THE LARGEST AND ALL SUBSEQUENT NUMBERS OF CONGRUENCES OF \(n\)-ELEMENT LATTICES
Gábor Czédli · Ural mathematical journal · 2025
For a positive integer \(n\), let SCL\((n)=\{|\)Con \((L)|: L\) is an \(n\)-element lattice\(\}\) stand for the set of Sizes of the Congruence Lattices of \(n\)-element lattices. The \(k\)-th Largest Number of Congruences of \(n\)-element lattices, denoted by lnc \((n, k)\), is the \(k\)-th largest member of SCL \( (n)\). Let \((n_1,\dots,n_6):=(1,4,5,6,6,7)\), and let \(n_k:=k\) for \(k\geq 7\). In 1997, R. Freese proved that for \(n\geq n_1=1\), lnc \( (n, 1)=2^{n-1}\). For \(n\geq n_2\), the present author gave lnc \((n, 2)\). For \(k=3,4,5\) and \(n\geq n_k\), C. Mureşan and J. Kulin determined lnc \((n, k)\) in their 2020 paper. For \(k\leq 5\) and \(n\geq n_k\), the above-mentioned authors described the \(n\)-element lattices witnessing lnc \((n, k)\), too. For all positive integers \(k\) and \(n \geq n_k\), this paper determines lnc \((n, k)\) and presents the lattices that witness it. It turns out that, for each fixed \(k\), the quotient lcd \((k):=\) lnc \((n, k)/\) lnc \((n, 1)\) does not depend on \(n\geq n_k\). Furthermore, lcd \((k)\) converges to \(1/8\) as \(k\) tends to infinity.