# Are there superexponential Pfaffian functions?

This question is motivated by model theory, but it's really an analysis question (which means it may have an easy analysis answer that I just don't have the background for). Here's the main question, followed by some definitions and motivation:

Is there a partial function $f:\mathbb R\to \mathbb R$, defined on an unbounded interval $(a,\infty)$, which is both superexponential and Pfaffian?

I suspect if there is a solution to the above, there is also a totally defined function, but that's the form of the question as it's useful to me.

A function $f$ is superexponential if, for all $n$, there is an $a$ where for all $x>a$, $f(x)>\exp(\exp(\cdots \exp(x)\cdots))$, where $\exp(x)$ is $e^x$, and there are $n$ iterates of $exp$ in the preceding expression.

Equivalently (for model theorists), a function is superexponential if and only if it is not eventually bounded by any function definable in $(\mathbb R, +,\cdot,0,1,\exp)$. It is possible to construct (real)-analytic superexponential functions without too much trouble; see this answer by fedja to a Math.StackExchange question to this effect.

A Pfaffian chain is a sequence of unary functions $(f_1,\ldots,f_n)$ such that for each $k$, there is a real polynomial $p_k(x,y_1,\ldots,y_k)$ where $f_k'=p_k(x,f_1,\ldots,f_k)$ on a tail of $\mathbb R$. A function is Pfaffian if it is part of a Pfaffian chain.

So for example, all polynomials are Pfaffian, as is the exponential function. However, $\sin$ and $\cos$ are not Pfaffians, since each one would need to "reference the other."

For motivation see another question, where this question was raised in a comment by Joel David Hamkins; if such a function exists, it would generate a superexponential o-minimal theory, resolving an interesting open question. However, a negative answer, while interesting, would not resolve that question. I fiddled with this formulation for a few months but eventually gave up.

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