Random k-Sat: A Tight Threshold For Moderately Growing k

We consider a random instance I of k-SAT with n variables and m clauses, where k=k(n) satisfies k—log2n→∞. Let m0=2knln2 and let ∈=∈(n)>0 be such that ∈n→∞. We prove that \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ {}^{{\lim }}_{{n \to \infty }} \Pr {\left( {I\;{\text{is}}\;{\text{satisfiable}}} \right)} = \left\{ {^{{1\;m \leqslant {\left( {1 - \in } \right)}m_{0} }}_{{0\;m \geqslant {\left( {1 + \in } \right)}m_{0} }} .} \right. $$\end{document}