How many Recursion Sequences?

Consider the sequence defined by the recurrence relation \ $b_{n+2} \;= \; 4b_{n+1} - b_n \hspace{2cm} b_0 = b_1 = 1$ It is clear that $b_n$ is an integer for all $n \ge 0$ . It is easy to solve this recurrence relation to see that $b_n \; = \; \frac{1}{2\sqrt{3}}\Big[(\sqrt{3}-1)(2+\sqrt{3})^n + (\sqrt{3}+1)(2-\sqrt{3})^n\Big] \hspace{2cm} n \ge 0$ From this we note that $3b_n^2 - 2 \; = \; \tfrac14\Big[(\sqrt{3}-1)(2+\sqrt{3})^n - (\sqrt{3}+1)(2-\sqrt{3})^n\Big]^2 \hspace{2cm} n \ge 0$ Now $\frac{(\sqrt{3}-1)(2+\sqrt{3})^n}{(\sqrt{3}+1)(2-\sqrt{3})^n} \; = \; \frac{2(2+\sqrt{3})^{2n}}{(\sqrt{3}+1)^2} \; \ge \; 2(1 + \sqrt{3})^{2n-2} \; > \; 1 \hspace{2cm} n \ge 1$ and hence $\sqrt{3b_n^2 - 2} \; =\; \tfrac12\Big[(\sqrt{3}-1)(2+\sqrt{3})^n - (\sqrt{3}+1)(2-\sqrt{3})^n\Big] \hspace{2cm} n\ge 1$ and hence $2b_n + \sqrt{3b_n^2 - 2} \; = \; \frac{1}{2\sqrt{3}}\Big[(\sqrt{3}-1)(2+\sqrt{3})^{n+1} - (\sqrt{3}+1)(2-\sqrt{3})^{n+1}\Big] \; = \; b_{n+1} \hspace{2cm} n \ge 1$ Thus we can show by induction that $a_n = b_n$ for all $n \ge 1$ , and hence all the terms $a_n$ are integers. Thus the answer is $\boxed{0}$ .