Solving a general two-term combinatorial recurrence relation What is known about explicit (not necessarily closed-form) solutions to the recurrence
$$R^n_k= (\alpha n) R^{n-1}_k + (\alpha' n + \beta' k) R^{n-1} _{k-1},$$
with initial condition $R_0^0 = 1$ and with $R^n_k = 0$ for $n < 0$ or $k < 0$?  Special cases of this are closely related to recurrences satisfied by some interesting combinatorial numbers, such as the binomial coefficients and the Stirling numbers.
The more general recurrence
$$R^n_k= (\alpha n + \beta k + \gamma) R^{n-1}_k + (\alpha' n + \beta' k + \gamma') R^{n-1} _{k-1},$$
is open Problem 6.94 in Concrete Mathematics (2nd edition, p. 319).
The closest published result I have found thus far is the following formula due to Neuwirth ("Recursively defined combinatorial functions: Extending Galton's board," Discrete Mathematics, 2001) for the case $\alpha' = 0$ of the Concrete Mathematics problem,
$$R^n_k = \prod_{i=1}^k (\beta' i + \gamma') \sum_{i=0}^n \sum_{j=0}^n  s^n_i \binom{i}{j} S^j_k \alpha^{n-i} (\gamma - \alpha)^{i-j} \beta^{j-k},$$
which, of course, gives me an answer to my question when $\alpha'=0$.    (Here, $s^n_i$ and $S^j_k$ are unsigned Stirling numbers of the first and second kinds, respectively.)
I have tried generating functions without any success thus far.  An answer like Neuwirth's that involves sums and binomial coefficients or Stirling numbers would be fine, as would a partial answer or just another idea to try.
 A: A general solution of the Graham-Knuth-Patashnik 6.94 problem 
$$
R^{n}_k=(\alpha \, n + \beta\, k + \gamma)\, R^{n−1}_k+ (\alpha′\, n+ \beta′\, k +\gamma′)\, R^{n−1}_{k−1} + \delta_{n,0}\delta_{k,0}\,, 
$$
with $R_{n}^k=0$ if $n<0$ or $k<0$, can be found in the paper ``Bivariate generating functions for a class of linear recurrences: General structure'', by J.F. Barbero G., J. Salas, and E.J.S. Villaseñor, published in J. Combin. Theory A 125 (2014) 146-165 (see also 
arXiv:1307.2010).
The solution was obtained by using generating functions.
A: By now it is not a general answer, but I hope it may help. The method is based on the one used in the Appendix A of arXiv:1009.1031.
To calculate $R^N_K$ one can introduce densities $\rho_k(t)$ accumulating the multiplier of $R^{N-t}_k$ after using the recurrence formula $2^{t}-1$ times. Then $R^N_K$ is just $\rho_0(N)$ for the initial condition $\rho_k(0)=\delta_{Kk}$.
The set of equations of their evolution reads:
$$\rho_k(t+1) = (\alpha n + \beta k + \gamma) \rho_k(t)+(\alpha' n + \beta' (k+1) + \gamma')\rho_{k+1}(t)$$
for integer $k$ and bearing in mind that $n = N-t$.
After introducing a generating function
$$G(t,z)=\sum_{k=-\infty}^\infty \rho_k(t)z^k,$$
with the initial condition $G(0,z)=z^K$, the set of equations is transformed into
$$G(t+1,z) = \left(
\gamma + \alpha n
+(\gamma'+\alpha'n)\frac{1}{z}+(\beta z+\beta')\frac{\partial }{\partial z}
\right)G(t,z).$$
So in general case $R^N_K$ is the constant term of
$$\left[  \prod_{n=1}^N \left(
\gamma + \alpha n
+(\gamma'+\alpha'n)\frac{1}{z}+(\beta z+\beta')\frac{\partial }{\partial z}
\right)\right]z^K.$$ 
The question is if it is possible to simplify it. As in general at different times (i.e. for different $n$) eigenvectors of the differential operators are different, one cannot use the same approach as in the paper.
However, for some special cases eigenvectors are the same for every $n$, that is for


*

*$\alpha' = 0$,

*$\beta = \beta' = 0$.


If decompose $z^M$ in the eigenfunctions $z^M = \sum_i a_i f_i(z)$, then instead of a product of operators one gets a product of eigenvalues
$$R^N_K = \sum_i a_i \times \text{[the constant term of $f_i(z)$]}\times\prod_{n=1}^N \lambda_i(n).$$
I keep writing the constant term of instead of $|_{z=0}$ as when $\gamma'\neq 0$ or $\alpha'\neq 0$ one needs to tackle negative powers of $z$. 
A: have a look at Migdal(2010), paper on analysis of Mafia game, there is a vey similar problem there with some solution
