TITLE: Exponential of a specific hypergeometric series
QUESTION [5 upvotes]: This is motivated by this question.
Let $f$ be the hypergeometric series
$ f(x) = 2 x \, _{4}F_3([1, 1, 4/3, 5/3], [2, 2, 2], 27 x) $
which is explictly given by
$ f(x) = \sum_{n \geq 1} \frac{(3n-1)!}{n!^3} x^n $
and appears in some studies of Mahler measures.

Is is true that $\exp(f)$ has only positive integer coefficients ?

The first few coefficients are $1, 2, 17, 218, 3404, 59644$.
If true, is there a combinatorial meaning for these coefficients ?
In the same vein, what can one say on the coefficients of $\exp(-f)$ which start
$1, -2, -13, -158, -2431, -42454, -802790$ ?

REPLY [2 votes]: Here's another approach. It resembles Ira's solutions, although less combinatorial. Maybe it will be of use.
Let $a_n := \frac{1}{3} \binom{3n}{n,n,n}$. ($a_1=1$ and $a_n \in \mathbb{Z}$, as $a_n = \binom{3n-1}{n-1,n,n}$.)
Note that $f(x) = \sum_{n \ge 1} \frac{a_n}{n} x^n$. Hence, the function we're dealing with is $\zeta_{\{a_n\}}(x) := \exp( \sum_{n \ge 1} \frac{a_n}{n} x^n)$, which we'll refer to as "the zeta function of $\{a_n\}$".
Lemma 1: Consider a sequence $\{ a_n \}_{n\ge1}$ of rational numbers, with $a_1=1$. $\zeta_{\{a_n\}}(x) \in \mathbb{Z}[[x]]$ iff $\forall n \in \mathbb{N}: n\mid \sum_{d\mid n} \mu(n/d) a_d$.
This lemma reduces our problem to proving a congruence: $\forall n \in \mathbb{N}: 3n\mid \sum_{d\mid n} \mu(n/d) \binom{3d}{d,d,d}$.
Lemma 2: The condition $\forall n \in \mathbb{N}: n\mid \sum_{d\mid n} \mu(n/d) a_d$ is equivalent to $a_{np^k} \equiv a_{np^{k-1}} \mod {p^k}$ for any prime $p$ and $n,k \in \mathbb{N}$.
So we need to prove the congruence $\binom{3np}{np,np,np} \equiv \binom{3n}{n,n,n} \mod {3np}\mathbb{Z}_{p}$ (note how I turned to $p$-adics, as it simplifies computations and presentation).
This congruence is a simple consequence of either:

A combinatorial argument, reminiscent of the combinatorial proof of (a weak version of) Wolstenholme's theorem.
Properties of the $p$-adic Gamma function. The congruence is equivalent to $\frac{\Gamma_{p}(3np)}{\Gamma_{p}^3(np)} \equiv 1 \mod {3np \mathbb{Z}_{p}}$. Kazandzidis congruences give something much stronger (although a small modification is necessary for the even prime and the second prime):
$$\frac{\Gamma_{p}(3np)}{\Gamma_{p}^3(np)} \equiv 1 \mod {n^3p^3 \mathbb{Z}_{p}}$$

Proofs of both lemmas are not hard. The first lemma follows by writing $\zeta_{a_n}(x)$ as $\prod_{n\ge1} (1-x^n)^{-\frac{b_n}{n}}$, where taking the logarithmic derivative shows $b_n = \sum_{d\mid n} \mu(n/d) a_d$. This gives a proof of one direction. For the other direction, notice that 
$$[x^n]\prod_{i=1}^{\infty} (1-x^i)^{-\frac{b_i}{i}} = [x^n]\prod_{i=1}^{n-1} (1-x^i)^{-\frac{b_i}{i}} + b_n$$
 and use an induction argument. The second lemma is just a "coupling" argument (note how $\mu(ap) = -\mu(a)$ for $p\nmid a$).