Timeline for Proof that $3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Nov 10, 2019 at 15:50 | comment | added | ReverseFlowControl | @SimonHenry: I found counter examples. It is subtle, but this problem is not exactly equivalent to the Collatz Conjecture; at least not as you describe. | |
| Nov 9, 2019 at 22:37 | answer | added | Max Alekseyev | timeline score: 0 | |
| Nov 3, 2019 at 22:46 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Better phrasing.
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| Nov 3, 2019 at 20:49 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
General form.
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| Nov 3, 2019 at 0:34 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Simplified to best form.
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| Nov 2, 2019 at 13:04 | answer | added | Gottfried Helms | timeline score: 4 | |
| Nov 2, 2019 at 9:08 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Fixed a typo.
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| Nov 1, 2019 at 23:34 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Reduced a fraction.
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| Nov 1, 2019 at 23:03 | history | edited | ReverseFlowControl |
Added a tag.
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| Nov 1, 2019 at 15:39 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Fixed typo.
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| Nov 1, 2019 at 5:53 | history | edited | ReverseFlowControl | CC BY-SA 4.0 |
Added the case for n=3 exhaustively.
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| Nov 1, 2019 at 0:39 | comment | added | ReverseFlowControl | @SimonHenry looking more closely at one of my proofs I just realized that this is exactly the 3n+1 problem. That is to say, any sequence satisfying this equation for an odd s exactly corresponds to meeting the condition I mentioned earlier. Sorry, had to go back and read my paper slowly. I did not realize this earlier. | |
| Oct 31, 2019 at 8:30 | comment | added | Gottfried Helms | @ReverseFlow - oh sorry for the not-recognizing of the/a difference. Well, perhaps pointing it out in the question, where exactly the by me mentioned representation differs from yours? You only mentioned some restrictions? But I think I'll leave this here in MO, and look forward to the process in MSE if there will happen some activity there. | |
| Oct 31, 2019 at 8:13 | comment | added | ReverseFlowControl | @GottfriedHelms thank you for the reference! Your notation and mine however are not equivalent. Also, the ideas behind the above formula are very much different from your work. | |
| Oct 31, 2019 at 7:54 | comment | added | Gottfried Helms | @ReverseFlow - have you ever seen my small essay on the question about cycles on my homepage go.helms-net.de/math/collatz/Collatz061102.pdf I had not yet much experience in writing such things (at 2006) and it is rather amateurish but I introduced a notation which shortens that big formula to refer to the exponents only and do a lot of things with this. Perhaps this might also be a good idea for your analyzes? (I surely should rewrite the whole essay soon...) | |
| Oct 31, 2019 at 7:48 | comment | added | Gottfried Helms | @ReverseFlow - you're correct, it is not in the wikipedia, sorry. I'm so used to that formula and have seen it so often elsewhere that I took the cycle-formula in wp for that one (of course, the cycle-formula is just a copy of your formula when the $3^ns + 3^{n-1} +... = 2^m $ is modified $3^ns + 3^{n-1} +... = 2^m s $ to reflect the cycling and then $(2^m-3^n)s = 3^{n-1}+...$ collected and then divided by $(2^m-3^n)$ . I think that formula is already in Crandall's 1978 article if not earlier and has been posed for questions multiple times in this or that small variation. | |
| Oct 31, 2019 at 4:42 | comment | added | ReverseFlowControl | @GottfriedHelms your first observation is correct. It’s hard to get any insight out of that if any though. As for your second observation, could you please provide the wikipedia link and section? The closest thing I found was a reference to rational representations which is not quite this, and not even an equation at that. The condition takes a bit to write down, but it remains an open problem to me whether the only solutions to this are exactly those that meet those restrictions; in that case, this problem is exactly equivalent to the collatz conjecture. I am not sure I could prove the link. | |
| Oct 30, 2019 at 22:29 | comment | added | Gottfried Helms | @SimonHenry: in the formulation $ {3^ns + ... \over 2^m }=1$ where $m>a_{n+1}$ it is the codification of the orbit of the odd positive number $s$ downto $1$ by the iterated Collatz-transformation (in the "Syracuse"-notation of the problem). Don't see the different restrictions on $a_k$ here against that of the Collatz-transformation, btw. ($a_0=0$, $a_{k+1}\gt a_k$ ). I think that formula is also in wikipedia. | |
| Oct 30, 2019 at 22:17 | comment | added | Gottfried Helms | At a short view it seems it is required that $m$ and $a_{n-1}$ must be such that $3 | 2^m - 2^{a_{n-1}} $ ... | |
| Oct 30, 2019 at 22:06 | comment | added | Gerry Myerson | Reminds me of a puzzle due to Erdos, from Mathematics Magazine, February 1994: prove that any positive integer can be written as the sum of terms of the form $2^a3^b$ where no summand divides another. See also Richard Blecksmith, Michael McCallum and J. L. Selfridge, 3-Smooth Representations of Integers, The American Mathematical Monthly Vol. 105, No. 6 (Jun. - Jul., 1998), pp. 529-543. | |
| Oct 30, 2019 at 18:34 | comment | added | ReverseFlowControl | The 3n+1 problem has a condition on $a_n$ which is not required here. | |
| Oct 30, 2019 at 18:29 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added the math.SE link
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| Oct 30, 2019 at 18:22 | history | edited | Martin Sleziak |
Removed the deprecated (abstract-algebra) tag - see the tag info: https://mathoverflow.net/tags/abstract-algebra/info (if there are some other suitable tags, choose them instead.)
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| Oct 30, 2019 at 18:19 | comment | added | Simon Henry | Isn't this a rephrasing of the 3n+1 problem ? (It really sounds like the type of expression you get after applying the 3n+1 iteration to s until you get 1... ) | |
| Oct 30, 2019 at 18:17 | history | edited | Asaf Karagila♦ | CC BY-SA 4.0 |
added 23 characters in body; edited title
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| Oct 30, 2019 at 17:58 | history | asked | ReverseFlowControl | CC BY-SA 4.0 |