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The are some linearly upper bounds on Hermite's constant $\gamma_d$, such as $\gamma_d \leq 2d/3$, $\gamma_d \leq d/4+1$. So we can claim that $\gamma_d=O(d)$. There is also a rather tight asymptotical bound for $\gamma_n$: $\frac{d}{2\pi e}+\frac{\log(\pi d)}{2\pi e}+o(1) \leq \gamma_d \leq \frac{1.744d}{2 \pi e}(1+o(1))$ (see page 34 of The LLL algorithm edited by Phong Q. Nguyen et al.). After this tight asymptotical bound, there is a sentence that "Thus, $\gamma_d$ is enssentially linear in $d$" in this book (also in page 34). My questions is that can we express $\gamma_d$ in the asymptotical form of $\Theta(d)$. I have searched some related papers, but I can't find the the corresponding expression.

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  • $\begingroup$ What you want seems to follow from the inequality you have linked? The first inequality implies that $\gamma_d = \Omega(d)$. The second inequality implies that $\gamma_d = O(d)$. Combined you have that $\gamma_d =\Theta(d)$. $\endgroup$
    – Mark
    Aug 2, 2023 at 5:39

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