>**Problem**: Let $x_1 = 1$ and $x_{n+1} = \frac{x_n}{n^2} + \frac{n^2}{x_n} + 2, \ n\ge 1$.
Find the third term in the asymptotic expansion of $x_n$.

I have posted it in MSE six months ago without solution for the third term
https://math.stackexchange.com/questions/3801405/the-limit-and-asymptotic-analysis-of-a-n2-n-from-a-n1-fraca-nn.

We have $\lim_{n\to \infty} (x_n - n) = \frac{1}{2}$ (see [1]; I also give a solution with the help of computer in the link above).
So the first two terms in the asymptotic expansion of $x_n$ are $x_n \sim n + \frac{1}{2}$.

For the third term, @Diger in MSE said $x_n \sim n + \frac{1}{2} + \frac{5}{8n}$ (see @Diger's answer in the link above).
However, I did some numerical experiment which does not support this result.
I am not convinced of the numerical evidence due to finite precision arithmetic.
I hope to prove or disprove it analytically.

*Numerical Experiment*: If $x_n \sim n + \frac{1}{2} + \frac{5}{8n}$, 
then it should hold $16n(x_{2n} - 2n - \frac{1}{2}) \approx 5$ and $8(2n+1)(x_{2n+1} - (2n+1) - \frac{1}{2}) \approx 5$
for large $n$. When $n=1500$, I use Maple to get
$16n(x_{2n} - 2n - \frac{1}{2}) \approx 4.368$ and $8(2n+1)(x_{2n+1} - (2n+1) - \frac{1}{2}) \approx 5.642$.
When $n$ is larger (e.g., $n=10000$), the numerical result seems unreliable. 

I ${\color{blue}{\textbf{GUESS}}}$ that
$$x_{2n} \sim 2n + \frac{1}{2} + \frac{q_1}{2n},$$
$$x_{2n+1} \sim (2n+1) + \frac{1}{2} + \frac{q_2}{2n+1}$$
where $q_1 + q_2 = \frac{5}{4}$ and $q_1 \ne q_2$ (if $q_1 = q_2$, then it is $x_n \sim n + \frac{1}{2} + \frac{5}{8n}$).
(Some numerical experiment shows $q_1 \approx \frac{61}{112}, q_2 \approx \frac{79}{112}$. But I am not convinced of it.)

Any comments and solutions are welcome and appreciated.


*Reference*

[1] Yuming Chen, Olaf Krafft and Martin Schaefer, “Variation of a Ukrainian Olympiad Problem: 10982”,
The American Mathematical Monthly, Vol. 111, No. 7 (Aug. - Sep., 2004), pp. 631-632