TITLE: Do we know any bound on $\operatorname{lcm}(2^1-1, 2^2-1,\dots,2^n-1)$?
QUESTION [9 upvotes]: $\DeclareMathOperator\lcm{lcm}$We know that $\operatorname{lcm}(1,\dotsc,n)$ is approximately $e^n$ and we also know that $\gcd(2^a-1, 2^b-1)=2^{\gcd(a,b)}-1$.
I wonder if there exists an upper bound/lower bound/approximation for $\operatorname{lcm}(2^1-1, 2^2-1,\dotsc,2^n-1)$.

REPLY [5 votes]: The Szymiczek paper that I mentioned in comments some time ago is now available on the web. From Theorems 7 and 8 in the paper, we get $$\log{\rm lcm}(2^1-1,\dots,2^{[x]}-1)={3\log2\over\pi^2}x^2+O(x\log x)$$