Over the last weekend, another question was asked and later disappeared on/from Math StakExchange. The sequence, from that question, is too nice to "let it go", thus, copy/pasting my answer here.
Let's define the following sequence
$a(n)=\left\lfloor n\cdot\varphi + \frac{1}{2}\right\rfloor$
where φ is the golden ratio. Is there a closed form for the following sequence?
$a^{\circ k}(n)=\underbrace{a(((...a(n)...)))}_{k \text{ times}}$
Calculating a few values of the sequence reveals that this is the A007067, with the property that
$a(a(n))=a(n)+n \tag{1}$
a result proved in 2006. Let's show it ...
Proposition 1.
$\left\lfloor \frac{a(n)}{\varphi} + \frac{1}{2}\right\rfloor=n$
Given \(x-1 < \lfloor x\rfloor \leq x\) we have
$n\varphi - \frac{1}{2} < a(n)\leq n\varphi + \frac{1}{2} \iff \\ n-\frac{1}{2\varphi} < \frac{a(n)}{\varphi}\leq n+\frac{1}{2\varphi} \iff \\ n+\frac{1}{2}-\frac{1}{2\varphi} < \frac{a(n)}{\varphi}+\frac{1}{2}\leq n+\frac{1}{2}+\frac{1}{2\varphi}$
Given \(1+\frac{1}{\phi}=\varphi\) and \(\frac{1}{2}-\frac{1}{2\varphi} > 0\) then
$n < n+\frac{1}{2}-\frac{1}{2\varphi} < \frac{a(n)}{\varphi}+\frac{1}{2}\leq n+\frac{\varphi}{2} < n+1$
or
$n < \frac{a(n)}{\varphi}+\frac{1}{2} < n+1 \Rightarrow \left\lfloor \frac{a(n)}{\varphi} + \frac{1}{2}\right\rfloor=n$
Proposition 2.
$a(a(n))=a(n)+n$
Obviously \(a(n)\in\mathbb{N}\), then
$a(a(n))= \left\lfloor a(n)\cdot\varphi + \frac{1}{2}\right\rfloor= \left\lfloor a(n)\cdot\left(1+\frac{1}{\varphi}\right) + \frac{1}{2}\right\rfloor=\\ a(n)+\left\lfloor \frac{a(n)}{\varphi} + \frac{1}{2}\right\rfloor = ...$
applying Proposition 1
$...= a(n)+n$
Now back to the original question, a few observations, using \((1)\)
$a^{\circ 2}(n)=a(a(n))=F_2\cdot a(n) + F_1\cdot n$
$a^{\circ 3}(n)=a(a(\color{red}{a(n)}))=a(\color{red}{a(n)})+\color{red}{a(n)}=\\ 2\cdot a(n)+n=F_3\cdot a(n) + F_2\cdot n$
$a^{\circ 4}(n)=a(a(a(\color{blue}{a(n)})))=2\cdot a(\color{blue}{a(n)})+\color{blue}{a(n)}=\\ 3\cdot a(n)+2n=F_4\cdot a(n) + F_3\cdot n$
It's easy to see, by induction, this is
$a^{\circ k}(n)=F_k\cdot a(n) + F_{k-1}\cdot n \tag{2}$
because
$a^{\circ (k+1)}(n)=a^{\circ k}(a(n))=F_k\cdot a(a(n)) + F_{k-1}\cdot a(n)=\\ (F_k+F_{k-1})\cdot a(n) + F_{k}\cdot n=F_{k+1}\cdot a(n) + F_{k}\cdot n$
