Two sequences and a limit

We note that $a_{n+1} = 2a_n + a_{n-1}$ with $a_1 = 1$ , $a_2=2$ . Solving this recurrence relation, we obtain $a_n \; = \; \frac{1}{2\sqrt{2}}\left[\big(1+\sqrt{2}\big)^n - \big(1-\sqrt{2}\big)^n\right]$ and hence $b_n \; = \; a_{n+1} - a_n \; = \; \frac{1}{2}\left[\big(1 + \sqrt{2}\big)^n + \big(1 - \sqrt{2}\big)^n\right]$ and hence $\frac{b_n}{a_n} \; = \; \sqrt{2}\frac{1 + u^n}{1 - u^n}$ where $u \; = \; \frac{1 - \sqrt{2}}{1 + \sqrt{2}}$ Since $-1 < u < 0$ , we deduce that $\lim_{n \to \infty}\frac{b_n}{a_n} \; = \; \sqrt{2}$ making the desired answer $\lfloor 10000\sqrt{2}\rfloor = \boxed{14142}$ .

It is worth noting that $\begin{aligned} 2a_n^2 - b^2 & = \; \frac14\left[\left(\big(1 + \sqrt{2}\big)^n - \big(1 - \sqrt{2}\big)^n\right)^n - \left(\big(1 + \sqrt{2}\big)^n + \big(1 - \sqrt{2}\big)^n\right)^n \right] \\ & = \; -\big(1 + \sqrt{2}\big)^n\big(1 - \sqrt{2}\big)^n \; = \; (-1)^{n+1} \end{aligned}$ as noted by @Chan Lye Lee . For that matter, $2a_{n+1}^2 - b_{n+1}^2 \; = \; 2(a_n + b_n)^2 - (2a_n + b_n)^2 \; = \; b_n^2 - 2a_n^2$ and so the fact that $2a_n^2 - b_n^2 = (-1)^{n+1}$ is a simple induction.

[Partial solution]

The table shows the first values of $a_n$ and $b_n$ .

It is interesting to note that $2a_n^2 = b_n^2 + (-1)^{n-1}$ . For example: $2a_3^2 = 2(25)=50=7^2+1=b_3^2+1$ . Hence $2a_n^2 \approx b_n^2$ . So $\frac{b_n}{a_n} \approx \sqrt{2}$ . Finally $\displaystyle\left\lfloor 10000\lim_{n \to \infty} \frac{b_n}{a_n} \right\rfloor = 10000\sqrt{2} =14142$ .

We can write $\left[\begin{array}{cc}a_{n+1}\\b_{n+1}\\\end{array}\right] =\left[\begin{array}{cc}1&1\\2&1\\\end{array}\right]\left[\begin{array}{cc}a_{n}\\b_{n}\\\end{array}\right]$ . The eigenvalues of the matrix are $1\pm\sqrt{2}$ , and an eigenvector associated with the dominant eigenvalue $1+\sqrt{2}$ is $\left[\begin{array}{cc}a\\b\\\end{array}\right]=\left[\begin{array}{cc}1\\\sqrt{2}\\\end{array}\right]$ . Thus $\lim_{n\to\infty}\frac{b_n}{a_n}=\frac{b}{a}=\sqrt{2}\approx1.4142$ and the answer is $\boxed{14142}$ .

It is interesting to note that the answer is independent of the initial values as long as $b_1\neq-\sqrt{2}a_1$ .

First, see that both sequence are increasing and

$L = \lim_{n \to \infty } \frac{b_{n}}{a_{n}} = \lim_{n \rightarrow \infty } \frac{b_{n+1}}{a_{n+1}}$

Now the fun part

$L = \lim_{n \rightarrow \infty } \frac{b_{n+1}}{a_{n+1}} = \lim_{n \rightarrow \infty } \frac{2a_{n}+b_{n}}{a_{n}+b_{n}} = \lim_{n \rightarrow \infty } (1+ \frac{1}{1+\frac{b_{n}}{a_{n}}})$

Ha...

$L=1+\frac{1}{L+1} \Rightarrow L^{2}-1=1 \Rightarrow L=\sqrt{2}$ $\lfloor10000 \times L\rfloor = \boxed{14142}$