First I get $ na_n-(n-1)^2a_{n-1}=\left\{ \begin{array}{ll} -n+2\ (n: odd) &\quad \\ -n+3\ (n: even) &\quad \end{array} \right.$

Prove by induction that \[ a_n=\frac{(n-1)!+1}n\] The rest is obvious.

$ a_{3}=\frac{2}{3} $. your formula does not work for $ n=1,3 $. your formula works for the sequence $ a_{1}=2,a_{2}=1 $...so it was a typo in the initial problem?

anonymouslonely wrote: $ a_{3}=\frac{2}{3} $. your formula does not work for $ n=1,3 $. your formula works for the sequence $ a_{1}=2,a_{2}=1 $...so it was a typo in the initial problem? No in the initial problem $ a_{1} $ is 2 and $ a_{2} $ is 1.Right!