Everything is in the title. I wonder if there is known good asymptotic (when $n\to+\infty$) for the quantity $LCM(\binom{2k}k)\_{1\le k\le n}$


Thanks in advance



Asymptotic of $Lcm(\binom{2k}k)_{1\le k\le n}$
Put $A\_n$ to be the lcm of $\binom{2k}{k}$ for $1\le k\le n$, and let $B\_n$ denote the lcm of the numbers at most $2n$ (thus $B\_n = \exp(\psi(2n))$). We claim that $A\_n = B\_n/2$ unless $n=2^r-1$ in which case $A\_n =B\_n$. Since asymptotics for the $\psi(2n) = \sum\_{r\le 2n} \Lambda(r)$ are well known (the prime number theorem), this solves the problem. 


Consider an odd prime $p$. The power of $p$ dividing $B\_n$ is the largest power $p^r$ lying below $2n$. Now, if $k\le n$, then the power of $p$ dividing $\binom{2k}{k}$ is 
$$ 
\sum\_{\ell=1}^{\infty}\Big( \Big\lfloor \frac{2k}{p^\ell} \Big\rfloor -2 \Big\lfloor \frac{k}{p^\ell} \Big\rfloor\Big). 
$$ 
Since only terms with $p^\ell \le 2k$ are relevant, and each term in parentheses is at most $1$, this exponent is at most $r$. Also if we take $2k=p^r+1$ (which is below $2n$, since $p^r$ is odd and at most $2n$) then 
$$ 
\Big\lfloor \frac{p^r+1}{p^\ell} \Big\rfloor -2 \Big\lfloor \frac{p^{r}+1}{2p^\ell} \Big\rfloor =p^{r-\ell} - 2\Big \lfloor \frac{p^{r}+1}{2p^{\ell}}\Big\rfloor = 1 
$$
for all $1\le \ell \le r$ (it must be odd, and therefore equal one). So the power of $p$ dividing $A\_n$ matches that dividing $B\_n$. 


The case for $2$ is similar. We check that the power of $2$ dividing $\binom{2k}{k}$ is strictly less than the largest power of $2$ at most $2k$, unless $k$ is one less than a power of $2$ in which case it equals the largest power of $2$ at most $2k$. 


