 2 within p+1.

Yes.

> You then move it over and leave yourself the non-distinct product of prime factors.

Yes! Just like you told me to!

> I got mixed up on your step 4 and assumed that you were further along, forgetting entirely that we are only trying to define G.
In that case, yes I think you are on the right path.

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.

> Hey man!  Glad to see you're still around here too,

Oh cool. Still, I should be clear about something. I don't think you have a proof. First, no amateur has a proof. A lot of really smart people have looked at the problem. If there's an elementary proof along the lines of doing arithmetic on prime factorizations, it would have been found.

Secondly, even if it is logically possible that you have a valid proof in your head, that is not enough. What you wrote isn't a proof. And your expository skills aren't sufficient to explain it to me. Perhaps you're right, maybe you need a real mathematician. If you're Ramanujan I'll perfectly well stipulate that I'm no Hardy!

Given that, I'll stick with this for a while more. I'm pleased I understood your G function. In fact I agree that your theorem 1 is correct, although I haven't worked out the detailed induction. So I want to push on a little more and see if I can understand the next step.

Edited by wtf

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