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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 \leq 2$$n \geq 2$. My other estimates are $p(n) \leq 2^{\frac{(n-1)(n-2)}{2}}$ (also $p(n) \leq 2^{\frac{(n-1)}{2}}$) and $p(n-1) < p(n)$ for $n \leq 4$$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)

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 $n \leq 2$. 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 $n \leq 4$. 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)

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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Gerry Myerson
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Let $p(n)$ denote count of laticceslattices on finite set $G$, $|G|=n$ (without isomoprhismisomorphism). 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 $n \leq 2$. 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 $n \leq 4$. 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)

Let $p(n)$ denote count of laticces on finite set $G$, $|G|=n$ (without isomoprhism). 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 $n \leq 2$. 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 $n \leq 4$. 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)

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 $n \leq 2$. 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 $n \leq 4$. 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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JBL
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