Let $s=max_{n}\frac{2^{\omega (n)}}{n^{1/4}}=max_{m}\frac{2^{m}}{(p_{1}p_{2}...p_{m})^{1/4}}=\frac{2^{\pi (2^{4})}}{\prod_{p\le 2^{4}}p^{1/4}}=\frac{64}{30030^{1/4}}=4.86173341$. Then $k=[s]+1=5$.

What is $\pi(n)$ ?

$\pi(n)$ is the number of prime $\leq n$

Rust wrote: $max_{m}\frac{2^{m}}{(p_{1}p_{2}...p_{m})^{1/4}}=\frac{2^{\pi (2^{4})}}{\prod_{p\le 2^{4}}p^{1/4}}$ Why?

N.T.TUAN wrote: Rust wrote: $max_{m}\frac{2^{m}}{(p_{1}p_{2}...p_{m})^{1/4}}=\frac{2^{\pi (2^{4})}}{\prod_{p\le 2^{4}}p^{1/4}}$ Why? If $p_{m}<2^{4}$, then $\frac{2}{p_{m}^{1/4}}>1$, therefore $\frac{2^{m}}{(p_{1}p_{2}...p_{m})^{1/4}}>\frac{2^{m-1}}{(p_{1}p_{2}...p_{m-1})^{1/4}}.$

Ok! Thanks!

My solution: Let $ p_0=2<p_1<p_2<\cdots $ be the sequence of all prime number. Then, there exist an index $ i $ such that \[ p_0p_1p_2\cdots p_{i-1}\leq n<p_0p_1p_2\cdots p_i \] We have $ \omega(n)\leq i $ and \[ \frac{2^{\omega(n)}}{\sqrt[4]{n}} \leq \frac{2^i}{\sqrt[4]{p_0p_1\cdots p_{i-1}}} \] Hence \[ k=\text{max} \{0,\lfloor \frac{2^2}{\sqrt[4]{2\cdot 3}} \rfloor\,\lfloor \frac{2^2}{\sqrt[4]{2\cdot 3\cdot 5}} \rfloor, \lfloor \frac{2^2}{\sqrt[4]{2\cdot 3\cdot 5\cdot 7}} \rfloor, \cdots \}=\lfloor \frac{2^2}{\sqrt[4]{2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13}} \rfloor=5 \] satisfied the given condition. Easy to see that $ k=4 $ doesn't satisfy with $ n=2\cdot 3\cdot 5\cdot 7=210 $