# Number of possible partial orderings on a finite set

I've been reading about lattices and partial orders (my reference: Applied Abstract Algebra by Lidl, Pilz) while this question struck me.

Let X be a finite set. Is there any way to determine the total number of non-isomorphic partial orderings of this set?

I searched math.SE a bit, but couldn't find any similar question. Any kind of response/help will be appreciated. Best regards.

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How do you define isomorphism between Hasse diagrams, so that two Hasse diagrams fail to be isomorphic even while the posets they represent are the same (i.e. isomorphic)? I know that two Hasse diagrams for the same partial ordering can look very different from each other, but I don't know what definition of isomorphism one would use. –  Michael Hardy Apr 26 '12 at 17:25
The question is phrased in a misleading way. What it seems to mean, if we are to judge by the O.P.'s comments and by his acceptance of an answer, is: "Let $X$ be a finite set. Is there any way to determine the total number of non-isomorphic partial orderings of this set?" –  Michael Hardy Apr 26 '12 at 18:09
@MichaelHardy : Yes, that is what I actually meant. I thought every partial order has a unique representation by Hasse diagram. I've changed the question accordingly. Thanks a million for pointing this out. –  Sayantan Apr 27 '12 at 0:55

That OEIS entry seems to be about the number of posets with $n$ vertices, not the number of Hasse diagrams corresponding to one poset. Two different Hasse diagrams corresponding to the same poset, representing the exact same partial ordering, may look quite different from each other. But the posets they represent are isomorphic to each other. I don't know what it means to say the Hasse diagrams themselves are or are not isomorphic to each other in such a situation. –  Michael Hardy Apr 26 '12 at 16:44