Minimal Least Common Multiplier for $n$ distinct integers Let $n\in\mathbb N^+$ be a positive integer.
I am trying to find a set of $n$ distinct positive integers with minimal LCM.


*

*What is the set of $n$ distinct integers with minimal LCM?

*What is the asymptotics of this LCM as a function of $n$?

Observe that we can choose the set to be $\{1,...,n\}$, which would give a bound of $2^{O(n)}$. However, I'm not sure that this is tight.
For example, if $n=9$ then LCM$(\{1,2,3,4,5,6,7,8,9\})=2520$, but we have 
LCM$(\{1,2,3,4,6,8,12,16,24\})=48$.
 A: ADDED: see Nicolas on highly composite 
Nicolas  Full List 
you can get a predictably small LCM by taking the first Superior Highly Composite Number with number of divisors at least your target number, you are calling that $n$ 
https://en.wikipedia.org/wiki/Superior_highly_composite_number
For your $7 \leq n \leq 12,$ you can get a small LCM by taking $n$ divisors of $60$ 
For your $13 \leq n \leq 16,$ you can get a small LCM by taking $n$ divisors of $120$ 
For your $17 \leq n \leq 24,$ you can get a small LCM by taking $n$ divisors of $360$ 
For your $25 \leq n \leq 48,$ you can get a small LCM by taking $n$ divisors of $2520$ 
A: HINT:
Given $n$ integers
$$
a_{\,1} ,a_{\,2} , \cdots ,a_{\,n} 
$$
having the prime factorization
$$
a_{\,k}  = p_{\,1} ^{\,b_{\,1,\;k} } p_{\,2} ^{\,b_{\,2,\;k} } \; \cdots \;p_{\,q} ^{\,b_{\,q,\;k} } 
$$
then the factorization of their LCM will have exponents which are the "envelope" , i.e.the maximum
of the exponents for each prime
$$
\eqalign{
  & LCM\left( {a_{\,1} ,a_{\,2} , \cdots ,a_{\,n} } \right) = p_{\,1} ^{\,\max \left( {b_{\,1,\;k} } \right)} p_{\,2} ^{\max \left( {\,b_{\,2,\;k} } \right)} \; 
  \cdots \;p_{\,q} ^{\max \left( {\,b_{\,q,\;k} } \right)}  =   \cr 
  &  = 2^{\,l_{\,1} \,m_{\,1}  + l_{\,2} \,m_{\,2}  + \; \cdots  + l_{\,q} \,m_{\,q} } \quad \left| \matrix{
  \;l_{\,j}  = \log _2 p_{\,j}  \hfill \cr 
  \;m_{\,j}  = \max \left( {\,b_{\,j,\;k} \left| {\;1 \le k \le n} \right.} \right) \hfill \cr}  \right. \cr} 
$$
Now, taking the integers to be consecutive powers of $2$ we have
$$
\left( {2^{\,0} ,2^{\,1} , \cdots ,2^{\,n - 1} } \right)\quad  \Rightarrow \quad \left\{ \matrix{
  d = No.\,of\,digits = n \hfill \cr 
  \log _2 LCM = n - 1 \hfill \cr}  \right.
$$
while adding a $3$
$$
\left( {\matrix{
   {2^{\,0} ,} & {2^{\,1} ,} & { \cdots ,} & {2^{\,n - 1} ,}  \cr 
   {2^{\,0} 3,} & {2^{\,1} 3} & { \cdots ,} & {2^{\,n - 1} 3}  \cr 
 } } \right)\quad  \Rightarrow \quad \left\{ \matrix{
  d = No.\,of\,digits = 2n \hfill \cr 
  \log _2 LCM = n - 1 + \log _2 3 \hfill \cr}  \right.
$$
That means that taking  the products of the first q primes,
with all the combinations of the exponents of each 
varying from $0$ to a given maximum $m_k$, we get
$$
Comb.\left( {p_{\,1} ^{\,0, \cdots ,m_{\,1} } p_{\,2} ^{\,0, \cdots ,m_{\,2} } \; \cdots \;p_{\,q} ^{\,0, \cdots ,m_{\,q} } } \right)\quad  \Rightarrow \quad \left\{ \matrix{
  D = \log _2 No.\,of\,digits = \log _2 n = \sum\limits_k {\log _2 \left( {1 + m_{\,k} } \right)}  \hfill \cr 
  L = \log _2 LCM = \sum\limits_k {l_{\,k} \,m_{\,k} }  \hfill \cr}  \right.
$$
and given $D$ to minimize $L$ (or v.v. given $L$ to maximize $D$), wrt  the variables $m_k$.
