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.)