Quadratic Recurrence

Define $S_n \; = \; \sum_{j=1}^n x_j^{-1}$ It is easy to calculate that $\lfloor S_1\rfloor = 0$ , $\lfloor S_2 \rfloor = 1$ and $\lfloor S_3 \rfloor =2$ .

It is also a simple induction to show that $x_n > 1$ for all $n \ge 1$ , and therefore that $x_{n+1}-x_n = (x_n-1)^2 > 0$ for all $n \ge 1$ , so that the sequence $x_n$ is increasing. Now note that $\frac{1}{x_n-1} - \frac{1}{x_{n+1}-1} \; = \; \frac{x_{n+1}-x_n}{(x_n-1)(x_{n+1}-1)} \; = \; \frac{(x_n-1)^2}{(x_n-1)x_n(x_n-1)} \; = \; \frac{1}{x_n}$ for all $n \ge 1$ , so that $S_n \; = \; \sum_{j=1}^n x_j^{-1} \; = \; \sum_{j=1}^n \left(\frac{1}{x_j-1} - \frac{1}{x_{j+1}-1}\right) \; = \; \frac{1}{x_1-1} - \frac{1}{x_{n+1}-1} \; = \; 3 - \frac{1}{x_{n+1}-1} \; < \; 3$ for all $n \ge 1$ . Thus $(S_n)_n$ is an increasing sequence that is bounded above, and so it converges. Hence $x_n^{-1}\to0$ as $n \to \infty$ , so that $x_n \to \infty$ as $n \to \infty$ . But this implies that $S_n \to 3$ as $n \to \infty$ . Since $S_n < 3$ for all $n\ge 1$ , with $\lim_{n \to\infty}S_n = 3$ , we deduce that the correct answer is $\boxed{\{0,1,2\}}$ .

As simple calculations show, the sequence $(x_n)_n$ diverges to $\infty$ extremely rapidly!

From $a_{n+1}={a_n}^{2}-a_n+1$ , we obtain $\frac{1}{a_{n+1}} \leq \frac{1}{a_{n+1}-1}=\frac{1}{a_{n-1}}-\frac{1}{a_n}$ .

This gives $\frac{1}{a_3} \leq \frac{1}{a_1}-\frac{1}{a_2}$ , $\frac{1}{a_4} \leq \frac{1}{a_2}-\frac{1}{a_3}$ etc, so taking summations we get $\sum_{i=1}^{n}\frac{1}{a_i} \leq \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_1}-\frac{1}{a_{n-1}}=\frac{29}{13}-\frac{1}{a_{n-1}} < 3$ for $n \geq 3$ .

It is easy to observe that the ${a_i}$ are always positive (in fact >1) from $a_{n+1}=a_n(a_n-1)+1$ . Therefore, $\lfloor\sum_{i=1}^{n}\frac{1}{a_i}\rfloor<3$ and so can only take the values ${0,1,2}$ .

Finally, it is easy to observe that $\lfloor\frac{1}{a_1}\rfloor=0$ , $\lfloor\frac{1}{a_1}+\frac{1}{a_2}\rfloor=1$ and $\lfloor\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\rfloor=2$ , hence the answer.