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.
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) of Enumerative Combinatorics, vol.1, 2nd ed.