Let $x_{1}=1$ and $x_{n+1}=x_{n}+\left\lfloor\frac{x_{n}}{n}\right\rfloor+2$ for $n=1,2,3, \ldots$, where $\lfloor x\rfloor$ denotes the largest integer not greater than $x$. Determine $x_{1997}$.
Answer: $x_{1997}=23913$.

Note that if $x_{n}=a n+b$ with $0 \leqslant b<n$, then

$$
x_{n+1}=x_{n}+a+2=a(n+1)+b+2 \text {. }
$$

Hence if $x_{N}=A N$ for some positive integers $A$ and $N$, then for $i=0,1, \ldots, N$ we have $x_{N+i}=A(N+i)+2 i$, and $x_{2 N}=(A+1) \cdot 2 N$. Since for $N=1$ the condition $x_{N}=A N$ holds with $A=1$, then for $N=2^{k}$ (where $k$ is any non-negative integer) it also holds with $A=k+1$. Now for $N=2^{10}=1024$ we have $A=11$ and $x_{N+i}=A(N+i)+2 i$, which for $i=973$ makes $x_{1997}=11 \cdot 1997+2 \cdot 973=23913$.