Timeline for Natural numbers in the form $\lfloor\frac{a^3+b^3}2+\frac{c^3+d^3}6\rfloor$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2021 at 11:39 | comment | added | Zhi-Wei Sun | I have extended the verification of the conjecture in the posting to $10^5$. | |
| Apr 12, 2021 at 11:36 | comment | added | Zhi-Wei Sun | I don't think individ's vague arguments are reasonable. Below $n$ one only has about $\root 3\of{n}$ cubes while one has about $\sqrt{n}$ squares. | |
| Apr 12, 2021 at 7:25 | comment | added | Yaakov Baruch | @individ: I don't completely follow. I see how you can reduce to two squares, but then from those cannot further reduce to one linear (due to $2/6$, or $3/5$ or $4/5$, not being square). Could you kindly clarify your comment? | |
| Apr 12, 2021 at 5:11 | comment | added | individ | There are four cubes. The transformation can be reduced to an equation with squares. And there will also be four squares. And as you know, four squares can make up any number. | |
| Apr 12, 2021 at 3:07 | comment | added | Zhi-Wei Sun | I also conjecture that each natural number can be written as $$\left\lfloor \frac{a^3+b^3}3\right\rfloor+\left\lfloor \frac{c^3+d^3}5\right\rfloor$$ with $a,b,c,d\in\mathbb N$. The pair $(3,5)$ may be replaced by some other pairs like $(4,5)$. | |
| Apr 12, 2021 at 2:52 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
added 37 characters in body
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| Apr 12, 2021 at 2:42 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |