20180303, 14:17  #12 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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) Last fiddled with by science_man_88 on 20180303 at 14:18 
20180303, 18:49  #13 
Feb 2018
2^{3} Posts 
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.

20180306, 18:05  #14 
Feb 2018
36_{10} Posts 

20180306, 18:08  #15 
Nov 2008
2·3^{3}·43 Posts 

20180306, 18:15  #16  
Feb 2018
44_{8} Posts 
Quote:


20180307, 12:47  #17  
Feb 2018
2^{3} Posts 
Quote:
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 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 is not following any alignment you mentioned (try to follow 27, and tell me what is the pattern to reach 1). Last fiddled with by Collag3n on 20180307 at 13:01 

20180307, 18:54  #18 
Feb 2018
2^{3} Posts 
By the way, if your goal is to show that proving only of the integer prove them all, I have better than an asymptotic density of :
If you only look at numbers with exactly factors of 2 (in a prime factorization sens: e.g. with 3 factors of 2: or ), i can tell you that proving with integer (or in other words, all numbers with exactly factors of 2) prove them all, whatever you choose. And with an asymptotic density close to 0 (by choosing an ), you will have hard time to beat this.... Last fiddled with by Collag3n on 20180307 at 19:52 
20180308, 05:21  #19 
Romulan Interpreter
Jun 2011
Thailand
2^{2}·3·5·163 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 20180308 at 05:23 
20180308, 06:50  #20 
Feb 2018
2^{3} 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. 
20180308, 07:39  #21 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×3×5×163 Posts 
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). Last fiddled with by LaurV on 20180308 at 07:41 
20180308, 08:18  #22 
Feb 2018
2^{3} 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 20180308 at 09:07 
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