This is a quick follow up on [R. Stanley's interesting post on MO](https://mathoverflow.net/questions/385303/the-square-root-of-a-graph) in a different direction, which might be easier.


For positive integers $a$, define the family of functions (infinite series) given by
$$F\_a(x)=\sum\_{n\geq0}a^{\binom{n+1}2}\frac{x^n}{n!}.$$
Then, $F\_2(x)$ becomes Stanley's $F(x)$.



> 
> **QUESTION.** Is it true that the coefficients of $\sqrt[a]{F\_a(x)}$ are all positive integers, as an exponential generating function?
> 
> 
> 



$a^{th}$-root of exponential generating functions
In fact, we can let $a$ be an indeterminate, and the coefficient of $x^n/n!$ in $\sqrt[a]{F\_a(x)}$ will be a polynomial with integer coefficients. This is because $a^{{n+1\choose 2}}$ counts graphs with edges colored with $a$ colors (including the color 0, which denotes no edge). Thus the number $c\_a(n)$ of connected such graphs on $n≥1$ vertices is a polynomial in $a$ with integer coefficients and is divisible by $a$. Hence the coefficient of $\frac{x^n}{n!}$ in $\exp \frac 1a\sum\_{n\geq 1}c\_a(n)\frac{x^n}{n!}$ is a polynomial in $a$ with integer coefficients.


**Addendum.** Let me explain why $c\_a(n)$ is divisible by $a$. Let
$G$ be a connected graph on $n$ vertices with no loops. If $G$ has
$q$ edges, then there are $(a-1)^q$ ways to color them. (The
nonedges, corresponding to the color 0, are already determined by
$G$.) At each vertex we are free to add a loop or not without
affecting connectedness. If we add a loop, we can color it in
$a-1$ ways, so $a$ possibilities in all for each vertex. Hence the
total number of coloring of all graphs obtained from $G$ by adding
loops is $(a-1)^q a^n$. This is divisible by $a$ (since $n\geq 1$),
and summing over all $G$ without loops will therefore give a
polynomial divisible by $a$, as claimed.


Now note that if $n\geq 2$, then $q\geq 1$ so $(a-1)^q a^n$ is
divisible by $a^2(a-1)$. Hence
$$ \exp \frac 1a\sum\_{n\geq 1}c\_a(n)\frac{x^n}{n!} =
 \exp\left( x+\sum\_{n\geq 2}a(a-1)d\_a(n)\frac{x^n}{n!}\right), $$
for some $d\_a(n)\in\mathbb{Z}[a]$. Taking this modulo $a(a-1)$ gives
$e^x=\sum\_{m\geq 0}\frac{x^m}{m!}$. It follows that the conjecture
in OEIS A178319 mentioned in the comment by Max Alekseyev is true.


