This question have been driving me crazy for months now. This comes from work on multiple integrals and convolutions but is phrased in terms of formal power series.

We start with a formal power series

$P(C) = \sum_{n=0}^\infty a_n C^{n+1}$

where $a_n = (-1)^n n!$

With these coefficients the formal power series can be expressed as a hypergeometric function

$P(C) = C \, _2F_0(1,1;;-C)$

I'm then interested in the formal power series $P_T(C)=\frac{P}{1-P}$ as well as, if possible, the series $P^n$ for arbitrary positive integer n (where this is the power series P raised to the nth power).

Specifically if

$P_T(C) = \sum_{n=0}^\infty b_n C^{n}$

then want to construct the function

$f(x) = \sum_{n=0}^\infty \frac{b_{n+1}}{(n!)^2} x^{n}$

which, from other results, should converge for all x. We can think of this as the doubly-exponential generating function for the $b_n$ sequence.

There are rules for multiplying and dividing formal power series (see here) and I've used these to get a recurrence relation for the coefficients in $P_T(C)$ (as well as P^n(C)) but I've been unable to solve these recurrence relations explicitly. They're in a form where each $b_n$ depends on all the previous $b_n$'s and I've not been able to make progress with them.

Explicitly the recurrence relation for the $b_n$ is $b_0 = 1$, $b_n = \sum_{k=1}^n b_{n-k} a_k$ (for n > 1). This looks simple enough but I don't think has a nice closed-form expression.

Nevertheless I do know what the $b_n$ are. They are the sequence A052186 (up to plus and minus signs). So

$P_T(C) = C+C^3-3 C^4+14 C^5-77 C^6+497 C^7-3676 C^8+\ldots$

and

$f(x) = 1 + \frac{1}{4}x^2 - \frac{1}{12}x^3 + \frac{7}{288}x^4 - \frac{77}{14400} x^5 + \frac{497}{518400}x^6 +\ldots$

The question is, is it possible to figure out the function $f(x)$!? Does it have a nice closed form? Perhaps in the form of a hypergeometric function? If it does, great, if it doesn't then at least I can stop searching for it!