Do we know any bound on $\operatorname{lcm}(2^1-1, 2^2-1,\dotsc\dots,2^n-1)$?

$\DeclareMathOperator\lcm{lcm}$We know that $\lcm(1,\dotsc,n)$$\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 $\lcm(2^1-1, 2^2-1,\dotsc,2^n-1)$$\operatorname{lcm}(2^1-1, 2^2-1,\dotsc,2^n-1)$.

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edited May 3, 2020 at 23:36

LSpice

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Do we know any bound on $\lcm$\operatorname{lcm}(2^1-1, 2^2-1,\dots\dotsc,2^n-1)$?

We$\DeclareMathOperator\lcm{lcm}$We know that $\lcm(1,\dots,n)$$\lcm(1,\dotsc,n)$ is approximately $e^n$ and and also we also know that $\gcd(2^a-1, 2^b-1)=2^{\gcd(a,b)}-1$.

I wonder if there exists an upperboundupper bound/lowerboundlower bound/approximation for $\lcm(2^1-1, 2^2-1,\dots,2^n-1)$$\lcm(2^1-1, 2^2-1,\dotsc,2^n-1)$.