TITLE: Real-rootedness, interlacing, root-bounds of a sequence of polynomials
QUESTION [11 upvotes]: Problem: the number $a(n,k)$ is defined by the following recurrence
\begin{equation}
a(n,k)=(k+1)(k+2)\, a(n-1, k)+\frac{(k+1)(k+2)(k+3)}{k} \,a(n-1, k-1),
\end{equation}
with $a(1,1)=1$ and $a(n,k)=0$ if $k<1$ or $k>n$.
For fixed $n$, the generating polynomial of $a(n,k)$ is defined as  $A_n(x)=\sum_{k=1}^n a(n,k)x^k$. The recurence above is equivalent to the following differential equation
$$
\frac{\mathrm{d}}{\mathrm{d}x} A_{n+1}(x) =24 A_n(x)+(36x+6)\frac{\mathrm{d}}{\mathrm{d}x} A_{n}(x)+(12x^2+6x)\frac{\mathrm{d}^2}{\mathrm{d}x^2} A_{n}(x)+(x^3+x^2)\frac{\mathrm{d}^3}{\mathrm{d}x^3} A_n(x).
$$
Question 1: all roots of $A_n(x)$ are real?
Question 2:  $A_n(x)$ interlaces $A_{n+1}(x)$? i.e.
$$
b_1 \leq a_1 \leq b_2 \leq a_2 \leq \cdots \leq b_n \leq a_n \leq b_{n+1},
$$
where $\{a_i\}$ and $\{b_j\}$ are roots of $A_n(x)$ and $A_{n+1}(x)$, respectively.
Question 3: all roots of $A_n(x)$ are located in $(-1,0]$?
These three statements are verified to be true for $n\leq 50$.
Some examples of $A_n(x)$ are given 
\begin{eqnarray*}
A_1(x) &=& x, \\
A_2(x) &=& 6x + 30 x^2,\\
A_3(x) &=& 36x + 540 x^2+ 1200 x^3,\\
A_4(x) &=& 216x + 7560 x^2+ 45600 x^3 +63000 x^4.
\end{eqnarray*}
Background of this problem: this number arises from certain graph enumeration problem.
One can easily prove the log-concavity of $\{a(n,k)\}_k$ by induction. Many literatures 
on real-rootedness or interlacing deal with recurrences with polynomial coefficients or
first and second order differential equations, not for this example. 
Any things about these numbers and polynomials would also be appreciated.
Progress:
as pointed out by Per Alexandersson (see below reply), the linear differential operator 
$p\mapsto 24 p +(36x+6)p' + (12x^2+6x)p'' + (x^3+x^2)p'''$
preserves real-rootedness. This show that if $A_n(x)$ has only real zeros,
so does the right hand side of the differential equation.
Notice that if a polynomial $f$ has only real zeros, the primitive integral $\int f dx$ of $f$ could have complex zeros in general. I am wondering, under what conditions of $f$, primitive integral $\int f dx$ has only real zeros.
(Assume that the constant term  of $\int f dx$ is zero.)

REPLY [3 votes]: Well, if you can show that the linear differential operator 
$p\mapsto 24 p +(36x+6)p' + (12x^2+6x)p'' + (x^3+x^2)p'''$
preserves real-rootedness, then you are a bit closer.
Now, this can be attacked by proving that the symbol of this operator is stable,
see
http://www.math.kth.se/~dirocco/ML2011/CIAMWORKSHOP/25/Branden.pdf
It will be some work; you end up with a 2-variable polynomial, and you want to show that this is non-zero whenever both variables have positive real part.
Now, it is quite easy to see that in "the limit", i.e, for large degrees of your polynomial $p$, this operator preserves real-rootedness, since then only the $(x^3+x^2)p'''$ term matters. And $(x^3+x^2)$ has roots 0 and -1, so it is not surprising that you get [-1,0] as the special interval.