In a recent question Proof related to Harmonic Progression it was shown that if $m_1, m_2 \ldots m_k$ are positive integers such that $m_1 < m_2 < \ldots < m_k$ and $$ \left\{\frac1{m_1}, \frac1{m_2}, \ldots ,\frac1{m_k} \right\} $$ form an arithmetic progression, then $$ m_1 \geq k-1 $$ For $k \in \{2,3,4\}$ this bound in fact is saturated. For example, $$ \left\{ \frac13, \frac14, \frac16,\frac1{12} \right\} $$ are an arithmetic progression, so $m_1 (k=4)$ can be $3$.
For a given sequence length $k$ let $m(k)$ be the smallest value of $m_1$ such that such an arithmetic progression of reciprocals of length $k$ can be formed starting from $\frac1{m_1}$. So for example, $$ m(2) = 1 \\ m(3) = 2 \\ m(4) = 3 \\ m(5) = m(6) = 10 \\ m(7) = 60\\ m(8) = 105 \\ \vdots $$
I'd like to know the asymptotic expression for $m(k)$ for large $k$. $m(k)$ will in most cases be the least common multiple of $1, 2, 3, \ldots k$ divided by $k$. in some cases such as $k=6$ $m(k)=m(k-1)$ but still the statement about
$$
m(k) = \frac{\operatorname{l.c.m.} (n\mid 1<n\leq k)}{k}
$$
almost holds true; for the lesser $k$ (for example, $k=5$) the denominator is $k+1$ instead of $k$.
But while I know the asymptotics of $k!$ (Stirling's approximation) and this g.c.d. is sort of related to $k!$, I don't know the asymptotics of this l.c.m.