Good morning. I have the following;
Let
$$\beta_k(x)=\sum_{n=0}^\infty{b(r;n,k)x^n}$$
Here, $b(r; n,k)$ are called Associated Sterling Numbers of the Second Kind, and are the number of ways to partition a set of elements into disjoint parts whose partition cardinalities are greater than r. For simplicity, I'm only interested in the case when $r=1$, and so from here we will use $b(1;n,k)=b(n,k)$.
It can be shown through integral transformations that this generating function can be written as a partial fraction decomposition
$$\beta_k(x)=\frac{(-1)^k}{x}\sum_{j=0}^k\sum_{i=0}^{k-j}\frac{(-1)^j}{i!j!}\left(\frac{x}{1-jx}\right)^{k-i-j+1}$$
For my purposes, I'm trying to write this as a single rational function, which can be done up to one exception. For all $k$, we have that
$$\beta_k(x)=\frac{x^{2k}P_k(x)}{(1-x)^k(1-2x)^{k-1}(1-3x)^{k-2}...(1-kx)}$$
Now i was able to come up with these polynomials $P_k(x)$ using brute force and mathematica, but I'm still trying to determine if there is a closed form for these. Does a closed form exist and if so what are the techniques used to determine such a form?
As the joke says, "If I wanted to go there I wouldn't start here". From Wikipedia we have (adjusting notation) $$b(r; n+1, k) = k b(r; n, k) + \binom{n}{r} b(r; n-r, k-1)$$
In particular, setting $r=1$, $$b(n+1, k) = k b(n, k) + n b(n-1, k-1)$$ yielding $$\beta_k(x) = \frac{x^2}{1 - kx} \left[ \beta_{k-1}(x) + x \frac{\textrm d}{\textrm dx} \beta_{k-1}(x)\right]$$ which has obvious similarities to the expression in the question. If we define $$Q_k(x) = \prod_{a=1}^k (1-ax)^{k+1-a}$$ and suppose that $$\beta_{k-1}(x) = \frac{x^{2(k-1)} P_{k-1}(x)}{Q_{k-1}(x)}$$ then
$$\begin{eqnarray}\beta_k(x) &=& \frac{x^2}{1 - kx} \left[ \frac{x^{2(k-1)} P_{k-1}(x)}{Q_{k-1}(x)} + x \frac{\textrm d}{\textrm dx} \frac{x^{2(k-1)} P_{k-1}(x)}{Q_{k-1}(x)}\right] \\ % &=& \frac{x^2}{1 - kx} \left[ \frac{x^{2(k-1)} P_{k-1}(x)}{Q_{k-1}(x)} + x \frac{x^{2(k-1)} Q_{k-1}(x) \frac{\textrm d}{\textrm dx} P_{k-1}(x) + P_{k-1}(x) Q_{k-1}(x) \frac{\textrm d}{\textrm dx} x^{2(k-1)} - x^{2(k-1)} P_{k-1}(x) \frac{\textrm d}{\textrm dx} Q_{k-1}(x)}{Q_{k-1}(x)^2}\right] \\ % &=& \frac{\left[x^{2k} + x^3 \frac{\textrm d}{\textrm dx} x^{2(k-1)} \right]P_{k-1}(x) Q_{k-1}(x) + x^{2k+1} Q_{k-1}(x) \frac{\textrm d}{\textrm dx} P_{k-1}(x) - x^{2k+1} P_{k-1}(x) \frac{\textrm d}{\textrm dx} Q_{k-1}(x)}{(1-kx)Q_{k-1}(x)^2}\\ % &=& \frac{(2k-1) x^{2k} P_{k-1}(x) Q_{k-1}(x) + x^{2k+1} \left[ Q_{k-1}(x) \frac{\textrm d}{\textrm dx} P_{k-1}(x) - P_{k-1}(x) \frac{\textrm d}{\textrm dx} Q_{k-1}(x) \right]}{(1-kx)Q_{k-1}(x)^2}\\ % &=& \frac{(2k-1) x^{2k} P_{k-1}(x) + x^{2k+1} \left[ \frac{\textrm d}{\textrm dx} P_{k-1}(x) - P_{k-1}(x) \sum_{a=1}^{k-1} \frac{a(a-k)}{(1 - ax)}\right]}{(1-kx)Q_{k-1}(x)}\\ % \end{eqnarray}$$
Clearly $\sum_{a=1}^{k-1} \frac{a(a-k)}{(1 - ax)}$ can be expressed as $\frac{R(x)}{\prod_{a=1}^{k-1} (1 - ax)}$ for some polynomial $R_k(x)$, so we have
$$\begin{eqnarray}\beta_k(x) &=& \frac{(2k-1) x^{2k} P_{k-1}(x) + x^{2k+1} \left[ \frac{\textrm d}{\textrm dx} P_{k-1}(x) - P_{k-1}(x) \frac{R_k(x)}{\prod_{a=1}^{k-1} (1 - ax)}\right]}{(1-kx)Q_{k-1}(x)}\\ % &=& \frac{(2k-1) x^{2k} P_{k-1}(x) \prod_{a=1}^{k-1} (1 - ax) + x^{2k+1} \left[ (\prod_{a=1}^{k-1} (1 - ax)) \frac{\textrm d}{\textrm dx} P_{k-1}(x) - P_{k-1}(x) R_k(x)\right]}{Q_k(x)}\\ \end{eqnarray}$$
and giving the recurrence
$$P_k(x) = P_{k-1}(x) \left( (2k-1) \prod_{a=1}^{k-1} (1 - ax) - xR_k(x)\right) + x\left(\prod_{a=1}^{k-1} (1 - ax)\right) \frac{\textrm d}{\textrm dx} P_{k-1}(x)$$
It's not all the way to a closed form, but I would consider it better than brute force. It allows for reasonably efficient computation with a CAS, as well as the simple observation that $\deg R_k(x) = k - 2$, so $\deg P_k(x) \le \deg P_{k-1}(x) + k - 1$