MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)

share|cite|improve this question
It's not clear what sort of lattice you're looking for. Lattices (posets) on unlabeled nodes are at -- on labeled nodes are $$ $$ I found this by putting the word "lattice" and the first few terms (computed by hand) into the OEIS: – 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
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.