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Let $G$ be a finite group. Define $t(G)$ as the minimal number, such that $\forall X \subset G$ if $|X| > t(G)$ and $\langle X \rangle = G$, then $XXX = G$. Is there some sort of formula for $t(S_n)$, for the symmetric group $S_n$?

Here $XXX$ stands for $\{abc| a, b, c \in X\}$.

I have asked a similar question without forcing $X$ to generate, and received an example of an $X$, such that $|X| = 2(n - 5)!$, but $XXX \neq \langle X \rangle$. It was $S_{n-5} \times \{0; 1\}$, which lies in the subgroup $S_{n - 5} \times C_5$. However, it was conjectured in the comments, that the largest counterexamples, that generate $G$, are very likely to be much smaller.

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    $\begingroup$ @PietroMajer, yes, I meant $t(G)$. $\tau(G)$ was in the previous question... $\endgroup$
    – Chain Markov
    Commented Jul 14, 2019 at 12:37
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    $\begingroup$ Check the references for theorem 3.3 in Breuillard-Lubotzky (ArXiv). It gives something such as $|t(S_n)|\le |S_n|^{1/(1+\delta_n)}$ for some $\delta_n>0$. Of course without stating what is $\delta_n$, this is empty information, but the references are likely to provide you with an explicit value ($\delta_n$ ought to tend to $0$ sufficiently slowly to give an interesting lower bound). $\endgroup$
    – YCor
    Commented Jul 14, 2019 at 12:45
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    $\begingroup$ The lower bound given in my answer still applies, because the $X$ I chose does generate. $\endgroup$
    – Geoff Robinson
    Commented Jul 14, 2019 at 12:51

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