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"You may prove that 3 cycles will bring an even input to a lower output, but that output may be odd, in which case the next 3 cycles will not necessarily produc

"You may prove that 3 cycles will bring an even input to a lower output, but that output may be odd, in which case the next 3 cycles will not necessarily produce a lower output." Your observation about long "worst-case" strings of "odd even odd even, ..." which grow the input arbitrarily doesn't change the property of evens as I have described to be decreasing. I'm not negating what you said, I follow your reasoning, but it's not clear to me how/if you are negating my claim for evens (and it's just a claim). I said: f(f(f(n:even):odd):even)=(3(n/2)+1)/2=(3/4)n+1/2 is less than the initial (n:even) for n>4 (actually 2 is sufficient) This would still occur in your strings. The fact that you can come up with strings that abitrarily begin with (or contain within them) these "worst case" alternations does not necessarily mean that it can cycle from it: Let's change one of the cases in my original comment (diff-style): < f(f(f(n:even):even):odd)=3((n/2)/2)+1=(3/4)n+1 < is less than the initial (n:even) for n>4 --- > f(f(f(f(n:even):even):odd):even)=(3((n/2)/2)+1)/2=((3/4)n+1)/2=(3/8)n+1/2 > is less than the initial (n:even) for n>4/5... so every valid input You cannot have two odds in a row so that's the only thing that CAN happen to that case, and it is now clearly anchored on an even, requiring that even to still be decreasing with respect to the initial even (now no more than 5 numbers prior) in every case. We can walk down these decreasing even steps until we are back in known territory (n<=8). Of course, I'm happy to hear counter arguments./thoughts.

waterhouse 14y ago
> "it's not clear to me how/if you are negating my claim for evens (and it's just a claim)." I assumed, from context and stuff, that you had thought you had a proof of the Collatz conjecture, or at least a claim that seemed to prove it, or at least a claim that might be tweaked and extended and refined a little to produce a proof. I aimed to show that all claims like yours were unsuitable for a proof of the Collatz conjecture, because either they could not be iterated or they could not show a decrease for all inputs. If all you are saying is "If n is 4 mod 8, then four successive applications of f will produce (3/8)n + 1/2; this is an isolated mathematical fact interesting by itself", and don't believe it's a proof of the Collatz conjecture, then I have no complaint. I've even written a few functions that will tell you what happens to n after k steps, given that n is x mod 2^m. Here is some output for anyone who is interested: http://pastebin.com/MjzahMFV http://pastebin.com/MjzahMFV > "The fact that you can come up with strings that abitrarily begin with (or contain within them) these "worst case" alternations does not necessarily mean that it can cycle from it:" Of course not. If I could produce a cycle that involved n getting bigger, then I would have disproved the Collatz conjecture. I wasn't attempting to do that, merely disproving a class of attempted proofs.
- vy8vWJlco 14y ago
  I'm still thinking about your modular perspective. I believe that is an accurate restatement of one of my cases, but it is not all of them and would not be sufficient alone. Whereas all the cases together, however, are exhaustive. To summarize, my above comments/claims boil down to: 1) any even number input taken will produce a sub-sequence ending in an even output, by exhaustion (see above), that is guaranteed to be smaller after a maximum number of steps 2) this new even can be taken again, as often as necessary, to reach the smallest integer in that sequence greater than 8 3) the next even number (at most 1 odd away) is by definition below 8 and, while it may not have a next even by my arguement of necessity, it will still be below 8, and all inputs below 8 terminate 4) there is nothing about this reasoning that limits it to a particular sequence/input (all initially-odd inputs adhere after 1 iteration), so it is true of every sequence Am I saying anything untrue or ambiguous? Thanks for taking the time!
  - vy8vWJlco 14y ago
    2 slight corrections, for clarity: ... 2) ... smallest even integer in that sequence greater than 8 ... 3) ... by definition less than or equal to 8 ... will still be less than or equal to 8, and all inputs less than or equal to 8 terminate ... Everything hinges on odds being as or less likely than evens which I believe I show initially "for 2n-1 (clearly odd), f(2n-1) = 3(2n-1)+1 = 2(3n-1), which is clearly even"; at least one even must follow every odd and, additionally, since there are only evens and odds, every odd must have been immediately preceded by an even. It's definitely an innocuous observation, but it justifies a walk all the way down to 8 by exhausting the (3) possible routes through the evens.
- vy8vWJlco 14y ago
  "I aimed to show that all claims like yours were unsuitable for a proof of the Collatz conjecture, because either they could not be iterated or they could not show a decrease for all inputs." I'm not sure I addressed your criticism fully. Earlier you noted: "For starters, that would only work if your argument applied to the entire set of outputs that you'd feed back in again as inputs." That is precisely the case I am describing. I begin with a universal property of any sequence, then proceed to show that all sequences beginning with an even terminate. I then note that the earlier property requires that all sequences beginning with odd numbers immediately follow this pattern after at most one iteration. A sequence that terminates after a finite number of steps + 1 still terminates after a finite number of steps. I have summarized this in a post above. The abmiguity you describe in the set of proofs you are disqualifying is not present in my arguement. I begin at the first value for all sequences (evens, and then odds, which is exhaustive of inputs, and outputs, as you note is necessary) and that is exhaustive of sequences. Is it a variation on the most naive approach that pretty much everyone would start with? Sure, and that's reason enough to be skeptical. But, does it work? It seems to...

waterhouse 14y ago

> "it's not clear to me how/if you are negating my claim for evens (and it's just a claim)." I assumed, from context and stuff, that you had thought you had a proof of the Collatz conjecture, or at least a claim that seemed to prove it, or at least a claim that might be tweaked and extended and refined a little to produce a proof. I aimed to show that all claims like yours were unsuitable for a proof of the Collatz conjecture, because either they could not be iterated or they could not show a decrease for all inputs. If all you are saying is "If n is 4 mod 8, then four successive applications of f will produce (3/8)n + 1/2; this is an isolated mathematical fact interesting by itself", and don't believe it's a proof of the Collatz conjecture, then I have no complaint. I've even written a few functions that will tell you what happens to n after k steps, given that n is x mod 2^m. Here is some output for anyone who is interested: http://pastebin.com/MjzahMFV http://pastebin.com/MjzahMFV > "The fact that you can come up with strings that abitrarily begin with (or contain within them) these "worst case" alternations does not necessarily mean that it can cycle from it:" Of course not. If I could produce a cycle that involved n getting bigger, then I would have disproved the Collatz conjecture. I wasn't attempting to do that, merely disproving a class of attempted proofs.

vy8vWJlco 14y ago
I'm still thinking about your modular perspective. I believe that is an accurate restatement of one of my cases, but it is not all of them and would not be sufficient alone. Whereas all the cases together, however, are exhaustive. To summarize, my above comments/claims boil down to: 1) any even number input taken will produce a sub-sequence ending in an even output, by exhaustion (see above), that is guaranteed to be smaller after a maximum number of steps 2) this new even can be taken again, as often as necessary, to reach the smallest integer in that sequence greater than 8 3) the next even number (at most 1 odd away) is by definition below 8 and, while it may not have a next even by my arguement of necessity, it will still be below 8, and all inputs below 8 terminate 4) there is nothing about this reasoning that limits it to a particular sequence/input (all initially-odd inputs adhere after 1 iteration), so it is true of every sequence Am I saying anything untrue or ambiguous? Thanks for taking the time!
- vy8vWJlco 14y ago
  2 slight corrections, for clarity: ... 2) ... smallest even integer in that sequence greater than 8 ... 3) ... by definition less than or equal to 8 ... will still be less than or equal to 8, and all inputs less than or equal to 8 terminate ... Everything hinges on odds being as or less likely than evens which I believe I show initially "for 2n-1 (clearly odd), f(2n-1) = 3(2n-1)+1 = 2(3n-1), which is clearly even"; at least one even must follow every odd and, additionally, since there are only evens and odds, every odd must have been immediately preceded by an even. It's definitely an innocuous observation, but it justifies a walk all the way down to 8 by exhausting the (3) possible routes through the evens.
vy8vWJlco 14y ago
"I aimed to show that all claims like yours were unsuitable for a proof of the Collatz conjecture, because either they could not be iterated or they could not show a decrease for all inputs." I'm not sure I addressed your criticism fully. Earlier you noted: "For starters, that would only work if your argument applied to the entire set of outputs that you'd feed back in again as inputs." That is precisely the case I am describing. I begin with a universal property of any sequence, then proceed to show that all sequences beginning with an even terminate. I then note that the earlier property requires that all sequences beginning with odd numbers immediately follow this pattern after at most one iteration. A sequence that terminates after a finite number of steps + 1 still terminates after a finite number of steps. I have summarized this in a post above. The abmiguity you describe in the set of proofs you are disqualifying is not present in my arguement. I begin at the first value for all sequences (evens, and then odds, which is exhaustive of inputs, and outputs, as you note is necessary) and that is exhaustive of sequences. Is it a variation on the most naive approach that pretty much everyone would start with? Sure, and that's reason enough to be skeptical. But, does it work? It seems to...