Ordinary or Rational Generating Function for Associated Stirling Numbers $b(n,k)$ I am trying to identify or find the ordinary or rational generating function (not the exponential generating function) for the Associated Stirling numbers of the Second kind, denoted $$b(1;n,k)=b(n,k)$$
These numbers are the number of ways to partition a set of $n$ elements into $k$ disjoint parts whose partition cardinality are greater than 1.
If there isn't already an explicit formula, how I do go about attempting to derive such a thing?
EDIT:
One thing that I do know is that 
$$b(n,j)=\sum_{k=0}^j(-1)^k\binom{n}{k}S(n-k,j-k)$$
where $S(n,k)$ are Stirling numbers of the second kind.  Therefore, I can say that 
$$\sum_{n=0}^\infty b(n,j)x^n=\sum_{n=0}^\infty \left(\sum_{k=0}^j(-1)^k\binom{n}{k}S(n-k,j-k)\right)x^n$$
which is quite ugly.  I also know that the Stirling Numbers of the Second kind have the ordinary generating function
$$\sum_{n=0}^\infty S(n,k)x^n=\frac{x^k}{(1-x)(1-2x)...(1-kx)}  $$
which I'm hoping to take advantage of after some brute force calculation...
EDIT 2:  Aftr some brute force calculation, it appears that I have a series generated, that looks like
$$b(n,j)=\sum_{k=0}^j(-1)^k\binom{n}{k}S(n-k,j-k)$$
$$\sum_{n=0}^\infty b(n,j)x^n=b(0,j)+b(1,j)+b(2,j)+b(3,j)+b(4,j)+...$$
Now
$$b(0,j)=1$$
$$b(1,j)=S(1,j)$$
$$b(2,j)=S(2,j)-2S(1,j-1)$$
$$b(3,j)=S(3,j)-3S(2,j-1)$$
$$b(4,j)=S(4,j)-4S(3,j-1)+6S(2,j-2)$$
$$b(5,j)=S(5,j)-5S(4,j-1)+10S(3,j-2)$$
$$b(6,j)=S(6,j)-6S(5,j-1)+15S(4,j-2)-20(3,j-3)$$
$$b(7,j)=S(7,j)-7S(6,j-1)+21S(5,j-2)-35(4,j-3)$$
$$... $$
Rearranging I can see that I have
$$\sum_{n=0}^\infty b(n,j)x^n=\sum_{n=0}^\infty S(n,j)x^n-\sum_{n=1}^\infty nS(n-1,j-1)x^n+\sum_{n=2}^\infty \frac{n(n-1)}{2!}S(n-2,j-2)x^n-\sum_{n=3}^\infty \frac{n(n-1)(n-2)}{3!}S(n-3,j-3)x^n+...$$
And it looks like a derivative is taking effect but my powers of $x$ are not changing, but it also looks like I'm running in circles.  I feel like my bounds are incorrect...Should, for example, I change
$$\sum_{n=1}^\infty nS(n-1,j-1)x^n=\sum_{n=0}^\infty (n+1)S(n,j-1)x^{n+1}$$
$$\sum_{n=2}^\infty \frac{n(n-1)}{2!}S(n-2,j-2)x^n=\sum_{n=0}^\infty \frac{(n+2)(n+1)}{2!}S(n,j-2)x^{n+2}$$
Any suggestions?
EDIT 3:  By comparing the first few numbers given $j$ values 2,3,4, there seems to be a trend, the generating function is structured 
$$\sum_{n=0}^{\infty}b(n,j)x^n=\frac{f(x)}{\prod_{k=1}^{j}(1-kx)^{j-k+1}}$$
The degree of the numerator is greater than the degree of the denominator in the cases where $j=2,3,4$ and the ratios of degree(numerator) to degree(denominator) for $j=2,3,4$ are $$\frac{4}{3}, \frac{9}{6}, \frac{14}{10}$$
 A: Using the recurrent relation $b(n+1,k) = k\cdot b(n,k) + n\cdot b(n-1,k-1)$ it is easy to get that the ordinary generating function $B(x,y) = \sum_{n,k} b(n,k)\cdot x^n\cdot y^k$ satisfies the following PDE:
$$\frac{\partial B(x,y)}{\partial y} + x^2\frac{\partial B(x,y)}{\partial x} = \frac{1-x^2y}{xy} B(x,y) - x.$$
Making substitution $z = \frac{1}{x}$ and $w(y,z)=B(1/z,y)$, it reduces to
$$\frac{\partial w(y,z)}{\partial y} - \frac{\partial w(y,z)}{\partial z} = \frac{z^2-y}{yz} w(y,z) - \frac{1}{z}.$$
While it is known how to solve this equation, its solution involves integral of the form
$$\int \frac{e^y dy}{y^u (u-y)^2},$$
which is not expressed in elementary (let alone, rational) functions. I suspect (but did not check carefully) that $w(y,z)$ (and thus $B(x,y)$) can be more or less easily expressed in terms of incomplete gamma function.
A: Yes , the series $\beta_k(x):=\sum_{n\ge0} b(n,k) x^n$  are  indeed rational functions, with the poles as you said. 
One step back. The Exponential Generating Function  of the polynomials $\big\{\sum_{k\ge0} b(n,k)t^k \big\}_{n\ge0}$ is
$$\sum_{n\ge0}\Big(\sum_{k\ge0}  b(n,k) t^k \Big)\frac{x^n}{n!} =e^{t(e^x-1-x)}$$
by the combinatorial meaning of the exponential of an  EGF. Hence also, for any $k\ge0$, 
$$ \sum_{n\ge0} b(n,k) \frac{x^n}{n!}=\frac{(e^x-1-x)^k}{k!}.$$
To convert the latter into an ordinary Generating Function, apply the usual transformation   $u(x)\mapsto \frac{1}{ x}\int_0^{\infty}u(s) e^{-\frac{s}{x}}ds$
$$\sum_{n\ge0} b(n,k) x^n =\frac{1}{k!x} \int_0^{\infty} (e^s-1-s)^k e^{-\frac{s}{x}}ds
,$$
which  clearly produces a rational function: if we expand the term $(e^s-1-s)^k$ and integrate with a linear change of variables we get a finite sum
$$\sum_{n\ge0} b(n,k) x^n =\frac{(-1)^k}{x}\sum_{i\ge0, j\ge0\atop i+j\le k} \frac{(-1)^j}{i!j!}\Big(\frac{x}{1-jx}\Big)^{k-i-j+1}
,$$
which is a rational function of the form you suggested in last edit. 
Rmk:  the above transformation can be applied formally, as all needed identities holds in a formal context; however, for $|x|<1/k$ the convergence of the integrals and series are ensured. 
$$*$$
[edit] One can show that $\beta_k(x)$ is a rational function, without computing it explicitly. The recursive relation of the coefficients  $b(n,k)$, translated into the sequence $\beta_k$ reads:
$$(1-kx)\beta_k=x^2(x\beta_{k-1})'$$
for $k\ge1$. 
Since $\beta_0(x):=1$, it follows by induction that $\beta_k$ are rational functions of the form
$$\beta_k(x)=\frac{x^{2k}P_k(x)}{(1-x)^k(1-2x)^{k-1}\cdots(1-kx)},$$
for  polynomials $P_k(x)$ of degree $\binom{k}{2}$. 
