The Dedekind number $M(n)$ is the number of antichains in the partial order of subsets of $\{1,\dotsc,n\}$. It is only known for $0 \leq n \leq 8$.
Question. What are some known upper and lower bounds for $M(9)$?
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Why is the section on asymptotics insufficient? – Mr Tsjolder from codidact Feb 04 '16 at 20:40
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Because asymptotics do not tell us anything about concrete values, right? – Martin Brandenburg Feb 04 '16 at 21:34
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does it not say that $2^{126} \leq M(9) \leq 1.106 \cdot 2^{126}$? These seem like concrete values... – Mr Tsjolder from codidact Feb 05 '16 at 07:38
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1@MrTsjolder: I do not see how this follows from the Wikipedia article, since the Landau symbols only tell us what happens for $n \to \infty$. This is obviously not the kind of mathematics that I am used to. Perhaps you can elaborate on this in a proper answer? – Martin Brandenburg Feb 05 '16 at 09:22
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As a naive computer scientist I tend to take the big O notation as upper boundary, but indeed this is not guaranteed because the upper boundary only counts as from a certain $c$ which is unknown (I did not read through the proof). The lower bound on the other hand is quite concrete... PS: safer bounds could be $2^{126} \leq M(9) \leq 2^{252}$ using monotonicity of the Dedekind numbers... – Mr Tsjolder from codidact Feb 05 '16 at 09:37
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Yes, but this estimate is obviously not very precise. – Martin Brandenburg Feb 05 '16 at 19:31
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$M(9)$ was computed exactly in $2023$ and is now known to be
$$M(9) = \boxed{ 286386577668298411128469151667598498812366 }.$$
This computation was done independently by Christian Jäkel in A computation of the ninth Dedekind Number and by Lennart Van Hirtum, Patrick De Causmaecker, Jens Goemaere, Tobias Kenter, Heinrich Riebler, Michael Lass, and Christian Plessl in A computation of D(9) using FPGA Supercomputing.
Incidentally, the lower bound $\log_2 M(n) \ge {n \choose \lfloor \frac{n}{2} \rfloor}$ is not just an asymptotic estimate, it is an exact inequality (it just counts the number of antichains consisting of any family of subsets of size $\lfloor \frac{n}{2} \rfloor$), and it gives $\log_2 M(9) \ge {9 \choose 4} = 126$. The exact value above satisfies $\log_2 M(9) = 137.7 \dots$ which is within the range of the asymptotic upper bound given on Wikipedia, which is "$126 \left( 1 + O \left( \frac{\log 9}{9} \right) \right)$" (so to speak). Here $126 \left( 1 + \frac{\log 9}{9} \right) = 156.8 \dots $.
Qiaochu Yuan
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Great! Thanks for reviving this. https://ih1.redbubble.net/image.2635601056.3192/raf,360x360,075,t,fafafa:ca443f4786.u2.jpg – Martin Brandenburg Aug 11 '24 at 11:01