2007 Putnam B3

Let $x_0 = 1$ and for $n \geq 0$, let $x_{n+1} = 3x_n + \lfloor x_n

\sqrt{5} \rfloor$. In particular,$x_1 = 5$,$x_2 = 26$,$x_3 = 136$,

$x_4 = 712$. Find a closed-form expression for $x_{2007}$. ($\lfloor a

\rfloor$means the largest integer$\leq a$.)

令$x_0 = 1$，对于$n \geq 0$，令$x_{n+1} = 3x_n + \lfloor x_n

\sqrt{5}\rfloor$。特别是，$x_1 = 5$，$x_2 = 26$，$x_3 = 136$，

$x_4 = 712$。求 $x_{2007}$ 的封闭式表达式。 ($\l地板a

\rfloor$表示最大整数$\leq a$。)