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Jan 10, 2022 at 18:55 comment added Jukka Kohonen If the problems are solved, it would be nice to wrap them up into a genuine answer so that the question is no longer "unanswered". (You can even answer your own question.)
Dec 28, 2021 at 23:34 comment added SY.Xu Thank you for all your comments sincerely! I think my problems are solved.
Dec 28, 2021 at 17:16 comment added Max Alekseyev See oeis.org/A025591
Dec 28, 2021 at 17:15 comment added Richard Stanley One reference is pages 67-71 of my book Algebraic Combinatorics, second ed. $S_n$ is the middle coefficient of the polynomial $(1+q)(1+q^2)\cdots (1+q^n)$. There is unlikely to be an explicit formula, but there is a known asymptotic formula $S_n \sim \sqrt{\frac{6}{\pi}}n^{-3/2}2^n$. See OEIS A025591.
Dec 28, 2021 at 14:27 comment added SY.Xu Thanks for comments. I think the poset $M(n)$ in Stanley's article "Some applications of algebra to combinatorics" is exactly what I described above. However I have a pool knowledge about the rank generating function, is there any way to get the maximal rank size from this function? There are only some conclusions about $M(n)$ in this article, is there any reference books which can help me understand better?
Dec 28, 2021 at 13:52 comment added Christian Gaetz I think your poset is isomorphic to the one in this question: mathoverflow.net/questions/76723/…. If So then it is known to be strongly Sperner by a result of Stanley, so the maximal antichain size is just the maximal rank size, and the rank generating function is known.
Dec 28, 2021 at 13:45 comment added Fedor Petrov It is not much less, at most by a $\sqrt{n}$ factor, since you may consider all subsets with given sum like $n(n+1)/4$, and by certain CLT there are about $\Theta(2^n/n)$ such subsets.
Dec 28, 2021 at 13:14 history edited SY.Xu CC BY-SA 4.0
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S Dec 28, 2021 at 13:05 review First questions
Dec 28, 2021 at 13:29
S Dec 28, 2021 at 13:05 history asked SY.Xu CC BY-SA 4.0