TITLE: A polynomial recurrence involving partial derivatives
QUESTION [12 upvotes]: Define recursively polynomials $f_n(a,b)$ by
         $$ f_0(a,b)=1,\ \ f_n(0,b)=0\ \mathrm{for}\ n>0 $$
      $$ \frac{\partial}{\partial a}f_n(a,b) = f_{n-1}(b-a,1-a). $$
   For instance,
   $$ f_1(a,b)  =  a,\ \
         f_2(a,b)  =  \frac 12(2ab-a^2) $$
   $$   f_3(a,b) = \frac 16(a^3-3a^2-3ab^2+6ab). $$
Is there a ``nice'' solution to this recurrence, e.g., a formula for
the generating function $\sum_{n\geq 0}f_n(a,b)x^n/n!$? What I am
really interested in is $f_n(1,1)$. For the motivation, see the
solution to Exercise~4.56(d) (pg. 645) of Enumerative
Combinatorics, vol.1, 2nd ed.

REPLY [3 votes]: There seems to be a PDE for $g(a,b,x)=\sum_{n\ge0}f_n(a,b)x^n$,
which can be thought of as a boundary value problem in the triangle $0\lt a\lt b\lt1$.
$$g_{aab}+g_{abb}+x^3g=0$$ 
($x$ is a parameter and  subscripts are derivatives)
with boundary values $g(0,b,x)=1$, $g_a(a,a,x) = x$, and 
$g_{ab}(a,1,x) = x^2$. This comes from iterating the $f_n$ recurrence,
after Pietro's remarks that $(a,b)\to(b-a,1-a)$ has period 3
suggested looking at third derivatives.
  Does that determine $g$ uniquely, nicely? I don't know yet.
[Edit: I wrongly wrote $g$ at first using $\frac{x^n}{n!}$.]