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Let $p(n)$ denote count of lattices on finite set $G$, $|G|=n$ (without isomorphism). It's know closed formula for $p(n)$?

It's clear, that $1 \leq p(n)$ and also that $p(n-1) \leq p(n)$ for $n \geq 2$. My other estimates are $p(n) \leq 2^{\frac{(n-1)(n-2)}{2}}$ (also $p(n) \leq 2^{\frac{(n-1)}{2}}$) and $p(n-1) < p(n)$ for $n \geq 4$. Better lower bound for $p(n)$ is $\min(1,n - 2) \leq p(n)$

If there are not closed formula for $p(n)$, what we are able say about that function?

Thanks for help. (Sorry for my bad English)

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  • $\begingroup$ It's not clear what sort of lattice you're looking for. Lattices (posets) on unlabeled nodes are at oeis.org/A006966 -- on labeled nodes are oeis.org/A055512 $$ $$ I found this by putting the word "lattice" and the first few terms (computed by hand) into the OEIS: oeis.org/Seis.html $\endgroup$
    – JBL
    Commented Oct 30, 2010 at 15:20
  • $\begingroup$ (Part of the reason it's not clear is that for either interpretation, at least one of the bounds you've written down is wrong.) $\endgroup$
    – JBL
    Commented Oct 30, 2010 at 15:30
  • $\begingroup$ Ah, I see, it's just that $n \leq 2$ and $n \leq 4$ are meant to be $n \geq 2$ and $n \geq 4$. $\endgroup$
    – JBL
    Commented Oct 30, 2010 at 18:12
  • $\begingroup$ I mean lattice as poset in which any two elements have a unique supremum and infimum. $\endgroup$
    – tomas.lang
    Commented Nov 1, 2010 at 13:37

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1, 1, 1, 1, 2, 5, 15, 53, 222, 1078, 5994, 37622, 262776, 2018305, 16873364, 152233518, 1471613387, 15150569446, 165269824761, ...

There is a lot of information in The On-Line Encyclopedia of Integer Sequences.

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