Collatz Conjecture Proof - Page 2

science_man_88 · 2018-03-03, 14:17

Quote:

Originally Posted by Steve One

You epitomise 'lack of reading'. 4+10n, not 4n+10

Since constants tend to come last in what I usually read, it's an easy mistake to make.

With 4+10n we have at least 2 cases:

n odd→ 2+5n is odd→ 15n+7 is even→15m+11 (2m+1=n) is either.
n even→2+5n is even→5m+1 is either(2m=n)

Collag3n · 2018-03-03, 18:49

Quote:

Originally Posted by science_man_88

His claim seems weaker, than a claim of all odd numbers go to one.

Indeed, there is nothing about odd reaching one and he didn't claim that (though). He mostly looked at even numbers where most of the people dealing with collatz just ignore them and go straight to odd numbers. I think even numbers can teach us things about collatz, but it would only be a beginning, certainly not a proof of the conjecture.

Steve One · 2018-03-06, 18:05

Quote:

Originally Posted by 10metreh

Am I correct in thinking you're saying that knowing 4, 14, 24, 34, 44 and 54 go down to 1 proves everything goes down to 1? If so, how does your proof show that 13 goes down to 1?

Try reading point 3.

10metreh · 2018-03-06, 18:08

Quote:

Originally Posted by Steve One

Try reading point 3.

Point 3 tells me that if 104 goes down to 1, then so does 13. How does knowing 4, 14, 24, 34, 44 and 54 go down to 1 prove that 104 does?

Steve One · 2018-03-06, 18:15

Quote:

Originally Posted by Collag3n

Indeed, there is nothing about odd reaching one and he didn't claim that (though). He mostly looked at even numbers where most of the people dealing with collatz just ignore them and go straight to odd numbers. I think even numbers can teach us things about collatz, but it would only be a beginning, certainly not a proof of the conjecture.

Yes. Only the beginning of the problem, which is why l put the point at the beginning. Then l go on to use an analogy using letters. Which part of the lettered analogy do you disagree with? If you don't, then you actually agree with me, without realising it. Please, which of 'that' part do you disagree with? I put it in small changes so as to recognise the lack of real change when extending the spacing of letters. The lack of the other odd/even numbers between, is covered by accepting the first comment.

Collag3n · 2018-03-07, 12:47

Quote:

Originally Posted by Steve One

Yes. Only the beginning of the problem, which is why l put the point at the beginning. Then l go on to use an analogy using letters. Which part of the lettered analogy do you disagree with? If you don't, then you actually agree with me, without realising it. Please, which of 'that' part do you disagree with? I put it in small changes so as to recognise the lack of real change when extending the spacing of letters. The lack of the other odd/even numbers between, is covered by accepting the first comment.

My point is that your sequences either ends on an odd number or on an even numbers.

What you show is that the one ending on an even numbers (e.g. the 64... sequence ending with 4) were already "treated" before (Who knows A, knows BA).

What you also show in your own way is that in the end, all those sequences with even numbers will reach an odd and that all odds (except multiple of 5) will be reached (7,17,27,... in second column, 1,11,21,....in third,.....3,13,....via 5n+x).

Ok, but first, don't we already know that all even numbers reach an odd number (all of them in the end) via the Collatz $\frac{n}{2}$ function?

What i say is that you are left with proving that odd numbers reaches 1, which is what everyone is trying first, knowing that even numbers will ALWAYS reach an odd number (no need to see that 224 leads to 14 which leads to 7). You can't prove 10n+4 reaches 1 without proving the odds.

So yes, your shortcut saying that 37 is proven by some alignment (based on even numbers) is a bit....short, and saying that if we can prove 74, 37 will be proven is a bit trivial, but useless since we need to prove 37 to prove 74.

Some explanations are needed here, because the Collatz function applied to these odd numbers $3n+1$ is not following any alignment you mentioned (try to follow 27, and tell me what is the pattern to reach 1).

Collag3n · 2018-03-07, 18:54

By the way, if your goal is to show that proving only $\frac{1}{10^{th}}$ of the integer prove them all, I have better than an asymptotic density of $\frac{1}{10}$ :
If you only look at numbers with exactly $n$ factors of 2 (in a prime factorization sens: e.g. with 3 factors of 2: $40=5\cdot2^3$ or $168=3\cdot 7 \cdot2^3$ ), i can tell you that proving $2^{n+1}\cdot m+2^n$ with $m$ integer (or in other words, all numbers with exactly $n$ factors of 2) prove them all, whatever $n$ you choose. And with an asymptotic density close to 0 (by choosing an $n\to\infty$ ), you will have hard time to beat this....

LaurV · 2018-03-08, 05:21

Well, if you take all numbers (mod 32) you only have to prove 4 in 32 (8 in 64). Or 13 in 128 if you take them (mod 128).

Going to 256 or 512, you already have much better odds than 1/10, you have to prove only 19 in 256 (or 38 in 512).

At (mod 1024), you must only prove 64 of them, which is one in 16, hehe... And this is getting much better at higher powers. Unfortunately this leads nowhere... There is always a significant fraction of the numbers which must be "proved".

Collag3n · 2018-03-08, 06:50

I don't think you need to look to so much numbers. Like I said, if you prove numbers of the form 64m+32, or 128m+64 or....(only one number modulo 64 or 128 or...), you proved them all
e.g. with 128m+64:
64 32 16 8 4 2 1
192 96 48 24 12 6 3
320 160 80 40 20 10 5
...
covering all odds in the end
But you are right, having a density of 0 does not mean you won't have to prove an infinity of numbers. If i am not mistaken, prime numbers have a density of 0, but we have an infinity of them.

LaurV · 2018-03-08, 07:39

Quote:

Originally Posted by Collag3n

64m+32, or 128m+64

I don't know what do you mean by that. For 32 (mod 64) or 64 (mod 128) you have nothing to "prove". They decrease in one step, and you can use descent. For (mod 64) you have to "prove" 7, 15, 27, 31, 39, 47, 59, and 63. Which is the same as "proving" 7, 15, 27 and 31 (mod 32), as there is no "collapse" from (mod 32) to (mod 64). For (mod 128), there are 3 "collapses" and the remaining classes are 27, 31, 39, 47, 63, 71, 79, 91, 95, 103, 111, 123, and 127.

I use quotes because it seems we understand different things when we say "prove". My understanding of the concept is that, for example, to "prove 27 (mod 64)", you have to show that the numbers of the form 64k+27 go to a number which is smaller than 64k+27, after a number of steps. If you can "prove" any of the "(mod x)" cases, then you prove the conjecture, you do not need to prove all of them, each of them is a better refinement of the former. For example, "7 (mod 64)" means "7 (mod 128) and 71 (mod 128)", but the "7 (mod 128)" will "collapse" after 11 steps into 5 (mod 81) which is smaller, so only the "71 (mod 128)" is left.

No one gives a dime about 64k+32, and so on... (i.e. even numbers).

Collag3n · 2018-03-08, 08:18

Well, this is because you are reasoning like everyone except Steve One. You try to prove that odds are reaching 1, And you are totally right.
Like I said, "proving that numbers of the form 10n+4 or 64m+32 reaches 1, prove them all" is correct, but nonsense because to prove that you need to prove odds reaches 1 anyway (read my remark about 74 and 37). That's why I told him that most don't even look at even numbers in the first place (especially the trivial cases you mentioned).
So when I say that you only need to prove 64m+32 or 128m+64 or ..., this is of course in the Steve One vision.
His reasoning seems to be that if you prove that 64m+32 reaches 1 (no idea how he plans to do this), then automatically, all odds are proven too (since 64m+32 pass through all odds). Well, in his case he uses 10n+4 and the fact that from there you reach all other numbers (5n+x, odd or even).

Yeah, I know, this is confusing....but I don't know, perhaps I am the one confused....

Note: 96 is 32 mod 64 and does not decrease in 1 step (not to "1" at least)

2018-03-07, 18:54	#18
Collag3n Feb 2018 2³ Posts	By the way, if your goal is to show that proving only $\frac{1}{10^{th}}$ of the integer prove them all, I have better than an asymptotic density of $\frac{1}{10}$ : If you only look at numbers with exactly $n$ factors of 2 (in a prime factorization sens: e.g. with 3 factors of 2: $40=5\cdot2^3$ or $168=3\cdot 7 \cdot2^3$ ), i can tell you that proving $2^{n+1}\cdot m+2^n$ with $m$ integer (or in other words, all numbers with exactly $n$ factors of 2) prove them all, whatever $n$ you choose. And with an asymptotic density close to 0 (by choosing an $n\to\infty$ ), you will have hard time to beat this.... Last fiddled with by Collag3n on 2018-03-07 at 19:52

2018-03-08, 05:21	#19
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 3·23·149 Posts	Well, if you take all numbers (mod 32) you only have to prove 4 in 32 (8 in 64). Or 13 in 128 if you take them (mod 128). Going to 256 or 512, you already have much better odds than 1/10, you have to prove only 19 in 256 (or 38 in 512). At (mod 1024), you must only prove 64 of them, which is one in 16, hehe... And this is getting much better at higher powers. Unfortunately this leads nowhere... There is always a significant fraction of the numbers which must be "proved". Last fiddled with by LaurV on 2018-03-08 at 05:23

2018-03-08, 08:18	#22
Collag3n Feb 2018 2³ Posts	Well, this is because you are reasoning like everyone except Steve One. You try to prove that odds are reaching 1, And you are totally right. Like I said, "proving that numbers of the form 10n+4 or 64m+32 reaches 1, prove them all" is correct, but nonsense because to prove that you need to prove odds reaches 1 anyway (read my remark about 74 and 37). That's why I told him that most don't even look at even numbers in the first place (especially the trivial cases you mentioned). So when I say that you only need to prove 64m+32 or 128m+64 or ..., this is of course in the Steve One vision. His reasoning seems to be that if you prove that 64m+32 reaches 1 (no idea how he plans to do this), then automatically, all odds are proven too (since 64m+32 pass through all odds). Well, in his case he uses 10n+4 and the fact that from there you reach all other numbers (5n+x, odd or even). Yeah, I know, this is confusing....but I don't know, perhaps I am the one confused.... Note: 96 is 32 mod 64 and does not decrease in 1 step (not to "1" at least) Last fiddled with by Collag3n on 2018-03-08 at 09:07

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2018-03-08, 06:50	#20
Collag3n Feb 2018 8₁₀ Posts	I don't think you need to look to so much numbers. Like I said, if you prove numbers of the form 64m+32, or 128m+64 or....(only one number modulo 64 or 128 or...), you proved them all e.g. with 128m+64: 64 32 16 8 4 2 1 192 96 48 24 12 6 3 320 160 80 40 20 10 5 ... covering all odds in the end But you are right, having a density of 0 does not mean you won't have to prove an infinity of numbers. If i am not mistaken, prime numbers have a density of 0, but we have an infinity of them.