Recursion for A014307

As $m\mapsto R(n,k,m)$ is a polynomial, let us make the change of notation $R_{n,k}(x)=R(n,k,x)$.

Define $P_n(x)=R_{n,0}(x)$. We will show that $$P_{n+1}(x)=P_{n}(x)+(x-1)\sum_{k=0}^{n-1}\binom{n}{k}P_{k+1}(x).$$

We have $R_{n,k+1}(x)-R_{n,k}(x)=-R_{n-1,k}(x)$, or a backwards recurrence, $R_{n,k}(x)=R_{n-1,k}(x)+R_{n,k+1}(x)$, for $k\ge 1$. We also have $R_{n,n}(x)=(x-1)R_{n,0}(x)$. This allows us to conclude by induction on $k$ that $$R_{n,n-k}(x)=(x-1)\sum_{j=0}^k\binom kjP_{n-j}(x).$$

Thus we have $$\begin{align*}P_{n+1}(x)&=\sum_{j=0}^n R_{n,j}(x)=P_n(x)+\sum_{k=0}^{n-1} R_{n,n-k}(x)\\ &=P_n(x)+(x-1)\sum_{k=0}^{n-1}\sum_{j=0}^k\binom kj P_{n-j}(x)\\ &=P_n(x)+(x-1)\sum_{k=0}^{n-1}P_{k+1}(x)\sum_{s=n-k-1}^{n-1}\binom{s}{n-k-1}\end{align*}$$ from which the desired recurrence results.

Now let $C_m(z)$ be the EGF for $(c_m)_n=P_n(m)$. Identifying the inner sum as a convolution we have $$C_m'=C_m+(m-1)C_m'(-1+\exp z)$$ $B_m(z)$ satisfies this differential equation, and therefore $b(n,m)=R(n,0,m)$ as desired.

Alternatively, we can explicitly solve the differential equation by integrating $\frac{1}{m-(m-1)\exp z}$. As $R_{0,0}(m)=1$, $\ln C_m(0)=0$ and the constant of integration is zero.