Let $a(n) = \lfloor n\phi + 0.5\rfloor$ for all $n \geqslant 1$, where $\phi$ is the golden ratio. Now let
$$a(n)^k = a(a(\ldots(a(n))))$$
where we have iterated $a(n)$, $k$ times in the RHS. I am looking for a nice closed form of $a(n)^k$. I am working on this problem for a long time but didn't get any satisfactory solution. Indeed, I was also able to find some sort of closed form but it again seems to be very painful to evaluate for some large $n$. I'm rather looking for a nice closed form.
I highhly appreciate your time and efforts. Thanks.
Further to my comments, a proof of $a(a(n))=a(n)+n$.
Given $x-1< \lfloor x\rfloor \leq x$, we have $$n\phi - \frac{1}{2} <a(n)\leq n\phi + \frac{1}{2} \iff\\ n-\frac{1}{2\phi} <\frac{a(n)}{\phi}\leq n+\frac{1}{2\phi} \iff\\ n+\frac{1}{2}-\frac{1}{2\phi} <\frac{a(n)}{\phi}+\frac{1}{2}\leq n+\frac{1}{2}+\frac{1}{2\phi}$$
Given $1+\frac{1}{\phi}=\phi$ and $\frac{1}{2}-\frac{1}{2\phi}>0$, then $$n<n+\frac{1}{2}-\frac{1}{2\phi} <\frac{a(n)}{\phi}+\frac{1}{2}\leq n+\frac{\phi}{2}<n+1$$
or $$n<\frac{a(n)}{\phi}+\frac{1}{2}<n+1 \Rightarrow \left\lfloor \frac{a(n)}{\phi} + \frac{1}{2}\right\rfloor=n$$
Obviously $a(n)\in\mathbb{N}$, then $$a(a(n))= \left\lfloor a(n)\phi + \frac{1}{2}\right\rfloor= \left\lfloor a(n)\left(1+\frac{1}{\phi}\right) + \frac{1}{2}\right\rfloor=\\ a(n)+\left\lfloor \frac{a(n)}{\phi} + \frac{1}{2}\right\rfloor = ...$$ applying Proposition 1 $$...= a(n)+n$$
Now $$a^{\circ 2}(n)=a(a(n))=F_2\cdot a(n) + F_1\cdot n$$ $$a^{\circ 3}(n)=a(a(\color{red}{a(n)}))=a(\color{red}{a(n)})+\color{red}{a(n)}=2\cdot a(n)+n=F_3\cdot a(n) + F_2\cdot n$$ $$a^{\circ 4}(n)=a(a(a(\color{blue}{a(n)})))=2\cdot a(\color{blue}{a(n)})+\color{blue}{a(n)}=3\cdot a(n)+2n=F_4\cdot a(n) + F_3\cdot n$$
By induction, this is $$a^{\circ k}(n)=F_k\cdot a(n) + F_{k-1}\cdot n \tag{1}$$ because $$a^{\circ (k+1)}(n)=a^{\circ k}(a(n))=F_k\cdot a(a(n)) + F_{k-1}\cdot a(n)=\\ (F_k+F_{k-1})\cdot a(n) + F_{k}\cdot n=F_{k+1}\cdot a(n) + F_{k}\cdot n$$