Timeline for Positive integers written as $\binom{w}2+\binom{x}4+\binom{y}6+\binom{z}8$ with $w,x,y,z\in\{2,3,\ldots\}$
Current License: CC BY-SA 4.0
37 events
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Mar 13, 2019 at 8:13 | comment | added | Yaakov Baruch | I really don't think it was necessary to edit (and bump up) the question for just an improved numerical check! In any case, if anything, I think the surprising rate at which numbers with a single representation keep showing up, makes it still quite possible that the conjecture fail for some very large $n$. | |
Mar 12, 2019 at 23:30 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
Update the verification record.
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Mar 12, 2019 at 22:51 | comment | added | Zhi-Wei Sun | @Yaakov Baruch I'm glad to hear that the 2-4-6-8 conjecture holds for $n\le 2\times 10^{12}$. Thank you very much for your verification. | |
Mar 12, 2019 at 19:40 | comment | added | Yaakov Baruch | @StefanKohl: up to $2\times 10^{12}$ there are no counterexamples, 19 numbers with one representation and 141 with 2. Besides the 10 numbers in my non-answer below, the other 9 with unique rep are: 52640483762, 110740757279, 151955228192, 232115559443, 281156393219, 802140167849, 953628522614, 1331706555617, 1436076907142. Their log-density does not seem to be decreasing, yet. The log-density of the numbers with 2 reps even has a clear concave up shape thus far. | |
Mar 9, 2019 at 21:53 | answer | added | Lucia | timeline score: 13 | |
Mar 4, 2019 at 13:21 | history | edited | Todd Trimble | CC BY-SA 4.0 |
Highlighted some possible questions in response to the linked MO meta discussion
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Mar 4, 2019 at 13:15 | comment | added | Todd Trimble | Meta discussion: meta.mathoverflow.net/questions/4125/… (I request voting up this comment for visibility). I took the liberty of editing the question to implement some suggestions there. | |
Mar 4, 2019 at 9:19 | comment | added | Zhi-Wei Sun | I think people here focus on math. rather than money! | |
Mar 3, 2019 at 11:39 | comment | added | Yaakov Baruch | That is actually a nice 370$! But I'm waiting for the 3-4-5-6-7 conjecture... | |
Mar 3, 2019 at 9:08 | comment | added | Zhi-Wei Sun | I think there is almost no hope to find a counterexample. If you can find a concrete example, I may offer 2468 RMB (Chinese dollars). | |
Feb 28, 2019 at 4:30 | comment | added | Zhi-Wei Sun | @YaakovBaruch Please note that in my 2-4-6-8 conjecture I require that the number $\binom w2$ is positive, so $0$ should not be considered as a uniquely represented number. | |
Feb 28, 2019 at 0:19 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
edited body
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Feb 27, 2019 at 23:49 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
added 266 characters in body
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Feb 27, 2019 at 21:01 | answer | added | Yaakov Baruch | timeline score: 9 | |
Feb 27, 2019 at 17:33 | comment | added | Stefan Kohl♦ | @YaakovBaruch: This suggests that if the conjecture really happens to be true and one finds a proof with methods from analytic number theory that it holds for sufficiently large numbers, then the bound for "sufficiently large" may well be big enough to turn completing the proof into a challenging computational problem ... . | |
Feb 27, 2019 at 17:26 | comment | added | Stefan Kohl♦ | @Zhi-WeiSun: I checked the latter one a bit further (to $10^8$), and found the counterexample 41215718. | |
Feb 27, 2019 at 13:58 | comment | added | Zhi-Wei Sun | To understand the curious 2-4-6-8 conjecture better, perhaps we should investigate a similar conjecture. I note that $10413917$ is the least positive integer not representable by $w^2+\binom x4+\binom y6+\binom z8$ with $w,x,y,z\in\mathbb N$. I also guess that each $n=0,1,2,\ldots$ can be written as $w(w+1)+\binom x4+\binom y6+\binom z8$ with $w,x,y,z\in\mathbb N$ (cf. oeis.org/A306571). I have verified this new one for $n\le 3\times10^7$, maybe some of you could check it further. | |
Feb 27, 2019 at 10:38 | comment | added | Yaakov Baruch | @StefanKohl: the list of uniquely represented numbers below $10^{10}$: $0, 23343989, 39866594, 54847142, 394239767, 1769927927, 2321530979$, $5022744494$ and $7969623044$. More than OP was expecting... | |
Feb 26, 2019 at 23:35 | comment | added | Yaakov Baruch | Verified up to $5\times 10^{11}$. That took 58G of memory on a machine with 52G, and 190 minutes, of which 23m system, presumably mostly swapping. So I think I reached my limit. Will be happy to give my code to anyone with better resources. | |
Feb 24, 2019 at 22:11 | comment | added | Zhi-Wei Sun | @StefanKohl: In my opinion, all those uniquely represented numbers should be below $10^9$. | |
Feb 24, 2019 at 17:03 | comment | added | Max Alekseyev | @StefanKohl: I did not count the number of representations, reserving just one bit for answering whether a number is representable or not. Accounting for the number of representations will multiply the memory usage by a factor, thus reducing the bound I can reach. | |
Feb 24, 2019 at 16:54 | comment | added | Stefan Kohl♦ | @MaxAlekseyev: How many integers have you found up to your search bound which can be represented only in 1, 2, 3 or 4 ways in the desired form? | |
Feb 24, 2019 at 14:13 | comment | added | Zhi-Wei Sun | I'm grateful to Prof. Max Alekseyev and Yaakov Baruch for checkng the conjecture. I'd like to offer 2468 US dollars as the prize for the first correct proof of the 2-4-6-8 conjecture. | |
Feb 24, 2019 at 13:54 | comment | added | Max Alekseyev | I've verified the conjecture for $n\leq 2\times 10^{11}$. | |
Feb 22, 2019 at 2:02 | comment | added | Zhi-Wei Sun | @Max Alsekseyev, thank you very much. Your verification result is exciting! | |
Feb 22, 2019 at 1:22 | comment | added | Max Alekseyev | I have verified that the conjecture holds for $n$ below $5\times 10^{10}$. | |
Feb 22, 2019 at 0:27 | comment | added | Yaakov Baruch | I got to $1.5\times10^9$ with no counterexamples... but was working in simple miserable awk. Perhaps tomorrow I'll convert to C and push beyond that. As an aside, I didn't always like prior questions from this OP, but I don't think this one deserved the down votes. | |
Feb 21, 2019 at 23:38 | comment | added | Zhi-Wei Sun | As I mentioned, $1/2+1/4+1/6+1/8$ is about $1.04$, just slightly larger than $1$. Thus we should not expect many ways of the required representions for numbers below $10^8$. Note that the first counterexample to the 2-4-6-9 problem is $1061619$ and the first counterexample to the 2-4-6-10 problem is $68286$. If the 2-4-6-8 conjecture holds for all natural numbers below $10^9$, it should be quite safe. I agree with Yaakov Baruch's analysis. If possible, I hope that Yaakov could extend his verification of the 2-4-6-8 conjecture to $10^9$. | |
Feb 21, 2019 at 20:32 | comment | added | Stefan Kohl♦ | @YaakovBaruch Given that the integers below $10^8$ which can be represented in the desired form in only one way are 1, 23343989, 39866594 and 54847142, and given that there are 10, 49, 166, 541 and 1344 numbers below $10^8$ which can be represented in precisely 2, 3, 4, 5, respectively 6 ways, I'd guess that the proposed 2-4-6-8 conjecture is likely false, and that one has quite reasonable chances of finding a counterexample when extending the search bound a bit (say, to $10^{12}$ or so -- perhaps even $10^{10}$ will do, but these estimates are of course speculation). | |
Feb 21, 2019 at 18:53 | comment | added | Yaakov Baruch | There goes that theory... $226687459$ and $310917242$ are also counterexamples to the 2-3-6-9 (up to $5\times 10^8$). | |
Feb 21, 2019 at 17:20 | comment | added | Yaakov Baruch | A somewhat vague, empiric observation: for the 2-4-6-9 conjecture the only counterexamples up to $2\times 10^8$ are $1061619, 1943709, 2009719, 3024382$ and $3044809$. This makes me think that this conjecture holds for small numbers due to the law of small numbers, and for large ones due to the redundancies caused by $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{9}>1$. Thus, having presumably pushed the check of the 2-4-6-8 conjecture far into its redundancy area, I'd be very surprised if it didn't hold. | |
Feb 21, 2019 at 1:11 | comment | added | Zhi-Wei Sun | Yaakov, thank you very much for your verification! I'd like to call the conjecture "the 2-4-6-8 conjecture". | |
Feb 20, 2019 at 23:39 | comment | added | Yaakov Baruch | I have verified the conjecture up to $5\times 10^8$. | |
Feb 19, 2019 at 17:05 | review | Close votes | |||
Feb 22, 2019 at 20:27 | |||||
Feb 19, 2019 at 16:48 | comment | added | Wlod AA | On the contrary, I feel that the present question is perfect for MathOverflow. | |
Feb 19, 2019 at 7:02 | comment | added | Zach Teitler | mathoverflow.net/help/on-topic "MathOverflow is not the right place to ask open problems.... If you want to contribute to (or view) a list of open problems, visit the Open Problem Garden." | |
Feb 19, 2019 at 6:46 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |