# Count of lattices on finite set

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)

• 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 – JBL Oct 30 '10 at 15:20
• (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.) – JBL Oct 30 '10 at 15:30
• Ah, I see, it's just that $n \leq 2$ and $n \leq 4$ are meant to be $n \geq 2$ and $n \geq 4$. – JBL Oct 30 '10 at 18:12
• I mean lattice as poset in which any two elements have a unique supremum and infimum. – tomas.lang Nov 1 '10 at 13:37