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Let $f:\mathbb N_+\to\mathbb N_+, f(n)=\min\{\max\{k!,(n-k)!,(n!-k!(n-k)!)\}|k\in\mathbb N_+, n-k> 0,\ n!-k!(n-k)!> 0 \}$.

What's the asypmtotic growth of $f(n)$?

Is it true that $f(n)=\Theta((n/2)!)$?

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  • $\begingroup$ "n!-k!(n-k)!>0" is always true as long as $0\lt k\lt n$. $\endgroup$ – Apass.Jack Jun 8 at 4:44
  • $\begingroup$ Why didn't you accept the answer given? $\endgroup$ – Viktor Glombik Aug 30 at 14:23
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It is not true that $f(n)=\Theta((n/2)!)$.

$$\begin{aligned} f(n)&=\min_{k\in\mathbb N_+, n-k> 0,\ n!-k!(n-k)!> 0}\{\max\{k!,\,(n-k)!,\,(n!-k!(n-k)!)\}\\ &=\min_{0\lt k\lt n}\{\max\{k!,\,(n-k)!,\,(n!-k!(n-k)!)\}\\ &=\min_{0\lt k\lt n}\{n!-k!(n-k)!\}\\ &=n!-\max_{0\lt k\lt n}\{k!(n-k)!\}\\ &=n!-1!(n-1)!\\ &=(n-1)(n-1)!\\ &=\omega((n/2)!) \end{aligned}$$

In fact, $f(n)\sim n!\sim\sqrt{2\pi n}\left(\dfrac ne\right)^n$ by Stirling's approximation.

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