A random permitting-context grammar is a context-free grammar $(N, \Sigma, P, S)$ equipped with a function $p : P \rightarrow 2^N$. The rule $A \rightarrow x$ can be applied to $uAw \Rightarrow uxw$ if every symbol in $p(A\rightarrow x)$ appears in $uw$. A forbidding-context grammar is similar, except  equipped with a function $f : P \rightarrow 2^N$, and a rule application is allowed if no symbol from $f(A\rightarrow x)$ appears in $uw$.$\{a^n b^m : 1 \le m \le 2^n\}$ is a random permitting-context language (RPCL), though $\{a^n b^m : m = 2^n\}$ is not (as can be proved by Ewert and Van der Walt's pumping lemma for RPCLs). In a permitting grammar we can allow the  nonterminals of some type to double, but we can't force it.It seems obvious to me that one can't get more than exponential  growth, but I can't see how to prove it (the pumping lemma certainly  doesn't help, since it will just increase the number of $a$'s). Is  there a known result that helps here?I'm also interested in superexponential growth in forbidding languages  (e.g. is $\{a^{2^{2^n}} : n \in \mathbb{N}\}$ an RFCL).
Is $\{a^n b^m : 1 \le m \le 2^{2^n}\}$ a permitting-context language?
fl.formal languages