Limit of recursive sequence with floor

$\def\peq{\mathrel{\phantom{=}}{}}$Note that$$ x_{n + 1} = \left[ x_n \frac{n + 309}{n + 307} \right] = x_n + \left[ \frac{2x_n}{n + 307} \right]. \quad \forall n \geqslant 0 $$ First, it will be proved by induction on $n$ that $x_n = 13n + 2015$ for $0 \leqslant n \leqslant 23$. For $n = 0$ it is obvious. Now suppose it holds for $n$. Since$$ 2(13n + 2015) - 13(n + 307) = 13n + 39 $$ and$$ 0 \leqslant n \leqslant 22 \Longrightarrow 0 \leqslant 13n + 39 < n + 307, $$ then$$ x_{n + 1} = (13n + 2015) + \left[ \frac{2(13n + 2015)}{n + 307} \right] = 13(n + 1) + 2015. $$ End of induction.

Next, it will be proved by induction on $n$ that$$ x_n = \begin{cases} (5m + 14)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2314; & n = 127m + k + 23,\ 0 \leqslant k \leqslant 24\\ (5m + 15)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2289; & n = 127m + k + 23,\ 25 \leqslant k \leqslant 50\\ (5m + 16)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2238; & n = 127m + k + 23,\ 51 \leqslant k \leqslant 75\\ (5m + 17)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2162; & n = 127m + k + 23,\ 76 \leqslant k \leqslant 100\\ (5m + 18)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2061; & n = 127m + k + 23,\ 101 \leqslant k \leqslant 126 \end{cases} $$ for $n \geqslant 23$, where $m$ and $k$ are integers. For $n = 23$, $x_{23} = 13 × 23 + 2015 = 2314$. Now suppose it holds for $n = 127m + k + 23$ where $0 \leqslant k \leqslant 126$.

If $0 \leqslant k \leqslant 24$, because\begin{align*} &\peq 2x_n - (5m + 14)(n + 307)\\ &= 2\left( (5m + 14)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2314 \right) - (5m + 14)(127m + k + 330)\\ &= (5m + 14)k + 3m + 8 \geqslant 0, \end{align*}$$ (5m + 14)k + 3m + 8 < 127m + k + 330 \Longleftrightarrow (5m + 13)k < 124m + 322, $$ and $0 \leqslant k \leqslant 24$ implies$$ (5m + 13)k \leqslant 120m + 312 < 124m + 322, $$ then\begin{align*} &\peq x_{n + 1} = x_n + \left[ \frac{2x_n}{n + 307} \right]\\ &= \left( (5m + 14)k + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2314 \right) + (5m + 14)\\ &= (5m + 14)(k + 1) + 635 · \dfrac{1}{2} m(m - 1) + 2033m + 2314. \end{align*} Note that $(5m + 14) × 25 + 2314 = (5m + 15) × 25 + 2289$, thus the proposition holds for $0 \leqslant k \leqslant 24$. If $25 \leqslant k \leqslant 50$, $51 \leqslant k \leqslant 75$, $76 \leqslant k \leqslant 100$ or $101 \leqslant k \leqslant 126$, analogous deduction with the fact that\begin{align*} &\peq (5m + 18) · 127 + 635 · \frac{1}{2} m(m - 1) + 2033m + 2061\\ &= 317.5 m^2 + 2350.5 m + 4347\\ &= (5(m + 1) + 14) · 0 + 635 · \frac{1}{2} (m + 1)m + 2033(m + 1) + 2314 \end{align*} shows that the proposition also holds. End of induction.

Now for $n \geqslant 23$, suppose $n = 127m + k + 23$, where $0 \leqslant k \leqslant 126$, then\begin{gather*} x_n \geqslant x_{127m + 23} = \frac{1}{2} · 635m(m - 1) + 2033m + 2314,\\ x_n \leqslant x_{127(m + 1) + 23} = \frac{1}{2} · 635m(m + 1) + 2033m + 4347,\\ 127m + 23 = y_{127m + 23} \leqslant y_n \leqslant y_{127(m + 1) + 23} = 127m + 150. \end{gather*} Because\begin{gather*} \lim_{m → ∞} \frac{\dfrac{1}{2} · 635m(m - 1) + 2033m + 2314}{(127m + 150)^2} = \frac{\dfrac{1}{2} · 635}{127^2} = \frac{5}{254},\\ \lim_{m → ∞} \frac{\dfrac{1}{2} · 635m(m + 1) + 2033m + 4347}{(127m + 23)^2} = \frac{5}{254}, \end{gather*} then$$ \lim_{n → ∞} \frac{x_n}{y_n^2} = \frac{5}{254} ≈ 0.01968504. $$