What [is the minimal] differential equation is satisfied by $y = \sum_{n=0}^\infty {x^n}(n!)^{-k}$?

Question

Specific Question

What [is the minimal] differential equation is satisfied by $y = \sum_{n=0}^\infty {x^n}(n!)^{-k}$?

That is, we are hunting for a monic differential operator of minimal order which annihilates our function. That is, plugging in the above function into our operator results in zero.

Exposition

Each function $ F_k(x) = \sum_{n=0}^\infty \frac{x^n}{(n!)^k} $ is a D-finite function, meaning it satisfies a linear differential equation with polynomial coefficients. Note that,

$F_1(x)=e^x$, the exponential function satisfies $F_1'=F_1$.

$F_2(x)=I_0(2\sqrt{x})$, a modified Bessel function satisfies $F_2= xF_2''+F_2'$.

$F_3(x)= F_{3}'\left(x\right)+3xF_{3}''\left(x\right)+x^{2}F_{3}'''\left(x\right)$

Motivation

Is $F_k(1)-1$ =$\sum_{n=1}^\infty 1 /(n!)^k$ demonstrably transcendental? Note that $F_1(1)=e$ and $F_2(1)=I_0(2)$ are both demonstrably transcendental but the proofs of these facts invoke differential equations.

Answer 1

$y(x)= \,_0F_{k-1}(;1,1,\ldots,1;x)$ satisfies the differential equation $$\left(x \frac{d}{dx}\right)^ky = xy$$ (cf. DLMF 16.8.3). But $$\left(x\frac{d}{dx}\right)^k =\sum_{j=0}^k\left\{\begin{array} & k\\ j\end{array}\right\}x^j\frac{d^j}{dx^j}\text{,}$$ where $\left\{\begin{array} & k\\ j\end{array}\right\}$ are the Stirling numbers of the second kind, an identity which reduces to the definition of the Stirling numbers when acting on $x^n$. For $k>0$, then, $$ \sum_{j=1}^k\left\{\begin{array} & k\\ j\end{array}\right\}x^{j-1}\frac{d^jy}{dx^j} = y\text{.}$$

Answer 2

The equation for $k=4$ is $$y=y'+7x y''+6x^2y^{(3)}+x^3y^{(4)},$$ from which one can assume that the coefficients are the Stirling numbers of the second kind, i.e. $$F_k(x) = \sum_{n=1}^{k} \left\{ \begin{matrix} k \\ n \end{matrix} \right\} x^{n-1} F_k^{(n)}(x)\tag1 $$

This hypothesis can be proved as follows.

Observe that for $k>0$ $$F_k(x)=\frac1x\left(x\frac d{dx}\right)^k F_k(x).\tag2$$

Indeed under this operation $$ \frac{x^n}{(n!)^k}\mapsto \frac{x^{n-1}}{((n-1)!)^k}, $$ as easily to check. Further: $$ \frac1x\left(x\frac d{dx}\right)^k=\sum_{k=1}^n\left\{ \begin{matrix} k \\ n \end{matrix} \right\} x^{n-1}\frac{d^n}{dx^n}\tag3, $$ as can be demonstrated by induction. Obviously $(3)$ holds for $k=1$. Assume it holds for some $k\ge1$. Then: $$\begin{aligned} \frac1x\left(x\frac d{dx}\right)^{k+1}&=\frac d{dx}\left(x\frac d{dx}\right)^{k}\\ &=\frac d{dx}\left(x\sum_{n=1}^k\left\{ \begin{matrix} k \\ n \end{matrix} \right\} x^{n-1}\frac{d^n}{dx^n}\right)\\ &=\frac d{dx}\left(\sum_{n=1}^k\left\{ \begin{matrix} k \\ n \end{matrix} \right\} x^{n}\frac{d^n}{dx^n}\right)\\ &=\sum_{n=1}^k\left\{ \begin{matrix} k \\ n \end{matrix} \right\} n x^{n-1}\frac{d^n}{dx^n}+\sum_{k=1}^n\left\{ \begin{matrix} k \\ n \end{matrix} \right\} x^{n}\frac{d^{n+1}}{dx^{n+1}}\\ &=\sum_{n=1}^{k+1}\left\{ \begin{matrix} k \\ n \end{matrix} \right\} n x^{n-1}\frac{d^n}{dx^n}+\sum_{n=1}^{k+1}\left\{ \begin{matrix} k \\ n-1 \end{matrix} \right\} x^{n-1}\frac{d^{n}}{dx^{n}}\\ &=\sum_{n=1}^{k+1}\left(\left\{ \begin{matrix} k \\ n \end{matrix} \right\} n +\left\{ \begin{matrix} k \\ n-1 \end{matrix} \right\}\right)x^{n-1}\frac{d^n}{dx^n}\\ &=\sum_{n=1}^{k+1}\left\{ \begin{matrix} k+1 \\ n \end{matrix} \right\} x^{n-1}\frac{d^n}{dx^n}.\\ \end{aligned}$$ QED.