Asymptotic of $Lcm(\binom{2k}k)_{1\le k\le n}$

Question

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

Answer 1

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 $2k$.

Answer 2

Kummer's theorem implies the formula: $$\mathrm{LCM}(\binom{2}{1},\binom{4}{2},\dots,\binom{2n}{n}) = 2^{\lfloor \log_2(n+1)\rfloor} \cdot \prod_{3\leq p\leq 2n} p^{\lfloor \log_p(2n-1)\rfloor},$$ where $p$ runs over primes.

In particular, $$\mathrm{LCM}(\binom{2}{1},\binom{4}{2},\dots,\binom{2n}{n}) \leq (n+1)\cdot (2n-1)^{\pi(2n)-1}$$ where the r.h.s. is larger than LCM in at most $(2n)\#$ (primorial) times.