$f(n) = \frac{n^2 + n + 4}{2}, g(f(n)) = f(g(n))$ such that $g(n)$ is an integer.

Question

Let $n$ be a strict positive integer.

Lets define an integer sequence $f(n)$ :

$$f(n) = \frac{n^2 + n + 4}{2}$$

so

$$f(1) = 3$$ $$f := {3,5,8,12,17,23,30,38,47,...}$$ $$f(17) = 155$$ etc

Notice

$$3+2=5$$ $$5+3=8$$ $$8+4=12$$ $$12+5=17$$ 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 :

https://oeis.org/A034329

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 ?

Answer 1

This started as an attempted proof, but ended up being more of demonstration of how hard the problem is. However, I still think it's worth sharing.

---

By definition, since $g(f(n))=f(g(n))$, then $$g\left(\frac{1}{2}(n^2+n+4)\right)=\frac{1}{2}(g^2(n)+g(n)+4).$$ What this means is that if we are looking for free parameters, then we must observe that $$\begin{align*} &n=0\implies g(2)\text{ depends on the value of }g(0),\\ &n=1\implies g(3)\text{ depends on the value of }g(1),\\ &n=2\implies g(5)\text{ depends on the value of }g(2),\\ &\vdots \end{align*}$$ Thus, $g(0),g(1),g(4),\cdots$ (and all such $g(k)$ such that $k$ is not of the form $\frac{1}{2}(n^2+n+1)$) do not depend on any other term (or more accurately, all other terms depend on them). So they are effectively our "free" parameters which define the rest of the sequence. Notice that this effectively creates an infinite family of "lineages of dependence".

$$\begin{align*} &g(0)\leftarrow g(2)\leftarrow g(5) \leftarrow g(6) \leftarrow g(17) \leftarrow \cdots\\ &g(1)\leftarrow g(3)\leftarrow g(8) \leftarrow g(38) \leftarrow g(743) \leftarrow \cdots\\ &g(4)\leftarrow g(12)\leftarrow g(80) \leftarrow g(3242) \leftarrow g(5256905) \leftarrow \cdots\\ &\vdots \end{align*}$$

So the difficulty here is choosing such $g(0),g(1),g(4),\cdots$ such that they obey: $$g(n+1)>g(n)\text{ and }f(n)+1<g(n)<f(f(n))-1\text{ for all }n.$$

The $g(n+1)>g(n)$ condition alone is not so difficult, as it can be achieved with some sort of greedy algorithm. Fortunately, we are guaranteed that $$g\left(\frac{1}{2}(n^2+n+4)\right)=\frac{1}{2}(g^2(n)+g(n)+4)>g(n),$$ by definition, at least, so our "lineages of dependences" will be ordered regardless of our choices of $g(0),g(1),g(4),\cdots$. Hence, as long as we choose our free parameters to obey $0<g(0)<g(1)<g(4)<\cdots$, then it is already a given that

$$\begin{align*} &g(0)< g(2)< g(5) < g(6) < g(17) < \cdots\\ &\wedge\\ &g(1)< g(3)< g(8) < g(38) < g(743) < \cdots\\ &\wedge\\ &g(4)< g(12) < g(80) < g(3242) < g(5256905) < \cdots\\ &\wedge\\ &\vdots \end{align*}$$

---

But this is now starting to look very category theoretic, which is far beyond my pay-grade, so I'll leave it here. Hopefully someone can pick up this line of reasoning to extend this into a proof (positive or negative).