How Do Floor Functions Affect the Results of Square Roots in Sequence Problems?

[anemone]

Here is this week's POTW:

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Compute \(\displaystyle \frac{\left\lfloor{\sqrt[4]{1}}\right\rfloor \cdot \left\lfloor{\sqrt[4]{3}}\right\rfloor \cdot\left\lfloor{\sqrt[4]{5}}\right\rfloor \cdots \left\lfloor{\sqrt[4]{2015}}\right\rfloor}{\left\lfloor{\sqrt[4]{2}}\right\rfloor \cdot \left\lfloor{\sqrt[4]{4}}\right\rfloor \cdot\left\lfloor{\sqrt[4]{6}}\right\rfloor \cdots \left\lfloor{\sqrt[4]{2016}}\right\rfloor}\).

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[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. castor28
3. Opalg
4. lfdahl
5. kaliprasad

Solution from castor28:

Spoiler

The expression can be written as:
$$
P=\prod_{n=0}^{1007}{\frac{\lfloor\sqrt[4]{2n+1}\rfloor}{\lfloor\sqrt[4]{2n+2}\rfloor}}
$$
Each fraction in the product is different from $1$ only when $2n+2=a^4$ for some integer $a$ (necessarily even). In that case, the fraction is equal to $\dfrac{a-1}{a}$.
As $6^4 < 2016 < 7^4$, this happens for $a=2,4,6$, and the expression is equal to:
$$
P = \frac12\times\frac34\times\frac56= \bf\frac{5}{16}
$$