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Let's prove that $a_3=a_1$. Note that the recurrence for $R$ translates to the following PDE for the generating $F(x,y):=\sum_{n,m\geq0} R(n,m,p,q) \frac{x^n}{n!}\frac{y^m}{m!}$: $$\frac{\partial}{\partial x} F(x,y) = \frac{\partial}{\partial y} \bigg(F(x,y) + \frac{p}q(e^{qy} - 1)F(x,y)\bigg)$$ with the boundary condition $F(0,y) = e^y$. This PDE is well solvable in CAS like Maple, it can be easily verified that $F(x,0)=\sum_{n\geq0} R(n,0,p,q) \frac{x^n}{n!}$ does indeed coincide with the derivative of the e.g.f for $a_1$ given in Ira's answer. QED