Timeline for On subsets of $\mathbb{N}$ reciprocally summable to $1$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 28, 2019 at 14:44 | comment | added | Dominic van der Zypen | @user142929 Thanks for your ideas - you don't disturb at all. If you want to contact me about this, use twitter.com/dominiczypen and write me a direct message | |
| Aug 28, 2019 at 8:28 | comment | added | user142929 | I wondered if it can be interesting to create different variants of the problem, for different definitions involving multiplicative functions $f(n)$ for the sums $\sum_{a\in A}\frac{1}{f(a)}$, for example I imagine write $f(n)=\operatorname{rad}(n)$, the product of distinct primes dividing $n>1$ (see the Wikipedia Radical of an integer) instead of your $f(n)=n$ in the denominators. If you think that it has a good mathematical content, feel free to study it. I hope don't disturb. | |
| Aug 19, 2019 at 7:05 | vote | accept | Dominic van der Zypen | ||
| Aug 18, 2019 at 22:56 | comment | added | Gabe Conant | Let $N=1+\max A$ and $r=1-\sum_{a\in A}\frac{1}{a}$. Then the Main Theorem of this paper by Croot implies that we can find $A'$ so that $\max A'<e^{r+o(1)}N$. | |
| Aug 18, 2019 at 20:36 | answer | added | Fedor Petrov | timeline score: 12 | |
| Aug 18, 2019 at 20:09 | comment | added | Gerhard Paseman | I believe yes, because for any finite extension E to bring the sum up to 1, if it over shoots, replace E by an allowed scaling factor, and try again. There should be some algorithms in the literature on Egyptian fractions. Gerhard "Go Sum Like An Egyptian" Paseman, 2019.08.18. | |
| Aug 18, 2019 at 19:52 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |