find a upper bound of a product of gcd

Question

Let $n,k \in \mathbb{N}$.

Find a upper bound for

$\gcd(n,2^n) \gcd(n+1, 2^{n+1}) \cdots \gcd(n+2^k, 2^{n+2^k})$.

Note that in the upper bound, there should be no $n$, but having $k$ is fine.

Attempt:

$\gcd(n,2^n) \gcd(n+1, 2^{n+1}) \cdots \gcd(n+2^k, 2^{n+2^k}) \leq \gcd(n, 2^{n+2^k}) \gcd(n+1, 2^{n+2^k}) \cdots \gcd(n+2^k, 2^{n+2^k}) = \frac{\gcd(1, 2^{n+2^k}) \gcd(2, 2^{n+2^k}) \cdots \gcd(n+2^k, 2^{n+2^k})}{\gcd(1, 2^{n+2^k}) \gcd(2, 2^{n+2^k}) \cdots \gcd(n-1, 2^{n+2^k})}$

Then I tried to eliminate the $n's$ by using How to calculate "gcd product" $\operatorname{gcdp}(n,m)=\gcd(n,1)\gcd(n,2)\cdots\gcd(n,m)$.

Then it just got really messy.

What I'm really trying to do is to find a bound without the $n$. But leaving $k$ is fine.

It would be even better to find the actually value...

Thank you.

Answer 1

The important thing to note is that there is no upper bound in terms of $k$ only.

For any $k \in \mathbb{N}$, let $n=2^N$ where we can choose $N$ to be as large as we like. Then the quantity that you are trying to bound contains the factor $$2^N=\gcd(2^N,2^{2^N})$$ independently of whatever $k$ is.