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\begin{document}${\cal B}_n$\end{document} be the set of all \documentclass[12pt]{minimal}
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\begin{document}$n\times n$\end{document} Boolean matrices. Let R(A) denote the row space of \documentclass[12pt]{minimal}
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\begin{document}$A\in{\cal B}_n$\end{document}, let \documentclass[12pt]{minimal}
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\begin{document}${\cal R}_n=\{r \mid r={\rm r}(A),\ A\in {\cal B}_n \}$\end{document}, and let \documentclass[12pt]{minimal}
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\begin{document}$a_n=\min\{q\ge 1\mid q\notin {\cal R}_n\}$\end{document}. By extensive computation we found that
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\begin{document}${\cal R}_9\cap[1,256]=[1,190]\cup [192,204]\cup\{206\}\cup[208,212]\cup\{214,216,220\}\cup\,[224,228]\cup\{230,232,236,240,248,256\},$\end{document} and therefore \documentclass[12pt]{minimal}
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\begin{document}$a_9=191$\end{document}. Furthermore, \documentclass[12pt]{minimal}
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\begin{document}$a_n\ge 5{\sqrt[11]{336}\,}^n$\end{document} for \documentclass[12pt]{minimal}
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\begin{document}$n\ge 31$\end{document}. We proved that if \documentclass[12pt]{minimal}
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\begin{document}$n\ge 7$\end{document}, then the set \documentclass[12pt]{minimal}
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\begin{document}${\cal R}_n\cap(2^{n-2}+2^{n-3},2^{n-1}]$\end{document} contains at least \documentclass[12pt]{minimal}
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\begin{document}$n^2-7n+14+\frac{1}{24}\left((n-8)(n-10)(2n-15)+3(n\bmod 2)\right)$\end{document}
elements.
