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
12 events
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| Mar 12, 2019 at 19:42 | comment | added | Yaakov Baruch | See comments to the original question for an updated search result (not based on any (mod 33) restriction). | |
| Mar 3, 2019 at 12:04 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Mar 3, 2019 at 1:13 | comment | added | Zhi-Wei Sun | @Yaakov Baruch In view of your observation, perhaps one should focus on checking the conjecture with $n\equiv 20\pmod{33}$. For $n>5\times10^{11}$ with $n\equiv20\pmod{33}$, do you have a quick way to check the conjecture for $n$? | |
| Mar 2, 2019 at 15:32 | comment | added | user21820 | Your facts presented in fact suggest the opposite (to your conclusion), namely that there is no deep reason for it to be true or false. The probabilistic heuristic and the empirical evidence suggests that the conjecture should be true for sufficiently large $n$. This means that whether or not there are counter-examples is a matter of small number coincidences. The congruence to $20$ mod $33$ should merely be an artifact of the residues of the binomial coefficients mod $33$, which less frequently sum to $20$. (Oh I just realized Stefan Kohl already stated my last point.) | |
| Mar 1, 2019 at 18:04 | comment | added | Stefan Kohl♦ | (continued:) 1450/35937, 1520/35937, 224/9801, 3364/107811, 460/11979, 248/9801, 3886/107811, 1180/35937, 208/9801, 232/11979, 370/11979, 100/3267, 4408/107811, 1120/35937, 232/9801, 1334/35937, 1240/35937, 268/9801, 3422/107811, 1040/35937, 16/1089, 1073/35937, 500/11979, 304/9801, 3248/107811, 1160/35937, 92/3267, 3596/107811, 1340/35937, 236/9801, 3016/107811, 80/3993, 74/3267. The minimum is at 20 modulo 33 (the value is 16/1089), which statistically explains that numbers having only one representation of the given form are typically congruent to 20 modulo 33. | |
| Mar 1, 2019 at 18:04 | comment | added | Stefan Kohl♦ | From congruence conditions, one finds the following relative distribution of the sums $\binom{w}{2}+\binom{x}{4}+\binom{y}{6}+\binom{z}{8}$ over the residue classes modulo 33 (read: 1450/35937 of the sums are congruent to 0 modulo 33, and so on): (to be continued ...) | |
| Mar 1, 2019 at 11:48 | comment | added | Yaakov Baruch | The 10 numbers are: $\binom{365}2+\binom{76}4+\binom{40}6+\binom{34}8$, $\binom{8167}2+\binom{58}4+\binom{43}6+\binom{9}8$, $\binom{8914}2+\binom{139}4+\binom{26}6+\binom{10}8$, $\binom{8693}2+\binom{240}4+\binom{43}6+\binom{45}8$, $\binom{57792}2+\binom{5}4+\binom{67}6+\binom{21}8$, $\binom{15456}2+\binom{341}4+\binom{93}6+\binom{53}8$, $\binom{99977}2+\binom{89}4+\binom{41}6+\binom{34}8$, $\binom{121487}2+\binom{346}4+\binom{39}6+\binom{16}8$, $\binom{155964}2+\binom{196}4+\binom{90}6+\binom{49}8$, $\binom{64714}2+\binom{584}4+\binom{134}6+\binom{31}8$. No visible patterns there... | |
| Feb 28, 2019 at 11:24 | comment | added | Zhi-Wei Sun | See oeis.org/A306477 for my definition of the number of representations of $n$. | |
| Feb 28, 2019 at 10:54 | comment | added | Yaakov Baruch | Truly $0$ should not be regarded as an exception, but less because of your picking a specific requirement, than because if the problem is posed over $\mathbb{Z}$ instead of $\mathbb{N}$, then $0$ has $2\times 4\times 6\times 8$ representations, while the numbers above all have $2^4$ (the minimum possible if the conjecture is true, given that all non-zero values of binomial polynomials $\binom{x}{a}$ have multiplicity $2$). | |
| Feb 28, 2019 at 10:44 | comment | added | Zhi-Wei Sun | 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 27, 2019 at 21:06 | history | edited | Yaakov Baruch | CC BY-SA 4.0 |
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| Feb 27, 2019 at 21:01 | history | answered | Yaakov Baruch | CC BY-SA 4.0 |