Timeline for How to solve a given combinatorial problem?

11 events

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May 18 at 10:57	comment	added	ploosu2		I got interested in the limiting behavior of this as $n$ and $k$ both go to infinity. I asked a new question: math.stackexchange.com/q/4918569/111594
May 16 at 18:53	vote	accept	LaVuna47
S May 16 at 13:51	history	suggested	Satana	CC BY-SA 4.0	Improved formatting
May 16 at 13:41	review	Suggested edits
S May 16 at 13:51
May 16 at 12:33	answer	added	ploosu2		timeline score: 3
May 15 at 12:12	comment	added	Henry		For $k\gg n$, I would have thought the sum equivalent to your $760$ would approach $n^{2k}\times n(1-(1-\frac1n)^n)$ asymptotically as $k$ increases. For $n=4, k=2$ this suggests $700$ rather than $760$, so not too far away even for such a low $k$. For $n=4, k=5$ this gives $2867200$ rather than the correct $2895232$, so getting much closer relatively.
May 15 at 11:59	comment	added	Henry		Good luck with your polynomial time ambition: you could see this as a Markov chain with states equivalent to the partitions of $n$, and there are almost $\frac {1} {4n\sqrt3} \exp\left({\pi \sqrt {\frac{2n}{3}}}\right)$ of those. But you could then do a recursion which was linear in $k$ if not $n$.
May 15 at 11:51	comment	added	Henry		Note that $28$ of your $40$ "four non-empty boxes" options are the starting position.
May 15 at 11:22	history	edited	Red Five	CC BY-SA 4.0	TeX edits
S May 15 at 11:11	review	First questions
May 15 at 11:14
S May 15 at 11:11	history	asked	LaVuna47	CC BY-SA 4.0