Integrality of a sequence formed by sums Consider the following sequence defined as a sum
$$a_n=\sum_{k=0}^{n-1}\frac{3^{3n-3k-1}\,(7k+8)\,(3k+1)!}{2^{2n-2k}\,k!\,(2k+3)!}.$$

QUESTION. For $n\geq1$, is the sequence of rational numbers $a_n$ always integral?


Let $A(x) = \sum_{n=1}^\infty a_n x^n$ and let
$$S(x) = \sum_{k=0}^\infty (7k+8)\frac{(3k+1)!}{k!\,(2k+3)!} x^k.$$
Then the formula for $a_n$ gives
$A(x) = R(x)S(x)$,
where
$$R(x) = \frac{1}{3}\biggl(\frac{1}{1-\frac{27}{4} x} -1\biggr).$$
A standard argument, for example by Lagrange inversion, gives
$$S\left(\frac{y}{(1+y)^3}\right)=\frac{4+y}{3(1+y)^2}.$$
A straightforward computation gives
$$R\left(\frac{y}{(1+y)^3}\right) = \frac{9y}{(4+y)(1-2y)^2}.$$
Thus
$$A\left(\frac{y}{(1+y)^3}\right)=\frac{3y}{(1-2y)^2(1+y)^2}.$$
Since the power series expansion of $y/(1+y)^3$ starts with $y$ and has integer coefficients, its compositional inverse has integer coefficients, so $A(x)$ does also.
