# First-order recursion problem with a special cubic equation: by Ji_Chen

Given a natural number $$n$$ and a real number $$d_{n},$$ for the least $$d_{n+1}$$ so that $$32d_{n}^{5}\!\left ( d_{n}+ 2 \right )\!=$$ $$= 2\left ( 5d_{n}^{4}+ 8d_{n}^{2}+ 8d_{n}+ 8 \right )d_{n+ 1}^{3}+ 4d_{n}\left ( 5d_{n}^{4}+ 11d_{n}^{3}- 2d_{n}^{2}+ 14d_{n}- 4 \right )d_{n+ 1}^{2}$$ $$+ 4d_{n}^{3}\left ( 5d_{n}^{3}+ 2d_{n}^{2}+ 24d_{n}- 16 \right )d_{n+ 1}.$$ Prove that if $$d_{2}= \frac{8}{5},$$ then $$\left \{ d_{n} \right \}_{n> 1}$$ is rational and $$\lim\frac{n^{2}}{\ln n}\left ( \frac{2}{n}- d_{n} \right )= 2$$ Source: AoPS/@Ji_Chen (still unsolved)

I'm into the related one here https://Artofproblemsolving.com/community/c6h318726p1713923 _I also see $$d_{2}= \!\frac{8}{5}.$$ Maybe they are same so I used discriminant to define the concrete value for $$d_{n}$$ but unsuccessfully, I need to the help.