A finite n-element semilattice is planar if it has at least 127 * 2^(n-8) subsemilattices, and this bound is sharp for n > 8 via an explicit non-planar counterexample with one fewer subsemilattice.
A note on lattices with many sublattices
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abstract
For every natural number $n\geq 5$, we prove that the number of subuniverses of an $n$-element lattice is $2^n$, $13\cdot 2^{n-4}$, $23\cdot 2^{n-5}$, or less than $23\cdot 2^{n-5}$. By a subuniverse, we mean a sublattice or the emptyset. Also, we describe the $n$-element lattices with exactly $2^n$, $13\cdot 2^{n-4}$, or $23\cdot 2^{n-5}$ subuniverses.
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math.RA 1years
2019 1verdicts
UNVERDICTED 1representative citing papers
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One hundred twenty-seven subsemilattices and planarity
A finite n-element semilattice is planar if it has at least 127 * 2^(n-8) subsemilattices, and this bound is sharp for n > 8 via an explicit non-planar counterexample with one fewer subsemilattice.