While evaluating some integrals I came across this system. I know that most of the times systems of non-linear recurrence relations are unsolvable, but looking at the outputs I got for different m it seems to me like a solution exists, just I'm not able to find it. Here is the problem:
$$ A_m=mA_{m-1}+B_{m-1}$$ $$ B_m=mB_{m-1}-A_{m-1} $$
Hint.
This recurrence is linear and can be written as
$$ \left(\matrix{A_m\\ B_m}\right) = \left(\matrix{m & 1 \\ -1 & m}\right)\left(\matrix{A_{m-1}\\ B_{m-1}}\right) $$
now $\left(\matrix{m & 1 \\ -1 & m}\right) = T\cdot \Lambda_m\cdot T^{-1}$ with
$$ T = \left( \begin{array}{cc} i & -i \\ 1 & 1 \\ \end{array} \right)\ \ \ \text{}\ \ \ \Lambda_m = \left( \begin{array}{cc} m-i & 0 \\ 0 & m+i \\ \end{array} \right) $$
so calling now $X_m = T^{-1}\cdot \left(\matrix{A_m\\ B_m}\right)$ we have the recast recurrence
$$ X_m = \Lambda_m\cdot X_{m-1} $$
with solution
$$ X_m = \left(\prod_{k=1}^m\Lambda_k\right)\cdot X_1 $$
etc.
NOTE
$$ \prod_{k=1}^m\Lambda_k = \left(\matrix{(m-i)!& 0\\ 0&(m+i)!}\right) $$
This result is not amenable to work with. Using generating functions we follow with
$$ \cases{ A_m x^m = m A_{m-1}x^m + B_{m-1}x^m\\ B_m x^m = -A_{m-1}x^m + m B_{m-1}x^m } $$
or
$$ \cases{ A(x) = x(x A(x))'+x B(x) + \phi_A(x)\\ B(x) = -x A(x) + x(x B(x))'+\phi_B(x) } $$
them we have
$$ \cases{ B'(x) = \frac{1-x}{x^2}-\frac 1x B(x) + \frac{1}{x^2}\phi_A(x)\\ A'(x) = \frac 1x A(x) + \frac{1-x}{x^2}B(x) + \frac{1}{x^2}\phi_B(x) } $$
Here $\phi_A(x), \phi_B(x)$ are initial condition polynomials. Assuming $\phi_A(x) = \phi_B(x) = 0$ we have the solution
$$ \cases{ A(x) = \frac{c_1 e^{-1/x} \cos (\log (x))}{x}-\frac{c_2 e^{-1/x} \sin (\log (x))}{x}\\ B(x) = \frac{c_1 e^{-1/x} \sin (\log (x))}{x}+\frac{c_2 e^{-1/x} \cos (\log(x))}{x} } $$
not expandable in series around $x=0$.