Timeline for A sum over partitions involving "subpartitions"
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2021 at 13:52 | comment | added | Christian Bertoni | Thank you for the very nice answer and the insight about the $x\neq 1$ case! | |
| Jul 20, 2021 at 23:30 | comment | added | Lev Borisov | Nah, don't see it even for the first derivative. | |
| Jul 20, 2021 at 23:19 | comment | added | Lev Borisov | Yeah, I don't see any pattern either. Perhaps there is a pattern for R_k'(u,1). | |
| Jul 20, 2021 at 22:52 | comment | added | Peter Taylor | sagecell.sagemath.org/… computes the first 12 and is easily extended, although I think the PowerSeriesRing constructor needs modification to use higher orders of $u$ if you want to get beyond about 18. I've spent a few hours with this and not seen anything which isn't easier to show directly. | |
| Jul 20, 2021 at 22:36 | comment | added | Lev Borisov | @PeterTaylor I would try to compute the first few $R_k(u,x)$ to see if there is a pattern. For example, it is easy to see that $R_k(u,x) = c_k(x) u^k + O(u^{k+1})$, so there is hope for some structure. | |
| Jul 20, 2021 at 14:37 | comment | added | Peter Taylor | @darijgrinberg, point conceded: the sum limits can be rewritten as $j \in \mathbb{N}^k$ to make this clearer. Lev, extending this to include $x$ I get $$R_k(u,x) = \exp(ux + \tfrac{u^2 x}2 + \cdots + \tfrac{u^kx}k) - \sum_{i=1}^{k} iS_{i-1}(x) \frac{\exp(\tfrac{u^ix}i) - 1}{ux} \exp(\tfrac{u^{i+1}x}{i+1}+\cdots+\tfrac{u^k x}k)$$ where $S_k(x) = [u^k] R_k(u,x)$ is the $S(k,x)$ of the question. The part outside the sum is easy and gives the upper bound $\binom{x+k-1}{k}$ already noted and improved upon, but the parts in the sum are less easy. | |
| Jul 20, 2021 at 10:22 | comment | added | darij grinberg | @PeterTaylor No, it would not be an empty sum. There is exactly one 0-tuple. | |
| Jul 20, 2021 at 9:58 | comment | added | Peter Taylor | @darijgrinberg, to start the induction at $k=0$ you need $R_0(u) = 1$, which is exceptional in that as defined it would be an empty sum. | |
| Jul 19, 2021 at 8:05 | history | bounty ended | Christian Bertoni | ||
| Jul 19, 2021 at 8:05 | vote | accept | Christian Bertoni | ||
| Jul 19, 2021 at 0:14 | comment | added | Lev Borisov | I am sure you are right :) | |
| Jul 18, 2021 at 21:50 | comment | added | darij grinberg | Nice proof, Lev! You can probably start your induction at $k=0$, though. | |
| Jul 18, 2021 at 21:45 | comment | added | darij grinberg | @MaxAlekseyev: Your "More generally" claim easily follows from Lev's proof, because if we set $f := \sum_{d=1}^k \dfrac{u^d}{d!}$ and $g := \sum_{d=k+1}^\infty \dfrac{u^d}{d!}$, then $f+g = -\log\left(1-u\right)$ and therefore $\exp\left(f+g\right) = \dfrac{1}{1-u}$, so that $\exp f \cdot \exp g = \exp\left(f+g\right) = \dfrac{1}{1-u}$ and therefore $\left(1-u\right) \exp f = \dfrac{1}{\exp g} = \exp\left(-g\right) \equiv 1-g \mod u^{2k+2}$ (since $g$ is divisible by $u^{k+1}$, and thus $\left(-g\right)^i \equiv 0 \mod u^{2k+2}$ for all $i \geq 2$). The rest you can figure out easily :) | |
| Jul 18, 2021 at 18:51 | comment | added | Lev Borisov | Yes, I noticed that. The point is that only "linear" terms in the exponentials contribute. | |
| Jul 18, 2021 at 16:04 | comment | added | Max Alekseyev | More generally, it seems that $$R_k(u) = \sum_{d=k}^{2k} \frac{u^d}{d+1} + O(u^{2k+1}).$$ | |
| Jul 17, 2021 at 23:15 | comment | added | Lev Borisov | I have not tried to work with the $S(n,x)$. Presumably, the method will still provide something, but I don't know how explicit it will be. | |
| Jul 17, 2021 at 20:22 | history | edited | Lev Borisov | CC BY-SA 4.0 |
added 97 characters in body
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| Jul 17, 2021 at 20:10 | history | answered | Lev Borisov | CC BY-SA 4.0 |