Limit of recursive sequence with floor 
Sequences $x_n$ and $y_n$ are defined as
  $$x_n=\left\lfloor x_{n-1}\frac{y_{n+1}}{y_{n-1}}\right\rfloor,\\
y_n=y_{n-1}+1,\\
x_0=2015,\ y_0=307.$$
  Compute $$\lim_{n\to\infty}\frac{x_n}{y_n^2}$$

My attempt: $y_n=307+n$ so $$x_n> x_{n-1} \frac{308+n}{306+n}>x_0\frac{(308+n)(309+n)}{306\cdot 307}\approx O(n^2),$$
so the limit is approximately $\dfrac{2015}{306\cdot 307}$. However, the answer is given as $\dfrac{2}{101}$. How to obtain this value?
 A: By your attempt we now that a limit exists, so we need to find just a limit of a subsequence of $\frac{x_{t(n)}}{y_{t(n)}^{2}}$ and that limit will be equal to the limit of the whole sequence.
So, it's easy to check numerically that
$$x_{127(k-1)+1}=740+k\frac{1941}{2}+k^2\frac{635}{2}$$
holds $\forall k \geq 1 \in \mathbb{N}$.
I'll try to demonstrate this equation later in this answare.
The following is a simple limit knowing that 
$$x(n)= 740 + n\frac{1941}{254}+n^2\frac{5}{254}$$
$$y(n) = 307 + n$$
So
$$\lim_{n\to\infty} \frac{x(n)}{y(n)^2}=
\frac{740 + n\frac{1941}{254}+n^2\frac{5}{254}}{94249 + 614 n + n^2}=\frac{5}{254}$$
Not really a demonstration... yet
First it easy to check that for $g(t)=A+t$
$$\frac{g(t+1)}{g(t-1)}=\frac{A+t+1}{A+t-1}=\frac{A+t-1+2}{A+t-1}=1+\frac{2}{A+t-1}=1+\frac{2}{g(t-1)}$$
holds for evry real number if $g(t-1)\neq0$.
So in our case the equality
$$x_n=\left\lfloor x_{n-1}\frac{y_{n+1}}{y_{n-1}}\right\rfloor = 
\left\lfloor x_{n-1} \left( 1+\frac{2}{y_{n-1}} \right)  \right\rfloor =
x_{n-1}+\left\lfloor \frac{2 x_{n-1}}{y_{n-1}}\right\rfloor
$$
holds $\forall n \in \mathbb{N}$.
Then is true that
$$x_{127(k-1)+1}=x_{127(k-1)}+\left\lfloor \frac{2 x_{127(k-1)}}{y_{127(k-1)}}\right\rfloor$$
I do not know how to go forward...
