(This is a follow-up to my previous question Can every finite graph be represented by an arithmetic sequence of natural numbers?)

Since it is obviously false that every finite graph can be represented by an arithmetic sequence (with an edge between two vertices/numbers $n_i, n_j$ iff GCD$(n_i, n_j)>1$) I'd like to reformulate my question:

Consider a family $F$ of parametrized computable functions $\lbrace f_{\alpha}:\mathbb{N}^k\mapsto\mathbb{N}\rbrace_{\alpha}$ with $\alpha \in \mathbb{N}^{n}$ for some $k,n \in \mathbb{N}$.

A sequence $(n_1,....,n_k,n_{k+1},...,n_{k+l})$ is an $F$-sequence if $n_{i+k} = f_\alpha(n_i,n_{i+1},...,n_{i+k-1})$ for some fixed $f_\alpha \in F$.

An arithmetic sequence is an $F$-sequence for $F = \lbrace f_\alpha(n) = n + \alpha\rbrace_{\alpha \in \mathbb{N}}$.

Question: Is there a family $F$ of functions $f_{\alpha}:\mathbb{N}^k\mapsto\mathbb{N}$ (as above) such that every graph on $n > k$ nodes can be represented by an $F$-sequence such that $n_i$ and $n_j$ are joined by an edge iff $n_i \neq n_j$ and GCD$(n_i, n_j) > 1$

My first question was genuinely "first-order" and admitted a straight-forward "first-order" answer (even though negative). This question is genuinely "second-order" and furthermore existential and probably doesn't admit a straight-forward definite answer. Sorry for that (and I tagged it as a "soft question"). But maybe someone can give a hint how to try to attack it?