Let $n$ be a strict positive integer and let's define an integer sequence $f(n)$ :
$$f(n) = \frac{n^2 + n + 4}{2}$$
so
$$
\begin{split}
f (\Bbb N)& \triangleq {3,5,8,12,17,23,30,38,47,\ldots}\\
f(1) & = 3 \\
&\:\,\vdots \\
f(17) & = 155 \\
&\:\,\vdots
\end{split}$$
etc.
Notice
$$
\begin{split}
3+2 &=5 \\
5+3 &=8 \\
8+4 &=12 \\
12+5 &=17 \\
&\:\,\vdots
\end{split}
$$
etc. and $f(R)$ has no real fixpoint for real $R$.
Now we want integer sequences $g(n)$ such that $g(n)$ grow faster than linear and
$$ g(n+1) > g(n)$$
$$f(g(n)) = g(f(n))$$
$$f(n) + 1 < g(n) < f(f(n)) - 1$$
And fixpoints are not allowed :
$$g(n) \neq n.$$
A possible solution seems
$$g :={5,9,17,29,47,73,109,155,...}$$
and you can check that $$f(g(n)) = g(f(n))$$ and the other conditions hold.
The sequence $g$ might not be unique and I assume no mistake has been made. But it is not so easy to see. (contradictions do not occur immediately but could happen for later values if one is not careful)
The sequence g resembles or equals this perhaps :
But it might not relate or be coincidence.
One of the ideas is that $g(n)$ is just floor$(f^{[r]}(n))$ ; In other words $g(n)$ is just the rounded down number of some real iterate of $f(n)$. But that is just a vague conjecture. Fractional iterations are not unique and the solutions for $g$ might not be unique either. So that idea is vague and nonconstructive, so its value is arguable.
I want solutions for $g(n)$. Is $g(n)$ close to a $3/2$th iterate of $f(n)$ ?? Does $g(n)$ grow like $O(n^3)$ because it is close to a $3/2$th iterate of $f(n)$ ? The integer conditions make this hard to see.
MAIN QUESTIONS
I want solutions for $g(n)$. How many solutions exist ? How many free parameters are there ?
