Floor functions go on and on

Consider

$\begin{aligned} \big \lfloor \underbrace{\sqrt{444 \cdots 444}}_{\text{Number of 4s}=2n} \big \rfloor & = 2 \big \lfloor \underbrace{\sqrt{111 \cdots 111}}_{\text{Number of 1s}=2n} \big \rfloor = 2 \left \lfloor \sqrt{\frac {10^{2n}-1}9} \right \rfloor \approx 2 \left \lfloor \frac {10^n}3 \right \rfloor = 2 \times \underbrace{333\cdots333}_{\text{Number of 3s}=n} = \underbrace{666\cdots666}_{\text{Number of 6s}=n}\end{aligned}$

For $2n=16$ , the answer $\boxed{66666666}$ .

I noticed the pattern:

For every two $4$ 's added, one $6$ is added onto the result.

So there are $16$ $4$ 's, therefore:

$\frac{16}{2}$ $= 8$

Therefore, there are $8$ $6$ 's, so the answer is: $\fbox{66666666}$

Proof