Let $n, k \in \mathbb{N}$.
Find $\gcd(n, 2^n)\cdot\gcd(n+1, 2^{n+1}) \cdots \gcd(n+k, 2^{n+k-1})$.
I've been thinking about this for a while. I tried to plug in $n=1$, but didn't find patterns. The answer seems to be $2^{k-1}$. Not sure if this has anything to do with the Euler Phi function.
Thanks!
Note that $2^k>k$ holds for all $k\in\mathbb{N} \implies V_{2}(2^k)>V_{2}(k) \implies gcd(k,2^k)=V_2(k)$
So what you're truly seeking is the value of $$\displaystyle\prod_{t=n}^{n+k}{V_{2}(t)}=V_2\Bigg({\displaystyle\prod_{t=n}^{n+k}}t\Bigg)$$
Which is equal to $$V_2\Bigg(\frac{(n+k)!}{(n-1)!}\Bigg)=V_2\bigg((n+k)!\bigg)-V_2\bigg((n-1)!\bigg)$$
But it's well known that $V_2(k!)=\displaystyle\sum_{i=1}^{\infty}{\bigg\lfloor\frac{k}{2^i}}\bigg\rfloor$ so finally we deduce that : $$\displaystyle\prod_{t=n}^{n+k}{gcd(t,2^t)}=2^s$$ Where $s$ is equal to $${\displaystyle\sum_{i=1}^{\infty}{\bigg\lfloor\frac{n+k}{2^i}}\bigg\rfloor-\displaystyle\sum_{i=1}^{\infty}{\bigg\lfloor\frac{n-1}{2^i}}\bigg\rfloor}={\displaystyle\sum_{i=1}^{\infty}{\Bigg(\scriptsize{\bigg\lfloor\frac{n+k}{2^i}}\bigg\rfloor-\bigg\lfloor\frac{n-1}{2^i}}\bigg\rfloor\Bigg)}$$