$p = \sqrt{1+\sqrt{1+\sqrt{1 + \cdots}}}$; $\sum_{k=2}^{\infty}{\frac{\lfloor p^k \rceil}{2^k}} = ? $ 
Let $p = \sqrt{1+\sqrt{1+\sqrt{1 + \cdots}}}$
  The sum $$\sum_{k=2}^{\infty}{\dfrac{\lfloor p^k \rceil}{2^k}}$$
  Can be expressed as $\frac{a}{b}$. Where $\lfloor \cdot \rceil$ denotes the nearest integer function. Find $a+b$.  

My Work
$p^2 = 1+p \implies p = \dfrac{1+\sqrt{5}}{2} = \phi$
And also $\phi^n = \dfrac{L_n + F_n\sqrt{5}}{2}$. Where $L_n$ is the $n-th$ Lucas-Number.
What to do with the next part?    
Note: Problem Collected from Brilliant.org 

For $n \ge 2$, the numbers $\lfloor \phi^k \rceil$ are equal to the Lucas numbers, which satisfy the Fibonacci recurrence but in which $L_1=1$ and $L_2=3$.  The Lucas numbers satisfy
$$L_k = \phi^k + (-1)^k \phi^{-k}$$
and thus have the generating function
$$L(x) = \frac{2-x}{1-x-x^2} $$
so the sum we want is
$$L \left (\frac12 \right ) - L_0 - \frac12 L_1 = \frac72$$ 
Note that $\phi^{-1}= \phi-1 = 0.618\ldots$ and exponentiation to the power of $k$ does not affect a rounding for $k \ge 2$.  
