Reach for the Summit - M-S3-A2

From the linear recurrence relation $a_n = 3a_{n-1} - a_{n-2} - 2a_{n-3}$ , the characteristic polynomial is as below:

$\begin{aligned} r^3 - 3r^2 + r + 2 & = 0 \\ (r-2)(r^2 - r -1) & = 0 \\ (r-2)(r-\varphi) \left(r + \varphi^{-1} \right) & = 0 & \small \blue{\text{where }\varphi = \frac {1+\sqrt 5}2 \text{ is the golden ratio.}} \end{aligned}$

Therefore $a_n = c_1 2^n + c_2 \varphi^n + c_3 (-\varphi^{-1})^n$ , where $c_1$ , $c_2$ , and $c_3$ are constants. Solving the simultaneous equations below:

$\begin{cases} a_1 = 2c_1 + c_2 \varphi - c_3 \varphi^{-1} = 1 \\ a_2 = 4c_1 + c_2 \varphi^2 + c_3 \varphi^{-2} = 3 \\ a_3 = 8c_1 + c_2 \varphi^3 + c_3 \varphi^{-3} = 6 \end{cases}$

we get $c_1 = 1$ , $c_2 = - \dfrac 1{\sqrt 5}$ , and $c_3 = \dfrac 1{\sqrt 5}$ . Therefore $a_n = 2^n - \dfrac {\varphi^n - (-\varphi)^{-n}}{\sqrt 5} = 2^n - F_n$ , where $F_n$ denotes that $n$ th Fibonacci number .

Then we have $a_n > \lambda \times 2^{n-2}$ and $\lambda < \dfrac {a_n}{2^{n-2}} = \dfrac {2^n - F_n}{2^{n-2}} = 4 - \dfrac {F_n}{2^{n-2}} < 4$ . Since $F_n < 2^{n-2}$ for all $n \ge 4$ , the right-hand side approaches the maximum of $4$ , as $n \to \infty$ . Therefor the maximum integer value of $\lambda$ is $\boxed 3$ .

$a_4 = 3(6) - (3) - 2(1) = 13$

And $13 > \lambda × 2^{4-2}$ so $\lambda < 3.25$ .

Since $\lambda$ is constant for all $a_n > \lambda × 2^{n-2}$ , the floor value of $\lambda$ will be the maximum (constant/non increasing afterwards) at $a_4$ , which is $\boxed{3}$ .

Let $x_{n}=a_{n} / 2^{n-2}$ our goal to find greatest integer lesser or equal then minimum of this sequence for $n\ge4$

$x_1=2, x_2=3, x_3=3,$ and $x_n = \frac{1}{4}(6x_{n-1}-x_{n-2}-x_{n-3})$

$x_n$ is increasing sequence.

It can be proven by induction:

base $x_1 \le x_2 \le x_3$ is given.

If $x_{n-1} \ge x_{n-2} \ge x_{n-3}$ then $x_n = \frac{1}{4}(6x_{n-1}-x_{n-2}-x_{n-3}) \ge \frac{1}{4}(6x_{n-1}-x_{n-1}-x_{n-1}) = x_{n-1}$

So minimum of $x_n$ for $n \ge 4$ is $x_4 = \frac{13}{4}$ and answer is $\lfloor \frac{13}{4} \rfloor = 3$