mathisreal wrote: Let $a_0,a_1,a_2,\dots$ be a periodic sequence of real numbers(that is, there is a fixed positive integer $k$ such that $a_n=a_{n+k}$ for every integer $n\geq 0$). The following equality is true, for all $n\geq 0$: $a_{n+2}=\frac{1}{n+2} (a_n - \frac{n+1}{a_{n+1}})$ if $a_0=2020$, determine the value of $a_1$. Note that $a_n\ne 0$ $\forall n$ Equation is $(n+2)a_{n+2}a_{n+1}=a_{n+1}a_n-(n+1)$ Which may be written $(n+2)!a_{n+2}a_{n+1}+(n+2)!=(n+1)!a_{n+1}a_n+(n+1)$ And so $(n+2)!a_{n+2}a_{n+1}+(n+2)!=a$ constant So $a_{n+2}=\frac 1{a_{n+1}}\left(\frac a{(n+2)!}-1\right)$ And so $a_{n+2}=a_n\frac{\frac a{(n+2)!}-1}{\frac a{(n+1)!}-1}$ If $a<0$, this implies $|a_{n+2}|<|a_n|$ from a given point and $a_n$ can not be periodic If $a>0$, this implies $|a_{n+2}|>|a_n|$ from a given point and $a_n$ can not be periodic So $a=0$ and $a_{n+1}a_{n}=-1$ And so $\boxed{a_1=-\frac 1{2020}}$